Optimal. Leaf size=397 \[ -\frac {\sqrt [4]{-a^2+b^2} e^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{3/2} d}-\frac {\sqrt [4]{-a^2+b^2} e^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{3/2} d}+\frac {2 e \sqrt {e \cos (c+d x)}}{b d}+\frac {2 a e^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt {e \cos (c+d x)}}-\frac {a \left (a^2-b^2\right ) e^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \cos (c+d x)}}-\frac {a \left (a^2-b^2\right ) e^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \cos (c+d x)}} \]
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Rubi [A]
time = 0.57, antiderivative size = 397, 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 = {2774, 2946,
2721, 2720, 2781, 2886, 2884, 335, 218, 214, 211} \begin {gather*} -\frac {e^{3/2} \sqrt [4]{b^2-a^2} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} d}-\frac {a e^2 \left (a^2-b^2\right ) \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 )}{b^2 d \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \cos (c+d x)}}-\frac {a e^2 \left (a^2-b^2\right ) \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 )}{b^2 d \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \cos (c+d x)}}-\frac {e^{3/2} \sqrt [4]{b^2-a^2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} d}+\frac {2 a 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 {2 e \sqrt {e \cos (c+d x)}}{b d} \end {gather*}
Antiderivative was successfully verified.
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Rule 211
Rule 214
Rule 218
Rule 335
Rule 2720
Rule 2721
Rule 2774
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)} \, dx &=\frac {2 e \sqrt {e \cos (c+d x)}}{b d}+\frac {e^2 \int \frac {b+a \sin (c+d x)}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))} \, dx}{b}\\ &=\frac {2 e \sqrt {e \cos (c+d x)}}{b d}+\frac {\left (a e^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{b^2}+\frac {\left (\left (-a^2+b^2\right ) e^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))} \, dx}{b^2}\\ &=\frac {2 e \sqrt {e \cos (c+d x)}}{b d}-\frac {\left (a \sqrt {-a^2+b^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}{2 b^2}-\frac {\left (a \sqrt {-a^2+b^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}{2 b^2}-\frac {\left (\left (a^2-b^2\right ) 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 )}{b d}+\frac {\left (a e^2 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{b^2 \sqrt {e \cos (c+d x)}}\\ &=\frac {2 e \sqrt {e \cos (c+d x)}}{b d}+\frac {2 a 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 {\left (2 \left (a^2-b^2\right ) 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 \sqrt {-a^2+b^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}{2 b^2 \sqrt {e \cos (c+d x)}}-\frac {\left (a \sqrt {-a^2+b^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}{2 b^2 \sqrt {e \cos (c+d x)}}\\ &=\frac {2 e \sqrt {e \cos (c+d x)}}{b d}+\frac {2 a 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 \sqrt {-a^2+b^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 )}{b^2 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {a \sqrt {-a^2+b^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 )}{b^2 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {\left (\sqrt {-a^2+b^2} 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 )}{b d}-\frac {\left (\sqrt {-a^2+b^2} 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 )}{b d}\\ &=-\frac {\sqrt [4]{-a^2+b^2} e^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{3/2} d}-\frac {\sqrt [4]{-a^2+b^2} e^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{3/2} d}+\frac {2 e \sqrt {e \cos (c+d x)}}{b d}+\frac {2 a 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 \sqrt {-a^2+b^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 )}{b^2 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {a \sqrt {-a^2+b^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 )}{b^2 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \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 = 26.74, size = 434, normalized size = 1.09 \begin {gather*} \frac {\left (\frac {1}{20}-\frac {i}{20}\right ) (e \cos (c+d x))^{3/2} \left ((4+4 i) a b^{3/2} 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 ) \cos ^{\frac {5}{2}}(c+d x)-5 \left (a^2-b^2\right ) \left (2 \sqrt [4]{-a^2+b^2} \tan ^{-1}\left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \sqrt [4]{-a^2+b^2} \tan ^{-1}\left (1+\frac {(1+i) \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )+(4+4 i) \sqrt {b} \sqrt {\cos (c+d x)}+\sqrt [4]{-a^2+b^2} \log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (c+d x)}+i b \cos (c+d x)\right )-\sqrt [4]{-a^2+b^2} \log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (c+d x)}+i b \cos (c+d x)\right )\right )\right ) \sin (c+d x) \left (a+b \sqrt {\sin ^2(c+d x)}\right )}{b^{3/2} \left (-a^2+b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) \sqrt {\sin ^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 = 9.62, size = 828, normalized size = 2.09
method | result | size |
default | \(\text {Expression too large to display}\) | \(828\) |
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 )}^{3/2}}{a+b\,\sin \left (c+d\,x\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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