Optimal. Leaf size=224 \[ \frac {4 e^3}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos (c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos ^3(c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {4 e}{7 a^2 d (e \sin (c+d x))^{7/2}}+\frac {16 e \cos (c+d x)}{77 a^2 d (e \sin (c+d x))^{7/2}}-\frac {4 \cos (c+d x)}{231 a^2 d e (e \sin (c+d x))^{3/2}}+\frac {4 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{231 a^2 d e^2 \sqrt {e \sin (c+d x)}} \]
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Rubi [A]
time = 0.46, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps
used = 17, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3957, 2954,
2952, 2647, 2716, 2721, 2720, 2644, 14} \begin {gather*} \frac {4 e^3}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos ^3(c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos (c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}+\frac {4 \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{231 a^2 d e^2 \sqrt {e \sin (c+d x)}}-\frac {4 e}{7 a^2 d (e \sin (c+d x))^{7/2}}+\frac {16 e \cos (c+d x)}{77 a^2 d (e \sin (c+d x))^{7/2}}-\frac {4 \cos (c+d x)}{231 a^2 d e (e \sin (c+d x))^{3/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 14
Rule 2644
Rule 2647
Rule 2716
Rule 2720
Rule 2721
Rule 2952
Rule 2954
Rule 3957
Rubi steps
\begin {align*} \int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{5/2}} \, dx &=\int \frac {\cos ^2(c+d x)}{(-a-a \cos (c+d x))^2 (e \sin (c+d x))^{5/2}} \, dx\\ &=\frac {e^4 \int \frac {\cos ^2(c+d x) (-a+a \cos (c+d x))^2}{(e \sin (c+d x))^{13/2}} \, dx}{a^4}\\ &=\frac {e^4 \int \left (\frac {a^2 \cos ^2(c+d x)}{(e \sin (c+d x))^{13/2}}-\frac {2 a^2 \cos ^3(c+d x)}{(e \sin (c+d x))^{13/2}}+\frac {a^2 \cos ^4(c+d x)}{(e \sin (c+d x))^{13/2}}\right ) \, dx}{a^4}\\ &=\frac {e^4 \int \frac {\cos ^2(c+d x)}{(e \sin (c+d x))^{13/2}} \, dx}{a^2}+\frac {e^4 \int \frac {\cos ^4(c+d x)}{(e \sin (c+d x))^{13/2}} \, dx}{a^2}-\frac {\left (2 e^4\right ) \int \frac {\cos ^3(c+d x)}{(e \sin (c+d x))^{13/2}} \, dx}{a^2}\\ &=-\frac {2 e^3 \cos (c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos ^3(c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {\left (2 e^2\right ) \int \frac {1}{(e \sin (c+d x))^{9/2}} \, dx}{11 a^2}-\frac {\left (6 e^2\right ) \int \frac {\cos ^2(c+d x)}{(e \sin (c+d x))^{9/2}} \, dx}{11 a^2}-\frac {\left (2 e^3\right ) \text {Subst}\left (\int \frac {1-\frac {x^2}{e^2}}{x^{13/2}} \, dx,x,e \sin (c+d x)\right )}{a^2 d}\\ &=-\frac {2 e^3 \cos (c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos ^3(c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}+\frac {16 e \cos (c+d x)}{77 a^2 d (e \sin (c+d x))^{7/2}}-\frac {10 \int \frac {1}{(e \sin (c+d x))^{5/2}} \, dx}{77 a^2}+\frac {12 \int \frac {1}{(e \sin (c+d x))^{5/2}} \, dx}{77 a^2}-\frac {\left (2 e^3\right ) \text {Subst}\left (\int \left (\frac {1}{x^{13/2}}-\frac {1}{e^2 x^{9/2}}\right ) \, dx,x,e \sin (c+d x)\right )}{a^2 d}\\ &=\frac {4 e^3}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos (c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos ^3(c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {4 e}{7 a^2 d (e \sin (c+d x))^{7/2}}+\frac {16 e \cos (c+d x)}{77 a^2 d (e \sin (c+d x))^{7/2}}-\frac {4 \cos (c+d x)}{231 a^2 d e (e \sin (c+d x))^{3/2}}-\frac {10 \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx}{231 a^2 e^2}+\frac {4 \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx}{77 