Optimal. Leaf size=277 \[ \frac {45 d^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b}-\frac {45 d^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{32 \sqrt {2} b}+\frac {45 d^{3/2} \log \left (\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{64 \sqrt {2} b}-\frac {45 d^{3/2} \log \left (\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{64 \sqrt {2} b}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{9/2}}{4 b d^3}+\frac {45 d \sqrt {d \tan (a+b x)}}{16 b}-\frac {9 \cos ^2(a+b x) (d \tan (a+b x))^{5/2}}{16 b d} \]
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Rubi [A] time = 0.20, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {2591, 288, 321, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac {45 d^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b}-\frac {45 d^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{32 \sqrt {2} b}+\frac {45 d^{3/2} \log \left (\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{64 \sqrt {2} b}-\frac {45 d^{3/2} \log \left (\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{64 \sqrt {2} b}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{9/2}}{4 b d^3}+\frac {45 d \sqrt {d \tan (a+b x)}}{16 b}-\frac {9 \cos ^2(a+b x) (d \tan (a+b x))^{5/2}}{16 b d} \]
Antiderivative was successfully verified.
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Rule 204
Rule 211
Rule 288
Rule 321
Rule 329
Rule 617
Rule 628
Rule 1162
Rule 1165
Rule 2591
Rubi steps
\begin {align*} \int \sin ^4(a+b x) (d \tan (a+b x))^{3/2} \, dx &=\frac {d \operatorname {Subst}\left (\int \frac {x^{11/2}}{\left (d^2+x^2\right )^3} \, dx,x,d \tan (a+b x)\right )}{b}\\ &=-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{9/2}}{4 b d^3}+\frac {(9 d) \operatorname {Subst}\left (\int \frac {x^{7/2}}{\left (d^2+x^2\right )^2} \, dx,x,d \tan (a+b x)\right )}{8 b}\\ &=-\frac {9 \cos ^2(a+b x) (d \tan (a+b x))^{5/2}}{16 b d}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{9/2}}{4 b d^3}+\frac {(45 d) \operatorname {Subst}\left (\int \frac {x^{3/2}}{d^2+x^2} \, dx,x,d \tan (a+b x)\right )}{32 b}\\ &=\frac {45 d \sqrt {d \tan (a+b x)}}{16 b}-\frac {9 \cos ^2(a+b x) (d \tan (a+b x))^{5/2}}{16 b d}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{9/2}}{4 b d^3}-\frac {\left (45 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (d^2+x^2\right )} \, dx,x,d \tan (a+b x)\right )}{32 b}\\ &=\frac {45 d \sqrt {d \tan (a+b x)}}{16 b}-\frac {9 \cos ^2(a+b x) (d \tan (a+b x))^{5/2}}{16 b d}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{9/2}}{4 b d^3}-\frac {\left (45 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{16 b}\\ &=\frac {45 d \sqrt {d \tan (a+b x)}}{16 b}-\frac {9 \cos ^2(a+b x) (d \tan (a+b x))^{5/2}}{16 b d}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{9/2}}{4 b d^3}-\frac {\left (45 d^2\right ) \operatorname {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{32 b}-\frac {\left (45 d^2\right ) \operatorname {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{32 b}\\ &=\frac {45 d \sqrt {d \tan (a+b x)}}{16 b}-\frac {9 \cos ^2(a+b x) (d \tan (a+b x))^{5/2}}{16 b d}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{9/2}}{4 b d^3}+\frac {\left (45 d^{3/2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}+2 x}{-d-\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b}+\frac {\left (45 d^{3/2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}-2 x}{-d+\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b}-\frac {\left (45 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{64 b}-\frac {\left (45 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{64 b}\\ &=\frac {45 d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b}-\frac {45 d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b}+\frac {45 d \sqrt {d \tan (a+b x)}}{16 b}-\frac {9 \cos ^2(a+b x) (d \tan (a+b x))^{5/2}}{16 b d}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{9/2}}{4 b d^3}-\frac {\left (45 d^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b}+\frac {\left (45 d^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b}\\ &=\frac {45 d^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b}-\frac {45 d^{3/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b}+\frac {45 d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b}-\frac {45 d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b}+\frac {45 d \sqrt {d \tan (a+b x)}}{16 b}-\frac {9 \cos ^2(a+b x) (d \tan (a+b x))^{5/2}}{16 b d}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{9/2}}{4 b d^3}\\ \end {align*}
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Mathematica [A] time = 0.77, size = 123, normalized size = 0.44 \[ -\frac {d \csc (a+b x) \sqrt {d \tan (a+b x)} \left (-143 \sin (a+b x)-14 \sin (3 (a+b x))+\sin (5 (a+b x))-45 \sqrt {\sin (2 (a+b x))} \sin ^{-1}(\cos (a+b x)-\sin (a+b x))+45 \sqrt {\sin (2 (a+b x))} \log \left (\sin (a+b x)+\sqrt {\sin (2 (a+b x))}+\cos (a+b x)\right )\right )}{64 b} \]
Antiderivative was successfully verified.
