Optimal. Leaf size=103 \[ \frac {\left (a^2 (A+2 C)+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a A b \tan (c+d x)}{d}+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}+2 a b C x-\frac {b^2 (A-2 C) \sin (c+d x)}{2 d} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.31, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3048, 3031, 3023, 2735, 3770} \[ \frac {\left (a^2 (A+2 C)+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a A b \tan (c+d x)}{d}+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}+2 a b C x-\frac {b^2 (A-2 C) \sin (c+d x)}{2 d} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 2735
Rule 3023
Rule 3031
Rule 3048
Rule 3770
Rubi steps
\begin {align*} \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx &=\frac {A (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int (a+b \cos (c+d x)) \left (2 A b+a (A+2 C) \cos (c+d x)-b (A-2 C) \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {a A b \tan (c+d x)}{d}+\frac {A (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}-\frac {1}{2} \int \left (-2 A b^2-a^2 (A+2 C)-4 a b C \cos (c+d x)+b^2 (A-2 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {b^2 (A-2 C) \sin (c+d x)}{2 d}+\frac {a A b \tan (c+d x)}{d}+\frac {A (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}-\frac {1}{2} \int \left (-2 A b^2-a^2 (A+2 C)-4 a b C \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=2 a b C x-\frac {b^2 (A-2 C) \sin (c+d x)}{2 d}+\frac {a A b \tan (c+d x)}{d}+\frac {A (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}-\frac {1}{2} \left (-2 A b^2-a^2 (A+2 C)\right ) \int \sec (c+d x) \, dx\\ &=2 a b C x+\frac {\left (2 A b^2+a^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {b^2 (A-2 C) \sin (c+d x)}{2 d}+\frac {a A b \tan (c+d x)}{d}+\frac {A (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}
________________________________________________________________________________________
Mathematica [B] time = 1.29, size = 249, normalized size = 2.42 \[ \frac {-2 \left (a^2 (A+2 C)+2 A b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \left (a^2 (A+2 C)+2 A b^2\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a^2 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {a^2 A}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {8 a A b \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {8 a A b \sin \left (\frac {1}{2} (c+d x)\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}+8 a b C (c+d x)+4 b^2 C \sin (c+d x)}{4 d} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.94, size = 139, normalized size = 1.35 \[ \frac {8 \, C a b d x \cos \left (d x + c\right )^{2} + {\left ({\left (A + 2 \, C\right )} a^{2} + 2 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (A + 2 \, C\right )} a^{2} + 2 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, C b^{2} \cos \left (d x + c\right )^{2} + 4 \, A a b \cos \left (d x + c\right ) + A a^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 0.49, size = 189, normalized size = 1.83 \[ \frac {4 \, {\left (d x + c\right )} C a b + \frac {4 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + {\left (A a^{2} + 2 \, C a^{2} + 2 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (A a^{2} + 2 \, C a^{2} + 2 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.27, size = 133, normalized size = 1.29 \[ \frac {a^{2} A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {2 a A b \tan \left (d x +c \right )}{d}+2 a b C x +\frac {2 C a b c}{d}+\frac {A \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{2} C \sin \left (d x +c \right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [A] time = 0.39, size = 140, normalized size = 1.36 \[ \frac {8 \, {\left (d x + c\right )} C a b - A a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, C b^{2} \sin \left (d x + c\right ) + 8 \, A a b \tan \left (d x + c\right )}{4 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 1.93, size = 188, normalized size = 1.83 \[ \frac {2\,\left (\frac {A\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+A\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+C\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+2\,C\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\right )}{d}+\frac {\frac {C\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{2}+\frac {C\,b^2\,\sin \left (c+d\,x\right )}{4}+A\,a\,b\,\sin \left (2\,c+2\,d\,x\right )}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )} \]
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]
________________________________________________________________________________________