a^2 e^2}\\ &=\frac {4 e^3}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos (c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos ^3(c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {4 e}{7 a^2 d (e \sin (c+d x))^{7/2}}+\frac {16 e \cos (c+d x)}{77 a^2 d (e \sin (c+d x))^{7/2}}-\frac {4 \cos (c+d x)}{231 a^2 d e (e \sin (c+d x))^{3/2}}-\frac {\left (10 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{231 a^2 e^2 \sqrt {e \sin (c+d x)}}+\frac {\left (4 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{77 a^2 e^2 \sqrt {e \sin (c+d x)}}\\ &=\frac {4 e^3}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos (c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos ^3(c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {4 e}{7 a^2 d (e \sin (c+d x))^{7/2}}+\frac {16 e \cos (c+d x)}{77 a^2 d (e \sin (c+d x))^{7/2}}-\frac {4 \cos (c+d x)}{231 a^2 d e (e \sin (c+d x))^{3/2}}+\frac {4 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{231 a^2 d e^2 \sqrt {e \sin (c+d x)}}\\ \end {align*}
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Mathematica [A]
time = 0.65, size = 113, normalized size = 0.50 \begin {gather*} -\frac {\csc \left (\frac {1}{2} (c+d x)\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \left (52+97 \cos (c+d x)+4 \cos (2 (c+d x))+\cos (3 (c+d x))+\csc ^4\left (\frac {1}{2} (c+d x)\right ) F\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right ) \sin ^{\frac {11}{2}}(c+d x)\right )}{1848 a^2 d e^2 \sqrt {e \sin (c+d x)}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.26, size = 160, normalized size = 0.71
method | result | size |
default | \(\frac {\frac {4 e^{3} \left (11 \left (\cos ^{2}\left (d x +c \right )\right )-4\right )}{77 a^{2} \left (e \sin \left (d x +c \right )\right )^{\frac {11}{2}}}-\frac {2 \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {13}{2}}\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \left (\sin ^{7}\left (d x +c \right )\right )+47 \left (\sin ^{5}\left (d x +c \right )\right )-87 \left (\sin ^{3}\left (d x +c \right )\right )+42 \sin \left (d x +c \right )\right )}{231 e^{2} a^{2} \sin \left (d x +c \right )^{6} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\) | \(160\) |
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 [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 0.74, size = 226, normalized size = 1.01 \begin {gather*} \frac {2 \, {\left (\sqrt {-i} {\left (\sqrt {2} \cos \left (d x + c\right )^{4} + 2 \, \sqrt {2} \cos \left (d x + c\right )^{3} - 2 \, \sqrt {2} \cos \left (d x + c\right ) - \sqrt {2}\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {i} {\left (\sqrt {2} \cos \left (d x + c\right )^{4} + 2 \, \sqrt {2} \cos \left (d x + c\right )^{3} - 2 \, \sqrt {2} \cos \left (d x + c\right ) - \sqrt {2}\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (2 \, \cos \left (d x + c\right )^{3} + 4 \, \cos \left (d x + c\right )^{2} + 47 \, \cos \left (d x + c\right ) + 24\right )} \sqrt {\sin \left (d x + c\right )}\right )}}{231 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} e^{\frac {5}{2}} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} e^{\frac {5}{2}} - 2 \, a^{2} d \cos \left (d x + c\right ) e^{\frac {5}{2}} - a^{2} d e^{\frac {5}{2}}\right )}} \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 {{\cos \left (c+d\,x\right )}^2}{a^2\,{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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