Integrand size = 21, antiderivative size = 270 \[ \int \frac {\sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2 b^4 \left (5 a^2-4 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^5 (a-b)^{3/2} (a+b)^{3/2} d}-\frac {b \left (a^2+4 b^2\right ) \text {arctanh}(\sin (c+d x))}{a^5 d}+\frac {\left (2 a^4+7 a^2 b^2-12 b^4\right ) \tan (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}-\frac {b \left (a^2-2 b^2\right ) \sec (c+d x) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2-4 b^2\right ) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \sec ^2(c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:
2*b^4*(5*a^2-4*b^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^5 /(a-b)^(3/2)/(a+b)^(3/2)/d-b*(a^2+4*b^2)*arctanh(sin(d*x+c))/a^5/d+1/3*(2* a^4+7*a^2*b^2-12*b^4)*tan(d*x+c)/a^4/(a^2-b^2)/d-b*(a^2-2*b^2)*sec(d*x+c)* tan(d*x+c)/a^3/(a^2-b^2)/d+1/3*(a^2-4*b^2)*sec(d*x+c)^2*tan(d*x+c)/a^2/(a^ 2-b^2)/d+b^2*sec(d*x+c)^2*tan(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))
Time = 6.78 (sec) , antiderivative size = 499, normalized size of antiderivative = 1.85 \[ \int \frac {\sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {2 b^4 \left (5 a^2-4 b^2\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{a^5 \left (a^2-b^2\right ) \sqrt {-a^2+b^2} d}+\frac {\left (a^2 b+4 b^3\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^5 d}+\frac {\left (-a^2 b-4 b^3\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^5 d}+\frac {a-6 b}{12 a^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{6 a^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{6 a^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {-a+6 b}{12 a^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 a^2 \sin \left (\frac {1}{2} (c+d x)\right )+9 b^2 \sin \left (\frac {1}{2} (c+d x)\right )}{3 a^4 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 a^2 \sin \left (\frac {1}{2} (c+d x)\right )+9 b^2 \sin \left (\frac {1}{2} (c+d x)\right )}{3 a^4 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {b^5 \sin (c+d x)}{a^4 (a-b) (a+b) d (a+b \cos (c+d x))} \] Input:
Integrate[Sec[c + d*x]^4/(a + b*Cos[c + d*x])^2,x]
Output:
(-2*b^4*(5*a^2 - 4*b^2)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2 ]])/(a^5*(a^2 - b^2)*Sqrt[-a^2 + b^2]*d) + ((a^2*b + 4*b^3)*Log[Cos[(c + d *x)/2] - Sin[(c + d*x)/2]])/(a^5*d) + ((-(a^2*b) - 4*b^3)*Log[Cos[(c + d*x )/2] + Sin[(c + d*x)/2]])/(a^5*d) + (a - 6*b)/(12*a^3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + Sin[(c + d*x)/2]/(6*a^2*d*(Cos[(c + d*x)/2] - Sin [(c + d*x)/2])^3) + Sin[(c + d*x)/2]/(6*a^2*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) + (-a + 6*b)/(12*a^3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^ 2) + (2*a^2*Sin[(c + d*x)/2] + 9*b^2*Sin[(c + d*x)/2])/(3*a^4*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (2*a^2*Sin[(c + d*x)/2] + 9*b^2*Sin[(c + d* x)/2])/(3*a^4*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) - (b^5*Sin[c + d*x] )/(a^4*(a - b)*(a + b)*d*(a + b*Cos[c + d*x]))
Time = 1.93 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.03, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.810, Rules used = {3042, 3281, 3042, 3534, 25, 3042, 3534, 27, 3042, 3534, 27, 3042, 3480, 3042, 3138, 218, 4257}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 3281 |
| \(\displaystyle \frac {\int \frac {\left (a^2-b \cos (c+d x) a-4 b^2+3 b^2 \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a^2-b \sin \left (c+d x+\frac {\pi }{2}\right ) a-4 b^2+3 b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {\frac {\int -\frac {\left (-2 b \left (a^2-4 b^2\right ) \cos ^2(c+d x)-a \left (2 a^2+b^2\right ) \cos (c+d x)+6 b \left (a^2-2 b^2\right )\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)}dx}{3 a}+\frac {\left (a^2-4 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}}{a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\left (a^2-4 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\int \frac {\left (-2 b \left (a^2-4 b^2\right ) \cos ^2(c+d x)-a \left (2 a^2+b^2\right ) \cos (c+d x)+6 b \left (a^2-2 b^2\right )\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)}dx}{3 a}}{a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (a^2-4 