Optimal. Leaf size=328 \[ -\frac{a \left (5 a^2-11 b^2\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{4 b^2 d \left (a^2-b^2\right )^2}-\frac{\left (-29 a^2 b^2+15 a^4+8 b^4\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^3 d \left (a^2-b^2\right )^2}-\frac{a \left (-38 a^2 b^2+15 a^4+35 b^4\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^3 d (a-b)^2 (a+b)^3}+\frac{\left (-29 a^2 b^2+15 a^4+8 b^4\right ) \sin (c+d x)}{4 b^3 d \left (a^2-b^2\right )^2 \sqrt{\cos (c+d x)}}-\frac{a^2 \left (5 a^2-11 b^2\right ) \sin (c+d x)}{4 b^2 d \left (a^2-b^2\right )^2 \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac{a^2 \sin (c+d x)}{2 b d \left (a^2-b^2\right ) \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^2} \]
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Rubi [A] time = 0.999823, antiderivative size = 328, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 11, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.478, Rules used = {4264, 3845, 4098, 4102, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ -\frac{a \left (5 a^2-11 b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^2 d \left (a^2-b^2\right )^2}-\frac{\left (-29 a^2 b^2+15 a^4+8 b^4\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^3 d \left (a^2-b^2\right )^2}-\frac{a \left (-38 a^2 b^2+15 a^4+35 b^4\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^3 d (a-b)^2 (a+b)^3}+\frac{\left (-29 a^2 b^2+15 a^4+8 b^4\right ) \sin (c+d x)}{4 b^3 d \left (a^2-b^2\right )^2 \sqrt{\cos (c+d x)}}-\frac{a^2 \left (5 a^2-11 b^2\right ) \sin (c+d x)}{4 b^2 d \left (a^2-b^2\right )^2 \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac{a^2 \sin (c+d x)}{2 b d \left (a^2-b^2\right ) \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 4264
Rule 3845
Rule 4098
Rule 4102
Rule 4106
Rule 3849
Rule 2805
Rule 3787
Rule 3771
Rule 2639
Rule 2641
Rubi steps
\begin{align*} \int \frac{1}{\cos ^{\frac{9}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec ^{\frac{9}{2}}(c+d x)}{(a+b \sec (c+d x))^3} \, dx\\ &=-\frac{a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^2}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x) \left (\frac{3 a^2}{2}-2 a b \sec (c+d x)-\frac{1}{2} \left (5 a^2-4 b^2\right ) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac{a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^2}-\frac{a^2 \left (5 a^2-11 b^2\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\sec (c+d x)} \left (-\frac{1}{4} a^2 \left (5 a^2-11 b^2\right )+a b \left (a^2-4 b^2\right ) \sec (c+d x)+\frac{1}{4} \left (15 a^4-29 a^2 b^2+8 b^4\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )^2}\\ &=\frac{\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)}}-\frac{a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^2}-\frac{a^2 \left (5 a^2-11 b^2\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{8} a \left (15 a^4-29 a^2 b^2+8 b^4\right )-\frac{1}{2} b \left (5 a^4-10 a^2 b^2+2 b^4\right ) \sec (c+d x)-\frac{3}{8} a \left (5 a^4-11 a^2 b^2+8 b^4\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{b^3 \left (a^2-b^2\right )^2}\\ &=\frac{\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)}}-\frac{a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^2}-\frac{a^2 \left (5 a^2-11 b^2\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{8} a^2 \left (15 a^4-29 a^2 b^2+8 b^4\right )-\left (\frac{1}{2} a b \left (5 a^4-10 a^2 b^2+2 b^4\right )-\frac{1}{8} a b \left (15 a^4-29 a^2 b^2+8 b^4\right )\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{a^2 b^3 \left (a^2-b^2\right )^2}-\frac{\left (a \left (15 a^4-38 a^2 b^2+35 b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{8 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)}}-\frac{a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^2}-\frac{a^2 \left (5 a^2-11 b^2\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac{\left (a \left (15 a^4-38 a^2 b^2+35 b^4\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{8 b^3 \left (a^2-b^2\right )^2}-\frac{\left (a \left (5 a^2-11 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\sec (c+d x)} \, dx}{8 b^2 \left (a^2-b^2\right )^2}-\frac{\left (\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{8 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac{a \left (15 a^4-38 a^2 b^2+35 b^4\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 (a-b)^2 b^3 (a+b)^3 d}+\frac{\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)}}-\frac{a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^2}-\frac{a^2 \left (5 a^2-11 b^2\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))}-\frac{\left (a \left (5 a^2-11 b^2\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{8 b^2 \left (a^2-b^2\right )^2}-\frac{\left (15 a^4-29 a^2 b^2+8 b^4\right ) \int \sqrt{\cos (c+d x)} \, dx}{8 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (15 a^4-29 a^2 b^2+8 b^4\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac{a \left (5 a^2-11 b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^2 \left (a^2-b^2\right )^2 d}-\frac{a \left (15 a^4-38 a^2 b^2+35 b^4\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 (a-b)^2 b^3 (a+b)^3 d}+\frac{\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)}}-\frac{a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^2}-\frac{a^2 \left (5 a^2-11 b^2\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))}\\ \end{align*}
Mathematica [A] time = 3.32688, size = 336, normalized size = 1.02 \[ \frac{4 \sqrt{\cos (c+d x)} \left (\frac{a^3 \sin (c+d x) \left (a \left (7 a^2-13 b^2\right ) \cos (c+d x)+9 a^2 b-15 b^3\right )}{\left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}+8 \tan (c+d x)\right )-\frac{\frac{8 b \left (-10 a^2 b^2+5 a^4+2 b^4\right ) \left (2 \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-\frac{2 b \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}\right )}{a}-\frac{2 \left (-29 a^2 b^2+15 a^4+8 b^4\right ) \sin (c+d x) \left (-2 b (a+b) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a b \sqrt{\sin ^2(c+d x)}}+\frac{2 \left (-95 a^3 b^2+45 a^5+56 a b^4\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}}{(a-b)^2 (a+b)^2}}{16 b^3 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 7.894, size = 2014, normalized size = 6.1 \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(-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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Fricas [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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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{1}{{\left (b \sec \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac{9}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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