Optimal. Leaf size=142 \[ -\frac{11 \cos ^5(a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac{77 \cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{30 b d^3}-\frac{77 \cos (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{20 b d^2 \sqrt{\sin (2 a+2 b x)}}-\frac{2 \cos ^5(a+b x)}{b d \sqrt{d \tan (a+b x)}} \]
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Rubi [A] time = 0.187095, antiderivative size = 142, normalized size of antiderivative = 1., 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 = {2609, 2612, 2615, 2572, 2639} \[ -\frac{11 \cos ^5(a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac{77 \cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{30 b d^3}-\frac{77 \cos (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{20 b d^2 \sqrt{\sin (2 a+2 b x)}}-\frac{2 \cos ^5(a+b x)}{b d \sqrt{d \tan (a+b x)}} \]
Antiderivative was successfully verified.
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Rule 2609
Rule 2612
Rule 2615
Rule 2572
Rule 2639
Rubi steps
\begin{align*} \int \frac{\cos ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx &=-\frac{2 \cos ^5(a+b x)}{b d \sqrt{d \tan (a+b x)}}-\frac{11 \int \cos ^5(a+b x) \sqrt{d \tan (a+b x)} \, dx}{d^2}\\ &=-\frac{2 \cos ^5(a+b x)}{b d \sqrt{d \tan (a+b x)}}-\frac{11 \cos ^5(a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac{77 \int \cos ^3(a+b x) \sqrt{d \tan (a+b x)} \, dx}{10 d^2}\\ &=-\frac{2 \cos ^5(a+b x)}{b d \sqrt{d \tan (a+b x)}}-\frac{77 \cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{30 b d^3}-\frac{11 \cos ^5(a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac{77 \int \cos (a+b x) \sqrt{d \tan (a+b x)} \, dx}{20 d^2}\\ &=-\frac{2 \cos ^5(a+b x)}{b d \sqrt{d \tan (a+b x)}}-\frac{77 \cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{30 b d^3}-\frac{11 \cos ^5(a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac{\left (77 \sqrt{\cos (a+b x)} \sqrt{d \tan (a+b x)}\right ) \int \sqrt{\cos (a+b x)} \sqrt{\sin (a+b x)} \, dx}{20 d^2 \sqrt{\sin (a+b x)}}\\ &=-\frac{2 \cos ^5(a+b x)}{b d \sqrt{d \tan (a+b x)}}-\frac{77 \cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{30 b d^3}-\frac{11 \cos ^5(a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac{\left (77 \cos (a+b x) \sqrt{d \tan (a+b x)}\right ) \int \sqrt{\sin (2 a+2 b x)} \, dx}{20 d^2 \sqrt{\sin (2 a+2 b x)}}\\ &=-\frac{2 \cos ^5(a+b x)}{b d \sqrt{d \tan (a+b x)}}-\frac{77 \cos (a+b x) E\left (\left .a-\frac{\pi }{4}+b x\right |2\right ) \sqrt{d \tan (a+b x)}}{20 b d^2 \sqrt{\sin (2 a+2 b x)}}-\frac{77 \cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{30 b d^3}-\frac{11 \cos ^5(a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}\\ \end{align*}
Mathematica [C] time = 0.838343, size = 89, normalized size = 0.63 \[ \frac{\sin (a+b x) \left (-308 \tan ^2(a+b x) \sqrt{\sec ^2(a+b x)} \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\tan ^2(a+b x)\right )+34 \cos (2 (a+b x))+3 \cos (4 (a+b x))-277\right )}{120 b (d \tan (a+b x))^{3/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.165, size = 536, normalized size = 3.8 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (b x + a\right )^{5}}{\left (d \tan \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d \tan \left (b x + a\right )} \cos \left (b x + a\right )^{5}}{d^{2} \tan \left (b x + a\right )^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (b x + a\right )^{5}}{\left (d \tan \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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