Optimal. Leaf size=215 \[ \frac {3 e^{5/2} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \tan ^{-1}\left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d \left (a^2 \sin (c+d x)+a^2 \cos (c+d x)+a^2\right )}+\frac {3 e^{5/2} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \sinh ^{-1}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right )}{d \left (a^2 \sin (c+d x)+a^2 \cos (c+d x)+a^2\right )}+\frac {e (e \cos (c+d x))^{3/2}}{a d \sqrt {a \sin (c+d x)+a}} \]
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Rubi [A] time = 0.29, antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2679, 2684, 2775, 203, 2833, 63, 215} \[ \frac {3 e^{5/2} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \tan ^{-1}\left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d \left (a^2 \sin (c+d x)+a^2 \cos (c+d x)+a^2\right )}+\frac {3 e^{5/2} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \sinh ^{-1}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right )}{d \left (a^2 \sin (c+d x)+a^2 \cos (c+d x)+a^2\right )}+\frac {e (e \cos (c+d x))^{3/2}}{a d \sqrt {a \sin (c+d x)+a}} \]
Antiderivative was successfully verified.
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Rule 63
Rule 203
Rule 215
Rule 2679
Rule 2684
Rule 2775
Rule 2833
Rubi steps
\begin {align*} \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^{3/2}} \, dx &=\frac {e (e \cos (c+d x))^{3/2}}{a d \sqrt {a+a \sin (c+d x)}}+\frac {\left (3 e^2\right ) \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx}{2 a}\\ &=\frac {e (e \cos (c+d x))^{3/2}}{a d \sqrt {a+a \sin (c+d x)}}+\frac {\left (3 e^3 \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}\right ) \int \frac {\sqrt {1+\cos (c+d x)}}{\sqrt {e \cos (c+d x)}} \, dx}{2 a (a+a \cos (c+d x)+a \sin (c+d x))}-\frac {\left (3 e^3 \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}\right ) \int \frac {\sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}} \, dx}{2 a (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=\frac {e (e \cos (c+d x))^{3/2}}{a d \sqrt {a+a \sin (c+d x)}}+\frac {\left (3 e^3 \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {e x} \sqrt {1+x}} \, dx,x,\cos (c+d x)\right )}{2 a d (a+a \cos (c+d x)+a \sin (c+d x))}-\frac {\left (3 e^3 \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{1+e x^2} \, dx,x,-\frac {\sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}}\right )}{a d (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=\frac {e (e \cos (c+d x))^{3/2}}{a d \sqrt {a+a \sin (c+d x)}}+\frac {3 e^{5/2} \tan ^{-1}\left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d \left (a^2+a^2 \cos (c+d x)+a^2 \sin (c+d x)\right )}+\frac {\left (3 e^2 \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{e}}} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{a d (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=\frac {e (e \cos (c+d x))^{3/2}}{a d \sqrt {a+a \sin (c+d x)}}+\frac {3 e^{5/2} \sinh ^{-1}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d \left (a^2+a^2 \cos (c+d x)+a^2 \sin (c+d x)\right )}+\frac {3 e^{5/2} \tan ^{-1}\left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d \left (a^2+a^2 \cos (c+d x)+a^2 \sin (c+d x)\right )}\\ \end {align*}
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Mathematica [C] time = 0.15, size = 80, normalized size = 0.37 \[ -\frac {2 \sqrt [4]{2} \sqrt {a (\sin (c+d x)+1)} (e \cos (c+d x))^{7/2} \, _2F_1\left (\frac {3}{4},\frac {7}{4};\frac {11}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{7 a^2 d e (\sin (c+d x)+1)^{9/4}} \]
Antiderivative was successfully verified.
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fricas [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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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 [A] time = 0.24, size = 232, normalized size = 1.08 \[ -\frac {\left (e \cos \left (d x +c \right )\right )^{\frac {5}{2}} \left (-3 \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sin \left (d x +c \right )-3 \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sin \left (d x +c \right )+2 \cos \left (d x +c \right ) \sin \left (d x +c \right )+2 \left (\cos ^{2}\left (d x +c \right )\right )-2 \cos \left (d x +c \right )\right )}{2 d \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )-\left (\cos ^{2}\left (d x +c \right )\right )-2 \sin \left (d x +c \right )-\cos \left (d x +c \right )+2\right ) \left (a \left (1+\sin \left (d x +c \right )\right )\right )^{\frac {3}{2}}} \]
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 \[ \int \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]
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 \[ \int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]
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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