Optimal. Leaf size=535 \[ \frac {3 \sqrt {b} \left (5 a^2+2 b^2\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 \left (-a^2+b^2\right )^{11/4} d \sqrt {e}}+\frac {3 \sqrt {b} \left (5 a^2+2 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 \left (-a^2+b^2\right )^{11/4} d \sqrt {e}}-\frac {7 a F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d \sqrt {e \sin (c+d x)}}+\frac {3 a \left (5 a^2+2 b^2\right ) \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{8 \left (a^2-b^2\right )^2 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}+\frac {3 a \left (5 a^2+2 b^2\right ) \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{8 \left (a^2-b^2\right )^2 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}-\frac {b \sqrt {e \sin (c+d x)}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac {7 a b \sqrt {e \sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))} \]
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Rubi [A]
time = 0.80, antiderivative size = 535, 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 = {2773, 2943,
2946, 2721, 2720, 2781, 2886, 2884, 335, 218, 214, 211} \begin {gather*} \frac {3 \sqrt {b} \left (5 a^2+2 b^2\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{8 d \sqrt {e} \left (b^2-a^2\right )^{11/4}}+\frac {3 \sqrt {b} \left (5 a^2+2 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{8 d \sqrt {e} \left (b^2-a^2\right )^{11/4}}-\frac {7 a b \sqrt {e \sin (c+d x)}}{4 d e \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac {b \sqrt {e \sin (c+d x)}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {7 a \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{4 d \left (a^2-b^2\right )^2 \sqrt {e \sin (c+d x)}}+\frac {3 a \left (5 a^2+2 b^2\right ) \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 )}{8 d \left (a^2-b^2\right )^2 \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \sin (c+d x)}}+\frac {3 a \left (5 a^2+2 b^2\right ) \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 )}{8 d \left (a^2-b^2\right )^2 \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \sin (c+d x)}} \end {gather*}
Antiderivative was successfully verified.
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Rule 211
Rule 214
Rule 218
Rule 335
Rule 2720
Rule 2721
Rule 2773
Rule 2781
Rule 2884
Rule 2886
Rule 2943
Rule 2946
Rubi steps
\begin {align*} \int \frac {1}{(a+b \cos (c+d x))^3 \sqrt {e \sin (c+d x)}} \, dx &=-\frac {b \sqrt {e \sin (c+d x)}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac {\int \frac {-2 a+\frac {3}{2} b \cos (c+d x)}{(a+b \cos (c+d x))^2 \sqrt {e \sin (c+d x)}} \, dx}{2 \left (a^2-b^2\right )}\\ &=-\frac {b \sqrt {e \sin (c+d x)}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac {7 a b \sqrt {e \sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))}+\frac {\int \frac {\frac {1}{2} \left (4 a^2+3 b^2\right )-\frac {7}{4} a b \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=-\frac {b \sqrt {e \sin (c+d x)}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac {7 a b \sqrt {e \sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))}-\frac {(7 a) \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx}{8 \left (a^2-b^2\right )^2}+\frac {\left (3 \left (5 a^2+2 b^2\right )\right ) \int \frac {1}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx}{8 \left (a^2-b^2\right )^2}\\ &=-\frac {b \sqrt {e \sin (c+d x)}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac {7 a b \sqrt {e \sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))}-\frac {\left (3 a \left (5 a^2+2 b^2\right )\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{16 \left (-a^2+b^2\right )^{5/2}}-\frac {\left (3 a \left (5 a^2+2 b^2\right )\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{16 \left (-a^2+b^2\right )^{5/2}}-\frac {\left (3 b \left (5 a^2+2 b^2\right ) e\right ) \text {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 )}{8 \left (a^2-b^2\right )^2 d}-\frac {\left (7 a \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{8 \left (a^2-b^2\right )^2 \sqrt {e \sin (c+d x)}}\\ &=-\frac {7 a F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d \sqrt {e \sin (c+d x)}}-\frac {b \sqrt {e \sin (c+d x)}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac {7 a b \sqrt {e \sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))}-\frac {\left (3 b \left (5 a^2+2 b^2\right ) e\right ) \text {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 )}{4 \left (a^2-b^2\right )^2 d}-\frac {\left (3 a \left (5 a^2+2 b^2\right ) \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}{16 \left (-a^2+b^2\right )^{5/2} \sqrt {e \sin (c+d