Optimal. Leaf size=218 \[ \frac {\left (4 a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{6 d \sqrt {a+b \sin (c+d x)}}+\frac {\sec ^3(c+d x) (a \sin (c+d x)+b) \sqrt {a+b \sin (c+d x)}}{3 d}-\frac {\sec (c+d x) (b-4 a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{6 d}-\frac {2 a \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}} \]
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Rubi [A] time = 0.44, antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2691, 2866, 2752, 2663, 2661, 2655, 2653} \[ \frac {\left (4 a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{6 d \sqrt {a+b \sin (c+d x)}}+\frac {\sec ^3(c+d x) (a \sin (c+d x)+b) \sqrt {a+b \sin (c+d x)}}{3 d}-\frac {\sec (c+d x) (b-4 a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{6 d}-\frac {2 a \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}} \]
Antiderivative was successfully verified.
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Rule 2653
Rule 2655
Rule 2661
Rule 2663
Rule 2691
Rule 2752
Rule 2866
Rubi steps
\begin {align*} \int \sec ^4(c+d x) (a+b \sin (c+d x))^{3/2} \, dx &=\frac {\sec ^3(c+d x) (b+a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{3 d}-\frac {1}{3} \int \frac {\sec ^2(c+d x) \left (-2 a^2+\frac {b^2}{2}-\frac {3}{2} a b \sin (c+d x)\right )}{\sqrt {a+b \sin (c+d x)}} \, dx\\ &=-\frac {\sec (c+d x) (b-4 a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{6 d}+\frac {\sec ^3(c+d x) (b+a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{3 d}+\frac {\int \frac {-\frac {1}{4} b^2 \left (a^2-b^2\right )-a b \left (a^2-b^2\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx}{3 \left (a^2-b^2\right )}\\ &=-\frac {\sec (c+d x) (b-4 a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{6 d}+\frac {\sec ^3(c+d x) (b+a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{3 d}-\frac {1}{3} a \int \sqrt {a+b \sin (c+d x)} \, dx+\frac {1}{12} \left (4 a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}} \, dx\\ &=-\frac {\sec (c+d x) (b-4 a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{6 d}+\frac {\sec ^3(c+d x) (b+a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{3 d}-\frac {\left (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 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\left (\left (4 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}{12 \sqrt {a+b \sin (c+d x)}}\\ &=-\frac {\sec (c+d x) (b-4 a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{6 d}+\frac {\sec ^3(c+d x) (b+a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{3 d}-\frac {2 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 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\left (4 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}}}{6 d \sqrt {a+b \sin (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 2.55, size = 211, normalized size = 0.97 \[ \frac {-4 \left (4 a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )+\sec ^3(c+d x) \left (12 a^2 \sin (c+d x)+4 a^2 \sin (3 (c+d x))-6 a b \cos (2 (c+d x))-2 a b \cos (4 (c+d x))+12 a b+7 b^2 \sin (c+d x)-b^2 \sin (3 (c+d x))\right )+16 a (a+b) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} E\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )}{24 d \sqrt {a+b \sin (c+d x)}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \sec \left (d x + c\right )^{4} \sin \left (d x + c\right ) + a \sec \left (d x + c\right )^{4}\right )} \sqrt {b \sin \left (d x + c\right ) + a}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-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 = 1.20, size = 938, normalized size = 4.30 \[ \frac {-\sqrt {\left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) b +\left (\cos ^{2}\left (d x +c \right )\right ) a}\, b \left (4 a^{2}-b^{2}\right ) \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )-2 \sqrt {\left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) b +\left (\cos ^{2}\left (d x +c \right )\right ) a}\, b \left (a^{2}+b^{2}\right ) \sin \left (d x +c \right )+4 \sqrt {\left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) b +\left (\cos ^{2}\left (d x +c \right )\right ) a}\, a \,b^{2} \left (\cos ^{4}\left (d x +c \right )\right )+\sqrt {\left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) b +\left (\cos ^{2}\left (d x +c \right )\right ) a}\, \left (4 \sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, \EllipticF \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} b -3 \EllipticF \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, a \,b^{2}-\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, \EllipticF \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{3}-4 \sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, \EllipticE \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{3}+4 \sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, \EllipticE \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a \,b^{2}-a \,b^{2}\right ) \left (\cos ^{2}\left (d x +c \right )\right )-4 \sqrt {\left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) b +\left (\cos ^{2}\left (d x +c \right )\right ) a}\, a \,b^{2}}{6 \sqrt {-\left (a +b \sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right ) \left (1+\sin \left (d x +c \right )\right )}\, \left (\sin \left (d x +c \right )-1\right ) \left (1+\sin \left (d x +c \right )\right ) 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 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{4}\,{d x} \]
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 \[ \int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^4} \,d x \]
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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