Optimal. Leaf size=78 \[ \frac {d^2 \sqrt {\sin (2 a+2 b x)} \sec (a+b x) F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{2 b \sqrt {d \tan (a+b x)}}-\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{b} \]
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Rubi [A] time = 0.10, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2610, 2614, 2573, 2641} \[ \frac {d^2 \sqrt {\sin (2 a+2 b x)} \sec (a+b x) F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{2 b \sqrt {d \tan (a+b x)}}-\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{b} \]
Antiderivative was successfully verified.
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Rule 2573
Rule 2610
Rule 2614
Rule 2641
Rubi steps
\begin {align*} \int \cos (a+b x) (d \tan (a+b x))^{3/2} \, dx &=-\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{b}+\frac {1}{2} d^2 \int \frac {\sec (a+b x)}{\sqrt {d \tan (a+b x)}} \, dx\\ &=-\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{b}+\frac {\left (d^2 \sqrt {\sin (a+b x)}\right ) \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}} \, dx}{2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\\ &=-\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{b}+\frac {\left (d^2 \sec (a+b x) \sqrt {\sin (2 a+2 b x)}\right ) \int \frac {1}{\sqrt {\sin (2 a+2 b x)}} \, dx}{2 \sqrt {d \tan (a+b x)}}\\ &=\frac {d^2 F\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sec (a+b x) \sqrt {\sin (2 a+2 b x)}}{2 b \sqrt {d \tan (a+b x)}}-\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{b}\\ \end {align*}
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Mathematica [C] time = 0.14, size = 58, normalized size = 0.74 \[ \frac {d \cos (a+b x) \sqrt {d \tan (a+b x)} \left (\sqrt {\sec ^2(a+b x)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\tan ^2(a+b x)\right )-1\right )}{b} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {d \tan \left (b x + a\right )} d \cos \left (b x + a\right ) \tan \left (b x + a\right ), x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.47, size = 196, normalized size = 2.51 \[ -\frac {\left (-1+\cos \left (b x +a \right )\right ) \left (\sin \left (b x +a \right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticF \left (\sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )+\left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}-\cos \left (b x +a \right ) \sqrt {2}\right ) \cos \left (b x +a \right ) \left (\cos \left (b x +a \right )+1\right )^{2} \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}} \sqrt {2}}{2 b \sin \left (b x +a \right )^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \cos \left (b x + a\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \cos \left (a+b\,x\right )\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2} \,d x \]
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 \[ \int \left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}} \cos {\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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