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fricas [B] time = 74.62, size = 1580, normalized size = 5.70 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.52, size = 252, normalized size = 0.91 \[ -\frac {1}{128} \, d {\left (\frac {90 \, \sqrt {2} \sqrt {{\left | d \right |}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{b} + \frac {90 \, \sqrt {2} \sqrt {{\left | d \right |}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{b} + \frac {45 \, \sqrt {2} \sqrt {{\left | d \right |}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{b} - \frac {45 \, \sqrt {2} \sqrt {{\left | d \right |}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{b} - \frac {256 \, \sqrt {d \tan \left (b x + a\right )}}{b} - \frac {8 \, {\left (17 \, \sqrt {d \tan \left (b x + a\right )} d^{4} \tan \left (b x + a\right )^{2} + 13 \, \sqrt {d \tan \left (b x + a\right )} d^{4}\right )}}{{\left (d^{2} \tan \left (b x + a\right )^{2} + d^{2}\right )}^{2} b}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.54, size = 702, normalized size = 2.53 \[ \frac {\left (-1+\cos \left (b x +a \right )\right ) \left (45 i \EllipticPi \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sin \left (b x +a \right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-45 i \EllipticPi \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sin \left (b x +a \right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-8 \sqrt {2}\, \left (\cos ^{5}\left (b x +a \right )\right )+8 \left (\cos ^{4}\left (b x +a \right )\right ) \sqrt {2}+45 \EllipticPi \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sin \left (b x +a \right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-90 \sin \left (b x +a \right ) \EllipticF \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}+45 \EllipticPi \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sin \left (b x +a \right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}+34 \left (\cos ^{3}\left (b x +a \right )\right ) \sqrt {2}-34 \left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}+64 \cos \left (b x +a \right ) \sqrt {2}-64 \sqrt {2}\right ) \cos \left (b x +a \right ) \left (\cos \left (b x +a \right )+1\right )^{2} \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}} \sqrt {2}}{64 b \sin \left (b x +a \right )^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.83, size = 235, normalized size = 0.85 \[ -\frac {90 \, \sqrt {2} d^{\frac {13}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + 90 \, \sqrt {2} d^{\frac {13}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + 45 \, \sqrt {2} d^{\frac {13}{2}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) - 45 \, \sqrt {2} d^{\frac {13}{2}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) - 256 \, \sqrt {d \tan \left (b x + a\right )} d^{6} - \frac {8 \, {\left (17 \, \left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}} d^{8} + 13 \, \sqrt {d \tan \left (b x + a\right )} d^{10}\right )}}{d^{4} \tan \left (b x + a\right )^{4} + 2 \, d^{4} \tan \left (b x + a\right )^{2} + d^{4}}}{128 \, b d^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\sin \left (a+b\,x\right )}^4\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2} \,d x \]
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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