Integrand size = 31, antiderivative size = 270 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {2 b^2 \left (4 a^2 A b-3 A b^3-3 a^3 B+2 a b^2 B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {\left (a^2 A+6 A b^2-4 a b B\right ) \text {arctanh}(\sin (c+d x))}{2 a^4 d}-\frac {\left (2 a^2 A b-3 A b^3-a^3 B+2 a b^2 B\right ) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2 A-3 A b^2+2 a b B\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac {b (A b-a B) \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \]
-2*b^2*(4*A*a^2*b-3*A*b^3-3*B*a^3+2*B*a*b^2)*arctan((a-b)^(1/2)*tan(1/2*d* x+1/2*c)/(a+b)^(1/2))/a^4/(a-b)^(3/2)/(a+b)^(3/2)/d+1/2*(A*a^2+6*A*b^2-4*B *a*b)*arctanh(sin(d*x+c))/a^4/d-(2*A*a^2*b-3*A*b^3-B*a^3+2*B*a*b^2)*tan(d* x+c)/a^3/(a^2-b^2)/d+1/2*(A*a^2-3*A*b^2+2*B*a*b)*sec(d*x+c)*tan(d*x+c)/a^2 /(a^2-b^2)/d+b*(A*b-B*a)*sec(d*x+c)*tan(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+ c))
Time = 6.90 (sec) , antiderivative size = 438, normalized size of antiderivative = 1.62 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {2 b^2 \left (-4 a^2 A b+3 A b^3+3 a^3 B-2 a b^2 B\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{a^4 \left (a^2-b^2\right ) \sqrt {-a^2+b^2} d}+\frac {\left (-a^2 A-6 A b^2+4 a b B\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 a^4 d}+\frac {\left (a^2 A+6 A b^2-4 a b B\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 a^4 d}+\frac {A}{4 a^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {A}{4 a^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {-2 A b \sin \left (\frac {1}{2} (c+d x)\right )+a B \sin \left (\frac {1}{2} (c+d x)\right )}{a^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {-2 A b \sin \left (\frac {1}{2} (c+d x)\right )+a B \sin \left (\frac {1}{2} (c+d x)\right )}{a^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {A b^4 \sin (c+d x)-a b^3 B \sin (c+d x)}{a^3 (a-b) (a+b) d (a+b \cos (c+d x))} \]
(-2*b^2*(-4*a^2*A*b + 3*A*b^3 + 3*a^3*B - 2*a*b^2*B)*ArcTanh[((a - b)*Tan[ (c + d*x)/2])/Sqrt[-a^2 + b^2]])/(a^4*(a^2 - b^2)*Sqrt[-a^2 + b^2]*d) + (( -(a^2*A) - 6*A*b^2 + 4*a*b*B)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/(2 *a^4*d) + ((a^2*A + 6*A*b^2 - 4*a*b*B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x )/2]])/(2*a^4*d) + A/(4*a^2*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) - A /(4*a^2*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (-2*A*b*Sin[(c + d*x) /2] + a*B*Sin[(c + d*x)/2])/(a^3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (-2*A*b*Sin[(c + d*x)/2] + a*B*Sin[(c + d*x)/2])/(a^3*d*(Cos[(c + d*x)/2 ] + Sin[(c + d*x)/2])) + (A*b^4*Sin[c + d*x] - a*b^3*B*Sin[c + d*x])/(a^3* (a - b)*(a + b)*d*(a + b*Cos[c + d*x]))
Time = 1.74 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.05, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {3042, 3479, 3042, 3534, 25, 3042, 3534, 25, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 3479 |
| \(\displaystyle \frac {\int \frac {\left (A a^2+2 b B a-(A b-a B) \cos (c+d x) a-3 A b^2+2 b (A b-a B) \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {A a^2+2 b B a-(A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right ) a-3 A b^2+2 b (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {\frac {\int -\frac {\left (-b \left (A a^2+2 b B a-3 A b^2\right ) \cos ^2(c+d x)-a \left (A a^2-2 b B a+A b^2\right ) \cos (c+d x)+2 \left (-B a^3+2 A b a^2+2 b^2 B a-3 A b^3\right )\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}+\frac {\left (a^2 A+2 a b B-3 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 a d}}{a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\left (a^2 A+2 a b B-3 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\int \frac {\left (-b \left (A a^2+2 b B a-3 A b^2\right ) \cos ^2(c+d x)-a \left (A a^2-2 b B a+A b^2\right ) \cos (c+d x)+2 \left (-B a^3+2 A b a^2+2 b^2 B a-3 A b^3\right )\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}}{a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (a^2 A+2 a b B-3 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\int \frac {-b \left (A a^2+2 b B a-3 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-a \left (A a^2-2 b B a+A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+2 \left (-B a^3+2 A b a^2+2 b^2 B a-3 A b^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a}}{a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {\frac {\left (a^2 A+2 a b B-3 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {\int -\frac {\left (\left (a^2-b^2\right ) \left (A a^2-4 b B a+6 A b^2\right )+a b \left (A a^2+2 b B a-3 A b^2\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {2 \left (a^3 (-B)+2 a^2 A b+2 a b^2 B-3 A b^3\right ) \tan (c+d x)}{a d}}{2 a}}{a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\left (a^2 A+2 a b B-3 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \left (a^3 (-B)+2 a^2 A b+2 a b^2 B-3 A b^3\right ) \tan (c+d x)}{a d}-\frac {\int \frac {\left (\left (a^2-b^2\right ) \left (A a^2-4 b B a+6 A b^2\right )+a b \left (A a^2+2 b B a-3 A b^2\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}}{2 a}}{a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (a^2 A+2 a b B-3 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \left (a^3 (-B)+2 a^2 A b+2 a b^2 B-3 A b^3\right ) \tan (c+d x)}{a d}-\frac {\int \frac {\left (a^2-b^2\right ) \left (A a^2-4 b B a+6 A b^2\right )+a b \left (A a^2+2 b B a-3 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{2 a}}{a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3480 |
| \(\displaystyle \frac {\frac {\left (a^2 A+2 a b B-3 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \left (a^3 (-B)+2 a^2 A b+2 a b^2 B-3 A b^3\right ) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2-b^2\right ) \left (a^2 A-4 a b B+6 A b^2\right ) \int \sec (c+d x)dx}{a}-\frac {2 b^2 \left (-3 a^3 B+4 a^2 A b+2 a b^2 B-3 A b^3\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a}}{a}}{2 a}}{a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (a^2 A+2 a b B-3 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \left (a^3 (-B)+2 a^2 A b+2 a b^2 B-3 A b^3\right ) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2-b^2\right ) \left (a^2 A-4 a b B+6 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b^2 \left (-3 a^3 B+4 a^2 A b+2 a b^2 B-3 A b^3\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a}}{2 a}}{a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {\left (a^2 A+2 a b B-3 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \left (a^3 (-B)+2 a^2 A b+2 a b^2 B-3 A b^3\right ) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2-b^2\right ) \left (a^2 A-4 a b B+6 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {4 b^2 \left (-3 a^3 B+4 a^2 A b+2 a b^2 B-3 A b^3\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}}{a}}{2 a}}{a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\left (a^2 A+2 a b B-3 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \left (a^3 (-B)+2 a^2 A b+2 a b^2 B-3 A b^3\right ) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2-b^2\right ) \left (a^2 A-4 a b B+6 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {4 b^2 \left (-3 a^3 B+4 a^2 A b+2 a b^2 B-3 A b^3\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{2 a}}{a \left (a^2-b^2\right )}+\frac {b (A b-a B) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {b (A b-a B) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\frac {\left (a^2 A+2 a b B-3 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \left (a^3 (-B)+2 a^2 A b+2 a b^2 B-3 A b^3\right ) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2-b^2\right ) \left (a^2 A-4 a b B+6 A b^2\right ) \text {arctanh}(\sin (c+d x))}{a d}-\frac {4 b^2 \left (-3 a^3 B+4 a^2 A b+2 a b^2 B-3 A b^3\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{2 a}}{a \left (a^2-b^2\right )}\) |
(b*(A*b - a*B)*Sec[c + d*x]*Tan[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])) + (((a^2*A - 3*A*b^2 + 2*a*b*B)*Sec[c + d*x]*Tan[c + d*x])/(2*a*d) - (-(((-4*b^2*(4*a^2*A*b - 3*A*b^3 - 3*a^3*B + 2*a*b^2*B)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + ((a^2 - b^2)*(a^2*A + 6*A*b^2 - 4*a*b*B)*ArcTanh[Sin[c + d*x]])/(a*d))/a) + (2*( 2*a^2*A*b - 3*A*b^3 - a^3*B + 2*a*b^2*B)*Tan[c + d*x])/(a*d))/(2*a))/(a*(a ^2 - b^2))
3.3.64.3.1 Defintions of rubi rules used
