Integrand size = 33, antiderivative size = 195 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {2 b^2 \left (3 a^2-4 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^5 \sqrt {a-b} \sqrt {a+b} d}+\frac {b \left (a^2-4 b^2\right ) \text {arctanh}(\sin (c+d x))}{a^5 d}-\frac {\left (a^2-12 b^2\right ) \tan (c+d x)}{3 a^4 d}-\frac {2 b \sec (c+d x) \tan (c+d x)}{a^3 d}+\frac {4 \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d}-\frac {\sec ^2(c+d x) \tan (c+d x)}{a d (a+b \cos (c+d x))} \] Output:
-2*b^2*(3*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)^(1/2)/(a+b)^(1/2)/d+b*(a^2-4*b^2)*arctanh(sin(d*x+c))/a^5/d-1/3*(a ^2-12*b^2)*tan(d*x+c)/a^4/d-2*b*sec(d*x+c)*tan(d*x+c)/a^3/d+4/3*sec(d*x+c) ^2*tan(d*x+c)/a^2/d-sec(d*x+c)^2*tan(d*x+c)/a/d/(a+b*cos(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(475\) vs. \(2(195)=390\).
Time = 6.67 (sec) , antiderivative size = 475, normalized size of antiderivative = 2.44 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2 b^2 \left (3 a^2-4 b^2\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{a^5 \sqrt {-a^2+b^2} d}+\frac {\left (-a^2 b+4 b^3\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^5 d}+\frac {\left (a^2 b-4 b^3\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^5 d}+\frac {a-6 b}{12 a^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{6 a^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{6 a^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {-a+6 b}{12 a^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {-a^2 \sin \left (\frac {1}{2} (c+d x)\right )+9 b^2 \sin \left (\frac {1}{2} (c+d x)\right )}{3 a^4 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {-a^2 \sin \left (\frac {1}{2} (c+d x)\right )+9 b^2 \sin \left (\frac {1}{2} (c+d x)\right )}{3 a^4 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {b^3 \sin (c+d x)}{a^4 d (a+b \cos (c+d x))} \] Input:
Integrate[((1 - Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + b*Cos[c + d*x])^2,x]
Output:
(2*b^2*(3*a^2 - 4*b^2)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2] ])/(a^5*Sqrt[-a^2 + b^2]*d) + ((-(a^2*b) + 4*b^3)*Log[Cos[(c + d*x)/2] - S in[(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) + (-(a^2*S in[(c + d*x)/2]) + 9*b^2*Sin[(c + d*x)/2])/(3*a^4*d*(Cos[(c + d*x)/2] - Si n[(c + d*x)/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^3*Sin[c + d*x])/(a^4*d*(a + b*Cos[c + d*x]))
Time = 1.84 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.38, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.515, Rules used = {3042, 3535, 3042, 3535, 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.
\(\displaystyle \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1-\sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3535 |
\(\displaystyle \frac {\int \frac {\left (4 \left (a^2-b^2\right )-3 \left (a^2-b^2\right ) \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^2(c+d x)}{a d (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {4 \left (a^2-b^2\right )-3 \left (a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^2(c+d x)}{a d (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3535 |
\(\displaystyle \frac {\frac {\int -\frac {\left (-8 b \left (a^2-b^2\right ) \cos ^2(c+d x)+a \left (a^2-b^2\right ) \cos (c+d x)+12 b \left (a^2-b^2\right )\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)}dx}{3 a}+\frac {4 \left (a^2-b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}}{a \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^2(c+d x)}{a d (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\int \frac {\left (-8 b \left (a^2-b^2\right ) \cos ^2(c+d x)+a \left (a^2-b^2\right ) \cos (c+d x)+12 b \left (a^2-b^2\right )\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)}dx}{3 a}}{a \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^2(c+d x)}{a d (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\int \frac {-8 b \left (a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a \left (a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+12 b \left (a^2-b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}}{a \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^2(c+d x)}{a d (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {\int \frac {2 \left (a^4-13 b^2 a^2-2 b \left (a^2-b^2\right ) \cos (c+d x) a+12 b^4+6 b^2 \left (a^2-b^2\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}+\frac {6 b \left (a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{a