Optimal. Leaf size=447 \[ \frac {2 a \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{3 d e^2 \left (a^2-b^2\right ) \sqrt {e \sin (c+d x)}}-\frac {a b^2 \sqrt {\sin (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d e^2 \left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \sin (c+d x)}}-\frac {a b^2 \sqrt {\sin (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d e^2 \left (a^2-b^2\right ) \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \sin (c+d x)}}+\frac {2 (b-a \cos (c+d x))}{3 d e \left (a^2-b^2\right ) (e \sin (c+d x))^{3/2}}+\frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d e^{5/2} \left (b^2-a^2\right )^{7/4}}+\frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d e^{5/2} \left (b^2-a^2\right )^{7/4}} \]
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Rubi [A] time = 1.01, antiderivative size = 447, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {2696, 2867, 2642, 2641, 2702, 2807, 2805, 329, 212, 208, 205} \[ \frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d e^{5/2} \left (b^2-a^2\right )^{7/4}}+\frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d e^{5/2} \left (b^2-a^2\right )^{7/4}}+\frac {2 a \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{3 d e^2 \left (a^2-b^2\right ) \sqrt {e \sin (c+d x)}}-\frac {a b^2 \sqrt {\sin (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d e^2 \left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \sin (c+d x)}}-\frac {a b^2 \sqrt {\sin (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d e^2 \left (a^2-b^2\right ) \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \sin (c+d x)}}+\frac {2 (b-a \cos (c+d x))}{3 d e \left (a^2-b^2\right ) (e \sin (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 205
Rule 208
Rule 212
Rule 329
Rule 2641
Rule 2642
Rule 2696
Rule 2702
Rule 2805
Rule 2807
Rule 2867
Rubi steps
\begin {align*} \int \frac {1}{(a+b \cos (c+d x)) (e \sin (c+d x))^{5/2}} \, dx &=\frac {2 (b-a \cos (c+d x))}{3 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{3/2}}-\frac {2 \int \frac {-\frac {a^2}{2}+\frac {3 b^2}{2}-\frac {1}{2} a b \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx}{3 \left (a^2-b^2\right ) e^2}\\ &=\frac {2 (b-a \cos (c+d x))}{3 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{3/2}}+\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx}{3 \left (a^2-b^2\right ) e^2}-\frac {b^2 \int \frac {1}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx}{\left (a^2-b^2\right ) e^2}\\ &=\frac {2 (b-a \cos (c+d x))}{3 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{3/2}}-\frac {\left (a b^2\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{2 \left (-a^2+b^2\right )^{3/2} e^2}-\frac {\left (a b^2\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{2 \left (-a^2+b^2\right )^{3/2} e^2}+\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (\left (a^2-b^2\right ) e^2+b^2 x^2\right )} \, dx,x,e \sin (c+d x)\right )}{\left (a^2-b^2\right ) d e}+\frac {\left (a \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{3 \left (a^2-b^2\right ) e^2 \sqrt {e \sin (c+d x)}}\\ &=\frac {2 (b-a \cos (c+d x))}{3 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{3/2}}+\frac {2 a F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 \left (a^2-b^2\right ) d e^2 \sqrt {e \sin (c+d x)}}+\frac {\left (2 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{\left (a^2-b^2\right ) e^2+b^2 x^4} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{\left (a^2-b^2\right ) d e}-\frac {\left (a b^2 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{2 \left (-a^2+b^2\right )^{3/2} e^2 \sqrt {e \sin (c+d x)}}-\frac {\left (a b^2 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{2 \left (-a^2+b^2\right )^{3/2} e^2 \sqrt {e \sin (c+d x)}}\\ &=\frac {2 (b-a \cos (c+d x))}{3 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{3/2}}+\frac {2 a F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 \left (a^2-b^2\right ) d e^2 \sqrt {e \sin (c+d x)}}+\frac {a b^2 \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{\left (-a^2+b^2\right )^{3/2} \left (b-\sqrt {-a^2+b^2}\right ) d e^2 \sqrt {e \sin (c+d x)}}-\frac {a b^2 \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{\left (-a^2+b^2\right )^{3/2} \left (b+\sqrt {-a^2+b^2}\right ) d e^2 \sqrt {e \sin (c+d x)}}+\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e-b x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{\left (-a^2+b^2\right )^{3/2} d e^2}+\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e+b x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{\left (-a^2+b^2\right )^{3/2} d e^2}\\ &=\frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\left (-a^2+b^2\right )^{7/4} d e^{5/2}}+\frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\left (-a^2+b^2\right )^{7/4} d e^{5/2}}+\frac {2 (b-a \cos (c+d x))}{3 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{3/2}}+\frac {2 a F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 \left (a^2-b^2\right ) d e^2 \sqrt {e \sin (c+d x)}}+\frac {a b^2 \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{\left (-a^2+b^2\right )^{3/2} \left (b-\sqrt {-a^2+b^2}\right ) d e^2 \sqrt {e \sin (c+d x)}}-\frac {a b^2 \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{\left (-a^2+b^2\right )^{3/2} \left (b+\sqrt {-a^2+b^2}\right ) d e^2 \sqrt {e \sin (c+d x)}}\\ \end {align*}
