Optimal. Leaf size=197 \[ \frac {a^3 \cos (c+d x)}{d}+\frac {a^3 \sec (c+d x)}{d}+\frac {9 a^2 b \tan (c+d x)}{2 d}-\frac {3 a^2 b \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac {9}{2} a^2 b x-\frac {a b^2 \cos ^3(c+d x)}{d}+\frac {6 a b^2 \cos (c+d x)}{d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {15 b^3 \tan (c+d x)}{8 d}-\frac {b^3 \sin ^4(c+d x) \tan (c+d x)}{4 d}-\frac {5 b^3 \sin ^2(c+d x) \tan (c+d x)}{8 d}-\frac {15 b^3 x}{8} \]
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Rubi [A] time = 0.26, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2912, 2590, 14, 2591, 288, 321, 203, 270} \[ \frac {9 a^2 b \tan (c+d x)}{2 d}-\frac {3 a^2 b \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac {9}{2} a^2 b x+\frac {a^3 \cos (c+d x)}{d}+\frac {a^3 \sec (c+d x)}{d}-\frac {a b^2 \cos ^3(c+d x)}{d}+\frac {6 a b^2 \cos (c+d x)}{d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {15 b^3 \tan (c+d x)}{8 d}-\frac {b^3 \sin ^4(c+d x) \tan (c+d x)}{4 d}-\frac {5 b^3 \sin ^2(c+d x) \tan (c+d x)}{8 d}-\frac {15 b^3 x}{8} \]
Antiderivative was successfully verified.
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Rule 14
Rule 203
Rule 270
Rule 288
Rule 321
Rule 2590
Rule 2591
Rule 2912
Rubi steps
\begin {align*} \int \sin (c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx &=\int \left (a^3 \sin (c+d x) \tan ^2(c+d x)+3 a^2 b \sin ^2(c+d x) \tan ^2(c+d x)+3 a b^2 \sin ^3(c+d x) \tan ^2(c+d x)+b^3 \sin ^4(c+d x) \tan ^2(c+d x)\right ) \, dx\\ &=a^3 \int \sin (c+d x) \tan ^2(c+d x) \, dx+\left (3 a^2 b\right ) \int \sin ^2(c+d x) \tan ^2(c+d x) \, dx+\left (3 a b^2\right ) \int \sin ^3(c+d x) \tan ^2(c+d x) \, dx+b^3 \int \sin ^4(c+d x) \tan ^2(c+d x) \, dx\\ &=-\frac {a^3 \operatorname {Subst}\left (\int \frac {1-x^2}{x^2} \, dx,x,\cos (c+d x)\right )}{d}+\frac {\left (3 a^2 b\right ) \operatorname {Subst}\left (\int \frac {x^4}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{d}-\frac {\left (3 a b^2\right ) \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^2}{x^2} \, dx,x,\cos (c+d x)\right )}{d}+\frac {b^3 \operatorname {Subst}\left (\int \frac {x^6}{\left (1+x^2\right )^3} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {3 a^2 b \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac {b^3 \sin ^4(c+d x) \tan (c+d x)}{4 d}-\frac {a^3 \operatorname {Subst}\left (\int \left (-1+\frac {1}{x^2}\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac {\left (9 a^2 b\right ) \operatorname {Subst}\left (\int \frac {x^2}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac {\left (3 a b^2\right ) \operatorname {Subst}\left (\int \left (-2+\frac {1}{x^2}+x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac {\left (5 b^3\right ) \operatorname {Subst}\left (\int \frac {x^4}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{4 d}\\ &=\frac {a^3 \cos (c+d x)}{d}+\frac {6 a b^2 \cos (c+d x)}{d}-\frac {a b^2 \cos ^3(c+d x)}{d}+\frac {a^3 \sec (c+d x)}{d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {9 a^2 b \tan (c+d x)}{2 d}-\frac {3 a^2 b \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac {5 b^3 \sin ^2(c+d x) \tan (c+d x)}{8 d}-\frac {b^3 \sin ^4(c+d x) \tan (c+d x)}{4 d}-\frac {\left (9 a^2 b\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left (15 b^3\right ) \operatorname {Subst}\left (\int \frac {x^2}{1+x^2} \, dx,x,\tan (c+d x)\right )}{8 d}\\ &=-\frac {9}{2} a^2 b x+\frac {a^3 \cos (c+d x)}{d}+\frac {6 a b^2 \cos (c+d x)}{d}-\frac {a b^2 \cos ^3(c+d x)}{d}+\frac {a^3 \sec (c+d x)}{d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {9 a^2 b \tan (c+d x)}{2 d}+\frac {15 b^3 \tan (c+d x)}{8 d}-\frac {3 a^2 b \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac {5 b^3 \sin ^2(c+d x) \tan (c+d x)}{8 d}-\frac {b^3 \sin ^4(c+d x) \tan (c+d x)}{4 d}-\frac {\left (15 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{8 d}\\ &=-\frac {9}{2} a^2 b x-\frac {15 b^3 x}{8}+\frac {a^3 \cos (c+d x)}{d}+\frac {6 a b^2 \cos (c+d x)}{d}-\frac {a b^2 \cos ^3(c+d x)}{d}+\frac {a^3 \sec (c+d x)}{d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {9 a^2 b \tan (c+d x)}{2 d}+\frac {15 b^3 \tan (c+d x)}{8 d}-\frac {3 a^2 b \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac {5 b^3 \sin ^2(c+d x) \tan (c+d x)}{8 d}-\frac {b^3 \sin ^4(c+d x) \tan (c+d x)}{4 d}\\ \end {align*}
