Optimal. Leaf size=221 \[ \frac {8 \left (4 a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 b^4 d \sqrt {a+b \sin (c+d x)}}-\frac {32 a \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 b^4 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {4 \cos (c+d x) (4 a+b \sin (c+d x))}{3 b^3 d \sqrt {a+b \sin (c+d x)}}-\frac {2 \cos ^3(c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}} \]
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Rubi [A] time = 0.32, antiderivative size = 221, 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 = {2693, 2863, 2752, 2663, 2661, 2655, 2653} \[ \frac {8 \left (4 a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 b^4 d \sqrt {a+b \sin (c+d x)}}-\frac {4 \cos (c+d x) (4 a+b \sin (c+d x))}{3 b^3 d \sqrt {a+b \sin (c+d x)}}-\frac {32 a \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 b^4 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {2 \cos ^3(c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 2653
Rule 2655
Rule 2661
Rule 2663
Rule 2693
Rule 2752
Rule 2863
Rubi steps
\begin {align*} \int \frac {\cos ^4(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx &=-\frac {2 \cos ^3(c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}-\frac {2 \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx}{b}\\ &=-\frac {2 \cos ^3(c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}-\frac {4 \cos (c+d x) (4 a+b \sin (c+d x))}{3 b^3 d \sqrt {a+b \sin (c+d x)}}+\frac {8 \int \frac {-\frac {b}{2}-2 a \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx}{3 b^3}\\ &=-\frac {2 \cos ^3(c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}-\frac {4 \cos (c+d x) (4 a+b \sin (c+d x))}{3 b^3 d \sqrt {a+b \sin (c+d x)}}-\frac {(16 a) \int \sqrt {a+b \sin (c+d x)} \, dx}{3 b^4}+\frac {\left (4 \left (4 a^2-b^2\right )\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}} \, dx}{3 b^4}\\ &=-\frac {2 \cos ^3(c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}-\frac {4 \cos (c+d x) (4 a+b \sin (c+d x))}{3 b^3 d \sqrt {a+b \sin (c+d x)}}-\frac {\left (16 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 b^4 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\left (4 \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}{3 b^4 \sqrt {a+b \sin (c+d x)}}\\ &=-\frac {2 \cos ^3(c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}-\frac {32 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 b^4 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {8 \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}}}{3 b^4 d \sqrt {a+b \sin (c+d x)}}-\frac {4 \cos (c+d x) (4 a+b \sin (c+d x))}{3 b^3 d \sqrt {a+b \sin (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 1.03, size = 174, normalized size = 0.79 \[ \frac {b \cos (c+d x) \left (-16 a^2-20 a b \sin (c+d x)+b^2 \cos (2 (c+d x))-3 b^2\right )-8 \left (4 a^2-b^2\right ) (a+b) \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{3/2} F\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )+32 a (a+b)^2 \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{3/2} E\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )}{3 b^4 d (a+b \sin (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.81, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {b \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{4}}{3 \, a b^{2} \cos \left (d x + c\right )^{2} - a^{3} - 3 \, a b^{2} + {\left (b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}, x\right ) \]
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 \[ \int \frac {\cos \left (d x + c\right )^{4}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.78, size = 1047, normalized size = 4.74 \[ -\frac {2 \left (10 a \,b^{3} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+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 {b}{a +b}}\, \sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, b \left (4 \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 ) a \,b^{2}-\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 \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 \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}\right ) \sin \left (d x +c \right )-b^{4} \left (\cos ^{4}\left (d x +c \right )\right )+\left (8 a^{2} b^{2}+2 b^{4}\right ) \left (\cos ^{2}\left (d x +c \right )\right )+16 \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 -12 \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}-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}}\, \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}-16 \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}+16 \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}\right )}{3 \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}} b^{5} \cos \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 \[ \int \frac {\cos \left (d x + c\right )^{4}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
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 \[ \int \frac {{\cos \left (c+d\,x\right )}^4}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-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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