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\int \frac {-2 b \left (a^2-4 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-a \left (2 a^2+b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+6 b \left (a^2-2 b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}}{a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {\frac {\left (a^2-4 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {\int -\frac {2 \left (2 a^4+7 b^2 a^2-b \left (a^2+2 b^2\right ) \cos (c+d x) a-12 b^4-3 b^2 \left (a^2-2 b^2\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}+\frac {3 b \left (a^2-2 b^2\right ) \tan (c+d x) \sec (c+d x)}{a d}}{3 a}}{a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\left (a^2-4 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \left (a^2-2 b^2\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\int \frac {\left (2 a^4+7 b^2 a^2-b \left (a^2+2 b^2\right ) \cos (c+d x) a-12 b^4-3 b^2 \left (a^2-2 b^2\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{a}}{3 a}}{a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (a^2-4 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \left (a^2-2 b^2\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\int \frac {2 a^4+7 b^2 a^2-b \left (a^2+2 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a-12 b^4-3 b^2 \left (a^2-2 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{3 a}}{a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {\frac {\left (a^2-4 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \left (a^2-2 b^2\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\frac {\int -\frac {3 \left (a \left (a^2-2 b^2\right ) \cos (c+d x) b^2+\left (a^4+3 b^2 a^2-4 b^4\right ) b\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {\left (2 a^4+7 a^2 b^2-12 b^4\right ) \tan (c+d x)}{a d}}{a}}{3 a}}{a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\left (a^2-4 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \left (a^2-2 b^2\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\frac {\left (2 a^4+7 a^2 b^2-12 b^4\right ) \tan (c+d x)}{a d}-\frac {3 \int \frac {\left (a \left (a^2-2 b^2\right ) \cos (c+d x) b^2+\left (a^4+3 b^2 a^2-4 b^4\right ) b\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}}{a}}{3 a}}{a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (a^2-4 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \left (a^2-2 b^2\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\frac {\left (2 a^4+7 a^2 b^2-12 b^4\right ) \tan (c+d x)}{a d}-\frac {3 \int \frac {a \left (a^2-2 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b^2+\left (a^4+3 b^2 a^2-4 b^4\right ) b}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{a}}{3 a}}{a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3480 |
| \(\displaystyle \frac {\frac {\left (a^2-4 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \left (a^2-2 b^2\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\frac {\left (2 a^4+7 a^2 b^2-12 b^4\right ) \tan (c+d x)}{a d}-\frac {3 \left (\frac {b \left (a^4+3 a^2 b^2-4 b^4\right ) \int \sec (c+d x)dx}{a}-\frac {b^4 \left (5 a^2-4 b^2\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a}\right )}{a}}{a}}{3 a}}{a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (a^2-4 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \left (a^2-2 b^2\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\frac {\left (2 a^4+7 a^2 b^2-12 b^4\right ) \tan (c+d x)}{a d}-\frac {3 \left (\frac {b \left (a^4+3 a^2 b^2-4 b^4\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {b^4 \left (5 a^2-4 b^2\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}\right )}{a}}{a}}{3 a}}{a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {\left (a^2-4 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \left (a^2-2 b^2\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\frac {\left (2 a^4+7 a^2 b^2-12 b^4\right ) \tan (c+d x)}{a d}-\frac {3 \left (\frac {b \left (a^4+3 a^2 b^2-4 b^4\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b^4 \left (5 a^2-4 b^2\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}\right )}{a}}{a}}{3 a}}{a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\left (a^2-4 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \left (a^2-2 b^2\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\frac {\left (2 a^4+7 a^2 b^2-12 b^4\right ) \tan (c+d x)}{a d}-\frac {3 \left (\frac {b \left (a^4+3 a^2 b^2-4 b^4\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b^4 \left (5 a^2-4 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}}{a}}{3 a}}{a \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {b^2 \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\frac {\left (a^2-4 b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \left (a^2-2 b^2\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\frac {\left (2 a^4+7 a^2 b^2-12 b^4\right ) \tan (c+d x)}{a d}-\frac {3 \left (\frac {b \left (a^4+3 a^2 b^2-4 b^4\right ) \text {arctanh}(\sin (c+d x))}{a d}-\frac {2 b^4 \left (5 a^2-4 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}}{a}}{3 a}}{a \left (a^2-b^2\right )}\) |
Input:
Int[Sec[c + d*x]^4/(a + b*Cos[c + d*x])^2,x]
Output:
(b^2*Sec[c + d*x]^2*Tan[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])) + (((a^2 - 4*b^2)*Sec[c + d*x]^2*Tan[c + d*x])/(3*a*d) - ((3*b*(a^2 - 2*b^2 )*Sec[c + d*x]*Tan[c + d*x])/(a*d) - ((-3*((-2*b^4*(5*a^2 - 4*b^2)*ArcTan[ (Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + (b*(a^4 + 3*a^2*b^2 - 4*b^4)*ArcTanh[Sin[c + d*x]])/(a*d)))/a + ((2*a^4 + 7*a^2*b^2 - 12*b^4)*Tan[c + d*x])/(a*d))/a)/(3*a))/(a*(a^2 - b^2))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2 ))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n + 3)*Si n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[ 2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2* n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A*b - a*B)/(b*c - a*d) Int[1/(a + b*Sin[e + f*x]), x], x] + Simp[(B*c - A*d)/ (b*c - a*d) Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int [(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A *b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ [n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) | | EqQ[a, 0])))
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 2.02 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.22
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{3 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-a -2 b}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {a^{2}+a b +3 b^{2}}{a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {b \left (a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{5}}-\frac {1}{3 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a +2 b}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {a^{2}+a b +3 b^{2}}{a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {b \left (a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{5}}+\frac {2 b^{4} \left (-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (5 a^{2}-4 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{5}}}{d}\) | \(330\) |
| default | \(\frac {-\frac {1}{3 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-a -2 b}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {a^{2}+a b +3 b^{2}}{a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {b \left (a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{5}}-\frac {1}{3 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a +2 b}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {a^{2}+a b +3 b^{2}}{a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {b \left (a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{5}}+\frac {2 b^{4} \left (-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (5 a^{2}-4 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{5}}}{d}\) | \(330\) |
| risch | \(\frac {2 i \left (3 a^{3} b^{2} {\mathrm e}^{7 i \left (d x +c \right )}-6 a \,b^{4} {\mathrm e}^{7 i \left (d x +c \right )}+6 a^{4} b \,{\mathrm e}^{6 i \left (d x +c \right )}+3 a^{2} b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-12 b^{5} {\mathrm e}^{6 i \left (d x +c \right )}+21 a^{3} b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-30 a \,b^{4} {\mathrm e}^{5 i \left (d x +c \right )}+6 a^{4} b \,{\mathrm e}^{4 i \left (d x +c \right )}+21 a^{2} b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-36 b^{5} {\mathrm e}^{4 i \left (d x +c \right )}+12 a^{5} {\mathrm e}^{3 i \left (d x +c \right )}+21 a^{3} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-42 a \,b^{4} {\mathrm e}^{3 i \left (d x +c \right )}+2 a^{4} b \,{\mathrm e}^{2 i \left (d x +c \right )}+25 a^{2} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-36 b^{5} {\mathrm e}^{2 i \left (d x +c \right )}+4 a^{5} {\mathrm e}^{i \left (d x +c \right )}+11 a^{3} b^{2} {\mathrm e}^{i \left (d x +c \right )}-18 a \,b^{4} {\mathrm e}^{i \left (d x +c \right )}+2 a^{4} b +7 a^{2} b^{3}-12 b^{5}\right )}{3 a^{4} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} \left (a^{2}-b^{2}\right ) \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{3} d}-\frac {4 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{5} d}-\frac {5 b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{3}}+\frac {4 b^{6} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{5}}+\frac {5 b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{3}}-\frac {4 b^{6} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{5}}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{3} d}+\frac {4 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{5} d}\) | \(811\) |
Input:
int(sec(d*x+c)^4/(a+cos(d*x+c)*b)^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-1/3/a^2/(tan(1/2*d*x+1/2*c)+1)^3-1/2*(-a-2*b)/a^3/(tan(1/2*d*x+1/2*c )+1)^2-(a^2+a*b+3*b^2)/a^4/(tan(1/2*d*x+1/2*c)+1)-b*(a^2+4*b^2)/a^5*ln(tan (1/2*d*x+1/2*c)+1)-1/3/a^2/(tan(1/2*d*x+1/2*c)-1)^3-1/2*(a+2*b)/a^3/(tan(1 /2*d*x+1/2*c)-1)^2-(a^2+a*b+3*b^2)/a^4/(tan(1/2*d*x+1/2*c)-1)+b*(a^2+4*b^2 )/a^5*ln(tan(1/2*d*x+1/2*c)-1)+2*b^4/a^5*(-a*b/(a^2-b^2)*tan(1/2*d*x+1/2*c )/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)+(5*a^2-4*b^2)/(a-b)/ (a+b)/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1 /2))))
Time = 0.49 (sec) , antiderivative size = 1001, normalized size of antiderivative = 3.71 \[ \int \frac {\sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:
integrate(sec(d*x+c)^4/(a+b*cos(d*x+c))^2,x, algorithm="fricas")
Output:
[-1/6*(3*((5*a^2*b^5 - 4*b^7)*cos(d*x + c)^4 + (5*a^3*b^4 - 4*a*b^6)*cos(d *x + c)^3)*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d* x + c)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2* b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) + 3*((a^6*b^2 + 2*a^ 4*b^4 - 7*a^2*b^6 + 4*b^8)*cos(d*x + c)^4 + (a^7*b + 2*a^5*b^3 - 7*a^3*b^5 + 4*a*b^7)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) - 3*((a^6*b^2 + 2*a^4*b^ 4 - 7*a^2*b^6 + 4*b^8)*cos(d*x + c)^4 + (a^7*b + 2*a^5*b^3 - 7*a^3*b^5 + 4 *a*b^7)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - 2*(a^8 - 2*a^6*b^2 + a^4* b^4 + (2*a^7*b + 5*a^5*b^3 - 19*a^3*b^5 + 12*a*b^7)*cos(d*x + c)^3 + 2*(a^ 8 + a^6*b^2 - 5*a^4*b^4 + 3*a^2*b^6)*cos(d*x + c)^2 - 2*(a^7*b - 2*a^5*b^3 + a^3*b^5)*cos(d*x + c))*sin(d*x + c))/((a^9*b - 2*a^7*b^3 + a^5*b^5)*d*c os(d*x + c)^4 + (a^10 - 2*a^8*b^2 + a^6*b^4)*d*cos(d*x + c)^3), 1/6*(6*((5 *a^2*b^5 - 4*b^7)*cos(d*x + c)^4 + (5*a^3*b^4 - 4*a*b^6)*cos(d*x + c)^3)*s qrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c)) ) - 3*((a^6*b^2 + 2*a^4*b^4 - 7*a^2*b^6 + 4*b^8)*cos(d*x + c)^4 + (a^7*b + 2*a^5*b^3 - 7*a^3*b^5 + 4*a*b^7)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) + 3*((a^6*b^2 + 2*a^4*b^4 - 7*a^2*b^6 + 4*b^8)*cos(d*x + c)^4 + (a^7*b + 2*a ^5*b^3 - 7*a^3*b^5 + 4*a*b^7)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) + 2*( a^8 - 2*a^6*b^2 + a^4*b^4 + (2*a^7*b + 5*a^5*b^3 - 19*a^3*b^5 + 12*a*b^7)* cos(d*x + c)^3 + 2*(a^8 + a^6*b^2 - 5*a^4*b^4 + 3*a^2*b^6)*cos(d*x + c)...
\[ \int \frac {\sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{2}}\, dx \] Input:
integrate(sec(d*x+c)**4/(a+b*cos(d*x+c))**2,x)
Output:
Integral(sec(c + d*x)**4/(a + b*cos(c + d*x))**2, x)
Exception generated. \[ \int \frac {\sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(sec(d*x+c)^4/(a+b*cos(d*x+c))^2,x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.56 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.36 \[ \int \frac {\sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {\frac {6 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - a^{4} b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}} + \frac {6 \, {\left (5 \, a^{2} b^{4} - 4 \, b^{6}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{7} - a^{5} b^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, {\left (a^{2} b + 4 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{5}} - \frac {3 \, {\left (a^{2} b + 4 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{5}} + \frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 18 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a^{4}}}{3 \, d} \] Input:
integrate(sec(d*x+c)^4/(a+b*cos(d*x+c))^2,x, algorithm="giac")
Output:
-1/3*(6*b^5*tan(1/2*d*x + 1/2*c)/((a^6 - a^4*b^2)*(a*tan(1/2*d*x + 1/2*c)^ 2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)) + 6*(5*a^2*b^4 - 4*b^6)*(pi*floor(1 /2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/((a^7 - a^5*b^2)*sqrt(a^2 - b^2 )) + 3*(a^2*b + 4*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^5 - 3*(a^2*b + 4*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^5 + 2*(3*a^2*tan(1/2*d*x + 1/ 2*c)^5 + 3*a*b*tan(1/2*d*x + 1/2*c)^5 + 9*b^2*tan(1/2*d*x + 1/2*c)^5 - 2*a ^2*tan(1/2*d*x + 1/2*c)^3 - 18*b^2*tan(1/2*d*x + 1/2*c)^3 + 3*a^2*tan(1/2* d*x + 1/2*c) - 3*a*b*tan(1/2*d*x + 1/2*c) + 9*b^2*tan(1/2*d*x + 1/2*c))/(( tan(1/2*d*x + 1/2*c)^2 - 1)^3*a^4))/d
Time = 49.01 (sec) , antiderivative size = 3843, normalized size of antiderivative = 14.23 \[ \int \frac {\sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:
int(1/(cos(c + d*x)^4*(a + b*cos(c + d*x))^2),x)
Output:
((2*tan(c/2 + (d*x)/2)^7*(a^5 - 2*a*b^4 + 4*b^5 - 3*a^2*b^3 + a^3*b^2))/(a ^4*(a + b)*(a - b)) + (2*tan(c/2 + (d*x)/2)^3*(6*a*b^4 - 8*a^4*b + a^5 + 3 6*b^5 - 19*a^2*b^3 - 7*a^3*b^2))/(3*a^4*(a + b)*(a - b)) + (2*tan(c/2 + (d *x)/2)^5*(6*a*b^4 + 8*a^4*b + a^5 - 36*b^5 + 19*a^2*b^3 - 7*a^3*b^2))/(3*a ^4*(a + b)*(a - b)) + (2*tan(c/2 + (d*x)/2)*(a^5 - 2*a*b^4 - 4*b^5 + 3*a^2 *b^3 + a^3*b^2))/(a^4*(a + b)*(a - b)))/(d*(a + b - tan(c/2 + (d*x)/2)^8*( a - b) - tan(c/2 + (d*x)/2)^2*(2*a + 4*b) + tan(c/2 + (d*x)/2)^6*(2*a - 4* b) + 6*b*tan(c/2 + (d*x)/2)^4)) + (b*atan(((b*(a^2 + 4*b^2)*((32*tan(c/2 + (d*x)/2)*(32*b^12 - 32*a*b^11 - 48*a^2*b^10 + 48*a^3*b^9 + 2*a^4*b^8 - 2* a^5*b^7 + 7*a^6*b^6 - 12*a^7*b^5 + 7*a^8*b^4 - 2*a^9*b^3 + a^10*b^2))/(a^1 0*b + a^11 - a^8*b^3 - a^9*b^2) + (b*(a^2 + 4*b^2)*((32*(a^17*b - 4*a^10*b ^8 + 2*a^11*b^7 + 9*a^12*b^6 - 4*a^13*b^5 - 5*a^14*b^4 + a^15*b^3))/(a^14* b + a^15 - a^12*b^3 - a^13*b^2) + (32*b*tan(c/2 + (d*x)/2)*(a^2 + 4*b^2)*( 2*a^15*b - 2*a^10*b^6 + 2*a^11*b^5 + 4*a^12*b^4 - 4*a^13*b^3 - 2*a^14*b^2) )/(a^5*(a^10*b + a^11 - a^8*b^3 - a^9*b^2))))/a^5)*1i)/a^5 + (b*(a^2 + 4*b ^2)*((32*tan(c/2 + (d*x)/2)*(32*b^12 - 32*a*b^11 - 48*a^2*b^10 + 48*a^3*b^ 9 + 2*a^4*b^8 - 2*a^5*b^7 + 7*a^6*b^6 - 12*a^7*b^5 + 7*a^8*b^4 - 2*a^9*b^3 + a^10*b^2))/(a^10*b + a^11 - a^8*b^3 - a^9*b^2) - (b*(a^2 + 4*b^2)*((32* (a^17*b - 4*a^10*b^8 + 2*a^11*b^7 + 9*a^12*b^6 - 4*a^13*b^5 - 5*a^14*b^4 + a^15*b^3))/(a^14*b + a^15 - a^12*b^3 - a^13*b^2) - (32*b*tan(c/2 + (d*...
Time = 0.19 (sec) , antiderivative size = 2128, normalized size of antiderivative = 7.88 \[ \int \frac {\sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^4/(a+b*cos(d*x+c))^2,x)
Output:
(30*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*a**3*b**4 - 24*sqrt(a**2 - b**2 )*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*a*b**6 - 30*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2 )*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a**3*b**4 + 24*s qrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a*b**6 - 30*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2) *a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*sin(c + d*x)**4*a**2*b**5 + 24 *sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a** 2 - b**2))*sin(c + d*x)**4*b**7 + 60*sqrt(a**2 - b**2)*atan((tan((c + d*x) /2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a**2*b**5 - 48*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( a**2 - b**2))*sin(c + d*x)**2*b**7 - 30*sqrt(a**2 - b**2)*atan((tan((c + d *x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a**2*b**5 + 24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2)) *b**7 + 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**7*b + 6*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**5*b**3 - 21*co s(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3*b**5 + 12*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**7 - 3*cos(c + d*x)*l og(tan((c + d*x)/2) - 1)*a**7*b - 6*cos(c + d*x)*log(tan((c + d*x)/2) -...