d}}{3 a}}{a \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^2(c+d x)}{a d (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {\int \frac {\left (a^4-13 b^2 a^2-2 b \left (a^2-b^2\right ) \cos (c+d x) a+12 b^4+6 b^2 \left (a^2-b^2\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {6 b \left (a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{a d}}{3 a}}{a \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^2(c+d x)}{a d (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {\int \frac {a^4-13 b^2 a^2-2 b \left (a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+12 b^4+6 b^2 \left (a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}+\frac {6 b \left (a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{a d}}{3 a}}{a \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^2(c+d x)}{a d (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {\frac {\int -\frac {3 \left (b \left (a^4-5 b^2 a^2+4 b^4\right )-2 a b^2 \left (a^2-b^2\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {\left (a^4-13 a^2 b^2+12 b^4\right ) \tan (c+d x)}{a d}}{a}+\frac {6 b \left (a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{a d}}{3 a}}{a \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^2(c+d x)}{a d (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {\frac {\left (a^4-13 a^2 b^2+12 b^4\right ) \tan (c+d x)}{a d}-\frac {3 \int \frac {\left (b \left (a^4-5 b^2 a^2+4 b^4\right )-2 a b^2 \left (a^2-b^2\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}}{a}+\frac {6 b \left (a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{a d}}{3 a}}{a \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^2(c+d x)}{a d (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {\frac {\left (a^4-13 a^2 b^2+12 b^4\right ) \tan (c+d x)}{a d}-\frac {3 \int \frac {b \left (a^4-5 b^2 a^2+4 b^4\right )-2 a b^2 \left (a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{a}+\frac {6 b \left (a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{a d}}{3 a}}{a \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^2(c+d x)}{a d (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3480 |
\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {\frac {\left (a^4-13 a^2 b^2+12 b^4\right ) \tan (c+d x)}{a d}-\frac {3 \left (\frac {b \left (a^4-5 a^2 b^2+4 b^4\right ) \int \sec (c+d x)dx}{a}-\frac {b^2 \left (3 a^4-7 a^2 b^2+4 b^4\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a}\right )}{a}}{a}+\frac {6 b \left (a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{a d}}{3 a}}{a \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^2(c+d x)}{a d (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {\frac {\left (a^4-13 a^2 b^2+12 b^4\right ) \tan (c+d x)}{a d}-\frac {3 \left (\frac {b \left (a^4-5 a^2 b^2+4 b^4\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {b^2 \left (3 a^4-7 a^2 b^2+4 b^4\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}\right )}{a}}{a}+\frac {6 b \left (a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{a d}}{3 a}}{a \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^2(c+d x)}{a d (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {\frac {\left (a^4-13 a^2 b^2+12 b^4\right ) \tan (c+d x)}{a d}-\frac {3 \left (\frac {b \left (a^4-5 a^2 b^2+4 b^4\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b^2 \left (3 a^4-7 a^2 b^2+4 b^4\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}\right )}{a}}{a}+\frac {6 b \left (a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{a d}}{3 a}}{a \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^2(c+d x)}{a d (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {\frac {\left (a^4-13 a^2 b^2+12 b^4\right ) \tan (c+d x)}{a d}-\frac {3 \left (\frac {b \left (a^4-5 a^2 b^2+4 b^4\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b^2 \left (3 a^4-7 a^2 b^2+4 b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}}{a}+\frac {6 b \left (a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{a d}}{3 a}}{a \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^2(c+d x)}{a d (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {6 b \left (a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{a d}+\frac {\frac {\left (a^4-13 a^2 b^2+12 b^4\right ) \tan (c+d x)}{a d}-\frac {3 \left (\frac {b \left (a^4-5 a^2 b^2+4 b^4\right ) \text {arctanh}(\sin (c+d x))}{a d}-\frac {2 b^2 \left (3 a^4-7 a^2 b^2+4 b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}}{a}}{3 a}}{a \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^2(c+d x)}{a d (a+b \cos (c+d x))}\) |
