Optimal. Leaf size=149 \[ \frac {2 a^2 \cos (c+d x)}{d}+\frac {9 a^2 \tan ^5(c+d x)}{10 d}-\frac {3 a^2 \tan ^3(c+d x)}{2 d}+\frac {9 a^2 \tan (c+d x)}{2 d}+\frac {2 a^2 \sec ^5(c+d x)}{5 d}-\frac {2 a^2 \sec ^3(c+d x)}{d}+\frac {6 a^2 \sec (c+d x)}{d}-\frac {a^2 \sin ^2(c+d x) \tan ^5(c+d x)}{2 d}-\frac {9 a^2 x}{2} \]
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Rubi [A] time = 0.17, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2710, 3473, 8, 2590, 270, 2591, 288, 302, 203} \[ \frac {2 a^2 \cos (c+d x)}{d}+\frac {9 a^2 \tan ^5(c+d x)}{10 d}-\frac {3 a^2 \tan ^3(c+d x)}{2 d}+\frac {9 a^2 \tan (c+d x)}{2 d}+\frac {2 a^2 \sec ^5(c+d x)}{5 d}-\frac {2 a^2 \sec ^3(c+d x)}{d}+\frac {6 a^2 \sec (c+d x)}{d}-\frac {a^2 \sin ^2(c+d x) \tan ^5(c+d x)}{2 d}-\frac {9 a^2 x}{2} \]
Antiderivative was successfully verified.
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Rule 8
Rule 203
Rule 270
Rule 288
Rule 302
Rule 2590
Rule 2591
Rule 2710
Rule 3473
Rubi steps
\begin {align*} \int (a+a \sin (c+d x))^2 \tan ^6(c+d x) \, dx &=\int \left (a^2 \tan ^6(c+d x)+2 a^2 \sin (c+d x) \tan ^6(c+d x)+a^2 \sin ^2(c+d x) \tan ^6(c+d x)\right ) \, dx\\ &=a^2 \int \tan ^6(c+d x) \, dx+a^2 \int \sin ^2(c+d x) \tan ^6(c+d x) \, dx+\left (2 a^2\right ) \int \sin (c+d x) \tan ^6(c+d x) \, dx\\ &=\frac {a^2 \tan ^5(c+d x)}{5 d}-a^2 \int \tan ^4(c+d x) \, dx+\frac {a^2 \operatorname {Subst}\left (\int \frac {x^8}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{d}-\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^3}{x^6} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {a^2 \tan ^3(c+d x)}{3 d}+\frac {a^2 \tan ^5(c+d x)}{5 d}-\frac {a^2 \sin ^2(c+d x) \tan ^5(c+d x)}{2 d}+a^2 \int \tan ^2(c+d x) \, dx-\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \left (-1+\frac {1}{x^6}-\frac {3}{x^4}+\frac {3}{x^2}\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac {\left (7 a^2\right ) \operatorname {Subst}\left (\int \frac {x^6}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac {2 a^2 \cos (c+d x)}{d}+\frac {6 a^2 \sec (c+d x)}{d}-\frac {2 a^2 \sec ^3(c+d x)}{d}+\frac {2 a^2 \sec ^5(c+d x)}{5 d}+\frac {a^2 \tan (c+d x)}{d}-\frac {a^2 \tan ^3(c+d x)}{3 d}+\frac {a^2 \tan ^5(c+d x)}{5 d}-\frac {a^2 \sin ^2(c+d x) \tan ^5(c+d x)}{2 d}-a^2 \int 1 \, dx+\frac {\left (7 a^2\right ) \operatorname {Subst}\left (\int \left (1-x^2+x^4-\frac {1}{1+x^2}\right ) \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-a^2 x+\frac {2 a^2 \cos (c+d x)}{d}+\frac {6 a^2 \sec (c+d x)}{d}-\frac {2 a^2 \sec ^3(c+d x)}{d}+\frac {2 a^2 \sec ^5(c+d x)}{5 d}+\frac {9 a^2 \tan (c+d x)}{2 d}-\frac {3 a^2 \tan ^3(c+d x)}{2 d}+\frac {9 a^2 \tan ^5(c+d x)}{10 d}-\frac {a^2 \sin ^2(c+d x) \tan ^5(c+d x)}{2 d}-\frac {\left (7 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-\frac {9 a^2 x}{2}+\frac {2 a^2 \cos (c+d x)}{d}+\frac {6 a^2 \sec (c+d x)}{d}-\frac {2 a^2 \sec ^3(c+d x)}{d}+\frac {2 a^2 \sec ^5(c+d x)}{5 d}+\frac {9 a^2 \tan (c+d x)}{2 d}-\frac {3 a^2 \tan ^3(c+d x)}{2 d}+\frac {9 a^2 \tan ^5(c+d x)}{10 d}-\frac {a^2 \sin ^2(c+d x) \tan ^5(c+d x)}{2 d}\\ \end {align*}
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Mathematica [A] time = 0.84, size = 174, normalized size = 1.17 \[ -\frac {a^2 \sec ^5(c+d x) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4 (250 \sin (c+d x)-720 c \sin (2 (c+d x))-720 d x \sin (2 (c+d x))-824 \sin (2 (c+d x))+351 \sin (3 (c+d x))+5 \sin (5 (c+d x))+10 (90 c+90 d x+103) \cos (c+d x)-544 \cos (2 (c+d x))-180 c \cos (3 (c+d x))-180 d x \cos (3 (c+d x))-206 \cos (3 (c+d x))+20 \cos (4 (c+d x))-500)}{160 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.46, size = 152, normalized size = 1.02 \[ -\frac {45 \, a^{2} d x \cos \left (d x + c\right )^{3} - 10 \, a^{2} \cos \left (d x + c\right )^{4} - 90 \, a^{2} d x \cos \left (d x + c\right ) + 78 \, a^{2} \cos \left (d x + c\right )^{2} - 4 \, a^{2} - {\left (5 \, a^{2} \cos \left (d x + c\right )^{4} - 90 \, a^{2} d x \cos \left (d x + c\right ) + 84 \, a^{2} \cos \left (d x + c\right )^{2} - 6 \, a^{2}\right )} \sin \left (d x + c\right )}{10 \, {\left (d \cos \left (d x + c\right )^{3} + 2 \, d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [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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maple [A] time = 0.31, size = 251, normalized size = 1.68 \[ \frac {a^{2} \left (\frac {\sin ^{9}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}-\frac {4 \left (\sin ^{9}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}+\frac {8 \left (\sin ^{9}\left (d x +c \right )\right )}{5 \cos \left (d x +c \right )}+\frac {8 \left (\sin ^{7}\left (d x +c \right )+\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{6}+\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}+\frac {35 \sin \left (d x +c \right )}{16}\right ) \cos \left (d x +c \right )}{5}-\frac {7 d x}{2}-\frac {7 c}{2}\right )+2 a^{2} \left (\frac {\sin ^{8}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}-\frac {\sin ^{8}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{3}}+\frac {\sin ^{8}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {16}{5}+\sin ^{6}\left (d x +c \right )+\frac {6 \left (\sin ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (d x +c \right )\right )}{5}\right ) \cos \left (d x +c \right )\right )+a^{2} \left (\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}+\tan \left (d x +c \right )-d x -c \right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.42, size = 152, normalized size = 1.02 \[ \frac {{\left (6 \, \tan \left (d x + c\right )^{5} - 20 \, \tan \left (d x + c\right )^{3} - 105 \, d x - 105 \, c + \frac {15 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} + 90 \, \tan \left (d x + c\right )\right )} a^{2} + 2 \, {\left (3 \, \tan \left (d x + c\right )^{5} - 5 \, \tan \left (d x + c\right )^{3} - 15 \, d x - 15 \, c + 15 \, \tan \left (d x + c\right )\right )} a^{2} + 12 \, a^{2} {\left (\frac {15 \, \cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{2} + 1}{\cos \left (d x + c\right )^{5}} + 5 \, \cos \left (d x + c\right )\right )}}{30 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 10.92, size = 392, normalized size = 2.63 \[ -\frac {9\,a^2\,x}{2}-\frac {\frac {9\,a^2\,\left (c+d\,x\right )}{2}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (18\,a^2\,\left (c+d\,x\right )-\frac {a^2\,\left (180\,c+180\,d\,x-422\right )}{10}\right )+\frac {174\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5}-\frac {a^2\,\left (45\,c+45\,d\,x-128\right )}{10}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (18\,a^2\,\left (c+d\,x\right )-\frac {a^2\,\left (180\,c+180\,d\,x-90\right )}{10}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (27\,a^2\,\left (c+d\,x\right )-\frac {a^2\,\left (270\,c+270\,d\,x-168\right )}{10}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {63\,a^2\,\left (c+d\,x\right )}{2}-\frac {a^2\,\left (315\,c+315\,d\,x-360\right )}{10}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (27\,a^2\,\left (c+d\,x\right )-\frac {a^2\,\left (270\,c+270\,d\,x-600\right )}{10}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (36\,a^2\,\left (c+d\,x\right )-\frac {a^2\,\left (360\,c+360\,d\,x-424\right )}{10}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {63\,a^2\,\left (c+d\,x\right )}{2}-\frac {a^2\,\left (315\,c+315\,d\,x-536\right )}{10}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (36\,a^2\,\left (c+d\,x\right )-\frac {a^2\,\left (360\,c+360\,d\,x-600\right )}{10}\right )}{d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^5\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int 2 \sin {\left (c + d x \right )} \tan ^{6}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \tan ^{6}{\left (c + d x \right )}\, dx + \int \tan ^{6}{\left (c + d x \right )}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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