Optimal. Leaf size=268 \[ \frac {a \cos (c+d x)}{d \left (a^2-b^2\right )}-\frac {3 b \tan (c+d x)}{2 d \left (a^2-b^2\right )}+\frac {a \sec (c+d x)}{d \left (a^2-b^2\right )}+\frac {b \sin ^2(c+d x) \tan (c+d x)}{2 d \left (a^2-b^2\right )}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 b d \left (a^2-b^2\right )}+\frac {3 b x}{2 \left (a^2-b^2\right )}-\frac {a^2 x \left (2 a^2+b^2\right )}{2 b^3 \left (a^2-b^2\right )}+\frac {2 a^5 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^3 d \left (a^2-b^2\right )^{3/2}}-\frac {a^3 \cos (c+d x)}{b^2 d \left (a^2-b^2\right )} \]
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Rubi [A] time = 0.40, antiderivative size = 268, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 13, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.448, Rules used = {2902, 2590, 14, 2591, 288, 321, 203, 2793, 3023, 2735, 2660, 618, 204} \[ -\frac {a^3 \cos (c+d x)}{b^2 d \left (a^2-b^2\right )}+\frac {a \cos (c+d x)}{d \left (a^2-b^2\right )}+\frac {2 a^5 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^3 d \left (a^2-b^2\right )^{3/2}}-\frac {3 b \tan (c+d x)}{2 d \left (a^2-b^2\right )}+\frac {a \sec (c+d x)}{d \left (a^2-b^2\right )}+\frac {b \sin ^2(c+d x) \tan (c+d x)}{2 d \left (a^2-b^2\right )}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 b d \left (a^2-b^2\right )}-\frac {a^2 x \left (2 a^2+b^2\right )}{2 b^3 \left (a^2-b^2\right )}+\frac {3 b x}{2 \left (a^2-b^2\right )} \]
Antiderivative was successfully verified.
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Rule 14
Rule 203
Rule 204
Rule 288
Rule 321
Rule 618
Rule 2590
Rule 2591
Rule 2660
Rule 2735
Rule 2793
Rule 2902
Rule 3023
Rubi steps
\begin {align*} \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {a \int \sin (c+d x) \tan ^2(c+d x) \, dx}{a^2-b^2}-\frac {a^2 \int \frac {\sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx}{a^2-b^2}-\frac {b \int \sin ^2(c+d x) \tan ^2(c+d x) \, dx}{a^2-b^2}\\ &=\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d}-\frac {a^2 \int \frac {a+b \sin (c+d x)-2 a \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{2 b \left (a^2-b^2\right )}-\frac {a \operatorname {Subst}\left (\int \frac {1-x^2}{x^2} \, dx,x,\cos (c+d x)\right )}{\left (a^2-b^2\right ) d}-\frac {b \operatorname {Subst}\left (\int \frac {x^4}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{\left (a^2-b^2\right ) d}\\ &=-\frac {a^3 \cos (c+d x)}{b^2 \left (a^2-b^2\right ) d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d}+\frac {b \sin ^2(c+d x) \tan (c+d x)}{2 \left (a^2-b^2\right ) d}-\frac {a^2 \int \frac {a b+\left (2 a^2+b^2\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )}-\frac {a \operatorname {Subst}\left (\int \left (-1+\frac {1}{x^2}\right ) \, dx,x,\cos (c+d x)\right )}{\left (a^2-b^2\right ) d}-\frac {(3 b) \operatorname {Subst}\left (\int \frac {x^2}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 \left (a^2-b^2\right ) d}\\ &=-\frac {a^2 \left (2 a^2+b^2\right ) x}{2 b^3 \left (a^2-b^2\right )}+\frac {a \cos (c+d x)}{\left (a^2-b^2\right ) d}-\frac {a^3 \cos (c+d x)}{b^2 \left (a^2-b^2\right ) d}+\frac {a \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d}-\frac {3 b \tan (c+d x)}{2 \left (a^2-b^2\right ) d}+\frac {b \sin ^2(c+d x) \tan (c+d x)}{2 \left (a^2-b^2\right ) d}+\frac {a^5 \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^3 \left (a^2-b^2\right )}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 \left (a^2-b^2\right ) d}\\ &=\frac {3 b x}{2 \left (a^2-b^2\right )}-\frac {a^2 \left (2 a^2+b^2\right ) x}{2 b^3 \left (a^2-b^2\right )}+\frac {a \cos (c+d x)}{\left (a^2-b^2\right ) d}-\frac {a^3 \cos (c+d x)}{b^2 \left (a^2-b^2\right ) d}+\frac {a \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d}-\frac {3 b \tan (c+d x)}{2 \left (a^2-b^2\right ) d}+\frac {b \sin ^2(c+d x) \tan (c+d x)}{2 \left (a^2-b^2\right ) d}+\frac {\left (2 a^5\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^3 \left (a^2-b^2\right ) d}\\ &=\frac {3 b x}{2 \left (a^2-b^2\right )}-\frac {a^2 \left (2 a^2+b^2\right ) x}{2 b^3 \left (a^2-b^2\right )}+\frac {a \cos (c+d x)}{\left (a^2-b^2\right ) d}-\frac {a^3 \cos (c+d x)}{b^2 \left (a^2-b^2\right ) d}+\frac {a \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d}-\frac {3 b \tan (c+d x)}{2 \left (a^2-b^2\right ) d}+\frac {b \sin ^2(c+d x) \tan (c+d x)}{2 \left (a^2-b^2\right ) d}-\frac {\left (4 a^5\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^3 \left (a^2-b^2\right ) d}\\ &=\frac {3 b x}{2 \left (a^2-b^2\right )}-\frac {a^2 \left (2 a^2+b^2\right ) x}{2 b^3 \left (a^2-b^2\right )}+\frac {2 a^5 \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^3 \left (a^2-b^2\right )^{3/2} d}+\frac {a \cos (c+d x)}{\left (a^2-b^2\right ) d}-\frac {a^3 \cos (c+d x)}{b^2 \left (a^2-b^2\right ) d}+\frac {a \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d}-\frac {3 b \tan (c+d x)}{2 \left (a^2-b^2\right ) d}+\frac {b \sin ^2(c+d x) \tan (c+d x)}{2 \left (a^2-b^2\right ) d}\\ \end {align*}
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Mathematica [A] time = 1.53, size = 221, normalized size = 0.82 \[ \frac {\frac {8 a^5 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^3 \left (a^2-b^2\right )^{3/2}}+\frac {4 a^4 (c+d x)+2 a^2 b^2 (c+d x)-4 a b^3-6 b^4 (c+d x)}{b^5-a^2 b^3}-\frac {4 a \cos (c+d x)}{b^2}+\frac {4 \sin \left (\frac {1}{2} (c+d x)\right )}{(a+b) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {4 \sin \left (\frac {1}{2} (c+d x)\right )}{(a-b) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\sin (2 (c+d x))}{b}}{4 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 1.04, size = 521, normalized size = 1.94 \[ \left [\frac {\sqrt {-a^{2} + b^{2}} a^{5} \cos \left (d x + c\right ) \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 2 \, a^{3} b^{3} - 2 \, a b^{5} - {\left (2 \, a^{6} - a^{4} b^{2} - 4 \, a^{2} b^{4} + 3 \, b^{6}\right )} d x \cos \left (d x + c\right ) - 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )^{2} - {\left (2 \, a^{2} b^{4} - 2 \, b^{6} - {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} d \cos \left (d x + c\right )}, -\frac {2 \, \sqrt {a^{2} - b^{2}} a^{5} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - 2 \, a^{3} b^{3} + 2 \, a b^{5} + {\left (2 \, a^{6} - a^{4} b^{2} - 4 \, a^{2} b^{4} + 3 \, b^{6}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, a^{2} b^{4} - 2 \, b^{6} - {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} d \cos \left (d x + c\right )}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.22, size = 208, normalized size = 0.78 \[ \frac {\frac {4 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{5}}{{\left (a^{2} b^{3} - b^{5}\right )} \sqrt {a^{2} - b^{2}}} + \frac {4 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}}{{\left (a^{2} - b^{2}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}} - \frac {{\left (2 \, a^{2} + 3 \, b^{2}\right )} {\left (d x + c\right )}}{b^{3}} - \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} b^{2}}}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.43, size = 283, normalized size = 1.06 \[ -\frac {64}{d \left (64 a +64 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {2 a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{3}}-\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d b}+\frac {64}{d \left (64 a -64 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {2 a^{5} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \left (a -b \right ) \left (a +b \right ) b^{3} \sqrt {a^{2}-b^{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 16.81, size = 2098, normalized size = 7.83 \[ \text {result too large to display} \]
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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