Input:
Int[((1 - Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + b*Cos[c + d*x])^2,x]
Output:
-((Sec[c + d*x]^2*Tan[c + d*x])/(a*d*(a + b*Cos[c + d*x]))) + ((4*(a^2 - b ^2)*Sec[c + d*x]^2*Tan[c + d*x])/(3*a*d) - ((6*b*(a^2 - b^2)*Sec[c + d*x]* Tan[c + d*x])/(a*d) + ((-3*((-2*b^2*(3*a^4 - 7*a^2*b^2 + 4*b^4)*ArcTan[(Sq rt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + (b*(a^4 - 5*a^2*b^2 + 4*b^4)*ArcTanh[Sin[c + d*x]])/(a*d)))/a + ((a^4 - 13 *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_)])/(((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[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S in[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*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d *(A*b^2 + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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 = 0.88 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.51
method | result | size |
derivativedivides | \(\frac {-\frac {1}{3 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a +2 b}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {b \left (a +3 b \right )}{a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {b \left (a^{2}-4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{5}}-\frac {2 b^{2} \left (-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}+\frac {\left (3 a^{2}-4 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{5}}-\frac {1}{3 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-2 b -a}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {b \left (a +3 b \right )}{a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {b \left (a^{2}-4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{5}}}{d}\) | \(295\) |
default | \(\frac {-\frac {1}{3 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a +2 b}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {b \left (a +3 b \right )}{a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {b \left (a^{2}-4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{5}}-\frac {2 b^{2} \left (-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}+\frac {\left (3 a^{2}-4 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{5}}-\frac {1}{3 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-2 b -a}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {b \left (a +3 b \right )}{a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {b \left (a^{2}-4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{5}}}{d}\) | \(295\) |
risch | \(\frac {2 i \left (6 a \,{\mathrm e}^{7 i \left (d x +c \right )} b^{2}+3 a^{2} {\mathrm e}^{6 i \left (d x +c \right )} b +12 \,{\mathrm e}^{6 i \left (d x +c \right )} b^{3}-6 a^{3} {\mathrm e}^{5 i \left (d x +c \right )}+30 a \,{\mathrm e}^{5 i \left (d x +c \right )} b^{2}-3 a^{2} {\mathrm e}^{4 i \left (d x +c \right )} b +36 \,{\mathrm e}^{4 i \left (d x +c \right )} b^{3}+42 a \,{\mathrm e}^{3 i \left (d x +c \right )} b^{2}-7 a^{2} {\mathrm e}^{2 i \left (d x +c \right )} b +36 \,{\mathrm e}^{2 i \left (d x +c \right )} b^{3}-2 a^{3} {\mathrm e}^{i \left (d x +c \right )}+18 a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-a^{2} b +12 b^{3}\right )}{3 a^{4} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{3} d}+\frac {4 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{5} d}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{3} d}-\frac {4 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{5} d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, d \,a^{3}}+\frac {4 b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,a^{5}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, d \,a^{3}}-\frac {4 b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,a^{5}}\) | \(629\) |
Input:
int((1-cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^2,x,method=_RETURNVERBO SE)
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-b*(a+3*b)/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^2/a^5*(-a*b*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)+(3*a^2-4*b^2)/((a+b)*(a-b))^(1/2)*arctan((a-b)*ta n(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2)))-1/3/a^2/(tan(1/2*d*x+1/2*c)+1)^3-1/ 2*(-2*b-a)/a^3/(tan(1/2*d*x+1/2*c)+1)^2-b*(a+3*b)/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))
Leaf count of result is larger than twice the leaf count of optimal. 391 vs. \(2 (182) = 364\).
