Optimal. Leaf size=104 \[ \frac {a^2 (3 A-2 B) \tan ^3(c+d x)}{15 d}+\frac {a^2 (3 A-2 B) \tan (c+d x)}{5 d}+\frac {a^2 (3 A-2 B) \sec ^3(c+d x)}{15 d}+\frac {(A+B) \sec ^5(c+d x) (a \sin (c+d x)+a)^2}{5 d} \]
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Rubi [A] time = 0.12, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2855, 2669, 3767} \[ \frac {a^2 (3 A-2 B) \tan ^3(c+d x)}{15 d}+\frac {a^2 (3 A-2 B) \tan (c+d x)}{5 d}+\frac {a^2 (3 A-2 B) \sec ^3(c+d x)}{15 d}+\frac {(A+B) \sec ^5(c+d x) (a \sin (c+d x)+a)^2}{5 d} \]
Antiderivative was successfully verified.
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Rule 2669
Rule 2855
Rule 3767
Rubi steps
\begin {align*} \int \sec ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx &=\frac {(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))^2}{5 d}+\frac {1}{5} (a (3 A-2 B)) \int \sec ^4(c+d x) (a+a \sin (c+d x)) \, dx\\ &=\frac {a^2 (3 A-2 B) \sec ^3(c+d x)}{15 d}+\frac {(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))^2}{5 d}+\frac {1}{5} \left (a^2 (3 A-2 B)\right ) \int \sec ^4(c+d x) \, dx\\ &=\frac {a^2 (3 A-2 B) \sec ^3(c+d x)}{15 d}+\frac {(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))^2}{5 d}-\frac {\left (a^2 (3 A-2 B)\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac {a^2 (3 A-2 B) \sec ^3(c+d x)}{15 d}+\frac {(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))^2}{5 d}+\frac {a^2 (3 A-2 B) \tan (c+d x)}{5 d}+\frac {a^2 (3 A-2 B) \tan ^3(c+d x)}{15 d}\\ \end {align*}
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Mathematica [A] time = 0.02, size = 178, normalized size = 1.71 \[ \frac {2 a^2 A \tan ^5(c+d x)}{5 d}+\frac {2 a^2 A \sec ^5(c+d x)}{5 d}+\frac {a^2 A \tan (c+d x) \sec ^4(c+d x)}{d}-\frac {a^2 A \tan ^3(c+d x) \sec ^2(c+d x)}{d}-\frac {4 a^2 B \tan ^5(c+d x)}{15 d}+\frac {a^2 B \sec ^5(c+d x)}{15 d}+\frac {a^2 B \tan ^2(c+d x) \sec ^3(c+d x)}{3 d}+\frac {2 a^2 B \tan ^3(c+d x) \sec ^2(c+d x)}{3 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.68, size = 113, normalized size = 1.09 \[ -\frac {4 \, {\left (3 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} - 3 \, {\left (2 \, A - 3 \, B\right )} a^{2} - {\left (2 \, {\left (3 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} - 3 \, {\left (3 \, A - 2 \, B\right )} a^{2}\right )} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + 2 \, d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.21, size = 192, normalized size = 1.85 \[ -\frac {\frac {15 \, {\left (A a^{2} - B a^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1} + \frac {105 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 15 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 270 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 360 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 40 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 210 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 50 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 63 \, A a^{2} - 7 \, B a^{2}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{5}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.64, size = 231, normalized size = 2.22 \[ \frac {a^{2} A \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )+B \,a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{15}\right )+\frac {2 a^{2} A}{5 \cos \left (d x +c \right )^{5}}+2 B \,a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )-a^{2} A \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+\frac {B \,a^{2}}{5 \cos \left (d x +c \right )^{5}}}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.48, size = 147, normalized size = 1.41 \[ \frac {{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{2} + {\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} A a^{2} + 2 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} B a^{2} - \frac {{\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} B a^{2}}{\cos \left (d x + c\right )^{5}} + \frac {6 \, A a^{2}}{\cos \left (d x + c\right )^{5}} + \frac {3 \, B a^{2}}{\cos \left (d x + c\right )^{5}}}{15 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 9.36, size = 175, normalized size = 1.68 \[ \frac {2\,a^2\,\left (\frac {5\,B\,\sin \left (c+d\,x\right )}{2}-\frac {15\,A\,\cos \left (c+d\,x\right )}{4}-\frac {5\,B\,\cos \left (c+d\,x\right )}{8}-\frac {15\,A\,\sin \left (c+d\,x\right )}{4}-\frac {5\,B}{2}-3\,A\,\cos \left (2\,c+2\,d\,x\right )+\frac {3\,A\,\cos \left (3\,c+3\,d\,x\right )}{4}+2\,B\,\cos \left (2\,c+2\,d\,x\right )+\frac {B\,\cos \left (3\,c+3\,d\,x\right )}{8}+3\,A\,\sin \left (2\,c+2\,d\,x\right )+\frac {3\,A\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {B\,\sin \left (2\,c+2\,d\,x\right )}{2}-\frac {B\,\sin \left (3\,c+3\,d\,x\right )}{2}\right )}{15\,d\,\left (\frac {\cos \left (3\,c+3\,d\,x\right )}{4}-\frac {5\,\cos \left (c+d\,x\right )}{4}+\sin \left (2\,c+2\,d\,x\right )\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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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.
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