Optimal. Leaf size=390 \[ \frac {3 a e^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{5/2} \sqrt [4]{-a^2+b^2} d}-\frac {3 a e^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{5/2} \sqrt [4]{-a^2+b^2} d}-\frac {3 e^2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt {\cos (c+d x)}}+\frac {3 a^2 e^3 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{2 b^3 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}+\frac {3 a^2 e^3 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{2 b^3 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {e (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))} \]
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Rubi [A]
time = 0.54, antiderivative size = 390, 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, 2719, 2780, 2886, 2884, 335, 304, 211, 214} \begin {gather*} \frac {3 a e^{5/2} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 b^{5/2} d \sqrt [4]{b^2-a^2}}-\frac {3 a e^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 b^{5/2} d \sqrt [4]{b^2-a^2}}+\frac {3 a^2 e^3 \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^3 d \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \cos (c+d x)}}+\frac {3 a^2 e^3 \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^3 d \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \cos (c+d x)}}-\frac {e (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))}-\frac {3 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{b^2 d \sqrt {\cos (c+d x)}} \end {gather*}
Antiderivative was successfully verified.
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Rule 211
Rule 214
Rule 304
Rule 335
Rule 2719
Rule 2721
Rule 2772
Rule 2780
Rule 2884
Rule 2886
Rule 2946
Rubi steps
\begin {align*} \int \frac {(e \cos (c+d x))^{5/2}}{(a+b \sin (c+d x))^2} \, dx &=-\frac {e (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))}-\frac {\left (3 e^2\right ) \int \frac {\sqrt {e \cos (c+d x)} \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{2 b}\\ &=-\frac {e (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))}-\frac {\left (3 e^2\right ) \int \sqrt {e \cos (c+d x)} \, dx}{2 b^2}+\frac {\left (3 a e^2\right ) \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)} \, dx}{2 b^2}\\ &=-\frac {e (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))}-\frac {\left (3 a^2 e^3\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^3}+\frac {\left (3 a^2 e^3\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^3}+\frac {\left (3 a e^3\right ) \text {Subst}\left (\int \frac {\sqrt {x}}{\left (a^2-b^2\right ) e^2+b^2 x^2} \, dx,x,e \cos (c+d x)\right )}{2 b d}-\frac {\left (3 e^2 \sqrt {e \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 b^2 \sqrt {\cos (c+d x)}}\\ &=-\frac {3 e^2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt {\cos (c+d x)}}-\frac {e (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))}+\frac {\left (3 a e^3\right ) \text {Subst}\left (\int \frac {x^2}{\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 (3 a^2 e^3 \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^3 \sqrt {e \cos (c+d x)}}+\frac {\left (3 a^2 e^3 \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^3 \sqrt {e \cos (c+d x)}}\\ &=-\frac {3 e^2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt {\cos (c+d x)}}+\frac {3 a^2 e^3 \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^3 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}+\frac {3 a^2 e^3 \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^3 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {e (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))}-\frac {\left (3 a e^3\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^2 d}+\frac {\left (3 a e^3\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^2 d}\\ &=\frac {3 a e^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{5/2} \sqrt [4]{-a^2+b^2} d}-\frac {3 a e^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{5/2} \sqrt [4]{-a^2+b^2} d}-\frac {3 e^2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt {\cos (c+d x)}}+\frac {3 a^2 e^3 \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^3 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}+\frac {3 a^2 e^3 \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^3 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {e (e \cos (c+d x))^{3/2}}{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 = 86.90, size = 371, normalized size = 0.95 \begin {gather*} \frac {(e \cos (c+d x))^{5/2} \left (-8 b^{3/2} \cos ^{\frac {3}{2}}(c+d x)-\frac {\left (8 b^{5/2} F_1\left (\frac {3}{4};-\frac {1}{2},1;\frac {7}{4};\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(c+d x)+3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \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 )\right ) \left (a+b \sqrt {\sin ^2(c+d x)}\right )}{a^2-b^2}\right )}{8 b^{5/2} d \cos ^{\frac {5}{2}}(c+d x) (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 = 23.79, size = 12860, normalized size = 32.97
method | result | size |
default | \(\text {Expression too large to display}\) | \(12860\) |
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(-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 {{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\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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