Optimal. Leaf size=207 \[ -\frac {16 a \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{15 d \sqrt {a+b \sin (c+d x)}}+\frac {2 \left (23 a^2+9 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{15 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {2 b \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 d}-\frac {16 a b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{15 d} \]
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Rubi [A] time = 0.26, antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2656, 2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac {16 a \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{15 d \sqrt {a+b \sin (c+d x)}}+\frac {2 \left (23 a^2+9 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{15 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {2 b \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 d}-\frac {16 a b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{15 d} \]
Antiderivative was successfully verified.
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Rule 2653
Rule 2655
Rule 2656
Rule 2661
Rule 2663
Rule 2752
Rule 2753
Rubi steps
\begin {align*} \int (a+b \sin (c+d x))^{5/2} \, dx &=-\frac {2 b \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 d}+\frac {2}{5} \int \sqrt {a+b \sin (c+d x)} \left (\frac {1}{2} \left (5 a^2+3 b^2\right )+4 a b \sin (c+d x)\right ) \, dx\\ &=-\frac {16 a b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{15 d}-\frac {2 b \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 d}+\frac {4}{15} \int \frac {\frac {1}{4} a \left (15 a^2+17 b^2\right )+\frac {1}{4} b \left (23 a^2+9 b^2\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx\\ &=-\frac {16 a b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{15 d}-\frac {2 b \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 d}-\frac {1}{15} \left (8 a \left (a^2-b^2\right )\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}} \, dx+\frac {1}{15} \left (23 a^2+9 b^2\right ) \int \sqrt {a+b \sin (c+d x)} \, dx\\ &=-\frac {16 a b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{15 d}-\frac {2 b \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 d}+\frac {\left (\left (23 a^2+9 b^2\right ) \sqrt {a+b \sin (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}} \, dx}{15 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {\left (8 a \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}} \, dx}{15 \sqrt {a+b \sin (c+d x)}}\\ &=-\frac {16 a b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{15 d}-\frac {2 b \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 d}+\frac {2 \left (23 a^2+9 b^2\right ) 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)}}{15 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {16 a \left (a^2-b^2\right ) 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}}}{15 d \sqrt {a+b \sin (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 0.99, size = 185, normalized size = 0.89 \[ \frac {b \cos (c+d x) \left (-22 a^2-28 a b \sin (c+d x)+3 b^2 \cos (2 (c+d x))-3 b^2\right )+16 a \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )-2 \left (23 a^3+23 a^2 b+9 a b^2+9 b^3\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} E\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )}{15 d \sqrt {a+b \sin (c+d x)}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}\right )} \sqrt {b \sin \left (d x + c\right ) + a}, x\right ) \]
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 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.22, size = 890, normalized size = 4.30 \[ \frac {2 a^{4} \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticF \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right )+\frac {16 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticF \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{3} b}{15}-\frac {4 a^{2} \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticF \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{2}}{5}-\frac {16 a \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticF \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{3}}{15}-\frac {6 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticF \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{4}}{5}-\frac {46 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticE \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{4}}{15}+\frac {28 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticE \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} b^{2}}{15}+\frac {6 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticE \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{4}}{5}+\frac {2 b^{4} \left (\sin ^{4}\left (d x +c \right )\right )}{5}+\frac {28 a \,b^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{15}+\frac {22 a^{2} b^{2} \left (\sin ^{2}\left (d x +c \right )\right )}{15}-\frac {2 b^{4} \left (\sin ^{2}\left (d x +c \right )\right )}{5}-\frac {28 a \,b^{3} \sin \left (d x +c \right )}{15}-\frac {22 a^{2} b^{2}}{15}}{b \cos \left (d x +c \right ) \sqrt {a +b \sin \left (d x +c \right )}\, d} \]
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 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+b\,\sin \left (c+d\,x\right )\right )}^{5/2} \,d x \]
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 \[ \int \left (a + b \sin {\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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