Optimal. Leaf size=203 \[ \frac {a \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {a+b \sin (c+d x)}}-\frac {\left (a^2+3 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {a b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{d}+\frac {\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{d} \]
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Rubi [A] time = 0.28, antiderivative size = 203, 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, 2753, 2752, 2663, 2661, 2655, 2653} \[ \frac {a \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {a+b \sin (c+d x)}}-\frac {\left (a^2+3 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {a b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{d}+\frac {\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{d} \]
Antiderivative was successfully verified.
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Rule 2653
Rule 2655
Rule 2661
Rule 2663
Rule 2691
Rule 2752
Rule 2753
Rubi steps
\begin {align*} \int \sec ^2(c+d x) (a+b \sin (c+d x))^{5/2} \, dx &=\frac {\sec (c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{d}-\int \sqrt {a+b \sin (c+d x)} \left (\frac {3 b^2}{2}+\frac {3}{2} a b \sin (c+d x)\right ) \, dx\\ &=\frac {a b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{d}+\frac {\sec (c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{d}-\frac {2}{3} \int \frac {3 a b^2+\frac {3}{4} b \left (a^2+3 b^2\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx\\ &=\frac {a b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{d}+\frac {\sec (c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{d}+\frac {1}{2} \left (a \left (a^2-b^2\right )\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}} \, dx-\frac {1}{2} \left (a^2+3 b^2\right ) \int \sqrt {a+b \sin (c+d x)} \, dx\\ &=\frac {a b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{d}+\frac {\sec (c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{d}-\frac {\left (\left (a^2+3 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}{2 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\left (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}{2 \sqrt {a+b \sin (c+d x)}}\\ &=\frac {a b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{d}+\frac {\sec (c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{d}-\frac {\left (a^2+3 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)}}{d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {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}}}{d \sqrt {a+b \sin (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 0.90, size = 203, normalized size = 1.00 \[ \frac {a^3 \tan (c+d x)-a \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )+2 a^2 b \sec (c+d x)+a^2 b \sin (c+d x) \tan (c+d x)+\left (a^3+a^2 b+3 a b^2+3 b^3\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} E\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )+3 a b^2 \tan (c+d x)+b^3 \sin (c+d x) \tan (c+d x)}{d \sqrt {a+b \sin (c+d x)}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (2 \, a b \sec \left (d x + c\right )^{2} \sin \left (d x + c\right ) - {\left (b^{2} \cos \left (d x + c\right )^{2} - a^{2} - b^{2}\right )} \sec \left (d x + c\right )^{2}\right )} \sqrt {b \sin \left (d x + c\right ) + a}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-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 = 0.88, size = 1042, normalized size = 5.13 \[ -\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}\, \left (\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}}\, \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}}\, a^{3} b +3 \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}}\, \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}}\, a^{2} b^{2}-\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}}\, \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}}\, a \,b^{3}-3 \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}}\, \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}}\, b^{4}-\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}}\, \EllipticE \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}}\, a^{4}-2 \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}}\, \EllipticE \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}}\, a^{2} b^{2}+3 \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}}\, \EllipticE \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}}\, b^{4}+a^{2} b^{2} \left (\cos ^{2}\left (d x +c \right )\right )+b^{4} \left (\cos ^{2}\left (d x +c \right )\right )-a^{3} b \sin \left (d x +c \right )-3 a \,b^{3} \sin \left (d x +c \right )-3 a^{2} b^{2}-b^{4}\right )}{b \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 )}\, \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 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{2}\,{d x} \]
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 \[ \int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^2} \,d x \]
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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