Optimal. Leaf size=182 \[ \frac {b \left (a^2 (4 A+6 C)+A b^2\right ) \tan (c+d x)}{2 d}+\frac {a \left (a^2 (3 A+4 C)+12 b^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \left (a^2 (3 A+4 C)+2 A b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^3}{4 d}+\frac {A b \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{4 d}+b^3 C x \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.60, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3048, 3047, 3031, 3021, 2735, 3770} \[ \frac {b \left (a^2 (4 A+6 C)+A b^2\right ) \tan (c+d x)}{2 d}+\frac {a \left (a^2 (3 A+4 C)+12 b^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \left (a^2 (3 A+4 C)+2 A b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {A b \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{4 d}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^3}{4 d}+b^3 C x \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 2735
Rule 3021
Rule 3031
Rule 3047
Rule 3048
Rule 3770
Rubi steps
\begin {align*} \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx &=\frac {A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \int (a+b \cos (c+d x))^2 \left (3 A b+a (3 A+4 C) \cos (c+d x)+4 b C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {A b (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{4 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{12} \int (a+b \cos (c+d x)) \left (3 \left (2 A b^2+a^2 (3 A+4 C)\right )+3 a b (5 A+8 C) \cos (c+d x)+12 b^2 C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {a \left (2 A b^2+a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {A b (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{4 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {1}{24} \int \left (-12 b \left (A b^2+a^2 (4 A+6 C)\right )-3 a \left (12 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) \cos (c+d x)-24 b^3 C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {b \left (A b^2+a^2 (4 A+6 C)\right ) \tan (c+d x)}{2 d}+\frac {a \left (2 A b^2+a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {A b (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{4 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {1}{24} \int \left (-3 a \left (12 b^2 (A+2 C)+a^2 (3 A+4 C)\right )-24 b^3 C \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=b^3 C x+\frac {b \left (A b^2+a^2 (4 A+6 C)\right ) \tan (c+d x)}{2 d}+\frac {a \left (2 A b^2+a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {A b (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{4 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{8} \left (a \left (12 b^2 (A+2 C)+a^2 (3 A+4 C)\right )\right ) \int \sec (c+d x) \, dx\\ &=b^3 C x+\frac {a \left (12 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b \left (A b^2+a^2 (4 A+6 C)\right ) \tan (c+d x)}{2 d}+\frac {a \left (2 A b^2+a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {A b (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{4 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.92, size = 127, normalized size = 0.70 \[ \frac {a \left (a^2 (3 A+4 C)+12 b^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))+8 a^2 A b \tan ^3(c+d x)+\tan (c+d x) \left (2 a^3 A \sec ^3(c+d x)+a \left (a^2 (3 A+4 C)+12 A b^2\right ) \sec (c+d x)+8 b \left (3 a^2 (A+C)+A b^2\right )\right )+8 b^3 C d x}{8 d} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.83, size = 199, normalized size = 1.09 \[ \frac {16 \, C b^{3} d x \cos \left (d x + c\right )^{4} + {\left ({\left (3 \, A + 4 \, C\right )} a^{3} + 12 \, {\left (A + 2 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (3 \, A + 4 \, C\right )} a^{3} + 12 \, {\left (A + 2 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, A a^{2} b \cos \left (d x + c\right ) + 2 \, A a^{3} + 8 \, {\left ({\left (2 \, A + 3 \, C\right )} a^{2} b + A b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left ({\left (3 \, A + 4 \, C\right )} a^{3} + 12 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [B] time = 1.05, size = 526, normalized size = 2.89 \[ \frac {8 \, {\left (d x + c\right )} C b^{3} + {\left (3 \, A a^{3} + 4 \, C a^{3} + 12 \, A a b^{2} + 24 \, C a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (3 \, A a^{3} + 4 \, C a^{3} + 12 \, A a b^{2} + 24 \, C a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (5 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 4 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 8 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 72 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, A b^{3} \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 )}^{4}}}{8 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.39, size = 267, normalized size = 1.47 \[ \frac {A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 A \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {C \,a^{3} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {2 A \,a^{2} b \tan \left (d x +c \right )}{d}+\frac {A \,a^{2} b \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {3 C \,a^{2} b \tan \left (d x +c \right )}{d}+\frac {3 A a \,b^{2} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {3 A a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {3 C a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {A \,b^{3} \tan \left (d x +c \right )}{d}+b^{3} C x +\frac {b^{3} C c}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [A] time = 0.53, size = 261, normalized size = 1.43 \[ \frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} b + 16 \, {\left (d x + c\right )} C b^{3} - A a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, A a b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C a b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, C a^{2} b \tan \left (d x + c\right ) + 16 \, A b^{3} \tan \left (d x + c\right )}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 3.51, size = 1547, normalized size = 8.50 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________