Optimal. Leaf size=452 \[ \frac {2 a \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{3 d e^2 \left (a^2-b^2\right ) \sqrt {e \sin (c+d x)}}+\frac {a b^2 \sqrt {\sin (c+d x)} \Pi \left (\frac {2 a}{a-\sqrt {a^2-b^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d e^2 \left (a^2-b^2\right ) \left (-a \sqrt {a^2-b^2}+a^2-b^2\right ) \sqrt {e \sin (c+d x)}}+\frac {a b^2 \sqrt {\sin (c+d x)} \Pi \left (\frac {2 a}{a+\sqrt {a^2-b^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d e^2 \left (a^2-b^2\right ) \left (a \sqrt {a^2-b^2}+a^2-b^2\right ) \sqrt {e \sin (c+d x)}}+\frac {2 (b-a \cos (c+d x))}{3 d e \left (a^2-b^2\right ) (e \sin (c+d x))^{3/2}}-\frac {a^{3/2} b \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{a^2-b^2}}\right )}{d e^{5/2} \left (a^2-b^2\right )^{7/4}}-\frac {a^{3/2} b \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{a^2-b^2}}\right )}{d e^{5/2} \left (a^2-b^2\right )^{7/4}} \]
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Rubi [A] time = 1.05, antiderivative size = 452, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3872, 2866, 2867, 2642, 2641, 2702, 2807, 2805, 329, 212, 208, 205} \[ -\frac {a^{3/2} b \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{a^2-b^2}}\right )}{d e^{5/2} \left (a^2-b^2\right )^{7/4}}-\frac {a^{3/2} b \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{a^2-b^2}}\right )}{d e^{5/2} \left (a^2-b^2\right )^{7/4}}+\frac {2 a \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{3 d e^2 \left (a^2-b^2\right ) \sqrt {e \sin (c+d x)}}+\frac {a b^2 \sqrt {\sin (c+d x)} \Pi \left (\frac {2 a}{a-\sqrt {a^2-b^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d e^2 \left (a^2-b^2\right ) \left (-a \sqrt {a^2-b^2}+a^2-b^2\right ) \sqrt {e \sin (c+d x)}}+\frac {a b^2 \sqrt {\sin (c+d x)} \Pi \left (\frac {2 a}{a+\sqrt {a^2-b^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d e^2 \left (a^2-b^2\right ) \left (a \sqrt {a^2-b^2}+a^2-b^2\right ) \sqrt {e \sin (c+d x)}}+\frac {2 (b-a \cos (c+d x))}{3 d e \left (a^2-b^2\right ) (e \sin (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 205
Rule 208
Rule 212
Rule 329
Rule 2641
Rule 2642
Rule 2702
Rule 2805
Rule 2807
Rule 2866
Rule 2867
Rule 3872
Rubi steps
\begin {align*} \int \frac {1}{(a+b \sec (c+d x)) (e \sin (c+d x))^{5/2}} \, dx &=-\int \frac {\cos (c+d x)}{(-b-a \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 {a b-\frac {1}{2} a^2 \cos (c+d x)}{(-b-a \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 {(a b) \int \frac {1}{(-b-a \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}-a \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}+a \sin (c+d x)\right )} \, dx}{2 \left (a^2-b^2\right )^{3/2} e^2}+\frac {\left (a^2 b\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (\left (-a^2+b^2\right ) e^2+a^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 a^2 b\right ) \operatorname {Subst}\left (\int \frac {1}{\left (-a^2+b^2\right ) e^2+a^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}-a \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}+a \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 a}{a-\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 (a-\sqrt {a^2-b^2}\right ) d e^2 \sqrt {e \sin (c+d x)}}+\frac {a b^2 \Pi \left (\frac {2 a}{a+\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 (a+\sqrt {a^2-b^2}\right ) d e^2 \sqrt {e \sin (c+d x)}}-\frac {\left (a^2 b\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a^2-b^2} e-a x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{\left (a^2-b^2\right )^{3/2} d e^2}-\frac {\left (a^2 b\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a^2-b^2} e+a x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{\left (a^2-b^2\right )^{3/2} d e^2}\\ &=-\frac {a^{3/2} b \tan ^{-1}\left (\frac {\sqrt {a} \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 {a^{3/2} b \tanh ^{-1}\left (\frac {\sqrt {a} \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 a}{a-\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 (a-\sqrt {a^2-b^2}\right ) d e^2 \sqrt {e \sin (c+d x)}}+\frac {a b^2 \Pi \left (\frac {2 a}{a+\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 (a+\sqrt {a^2-b^2}\right ) d e^2 \sqrt {e \sin (c+d x)}}\\ \end {align*}
