Optimal. Leaf size=166 \[ \frac {a^{5/2} (c-d) (3 c+5 d) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{d^{5/2} f (c+d)^{3/2}}-\frac {a^3 (3 c+d) \cos (e+f x)}{d^2 f (c+d) \sqrt {a \sin (e+f x)+a}}+\frac {a^2 (c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{d f (c+d) (c+d \sin (e+f x))} \]
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Rubi [A] time = 0.39, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2762, 2981, 2773, 208} \[ -\frac {a^3 (3 c+d) \cos (e+f x)}{d^2 f (c+d) \sqrt {a \sin (e+f x)+a}}+\frac {a^{5/2} (c-d) (3 c+5 d) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{d^{5/2} f (c+d)^{3/2}}+\frac {a^2 (c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{d f (c+d) (c+d \sin (e+f x))} \]
Antiderivative was successfully verified.
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Rule 208
Rule 2762
Rule 2773
Rule 2981
Rubi steps
\begin {align*} \int \frac {(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^2} \, dx &=\frac {a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{d (c+d) f (c+d \sin (e+f x))}-\frac {a \int \frac {\sqrt {a+a \sin (e+f x)} \left (\frac {1}{2} a (c-5 d)-\frac {1}{2} a (3 c+d) \sin (e+f x)\right )}{c+d \sin (e+f x)} \, dx}{d (c+d)}\\ &=-\frac {a^3 (3 c+d) \cos (e+f x)}{d^2 (c+d) f \sqrt {a+a \sin (e+f x)}}+\frac {a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{d (c+d) f (c+d \sin (e+f x))}-\frac {\left (a^2 (c-d) (3 c+5 d)\right ) \int \frac {\sqrt {a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{2 d^2 (c+d)}\\ &=-\frac {a^3 (3 c+d) \cos (e+f x)}{d^2 (c+d) f \sqrt {a+a \sin (e+f x)}}+\frac {a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{d (c+d) f (c+d \sin (e+f x))}+\frac {\left (a^3 (c-d) (3 c+5 d)\right ) \operatorname {Subst}\left (\int \frac {1}{a c+a d-d x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{d^2 (c+d) f}\\ &=\frac {a^{5/2} (c-d) (3 c+5 d) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{d^{5/2} (c+d)^{3/2} f}-\frac {a^3 (3 c+d) \cos (e+f x)}{d^2 (c+d) f \sqrt {a+a \sin (e+f x)}}+\frac {a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{d (c+d) f (c+d \sin (e+f x))}\\ \end {align*}
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Mathematica [B] time = 4.09, size = 350, normalized size = 2.11 \[ \frac {(a (\sin (e+f x)+1))^{5/2} \left (\frac {\left (-3 c^2-2 c d+5 d^2\right ) \left (2 \log \left (\sqrt {d} \sqrt {c+d} \left (\tan ^2\left (\frac {1}{4} (e+f x)\right )+2 \tan \left (\frac {1}{4} (e+f x)\right )-1\right )+(c+d) \sec ^2\left (\frac {1}{4} (e+f x)\right )\right )-2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right )\right )+e+f x\right )}{(c+d)^{3/2}}+\frac {\left (3 c^2+2 c d-5 d^2\right ) \left (2 \log \left (-\sec ^2\left (\frac {1}{4} (e+f x)\right ) \left (-\sqrt {d} \sqrt {c+d} \sin \left (\frac {1}{2} (e+f x)\right )+\sqrt {d} \sqrt {c+d} \cos \left (\frac {1}{2} (e+f x)\right )+c+d\right )\right )-2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right )\right )+e+f x\right )}{(c+d)^{3/2}}-\frac {4 \sqrt {d} (c-d)^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )}{(c+d) (c+d \sin (e+f x))}+8 \sqrt {d} \sin \left (\frac {1}{2} (e+f x)\right )-8 \sqrt {d} \cos \left (\frac {1}{2} (e+f x)\right )\right )}{4 d^{5/2} f \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.63, size = 1322, normalized size = 7.96 \[ \text {result too large to display} \]
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 = 1.48, size = 393, normalized size = 2.37 \[ -\frac {a^{2} \left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (-\sin \left (f x +e \right ) d \left (3 \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a \,c^{2}+2 \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a c d -5 \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a \,d^{2}-2 \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, c -2 \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, d \right )-3 \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a \,c^{3}-2 \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a \,c^{2} d +5 \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a c \,d^{2}+3 \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, c^{2}+\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a \left (c +d \right ) d}\, d^{2}\right )}{d^{2} \left (c +d \right ) \left (c +d \sin \left (f x +e \right )\right ) \sqrt {a \left (c +d \right ) d}\, \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f} \]
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 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^2} \,d x \]
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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