Optimal. Leaf size=445 \[ -\frac {3 a \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 d \sqrt {e} \left (b^2-a^2\right )^{7/4}}-\frac {3 a \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 d \sqrt {e} \left (b^2-a^2\right )^{7/4}}-\frac {b \sqrt {e \sin (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d \left (a^2-b^2\right ) \sqrt {e \sin (c+d x)}}+\frac {3 a^2 \sqrt {\sin (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{2 d \left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \sin (c+d x)}}+\frac {3 a^2 \sqrt {\sin (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{2 d \left (a^2-b^2\right ) \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \sin (c+d x)}} \]
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Rubi [A] time = 0.93, antiderivative size = 445, 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 = {2694, 2867, 2642, 2641, 2702, 2807, 2805, 329, 212, 208, 205} \[ -\frac {3 a \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 d \sqrt {e} \left (b^2-a^2\right )^{7/4}}-\frac {3 a \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 d \sqrt {e} \left (b^2-a^2\right )^{7/4}}-\frac {b \sqrt {e \sin (c+d x)}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d \left (a^2-b^2\right ) \sqrt {e \sin (c+d x)}}+\frac {3 a^2 \sqrt {\sin (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{2 d \left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \sin (c+d x)}}+\frac {3 a^2 \sqrt {\sin (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{2 d \left (a^2-b^2\right ) \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \sin (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 205
Rule 208
Rule 212
Rule 329
Rule 2641
Rule 2642
Rule 2694
Rule 2702
Rule 2805
Rule 2807
Rule 2867
Rubi steps
\begin {align*} \int \frac {1}{(a+b \cos (c+d x))^2 \sqrt {e \sin (c+d x)}} \, dx &=-\frac {b \sqrt {e \sin (c+d x)}}{\left (a^2-b^2\right ) d e (a+b \cos (c+d x))}+\frac {\int \frac {-a+\frac {1}{2} b \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx}{-a^2+b^2}\\ &=-\frac {b \sqrt {e \sin (c+d x)}}{\left (a^2-b^2\right ) d e (a+b \cos (c+d x))}-\frac {\int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx}{2 \left (a^2-b^2\right )}+\frac {(3 a) \int \frac {1}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx}{2 \left (a^2-b^2\right )}\\ &=-\frac {b \sqrt {e \sin (c+d x)}}{\left (a^2-b^2\right ) d e (a+b \cos (c+d x))}+\frac {\left (3 a^2\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{4 \left (-a^2+b^2\right )^{3/2}}+\frac {\left (3 a^2\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{4 \left (-a^2+b^2\right )^{3/2}}-\frac {(3 a b e) \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 )}{2 \left (a^2-b^2\right ) d}-\frac {\sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{2 \left (a^2-b^2\right ) \sqrt {e \sin (c+d x)}}\\ &=-\frac {F\left (\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 ) d \sqrt {e \sin (c+d x)}}-\frac {b \sqrt {e \sin (c+d x)}}{\left (a^2-b^2\right ) d e (a+b \cos (c+d x))}-\frac {(3 a b e) \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}+\frac {\left (3 a^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}{4 \left (-a^2+b^2\right )^{3/2} \sqrt {e \sin (c+d x)}}+\frac {\left (3 a^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}{4 \left (-a^2+b^2\right )^{3/2} \sqrt {e \sin (c+d x)}}\\ &=-\frac {F\left (\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 ) d \sqrt {e \sin (c+d x)}}-\frac {3 a^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)}}{2 \left (-a^2+b^2\right )^{3/2} \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {3 a^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)}}{2 \left (-a^2+b^2\right )^{3/2} \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {b \sqrt {e \sin (c+d x)}}{\left (a^2-b^2\right ) d e (a+b \cos (c+d x))}-\frac {(3 a b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e-b x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{2 \left (-a^2+b^2\right )^{3/2} d}-\frac {(3 a b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e+b x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{2 \left (-a^2+b^2\right )^{3/2} d}\\ &=-\frac {3 a \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 \left (-a^2+b^2\right )^{7/4} d \sqrt {e}}-\frac {3 a \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 \left (-a^2+b^2\right )^{7/4} d \sqrt {e}}-\frac {F\left (\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 ) d \sqrt {e \sin (c+d x)}}-\frac {3 a^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)}}{2 \left (-a^2+b^2\right )^{3/2} \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {3 a^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)}}{2 \left (-a^2+b^2\right )^{3/2} \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {b \sqrt {e \sin (c+d x)}}{\left (a^2-b^2\right ) d e (a+b \cos (c+d x))}\\ \end {align*}
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Mathematica [C] time = 10.12, size = 1182, normalized size = 2.66 \[ \frac {\sqrt {\sin (c+d x)} \left (\frac {4 a \cos (c+d x) \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 )}{(a+b \cos (c+d x)) \sqrt {1-\sin ^2(c+d x)}}-\frac {2 b \cos ^2(c+d x) \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 )}{(a+b \cos (c+d x)) \left (1-\sin ^2(c+d x)\right )}\right )}{2 (a-b) (a+b) d \sqrt {e \sin (c+d x)}}-\frac {b \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}} \]
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 \[ \int \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sqrt {e \sin \left (d x + c\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.26, size = 1351, normalized size = 3.04 \[ \text {result too large to display} \]
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 \[ \int \frac {1}{\sqrt {e\,\sin \left (c+d\,x\right )}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]
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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