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Mathematica [C] time = 12.71, size = 1233, normalized size = 2.73 \[ -\frac {a (b+a \cos (c+d x)) \sec (c+d x) \left (\frac {4 b \cos (c+d x) \left (\sqrt {1-\sin ^2(c+d x)} a+b\right ) \left (\frac {5 b \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 {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sqrt {\sin (c+d x)}}{\sqrt {1-\sin ^2(c+d x)} \left (2 \left (2 F_1\left (\frac {5}{4};\frac {1}{2},2;\frac {9}{4};\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) a^2+\left (a^2-b^2\right ) F_1\left (\frac {5}{4};\frac {3}{2},1;\frac {9}{4};\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^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 {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )\right ) \left (\left (\sin ^2(c+d x)-1\right ) a^2+b^2\right )}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \sqrt {a} \left (2 \tan ^{-1}\left (1-\frac {(1+i) \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \tan ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}+1\right )+\log \left (i a \sin (c+d x)-(1+i) \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+\sqrt {a^2-b^2}\right )-\log \left (i a \sin (c+d x)+(1+i) \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+\sqrt {a^2-b^2}\right )\right )}{\left (a^2-b^2\right )^{3/4}}\right )}{(b+a \cos (c+d x)) \sqrt {1-\sin ^2(c+d x)}}-\frac {2 a \cos ^2(c+d x) \left (\sqrt {1-\sin ^2(c+d x)} a+b\right ) \left (\frac {b \left (-2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )+2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}+1\right )-\log \left (a \sin (c+d x)-\sqrt {2} \sqrt {a} \sqrt [4]{b^2-a^2} \sqrt {\sin (c+d x)}+\sqrt {b^2-a^2}\right )+\log \left (a \sin (c+d x)+\sqrt {2} \sqrt {a} \sqrt [4]{b^2-a^2} \sqrt {\sin (c+d x)}+\sqrt {b^2-a^2}\right )\right )}{4 \sqrt {2} \sqrt {a} \left (b^2-a^2\right )^{3/4}}-\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 {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sqrt {\sin (c+d x)} \sqrt {1-\sin ^2(c+d x)}}{\left (2 \left (2 F_1\left (\frac {5}{4};-\frac {1}{2},2;\frac {9}{4};\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) a^2+\left (b^2-a^2\right ) F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^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 {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )\right ) \left (\left (\sin ^2(c+d x)-1\right ) a^2+b^2\right )}\right )}{(b+a \cos (c+d x)) \left (1-\sin ^2(c+d x)\right )}\right ) \sin ^{\frac {5}{2}}(c+d x)}{3 (a-b) (a+b) d (a+b \sec (c+d x)) (e \sin (c+d x))^{5/2}}-\frac {2 (b-a \cos (c+d x)) (b+a \cos (c+d x)) \tan (c+d x)}{3 \left (b^2-a^2\right ) d (a+b \sec (c+d x)) (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 \[ \int \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )} \left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 11.40, size = 681, normalized size = 1.51 \[ \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 \,a^{2} \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}} \ln \left (\frac {\sqrt {e \sin \left (d x +c \right )}+\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}{\sqrt {e \sin \left (d x +c \right )}-\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}\right )}{2 d e \left (a -b \right ) \left (a +b \right ) \left (-a^{2} e^{2}+b^{2} e^{2}\right )}+\frac {b \,a^{2} \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}\right )}{d e \left (a -b \right ) \left (a +b \right ) \left (-a^{2} e^{2}+b^{2} e^{2}\right )}+\frac {b^{2} \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}}}{a}}, \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}}}{a}\right )}-\frac {b^{2} \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}}}{a}}, \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}}}{a}\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 \[ \int \frac {\cos \left (c+d\,x\right )}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}\,\left (b+a\,\cos \left (c+d\,x\right )\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-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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