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_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(-(A*b^2 - a*b*B))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin [e + f*x])^(1 + n)/(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*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m + n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B) *(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n }, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Rat ionalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(I ntegerQ[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 = 2.06 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.21
| method | result | size |
| derivativedivides | \(\frac {-\frac {A}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-a A -4 A b +2 B a}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (A \,a^{2}+6 A \,b^{2}-4 B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{4}}+\frac {A}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-a A -4 A b +2 B a}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-A \,a^{2}-6 A \,b^{2}+4 B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{4}}-\frac {2 b^{2} \left (-\frac {b a \left (A b -B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}+\frac {\left (4 A \,a^{2} b -3 A \,b^{3}-3 B \,a^{3}+2 B a \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4}}}{d}\) | \(326\) |
| default | \(\frac {-\frac {A}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-a A -4 A b +2 B a}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (A \,a^{2}+6 A \,b^{2}-4 B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{4}}+\frac {A}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-a A -4 A b +2 B a}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-A \,a^{2}-6 A \,b^{2}+4 B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{4}}-\frac {2 b^{2} \left (-\frac {b a \left (A b -B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}+\frac {\left (4 A \,a^{2} b -3 A \,b^{3}-3 B \,a^{3}+2 B a \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4}}}{d}\) | \(326\) |
| risch | \(\text {Expression too large to display}\) | \(1276\) |
1/d*(-1/2*A/a^2/(tan(1/2*d*x+1/2*c)+1)^2-1/2*(-A*a-4*A*b+2*B*a)/a^3/(tan(1 /2*d*x+1/2*c)+1)+1/2*(A*a^2+6*A*b^2-4*B*a*b)/a^4*ln(tan(1/2*d*x+1/2*c)+1)+ 1/2*A/a^2/(tan(1/2*d*x+1/2*c)-1)^2-1/2*(-A*a-4*A*b+2*B*a)/a^3/(tan(1/2*d*x +1/2*c)-1)+1/2/a^4*(-A*a^2-6*A*b^2+4*B*a*b)*ln(tan(1/2*d*x+1/2*c)-1)-2*b^2 /a^4*(-b*a*(A*b-B*a)/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a- b*tan(1/2*d*x+1/2*c)^2+a+b)+(4*A*a^2*b-3*A*b^3-3*B*a^3+2*B*a*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 ))))
Leaf count of result is larger than twice the leaf count of optimal. 630 vs. \(2 (256) = 512\).
Time = 12.98 (sec) , antiderivative size = 1329, normalized size of antiderivative = 4.92 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \]
[-1/4*(2*((3*B*a^3*b^3 - 4*A*a^2*b^4 - 2*B*a*b^5 + 3*A*b^6)*cos(d*x + c)^3 + (3*B*a^4*b^2 - 4*A*a^3*b^3 - 2*B*a^2*b^4 + 3*A*a*b^5)*cos(d*x + c)^2)*s qrt(-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*co s(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) - ((A*a^6*b - 4*B*a^5*b^2 + 4*A* a^4*b^3 + 8*B*a^3*b^4 - 11*A*a^2*b^5 - 4*B*a*b^6 + 6*A*b^7)*cos(d*x + c)^3 + (A*a^7 - 4*B*a^6*b + 4*A*a^5*b^2 + 8*B*a^4*b^3 - 11*A*a^3*b^4 - 4*B*a^2 *b^5 + 6*A*a*b^6)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + ((A*a^6*b - 4*B* a^5*b^2 + 4*A*a^4*b^3 + 8*B*a^3*b^4 - 11*A*a^2*b^5 - 4*B*a*b^6 + 6*A*b^7)* cos(d*x + c)^3 + (A*a^7 - 4*B*a^6*b + 4*A*a^5*b^2 + 8*B*a^4*b^3 - 11*A*a^3 *b^4 - 4*B*a^2*b^5 + 6*A*a*b^6)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2 *(A*a^7 - 2*A*a^5*b^2 + A*a^3*b^4 + 2*(B*a^6*b - 2*A*a^5*b^2 - 3*B*a^4*b^3 + 5*A*a^3*b^4 + 2*B*a^2*b^5 - 3*A*a*b^6)*cos(d*x + c)^2 + (2*B*a^7 - 3*A* a^6*b - 4*B*a^5*b^2 + 6*A*a^4*b^3 + 2*B*a^3*b^4 - 3*A*a^2*b^5)*cos(d*x + c ))*sin(d*x + c))/((a^8*b - 2*a^6*b^3 + a^4*b^5)*d*cos(d*x + c)^3 + (a^9 - 2*a^7*b^2 + a^5*b^4)*d*cos(d*x + c)^2), 1/4*(4*((3*B*a^3*b^3 - 4*A*a^2*b^4 - 2*B*a*b^5 + 3*A*b^6)*cos(d*x + c)^3 + (3*B*a^4*b^2 - 4*A*a^3*b^3 - 2*B* a^2*b^4 + 3*A*a*b^5)*cos(d*x + c)^2)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) + ((A*a^6*b - 4*B*a^5*b^2 + 4*A*a^ 4*b^3 + 8*B*a^3*b^4 - 11*A*a^2*b^5 - 4*B*a*b^6 + 6*A*b^7)*cos(d*x + c)^...
