Optimal. Leaf size=265 \[ -\frac {2 \sqrt {2} a \left (4 a^2-b^2\right ) \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}} F\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{15 d \sqrt {2 a+b \sin (2 c+2 d x)}}+\frac {\left (92 a^2+9 b^2\right ) \sqrt {2 a+b \sin (2 c+2 d x)} E\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{60 \sqrt {2} d \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}-\frac {2 \sqrt {2} a b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{15 d} \]
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Rubi [A] time = 0.37, antiderivative size = 265, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2666, 2656, 2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac {2 \sqrt {2} a \left (4 a^2-b^2\right ) \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}} F\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{15 d \sqrt {2 a+b \sin (2 c+2 d x)}}+\frac {\left (92 a^2+9 b^2\right ) \sqrt {2 a+b \sin (2 c+2 d x)} E\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{60 \sqrt {2} d \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}-\frac {2 \sqrt {2} a b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{15 d} \]
Antiderivative was successfully verified.
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Rule 2653
Rule 2655
Rule 2656
Rule 2661
Rule 2663
Rule 2666
Rule 2752
Rule 2753
Rubi steps
\begin {align*} \int (a+b \cos (c+d x) \sin (c+d x))^{5/2} \, dx &=\int \left (a+\frac {1}{2} b \sin (2 c+2 d x)\right )^{5/2} \, dx\\ &=-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}+\frac {2}{5} \int \sqrt {a+\frac {1}{2} b \sin (2 c+2 d x)} \left (\frac {1}{8} \left (20 a^2+3 b^2\right )+2 a b \sin (2 c+2 d x)\right ) \, dx\\ &=-\frac {2 \sqrt {2} a b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{15 d}-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}+\frac {4}{15} \int \frac {\frac {1}{16} a \left (60 a^2+17 b^2\right )+\frac {1}{32} b \left (92 a^2+9 b^2\right ) \sin (2 c+2 d x)}{\sqrt {a+\frac {1}{2} b \sin (2 c+2 d x)}} \, dx\\ &=-\frac {2 \sqrt {2} a b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{15 d}-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}-\frac {1}{15} \left (2 a \left (4 a^2-b^2\right )\right ) \int \frac {1}{\sqrt {a+\frac {1}{2} b \sin (2 c+2 d x)}} \, dx+\frac {1}{60} \left (92 a^2+9 b^2\right ) \int \sqrt {a+\frac {1}{2} b \sin (2 c+2 d x)} \, dx\\ &=-\frac {2 \sqrt {2} a b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{15 d}-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}+\frac {\left (\left (92 a^2+9 b^2\right ) \sqrt {a+\frac {1}{2} b \sin (2 c+2 d x)}\right ) \int \sqrt {\frac {a}{a+\frac {b}{2}}+\frac {b \sin (2 c+2 d x)}{2 \left (a+\frac {b}{2}\right )}} \, dx}{60 \sqrt {\frac {a+\frac {1}{2} b \sin (2 c+2 d x)}{a+\frac {b}{2}}}}-\frac {\left (2 a \left (4 a^2-b^2\right ) \sqrt {\frac {a+\frac {1}{2} b \sin (2 c+2 d x)}{a+\frac {b}{2}}}\right ) \int \frac {1}{\sqrt {\frac {a}{a+\frac {b}{2}}+\frac {b \sin (2 c+2 d x)}{2 \left (a+\frac {b}{2}\right )}}} \, dx}{15 \sqrt {a+\frac {1}{2} b \sin (2 c+2 d x)}}\\ &=-\frac {2 \sqrt {2} a b \cos (2 c+2 d x) \sqrt {2 a+b \sin (2 c+2 d x)}}{15 d}-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt {2} d}+\frac {\left (92 a^2+9 b^2\right ) E\left (c-\frac {\pi }{4}+d x|\frac {2 b}{2 a+b}\right ) \sqrt {2 a+b \sin (2 c+2 d x)}}{60 \sqrt {2} d \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}-\frac {2 \sqrt {2} a \left (4 a^2-b^2\right ) F\left (c-\frac {\pi }{4}+d x|\frac {2 b}{2 a+b}\right ) \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}{15 d \sqrt {2 a+b \sin (2 c+2 d x)}}\\ \end {align*}
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Mathematica [A] time = 1.85, size = 202, normalized size = 0.76 \[ \frac {-32 a \left (4 a^2-b^2\right ) \sqrt {\frac {2 a+b \sin (2 (c+d x))}{2 a+b}} F\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )-b \left (88 a^2 \cos (2 (c+d x))+b \sin (4 (c+d x)) (28 a+3 b \sin (2 (c+d x)))\right )+2 \left (184 a^3+92 a^2 b+18 a b^2+9 b^3\right ) \sqrt {\frac {2 a+b \sin (2 (c+d x))}{2 a+b}} E\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{120 d \sqrt {4 a+2 b \sin (2 (c+d x))}} \]
Antiderivative was successfully verified.
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fricas [F] time = 2.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (b^{2} \cos \left (d x + c\right )^{4} - b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - a^{2}\right )} \sqrt {b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.52, size = 1138, normalized size = 4.29 \[ \text {result too large to display} \]
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 \[ \int {\left (b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]
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 \[ \int {\left (a+b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\right )}^{5/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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