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Mathematica [A] time = 0.76, size = 147, normalized size = 0.75 \[ \frac {\sec (c+d x) \left (32 \left (a^3+5 a b^2\right ) \cos (2 (c+d x))+96 a^3-24 b \left (12 a^2+5 b^2\right ) (c+d x) \cos (c+d x)+216 a^2 b \sin (c+d x)+24 a^2 b \sin (3 (c+d x))-8 a b^2 \cos (4 (c+d x))+360 a b^2+80 b^3 \sin (c+d x)+15 b^3 \sin (3 (c+d x))-b^3 \sin (5 (c+d x))\right )}{64 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.44, size = 135, normalized size = 0.69 \[ -\frac {8 \, a b^{2} \cos \left (d x + c\right )^{4} + 3 \, {\left (12 \, a^{2} b + 5 \, b^{3}\right )} d x \cos \left (d x + c\right ) - 8 \, a^{3} - 24 \, a b^{2} - 8 \, {\left (a^{3} + 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, b^{3} \cos \left (d x + c\right )^{4} - 24 \, a^{2} b - 8 \, b^{3} - 3 \, {\left (4 \, a^{2} b + 3 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{8 \, d \cos \left (d x + c\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.24, size = 336, normalized size = 1.71 \[ -\frac {3 \, {\left (12 \, a^{2} b + 5 \, b^{3}\right )} {\left (d x + c\right )} + \frac {16 \, {\left (3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3} + 3 \, a b^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 7 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 8 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 24 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 136 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 7 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, a^{3} - 40 \, a b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{8 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.64, size = 214, normalized size = 1.09 \[ \frac {a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+3 a^{2} b \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+3 a \,b^{2} \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+b^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.48, size = 164, normalized size = 0.83 \[ -\frac {12 \, {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a^{2} b + 8 \, {\left (\cos \left (d x + c\right )^{3} - \frac {3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} a b^{2} + {\left (15 \, d x + 15 \, c - \frac {9 \, \tan \left (d x + c\right )^{3} + 7 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1} - 8 \, \tan \left (d x + c\right )\right )} b^{3} - 8 \, a^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{8 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 16.02, size = 323, normalized size = 1.64 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (9\,a^2\,b+\frac {15\,b^3}{4}\right )+4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,a\,b^2+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (12\,a^3+32\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (12\,a^3+48\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (24\,a^2\,b+10\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (9\,a^2\,b+\frac {15\,b^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (24\,a^2\,b+10\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (30\,a^2\,b+\frac {9\,b^3}{2}\right )+4\,a^3}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {3\,b\,\mathrm {atan}\left (\frac {3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (12\,a^2+5\,b^2\right )}{36\,a^2\,b+15\,b^3}\right )\,\left (12\,a^2+5\,b^2\right )}{4\,d} \]
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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