Optimal. Leaf size=136 \[ -\frac {4 d^2 \sqrt {\sin (2 a+2 b x)} \sec (a+b x) F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{77 b \sqrt {d \tan (a+b x)}}+\frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {2 d \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{77 b}-\frac {4 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{77 b} \]
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Rubi [A] time = 0.18, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2611, 2613, 2614, 2573, 2641} \[ -\frac {4 d^2 \sqrt {\sin (2 a+2 b x)} \sec (a+b x) F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{77 b \sqrt {d \tan (a+b x)}}+\frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {2 d \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{77 b}-\frac {4 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{77 b} \]
Antiderivative was successfully verified.
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Rule 2573
Rule 2611
Rule 2613
Rule 2614
Rule 2641
Rubi steps
\begin {align*} \int \sec ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx &=\frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {1}{11} d^2 \int \frac {\sec ^5(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx\\ &=-\frac {2 d \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{77 b}+\frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {1}{77} \left (6 d^2\right ) \int \frac {\sec ^3(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx\\ &=-\frac {4 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{77 b}-\frac {2 d \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{77 b}+\frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {1}{77} \left (4 d^2\right ) \int \frac {\sec (a+b x)}{\sqrt {d \tan (a+b x)}} \, dx\\ &=-\frac {4 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{77 b}-\frac {2 d \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{77 b}+\frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {\left (4 d^2 \sqrt {\sin (a+b x)}\right ) \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}} \, dx}{77 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\\ &=-\frac {4 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{77 b}-\frac {2 d \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{77 b}+\frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {\left (4 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}{77 \sqrt {d \tan (a+b x)}}\\ &=-\frac {4 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)}}{77 b \sqrt {d \tan (a+b x)}}-\frac {4 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{77 b}-\frac {2 d \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{77 b}+\frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}\\ \end {align*}
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Mathematica [C] time = 0.82, size = 90, normalized size = 0.66 \[ -\frac {d \sec ^5(a+b x) \sqrt {d \tan (a+b x)} \left (16 \cos ^6(a+b x) \sqrt {\sec ^2(a+b x)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\tan ^2(a+b x)\right )+6 \cos (2 (a+b x))+\cos (4 (a+b x))-23\right )}{154 b} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {d \tan \left (b x + a\right )} d \sec \left (b x + a\right )^{5} \tan \left (b x + a\right ), x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \sec \left (b x + a\right )^{5}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.59, size = 251, normalized size = 1.85 \[ \frac {\left (-1+\cos \left (b x +a \right )\right ) \left (4 \sin \left (b x +a \right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \left (\cos ^{5}\left (b x +a \right )\right ) \EllipticF \left (\sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )-2 \sqrt {2}\, \left (\cos ^{5}\left (b x +a \right )\right )+2 \left (\cos ^{4}\left (b x +a \right )\right ) \sqrt {2}-\left (\cos ^{3}\left (b x +a \right )\right ) \sqrt {2}+\left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}+7 \cos \left (b x +a \right ) \sqrt {2}-7 \sqrt {2}\right ) \left (\cos \left (b x +a \right )+1\right )^{2} \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}} \sqrt {2}}{77 b \sin \left (b x +a \right )^{5} \cos \left (b x +a \right )^{4}} \]
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 \[ \int \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \sec \left (b x + a\right )^{5}\,{d x} \]
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 \[ \int \frac {{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2}}{{\cos \left (a+b\,x\right )}^5} \,d x \]
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 \[ \int \left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}} \sec ^{5}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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