\[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]
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 = 0.32 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.40 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {\frac {4 \, {\left (3 \, B a^{3} b^{2} - 4 \, A a^{2} b^{3} - 2 \, B a b^{4} + 3 \, A b^{5}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - a^{4} b^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {4 \, {\left (B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{5} - a^{3} b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}} - \frac {{\left (A a^{2} - 4 \, B a b + 6 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} + \frac {{\left (A a^{2} - 4 \, B a b + 6 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac {2 \, {\left (A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}}}{2 \, d} \]
-1/2*(4*(3*B*a^3*b^2 - 4*A*a^2*b^3 - 2*B*a*b^4 + 3*A*b^5)*(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*ta n(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/((a^6 - a^4*b^2)*sqrt(a^2 - b^2)) + 4*(B*a*b^3*tan(1/2*d*x + 1/2*c) - A*b^4*tan(1/2*d*x + 1/2*c))/((a^5 - a^3* b^2)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)) - (A*a ^2 - 4*B*a*b + 6*A*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 + (A*a^2 - 4*B*a*b + 6*A*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 - 2*(A*a*tan(1/2 *d*x + 1/2*c)^3 - 2*B*a*tan(1/2*d*x + 1/2*c)^3 + 4*A*b*tan(1/2*d*x + 1/2*c )^3 + A*a*tan(1/2*d*x + 1/2*c) + 2*B*a*tan(1/2*d*x + 1/2*c) - 4*A*b*tan(1/ 2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3))/d
Time = 9.65 (sec) , antiderivative size = 6692, normalized size of antiderivative = 24.79 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \]
(atan(-((((8*tan(c/2 + (d*x)/2)*(A^2*a^10 + 72*A^2*b^10 - 72*A^2*a*b^9 - 2 *A^2*a^9*b - 120*A^2*a^2*b^8 + 120*A^2*a^3*b^7 + 17*A^2*a^4*b^6 - 26*A^2*a ^5*b^5 + 23*A^2*a^6*b^4 - 20*A^2*a^7*b^3 + 11*A^2*a^8*b^2 + 32*B^2*a^2*b^8 - 32*B^2*a^3*b^7 - 64*B^2*a^4*b^6 + 64*B^2*a^5*b^5 + 20*B^2*a^6*b^4 - 32* B^2*a^7*b^3 + 16*B^2*a^8*b^2 - 96*A*B*a*b^9 - 8*A*B*a^9*b + 96*A*B*a^2*b^8 + 176*A*B*a^3*b^7 - 176*A*B*a^4*b^6 - 40*A*B*a^5*b^5 + 64*A*B*a^6*b^4 - 4 0*A*B*a^7*b^3 + 16*A*B*a^8*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (((8* (2*A*a^15 - 12*A*a^8*b^7 + 6*A*a^9*b^6 + 28*A*a^10*b^5 - 14*A*a^11*b^4 - 1 6*A*a^12*b^3 + 6*A*a^13*b^2 + 8*B*a^9*b^6 - 4*B*a^10*b^5 - 20*B*a^11*b^4 + 12*B*a^12*b^3 + 12*B*a^13*b^2 - 8*B*a^14*b))/(a^11*b + a^12 - a^9*b^3 - a ^10*b^2) - (4*tan(c/2 + (d*x)/2)*(A*a^2 + 6*A*b^2 - 4*B*a*b)*(8*a^13*b - 8 *a^8*b^6 + 8*a^9*b^5 + 16*a^10*b^4 - 16*a^11*b^3 - 8*a^12*b^2))/(a^4*(a^8* b + a^9 - a^6*b^3 - a^7*b^2)))*(A*a^2 + 6*A*b^2 - 4*B*a*b))/(2*a^4))*(A*a^ 2 + 6*A*b^2 - 4*B*a*b)*1i)/(2*a^4) + (((8*tan(c/2 + (d*x)/2)*(A^2*a^10 + 7 2*A^2*b^10 - 72*A^2*a*b^9 - 2*A^2*a^9*b - 120*A^2*a^2*b^8 + 120*A^2*a^3*b^ 7 + 17*A^2*a^4*b^6 - 26*A^2*a^5*b^5 + 23*A^2*a^6*b^4 - 20*A^2*a^7*b^3 + 11 *A^2*a^8*b^2 + 32*B^2*a^2*b^8 - 32*B^2*a^3*b^7 - 64*B^2*a^4*b^6 + 64*B^2*a ^5*b^5 + 20*B^2*a^6*b^4 - 32*B^2*a^7*b^3 + 16*B^2*a^8*b^2 - 96*A*B*a*b^9 - 8*A*B*a^9*b + 96*A*B*a^2*b^8 + 176*A*B*a^3*b^7 - 176*A*B*a^4*b^6 - 40*A*B *a^5*b^5 + 64*A*B*a^6*b^4 - 40*A*B*a^7*b^3 + 16*A*B*a^8*b^2))/(a^8*b + ...