Optimal. Leaf size=404 \[ -\frac {a e^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{3/2} \left (-a^2+b^2\right )^{3/4} d}-\frac {a e^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{3/2} \left (-a^2+b^2\right )^{3/4} d}-\frac {e^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt {e \cos (c+d x)}}+\frac {a^2 e^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{2 b^2 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \cos (c+d x)}}+\frac {a^2 e^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{2 b^2 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \cos (c+d x)}}-\frac {e \sqrt {e \cos (c+d x)}}{b d (a+b \sin (c+d x))} \]
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Rubi [A]
time = 0.60, antiderivative size = 404, 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 = {2772, 2946,
2721, 2720, 2781, 2886, 2884, 335, 218, 214, 211} \begin {gather*} -\frac {a e^{3/2} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} d \left (b^2-a^2\right )^{3/4}}+\frac {a^2 e^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{2 b^2 d \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \cos (c+d x)}}+\frac {a^2 e^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{2 b^2 d \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \cos (c+d x)}}-\frac {a e^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} d \left (b^2-a^2\right )^{3/4}}-\frac {e \sqrt {e \cos (c+d x)}}{b d (a+b \sin (c+d x))}-\frac {e^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt {e \cos (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 2772
Rule 2781
Rule 2884
Rule 2886
Rule 2946
Rubi steps
\begin {align*} \int \frac {(e \cos (c+d x))^{3/2}}{(a+b \sin (c+d x))^2} \, dx &=-\frac {e \sqrt {e \cos (c+d x)}}{b d (a+b \sin (c+d x))}-\frac {e^2 \int \frac {\sin (c+d x)}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))} \, dx}{2 b}\\ &=-\frac {e \sqrt {e \cos (c+d x)}}{b d (a+b \sin (c+d x))}-\frac {e^2 \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{2 b^2}+\frac {\left (a e^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))} \, dx}{2 b^2}\\ &=-\frac {e \sqrt {e \cos (c+d x)}}{b d (a+b \sin (c+d x))}-\frac {\left (a^2 e^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {-a^2+b^2}-b \cos (c+d x)\right )} \, dx}{4 b^2 \sqrt {-a^2+b^2}}-\frac {\left (a^2 e^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {-a^2+b^2}+b \cos (c+d x)\right )} \, dx}{4 b^2 \sqrt {-a^2+b^2}}+\frac {\left (a e^3\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 \cos (c+d x)\right )}{2 b d}-\frac {\left (e^2 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{2 b^2 \sqrt {e \cos (c+d x)}}\\ &=-\frac {e^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt {e \cos (c+d x)}}-\frac {e \sqrt {e \cos (c+d x)}}{b d (a+b \sin (c+d x))}+\frac {\left (a e^3\right ) \text {Subst}\left (\int \frac {1}{\left (a^2-b^2\right ) e^2+b^2 x^4} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{b d}-\frac {\left (a^2 e^2 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \left (\sqrt {-a^2+b^2}-b \cos (c+d x)\right )} \, dx}{4 b^2 \sqrt {-a^2+b^2} \sqrt {e \cos (c+d x)}}-\frac {\left (a^2 e^2 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \left (\sqrt {-a^2+b^2}+b \cos (c+d x)\right )} \, dx}{4 b^2 \sqrt {-a^2+b^2} \sqrt {e \cos (c+d x)}}\\ &=-\frac {e^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt {e \cos (c+d x)}}+\frac {a^2 e^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{2 b^2 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \cos (c+d x)}}-\frac {a^2 e^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{2 b^2 \sqrt {-a^2+b^2} \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {e \sqrt {e \cos (c+d x)}}{b d (a+b \sin (c+d x))}-\frac {\left (a e^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e-b x^2} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{2 b \sqrt {-a^2+b^2} d}-\frac {\left (a e^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e+b x^2} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{2 b \sqrt {-a^2+b^2} d}\\ &=-\frac {a e^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{3/2} \left (-a^2+b^2\right )^{3/4} d}-\frac {a e^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{3/2} \left (-a^2+b^2\right )^{3/4} d}-\frac {e^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt {e \cos (c+d x)}}+\frac {a^2 e^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{2 b^2 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \cos (c+d x)}}-\frac {a^2 e^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{2 b^2 \sqrt {-a^2+b^2} \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {e \sqrt {e \cos (c+d x)}}{b d (a+b \sin (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 = 23.29, size = 614, normalized size = 1.52 \begin {gather*} -\frac {(e \cos (c+d x))^{3/2} \sec (c+d x)}{b d (a+b \sin (c+d x))}+\frac {(e \cos (c+d x))^{3/2} \left (a+b \sqrt {1-\cos ^2(c+d x)}\right ) \left (\frac {5 b \left (a^2-b^2\right ) F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right ) \sqrt {\cos (c+d x)} \sqrt {1-\cos ^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};\cos ^2(c+d x),\frac {b^2 \cos ^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};\cos ^2(c+d x),\frac {b^2 \cos ^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};\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right )\right ) \cos ^2(c+d x)\right ) \left (a^2+b^2 \left (-1+\cos ^2(c+d x)\right )\right )}+\frac {a \left (-2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (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 {\cos (c+d x)}+b \cos (c+d x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (c+d x)}+b \cos (c+d x)\right )\right )}{4 \sqrt {2} \sqrt {b} \left (a^2-b^2\right )^{3/4}}\right ) \sin ^2(c+d x)}{b d \cos ^{\frac {3}{2}}(c+d x) \left (1-\cos ^2(c+d x)\right ) (a+b \sin (c+d x))} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 24.52, size = 7936, normalized size = 19.64
method | result | size |
default | \(\text {Expression too large to display}\) | \(7936\) |
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 \begin {gather*} \text {Failed to integrate} \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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \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 {{\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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