Time = 0.22 (sec) , antiderivative size = 851, normalized size of antiderivative = 4.36 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:
integrate((1-cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^2,x, algorithm="f ricas")
Output:
[1/6*(3*((3*a^2*b^3 - 4*b^5)*cos(d*x + c)^4 + (3*a^3*b^2 - 4*a*b^4)*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^4*b^2 - 5*a^2 *b^4 + 4*b^6)*cos(d*x + c)^4 + (a^5*b - 5*a^3*b^3 + 4*a*b^5)*cos(d*x + c)^ 3)*log(sin(d*x + c) + 1) - 3*((a^4*b^2 - 5*a^2*b^4 + 4*b^6)*cos(d*x + c)^4 + (a^5*b - 5*a^3*b^3 + 4*a*b^5)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) + 2*(a^6 - a^4*b^2 - (a^5*b - 13*a^3*b^3 + 12*a*b^5)*cos(d*x + c)^3 - (a^6 - 7*a^4*b^2 + 6*a^2*b^4)*cos(d*x + c)^2 - 2*(a^5*b - a^3*b^3)*cos(d*x + c)) *sin(d*x + c))/((a^7*b - a^5*b^3)*d*cos(d*x + c)^4 + (a^8 - a^6*b^2)*d*cos (d*x + c)^3), -1/6*(6*((3*a^2*b^3 - 4*b^5)*cos(d*x + c)^4 + (3*a^3*b^2 - 4 *a*b^4)*cos(d*x + c)^3)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt (a^2 - b^2)*sin(d*x + c))) - 3*((a^4*b^2 - 5*a^2*b^4 + 4*b^6)*cos(d*x + c) ^4 + (a^5*b - 5*a^3*b^3 + 4*a*b^5)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) + 3*((a^4*b^2 - 5*a^2*b^4 + 4*b^6)*cos(d*x + c)^4 + (a^5*b - 5*a^3*b^3 + 4* a*b^5)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - 2*(a^6 - a^4*b^2 - (a^5*b - 13*a^3*b^3 + 12*a*b^5)*cos(d*x + c)^3 - (a^6 - 7*a^4*b^2 + 6*a^2*b^4)*co s(d*x + c)^2 - 2*(a^5*b - a^3*b^3)*cos(d*x + c))*sin(d*x + c))/((a^7*b - a ^5*b^3)*d*cos(d*x + c)^4 + (a^8 - a^6*b^2)*d*cos(d*x + c)^3)]
\[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=- \int \left (- \frac {\sec ^{4}{\left (c + d x \right )}}{a^{2} + 2 a b \cos {\left (c + d x \right )} + b^{2} \cos ^{2}{\left (c + d x \right )}}\right )\, dx - \int \frac {\cos ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}}{a^{2} + 2 a b \cos {\left (c + d x \right )} + b^{2} \cos ^{2}{\left (c + d x \right )}}\, dx \] Input:
integrate((1-cos(d*x+c)**2)*sec(d*x+c)**4/(a+b*cos(d*x+c))**2,x)
Output:
-Integral(-sec(c + d*x)**4/(a**2 + 2*a*b*cos(c + d*x) + b**2*cos(c + d*x)* *2), x) - Integral(cos(c + d*x)**2*sec(c + d*x)**4/(a**2 + 2*a*b*cos(c + d *x) + b**2*cos(c + d*x)**2), x)
Exception generated. \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((1-cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^2,x, algorithm="m axima")
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 = 0.20 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.62 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {\frac {6 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )} a^{4}} + \frac {3 \, {\left (a^{2} b - 4 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{5}} - \frac {3 \, {\left (a^{2} b - 4 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{5}} + \frac {6 \, {\left (3 \, a^{2} b^{2} - 4 \, b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{5}} - \frac {2 \, {\left (3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 18 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a^{4}}}{3 \, d} \] Input:
integrate((1-cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^2,x, algorithm="g iac")
Output:
1/3*(6*b^3*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)*a^4) + 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 + 6* (3*a^2*b^2 - 4*b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + ar ctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/ (sqrt(a^2 - b^2)*a^5) - 2*(3*a*b*tan(1/2*d*x + 1/2*c)^5 + 9*b^2*tan(1/2*d* x + 1/2*c)^5 + 4*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*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 = 1.76 (sec) , antiderivative size = 1650, normalized size of antiderivative = 8.46 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:
int(-(cos(c + d*x)^2 - 1)/(cos(c + d*x)^4*(a + b*cos(c + d*x))^2),x)
Output:
((2*tan(c/2 + (d*x)/2)*(2*a*b^2 - a^2*b + 4*b^3))/a^4 + (2*tan(c/2 + (d*x) /2)^7*(2*a*b^2 + a^2*b - 4*b^3))/a^4 - (2*tan(c/2 + (d*x)/2)^3*(6*a*b^2 - a^2*b - 4*a^3 + 36*b^3))/(3*a^4) - (2*tan(c/2 + (d*x)/2)^5*(6*a*b^2 + a^2* b - 4*a^3 - 36*b^3))/(3*a^4))/(d*(a + b - tan(c/2 + (d*x)/2)^8*(a - b) - t an(c/2 + (d*x)/2)^2*(2*a + 4*b) + tan(c/2 + (d*x)/2)^6*(2*a - 4*b) + 6*b*t an(c/2 + (d*x)/2)^4)) + (2*b*atanh((256*b^5*tan(c/2 + (d*x)/2))/(64*a*b^4 + 256*b^5 - 64*a^2*b^3 - (256*b^6)/a) - (64*b^4*tan(c/2 + (d*x)/2))/(64*a* b^3 - 64*b^4 - (256*b^5)/a + (256*b^6)/a^2) - (256*b^6*tan(c/2 + (d*x)/2)) /(256*a*b^5 - 256*b^6 + 64*a^2*b^4 - 64*a^3*b^3) + (64*b^3*tan(c/2 + (d*x) /2))/(64*b^3 - (64*b^4)/a - (256*b^5)/a^2 + (256*b^6)/a^3))*(a^2 - 4*b^2)) /(a^5*d) + (b^2*atan(((b^2*(-(a + b)*(a - b))^(1/2)*(3*a^2 - 4*b^2)*((32*t an(c/2 + (d*x)/2)*(64*a*b^8 - 32*b^9 - 16*a^2*b^7 - 32*a^3*b^6 + 14*a^4*b^ 5 + 4*a^5*b^4 - 3*a^6*b^3 + a^7*b^2))/a^8 + (b^2*(-(a + b)*(a - b))^(1/2)* (3*a^2 - 4*b^2)*((32*(a^14*b - 4*a^10*b^5 + 6*a^11*b^4 + a^12*b^3 - 4*a^13 *b^2))/a^12 + (32*b^2*tan(c/2 + (d*x)/2)*(-(a + b)*(a - b))^(1/2)*(3*a^2 - 4*b^2)*(2*a^12*b + 2*a^10*b^3 - 4*a^11*b^2))/(a^8*(a^7 - a^5*b^2))))/(a^7 - a^5*b^2))*1i)/(a^7 - a^5*b^2) + (b^2*(-(a + b)*(a - b))^(1/2)*(3*a^2 - 4*b^2)*((32*tan(c/2 + (d*x)/2)*(64*a*b^8 - 32*b^9 - 16*a^2*b^7 - 32*a^3*b^ 6 + 14*a^4*b^5 + 4*a^5*b^4 - 3*a^6*b^3 + a^7*b^2))/a^8 - (b^2*(-(a + b)*(a - b))^(1/2)*(3*a^2 - 4*b^2)*((32*(a^14*b - 4*a^10*b^5 + 6*a^11*b^4 + a...
Time = 0.19 (sec) , antiderivative size = 1726, normalized size of antiderivative = 8.85 \[ \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:
int((1-cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^2,x)
Output:
( - 18*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sq rt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*a**3*b**2 + 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**4 + 18*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**2 - 2 4*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*b**4 + 18*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**3 - 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**5 - 36*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** 3 + 48*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sq rt(a**2 - b**2))*sin(c + d*x)**2*b**5 + 18*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**3 - 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**5 - 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**5*b + 15*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3*b**3 - 1 2*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**5 + 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**5*b - 15*cos(c + d*x)*log(tan((c + d*x )/2) - 1)*a**3*b**3 + 12*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a*b**5 ...