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Mathematica [C] time = 11.25, size = 1192, normalized size = 2.67 \[ \frac {\sin ^{\frac {5}{2}}(c+d x) \left (\frac {2 a b \left (a+b \sqrt {1-\sin ^2(c+d x)}\right ) \left (\frac {5 b \left (a^2-b^2\right ) \sqrt {\sin (c+d x)} \sqrt {1-\sin ^2(c+d x)} F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{b^2-a^2}\right )}{\left (2 \left (2 F_1\left (\frac {5}{4};-\frac {1}{2},2;\frac {9}{4};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{b^2-a^2}\right ) b^2+\left (a^2-b^2\right ) F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{b^2-a^2}\right )\right ) \sin ^2(c+d x)-5 \left (a^2-b^2\right ) F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{b^2-a^2}\right )\right ) \left (a^2+b^2 \left (\sin ^2(c+d x)-1\right )\right )}+\frac {a \left (-2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}+1\right )-\log \left (b \sin (c+d x)-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+\sqrt {a^2-b^2}\right )+\log \left (b \sin (c+d x)+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+\sqrt {a^2-b^2}\right )\right )}{4 \sqrt {2} \sqrt {b} \left (a^2-b^2\right )^{3/4}}\right ) \cos ^2(c+d x)}{(a+b \cos (c+d x)) \left (1-\sin ^2(c+d x)\right )}+\frac {2 \left (a^2-3 b^2\right ) \left (a+b \sqrt {1-\sin ^2(c+d x)}\right ) \left (\frac {5 a \left (a^2-b^2\right ) F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{b^2-a^2}\right ) \sqrt {\sin (c+d x)}}{\sqrt {1-\sin ^2(c+d x)} \left (5 \left (a^2-b^2\right ) F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{b^2-a^2}\right )-2 \left (2 F_1\left (\frac {5}{4};\frac {1}{2},2;\frac {9}{4};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{b^2-a^2}\right ) b^2+\left (b^2-a^2\right ) F_1\left (\frac {5}{4};\frac {3}{2},1;\frac {9}{4};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{b^2-a^2}\right )\right ) \sin ^2(c+d x)\right ) \left (a^2+b^2 \left (\sin ^2(c+d x)-1\right )\right )}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \sqrt {b} \left (2 \tan ^{-1}\left (1-\frac {(1+i) \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \tan ^{-1}\left (\frac {(1+i) \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}+1\right )+\log \left (i b \sin (c+d x)-(1+i) \sqrt {b} \sqrt [4]{b^2-a^2} \sqrt {\sin (c+d x)}+\sqrt {b^2-a^2}\right )-\log \left (i b \sin (c+d x)+(1+i) \sqrt {b} \sqrt [4]{b^2-a^2} \sqrt {\sin (c+d x)}+\sqrt {b^2-a^2}\right )\right )}{\left (b^2-a^2\right )^{3/4}}\right ) \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt {1-\sin ^2(c+d x)}}\right )}{3 (a-b) (a+b) d (e \sin (c+d x))^{5/2}}-\frac {2 (a \cos (c+d x)-b) \sin (c+d x)}{3 \left (a^2-b^2\right ) d (e \sin (c+d x))^{5/2}} \]
Warning: Unable to verify antiderivative.
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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] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )} \left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.98, size = 845, normalized size = 1.89 \[ \frac {2 b}{3 d e \left (a^{2}-b^{2}\right ) \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {b^{3} \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {e \sin \left (d x +c \right )+\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}} \sqrt {e \sin \left (d x +c \right )}\, \sqrt {2}+\sqrt {\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}}}{e \sin \left (d x +c \right )-\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}} \sqrt {e \sin \left (d x +c \right )}\, \sqrt {2}+\sqrt {\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}}}\right )}{4 d e \left (a -b \right ) \left (a +b \right ) \left (a^{2} e^{2}-b^{2} e^{2}\right )}+\frac {b^{3} \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \sin \left (d x +c \right )}}{\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}}}+1\right )}{2 d e \left (a -b \right ) \left (a +b \right ) \left (a^{2} e^{2}-b^{2} e^{2}\right )}+\frac {b^{3} \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \sin \left (d x +c \right )}}{\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}}}-1\right )}{2 d e \left (a -b \right ) \left (a +b \right ) \left (a^{2} e^{2}-b^{2} e^{2}\right )}+\frac {a b \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticPi \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {1}{1-\frac {\sqrt {-a^{2}+b^{2}}}{b}}, \frac {\sqrt {2}}{2}\right )}{2 d \,e^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, \left (a -b \right ) \left (a +b \right ) \sqrt {-a^{2}+b^{2}}\, \left (1-\frac {\sqrt {-a^{2}+b^{2}}}{b}\right )}-\frac {a b \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticPi \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {1}{1+\frac {\sqrt {-a^{2}+b^{2}}}{b}}, \frac {\sqrt {2}}{2}\right )}{2 d \,e^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, \left (a -b \right ) \left (a +b \right ) \sqrt {-a^{2}+b^{2}}\, \left (1+\frac {\sqrt {-a^{2}+b^{2}}}{b}\right )}+\frac {a \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {5}{2}}\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )}{3 d \,e^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, \left (a^{2}-b^{2}\right ) \left (\cos ^{2}\left (d x +c \right )-1\right )}+\frac {2 a \cos \left (d x +c \right ) \sin \left (d x +c \right )}{3 d \,e^{2} \sqrt {e \sin \left (d x +c \right )}\, \left (a^{2}-b^{2}\right ) \left (\cos ^{2}\left (d x +c \right )-1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [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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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}\,\left (a+b\,\cos \left (c+d\,x\right )\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e \sin {\left (c + d x \right )}\right )^{\frac {5}{2}} \left (a + b \cos {\left (c + d x \right )}\right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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