Optimal. Leaf size=227 \[ -\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b d^{3/2}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{4 \sqrt {2} b d^{3/2}}+\frac {\log \left (\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{8 \sqrt {2} b d^{3/2}}-\frac {\log \left (\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{8 \sqrt {2} b d^{3/2}}+\frac {\cos ^2(a+b x) (d \tan (a+b x))^{3/2}}{2 b d^3} \]
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Rubi [A] time = 0.16, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2591, 290, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b d^{3/2}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{4 \sqrt {2} b d^{3/2}}+\frac {\log \left (\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{8 \sqrt {2} b d^{3/2}}-\frac {\log \left (\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{8 \sqrt {2} b d^{3/2}}+\frac {\cos ^2(a+b x) (d \tan (a+b x))^{3/2}}{2 b d^3} \]
Antiderivative was successfully verified.
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Rule 204
Rule 290
Rule 297
Rule 329
Rule 617
Rule 628
Rule 1162
Rule 1165
Rule 2591
Rubi steps
\begin {align*} \int \frac {\sin ^2(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx &=\frac {d \operatorname {Subst}\left (\int \frac {\sqrt {x}}{\left (d^2+x^2\right )^2} \, dx,x,d \tan (a+b x)\right )}{b}\\ &=\frac {\cos ^2(a+b x) (d \tan (a+b x))^{3/2}}{2 b d^3}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {x}}{d^2+x^2} \, dx,x,d \tan (a+b x)\right )}{4 b d}\\ &=\frac {\cos ^2(a+b x) (d \tan (a+b x))^{3/2}}{2 b d^3}+\frac {\operatorname {Subst}\left (\int \frac {x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{2 b d}\\ &=\frac {\cos ^2(a+b x) (d \tan (a+b x))^{3/2}}{2 b d^3}-\frac {\operatorname {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{4 b d}+\frac {\operatorname {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{4 b d}\\ &=\frac {\cos ^2(a+b x) (d \tan (a+b x))^{3/2}}{2 b d^3}+\frac {\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 )}{8 \sqrt {2} b d^{3/2}}+\frac {\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 )}{8 \sqrt {2} b d^{3/2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{8 b d}+\frac {\operatorname {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{8 b d}\\ &=\frac {\log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b d^{3/2}}-\frac {\log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b d^{3/2}}+\frac {\cos ^2(a+b x) (d \tan (a+b x))^{3/2}}{2 b d^3}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b d^{3/2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b d^{3/2}}\\ &=-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b d^{3/2}}+\frac {\tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b d^{3/2}}+\frac {\log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b d^{3/2}}-\frac {\log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b d^{3/2}}+\frac {\cos ^2(a+b x) (d \tan (a+b x))^{3/2}}{2 b d^3}\\ \end {align*}
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Mathematica [A] time = 0.25, size = 105, normalized size = 0.46 \[ -\frac {\sqrt {\sin (2 (a+b x))} \sqrt {d \tan (a+b x)} \left (-2 \sqrt {\sin (2 (a+b x))}+\csc (a+b x) \sin ^{-1}(\cos (a+b x)-\sin (a+b x))+\csc (a+b x) \log \left (\sin (a+b x)+\sqrt {\sin (2 (a+b x))}+\cos (a+b x)\right )\right )}{8 b d^2} \]
Antiderivative was successfully verified.
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fricas [B] time = 104.13, size = 1856, normalized size = 8.18 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.02, size = 228, normalized size = 1.00 \[ \frac {\frac {8 \, \sqrt {d \tan \left (b x + a\right )} d \tan \left (b x + a\right )}{{\left (d^{2} \tan \left (b x + a\right )^{2} + d^{2}\right )} b} + \frac {2 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{b d^{2}} + \frac {2 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{b d^{2}} - \frac {\sqrt {2} {\left | d \right |}^{\frac {3}{2}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{b d^{2}} + \frac {\sqrt {2} {\left | d \right |}^{\frac {3}{2}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{b d^{2}}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.46, size = 522, normalized size = 2.30 \[ \frac {\left (-1+\cos \left (b x +a \right )\right ) \left (-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 ) \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 )}}+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 ) \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 )}}+\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 ) \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 )}}+\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 ) \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 )}}+2 \left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}-2 \cos \left (b x +a \right ) \sqrt {2}\right ) \left (\cos \left (b x +a \right )+1\right )^{2} \sqrt {2}}{8 b \cos \left (b x +a \right )^{2} \sin \left (b x +a \right ) \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.84, size = 193, normalized size = 0.85 \[ \frac {d^{2} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\sqrt {2} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} + \frac {8 \, \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} d^{2}}{d^{2} \tan \left (b x + a\right )^{2} + d^{2}}}{16 \, b d^{3}} \]
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 \[ \int \frac {{\sin \left (a+b\,x\right )}^2}{{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{2}{\left (a + b x \right )}}{\left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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