Optimal. Leaf size=292 \[ \frac {2 b \left (23 a^2+9 b^2\right ) \cos (c+d x)}{15 d \left (a^2-b^2\right )^3 \sqrt {a+b \sin (c+d x)}}+\frac {16 a b \cos (c+d x)}{15 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{3/2}}+\frac {2 b \cos (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^{5/2}}-\frac {16 a \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{15 d \left (a^2-b^2\right )^2 \sqrt {a+b \sin (c+d x)}}+\frac {2 \left (23 a^2+9 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{15 d \left (a^2-b^2\right )^3 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}} \]
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Rubi [A] time = 0.35, antiderivative size = 292, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2664, 2754, 2752, 2663, 2661, 2655, 2653} \[ \frac {2 b \left (23 a^2+9 b^2\right ) \cos (c+d x)}{15 d \left (a^2-b^2\right )^3 \sqrt {a+b \sin (c+d x)}}+\frac {16 a b \cos (c+d x)}{15 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{3/2}}+\frac {2 b \cos (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^{5/2}}-\frac {16 a \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{15 d \left (a^2-b^2\right )^2 \sqrt {a+b \sin (c+d x)}}+\frac {2 \left (23 a^2+9 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{15 d \left (a^2-b^2\right )^3 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}} \]
Antiderivative was successfully verified.
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Rule 2653
Rule 2655
Rule 2661
Rule 2663
Rule 2664
Rule 2752
Rule 2754
Rubi steps
\begin {align*} \int \frac {1}{(a+b \sin (c+d x))^{7/2}} \, dx &=\frac {2 b \cos (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{5/2}}-\frac {2 \int \frac {-\frac {5 a}{2}+\frac {3}{2} b \sin (c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx}{5 \left (a^2-b^2\right )}\\ &=\frac {2 b \cos (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{5/2}}+\frac {16 a b \cos (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}+\frac {4 \int \frac {\frac {3}{4} \left (5 a^2+3 b^2\right )-2 a b \sin (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx}{15 \left (a^2-b^2\right )^2}\\ &=\frac {2 b \cos (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{5/2}}+\frac {16 a b \cos (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}+\frac {2 b \left (23 a^2+9 b^2\right ) \cos (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt {a+b \sin (c+d x)}}-\frac {8 \int \frac {-\frac {1}{8} a \left (15 a^2+17 b^2\right )-\frac {1}{8} b \left (23 a^2+9 b^2\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx}{15 \left (a^2-b^2\right )^3}\\ &=\frac {2 b \cos (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{5/2}}+\frac {16 a b \cos (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}+\frac {2 b \left (23 a^2+9 b^2\right ) \cos (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt {a+b \sin (c+d x)}}-\frac {(8 a) \int \frac {1}{\sqrt {a+b \sin (c+d x)}} \, dx}{15 \left (a^2-b^2\right )^2}+\frac {\left (23 a^2+9 b^2\right ) \int \sqrt {a+b \sin (c+d x)} \, dx}{15 \left (a^2-b^2\right )^3}\\ &=\frac {2 b \cos (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{5/2}}+\frac {16 a b \cos (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}+\frac {2 b \left (23 a^2+9 b^2\right ) \cos (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt {a+b \sin (c+d x)}}+\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 \left (a^2-b^2\right )^3 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {\left (8 a \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 \left (a^2-b^2\right )^2 \sqrt {a+b \sin (c+d x)}}\\ &=\frac {2 b \cos (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{5/2}}+\frac {16 a b \cos (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}+\frac {2 b \left (23 a^2+9 b^2\right ) \cos (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt {a+b \sin (c+d x)}}+\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 \left (a^2-b^2\right )^3 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {16 a 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 \left (a^2-b^2\right )^2 d \sqrt {a+b \sin (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 1.42, size = 198, normalized size = 0.68 \[ \frac {2 \left (\frac {b \cos (c+d x) \left (34 a^4+b^2 \left (23 a^2+9 b^2\right ) \sin ^2(c+d x)+2 a b \left (27 a^2+5 b^2\right ) \sin (c+d x)-5 a^2 b^2+3 b^4\right )}{\left (a^2-b^2\right )^3}-\frac {\left (\frac {a+b \sin (c+d x)}{a+b}\right )^{5/2} \left (\left (23 a^2+9 b^2\right ) E\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )+8 a (b-a) F\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )\right )}{(a-b)^3}\right )}{15 d (a+b \sin (c+d x))^{5/2}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sin \left (d x + c\right ) + a}}{b^{4} \cos \left (d x + c\right )^{4} + a^{4} + 6 \, a^{2} b^{2} + b^{4} - 2 \, {\left (3 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left (a b^{3} \cos \left (d x + c\right )^{2} - a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )}, x\right ) \]
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 \[ \int \frac {1}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.61, size = 584, normalized size = 2.00 \[ \frac {\sqrt {-\left (-b \sin \left (d x +c \right )-a \right ) \left (\cos ^{2}\left (d x +c \right )\right )}\, \left (\frac {2 \sqrt {-\left (-b \sin \left (d x +c \right )-a \right ) \left (\cos ^{2}\left (d x +c \right )\right )}}{5 b^{2} \left (a^{2}-b^{2}\right ) \left (\sin \left (d x +c \right )+\frac {a}{b}\right )^{3}}+\frac {16 a \sqrt {-\left (-b \sin \left (d x +c \right )-a \right ) \left (\cos ^{2}\left (d x +c \right )\right )}}{15 \left (a^{2}-b^{2}\right )^{2} b \left (\sin \left (d x +c \right )+\frac {a}{b}\right )^{2}}+\frac {2 b \left (\cos ^{2}\left (d x +c \right )\right ) \left (23 a^{2}+9 b^{2}\right )}{15 \left (a^{2}-b^{2}\right )^{3} \sqrt {-\left (-b \sin \left (d x +c \right )-a \right ) \left (\cos ^{2}\left (d x +c \right )\right )}}+\frac {2 \left (15 a^{3}+17 a \,b^{2}\right ) \left (\frac {a}{b}-1\right ) \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {\frac {b \left (1-\sin \left (d x +c \right )\right )}{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 )}{\left (15 a^{6}-45 a^{4} b^{2}+45 a^{2} b^{4}-15 b^{6}\right ) \sqrt {-\left (-b \sin \left (d x +c \right )-a \right ) \left (\cos ^{2}\left (d x +c \right )\right )}}+\frac {2 b \left (23 a^{2}+9 b^{2}\right ) \left (\frac {a}{b}-1\right ) \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {\frac {b \left (1-\sin \left (d x +c \right )\right )}{a +b}}\, \sqrt {\frac {\left (-1-\sin \left (d x +c \right )\right ) b}{a -b}}\, \left (\left (-\frac {a}{b}-1\right ) \EllipticE \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right )+\EllipticF \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right )\right )}{15 \left (a^{2}-b^{2}\right )^{3} \sqrt {-\left (-b \sin \left (d x +c \right )-a \right ) \left (\cos ^{2}\left (d x +c \right )\right )}}\right )}{\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 \[ \int \frac {1}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]
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 \[ \int \frac {1}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{7/2}} \,d x \]
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 \[ \int \frac {1}{\left (a + b \sin {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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