x)}}-\frac {\left (3 a \left (5 a^2+2 b^2\right ) \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}{16 \left (-a^2+b^2\right )^{5/2} \sqrt {e \sin (c+d x)}}\\ &=-\frac {7 a F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d \sqrt {e \sin (c+d x)}}+\frac {3 a \left (5 a^2+2 b^2\right ) \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)}}{8 \left (-a^2+b^2\right )^{5/2} \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {3 a \left (5 a^2+2 b^2\right ) \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)}}{8 \left (-a^2+b^2\right )^{5/2} \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {b \sqrt {e \sin (c+d x)}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac {7 a b \sqrt {e \sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))}+\frac {\left (3 b \left (5 a^2+2 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e-b x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{8 \left (-a^2+b^2\right )^{5/2} d}+\frac {\left (3 b \left (5 a^2+2 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e+b x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{8 \left (-a^2+b^2\right )^{5/2} d}\\ &=\frac {3 \sqrt {b} \left (5 a^2+2 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 \left (-a^2+b^2\right )^{11/4} d \sqrt {e}}+\frac {3 \sqrt {b} \left (5 a^2+2 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 \left (-a^2+b^2\right )^{11/4} d \sqrt {e}}-\frac {7 a F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d \sqrt {e \sin (c+d x)}}+\frac {3 a \left (5 a^2+2 b^2\right ) \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)}}{8 \left (-a^2+b^2\right )^{5/2} \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {3 a \left (5 a^2+2 b^2\right ) \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)}}{8 \left (-a^2+b^2\right )^{5/2} \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {b \sqrt {e \sin (c+d x)}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac {7 a b \sqrt {e \sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in
optimal.
time = 29.75, size = 1226, normalized size = 2.29 \begin {gather*} \frac {\left (-\frac {b}{2 \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {7 a b}{4 \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}\right ) \sin (c+d x)}{d \sqrt {e \sin (c+d x)}}+\frac {\sqrt {\sin (c+d x)} \left (-\frac {14 a b \cos ^2(c+d x) \left (a+b \sqrt {1-\sin ^2(c+d x)}\right ) \left (\frac {a \left (-2 \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+b \sin (c+d x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+b \sin (c+d x)\right )\right )}{4 \sqrt {2} \sqrt {b} \left (a^2-b^2\right )^{3/4}}+\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 {b^2 \sin ^2(c+d x)}{-a^2+b^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)}{-a^2+b^2}\right )+2 \left (2 b^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)}{-a^2+b^2}\right )+\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)}{-a^2+b^2}\right )\right ) \sin ^2(c+d x)\right ) \left (a^2+b^2 \left (-1+\sin ^2(c+d x)\right )\right )}\right )}{(a+b \cos (c+d x)) \left (1-\sin ^2(c+d x)\right )}+\frac {2 \left (8 a^2+6 b^2\right ) \cos (c+d x) \left (a+b \sqrt {1-\sin ^2(c+d x)}\right ) \left (-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \sqrt {b} \left (2 \text {ArcTan}\left (1-\frac {(1+i) \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \text {ArcTan}\left (1+\frac {(1+i) \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )+\log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+i b \sin (c+d x)\right )-\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+i b \sin (c+d x)\right )\right )}{\left (-a^2+b^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 {b^2 \sin ^2(c+d x)}{-a^2+b^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)}{-a^2+b^2}\right )-2 \left (2 b^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)}{-a^2+b^2}\right )+\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 {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right )\right ) \sin ^2(c+d x)\right ) \left (a^2+b^2 \left (-1+\sin ^2(c+d x)\right )\right )}\right )}{(a+b \cos (c+d x)) \sqrt {1-\sin ^2(c+d x)}}\right )}{8 (a-b)^2 (a+b)^2 d \sqrt {e \sin (c+d x)}} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(2837\) vs.
\(2(559)=1118\).
time = 0.59, size = 2838, normalized size = 5.30
method | result | size |
default | \(\text {Expression too large to display}\) | \(2838\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {1}{\sqrt {e\,\sin \left (c+d\,x\right )}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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