Optimal. Leaf size=231 \[ \frac{8 a b \cos (c+d x)}{3 d \left (a^2-b^2\right )^2 \sqrt{a+b \sin (c+d x)}}+\frac{2 b \cos (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}-\frac{2 \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 d \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}+\frac{8 a \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 d \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}} \]
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Rubi [A] time = 0.228373, antiderivative size = 231, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {2664, 2754, 2752, 2663, 2661, 2655, 2653} \[ \frac{8 a b \cos (c+d x)}{3 d \left (a^2-b^2\right )^2 \sqrt{a+b \sin (c+d x)}}+\frac{2 b \cos (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}-\frac{2 \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 d \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}+\frac{8 a \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 d \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}} \]
Antiderivative was successfully verified.
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Rule 2664
Rule 2754
Rule 2752
Rule 2663
Rule 2661
Rule 2655
Rule 2653
Rubi steps
\begin{align*} \int \frac{1}{(a+b \sin (c+d x))^{5/2}} \, dx &=\frac{2 b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}-\frac{2 \int \frac{-\frac{3 a}{2}+\frac{1}{2} b \sin (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx}{3 \left (a^2-b^2\right )}\\ &=\frac{2 b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac{8 a b \cos (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt{a+b \sin (c+d x)}}+\frac{4 \int \frac{\frac{1}{4} \left (3 a^2+b^2\right )+a b \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx}{3 \left (a^2-b^2\right )^2}\\ &=\frac{2 b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac{8 a b \cos (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt{a+b \sin (c+d x)}}+\frac{(4 a) \int \sqrt{a+b \sin (c+d x)} \, dx}{3 \left (a^2-b^2\right )^2}-\frac{\int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx}{3 \left (a^2-b^2\right )}\\ &=\frac{2 b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac{8 a b \cos (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt{a+b \sin (c+d x)}}+\frac{\left (4 a \sqrt{a+b \sin (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}} \, dx}{3 \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{\sqrt{\frac{a+b \sin (c+d x)}{a+b}} \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}}} \, dx}{3 \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}\\ &=\frac{2 b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac{8 a b \cos (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt{a+b \sin (c+d x)}}+\frac{8 a E\left (\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )|\frac{2 b}{a+b}\right ) \sqrt{a+b \sin (c+d x)}}{3 \left (a^2-b^2\right )^2 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{2 F\left (\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )|\frac{2 b}{a+b}\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}{3 \left (a^2-b^2\right ) d \sqrt{a+b \sin (c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.904899, size = 166, normalized size = 0.72 \[ \frac{2 \left (b \cos (c+d x) \left (5 a^2+4 a b \sin (c+d x)-b^2\right )+(a-b) (a+b)^2 \left (\frac{a+b \sin (c+d x)}{a+b}\right )^{3/2} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )-4 a (a+b)^2 \left (\frac{a+b \sin (c+d x)}{a+b}\right )^{3/2} E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )\right )}{3 d (a-b)^2 (a+b)^2 (a+b \sin (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.339, size = 497, normalized size = 2.2 \begin{align*}{\frac{1}{d\cos \left ( dx+c \right ) }\sqrt{- \left ( -b\sin \left ( dx+c \right ) -a \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}} \left ({\frac{2}{ \left ( 3\,{a}^{2}-3\,{b}^{2} \right ) b}\sqrt{- \left ( -b\sin \left ( dx+c \right ) -a \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}} \left ( \sin \left ( dx+c \right ) +{\frac{a}{b}} \right ) ^{-2}}+{\frac{8\,b \left ( \cos \left ( dx+c \right ) \right ) ^{2}a}{3\, \left ({a}^{2}-{b}^{2} \right ) ^{2}}{\frac{1}{\sqrt{- \left ( -b\sin \left ( dx+c \right ) -a \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}}}+2\,{\frac{3\,{a}^{2}+{b}^{2}}{ \left ( 3\,{a}^{4}-6\,{a}^{2}{b}^{2}+3\,{b}^{4} \right ) \sqrt{- \left ( -b\sin \left ( dx+c \right ) -a \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}} \left ({\frac{a}{b}}-1 \right ) \sqrt{{\frac{a+b\sin \left ( dx+c \right ) }{a-b}}}\sqrt{{\frac{b \left ( 1-\sin \left ( dx+c \right ) \right ) }{a+b}}}\sqrt{{\frac{ \left ( -1-\sin \left ( dx+c \right ) \right ) b}{a-b}}}{\it EllipticF} \left ( \sqrt{{\frac{a+b\sin \left ( dx+c \right ) }{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ) }+{\frac{8\,ab}{3\, \left ({a}^{2}-{b}^{2} \right ) ^{2}} \left ({\frac{a}{b}}-1 \right ) \sqrt{{\frac{a+b\sin \left ( dx+c \right ) }{a-b}}}\sqrt{{\frac{b \left ( 1-\sin \left ( dx+c \right ) \right ) }{a+b}}}\sqrt{{\frac{ \left ( -1-\sin \left ( dx+c \right ) \right ) b}{a-b}}} \left ( \left ( -{\frac{a}{b}}-1 \right ){\it EllipticE} \left ( \sqrt{{\frac{a+b\sin \left ( dx+c \right ) }{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ) +{\it EllipticF} \left ( \sqrt{{\frac{a+b\sin \left ( dx+c \right ) }{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ) \right ){\frac{1}{\sqrt{- \left ( -b\sin \left ( dx+c \right ) -a \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}}} \right ){\frac{1}{\sqrt{a+b\sin \left ( dx+c \right ) }}}} \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{1}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{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{b \sin \left (d x + c\right ) + a}}{3 \, a b^{2} \cos \left (d x + c\right )^{2} - a^{3} - 3 \, a b^{2} +{\left (b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \sin{\left (c + d x \right )}\right )^{\frac{5}{2}}}\, dx \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 \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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