Optimal. Leaf size=369 \[ \frac {i e^{5/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} d}-\frac {i e^{5/2} \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} d}-\frac {i e^{5/2} \log \left (a-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{2 \sqrt {2} \sqrt {a} d}+\frac {i e^{5/2} \log \left (a+\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{2 \sqrt {2} \sqrt {a} d}-\frac {i e^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}{a d} \]
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Rubi [A]
time = 0.23, antiderivative size = 369, normalized size of antiderivative = 1.00, number of steps
used = 11, number of rules used = 8, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3582, 3576,
303, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {i e^{5/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} d}-\frac {i e^{5/2} \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} d}-\frac {i e^{5/2} \log \left (-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{2 \sqrt {2} \sqrt {a} d}+\frac {i e^{5/2} \log \left (\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{2 \sqrt {2} \sqrt {a} d}-\frac {i e^2 \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}{a d} \end {gather*}
Antiderivative was successfully verified.
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Rule 210
Rule 303
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 3576
Rule 3582
Rubi steps
\begin {align*} \int \frac {(e \sec (c+d x))^{5/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx &=-\frac {i e^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}{a d}+\frac {e^2 \int \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx}{2 a}\\ &=-\frac {i e^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}{a d}-\frac {\left (2 i e^4\right ) \text {Subst}\left (\int \frac {x^2}{a^2+e^2 x^4} \, dx,x,\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{d}\\ &=-\frac {i e^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}{a d}+\frac {\left (i e^3\right ) \text {Subst}\left (\int \frac {a-e x^2}{a^2+e^2 x^4} \, dx,x,\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{d}-\frac {\left (i e^3\right ) \text {Subst}\left (\int \frac {a+e x^2}{a^2+e^2 x^4} \, dx,x,\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{d}\\ &=-\frac {i e^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}{a d}-\frac {\left (i e^2\right ) \text {Subst}\left (\int \frac {1}{\frac {a}{e}-\frac {\sqrt {2} \sqrt {a} x}{\sqrt {e}}+x^2} \, dx,x,\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{2 d}-\frac {\left (i e^2\right ) \text {Subst}\left (\int \frac {1}{\frac {a}{e}+\frac {\sqrt {2} \sqrt {a} x}{\sqrt {e}}+x^2} \, dx,x,\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{2 d}-\frac {\left (i e^{5/2}\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {a}}{\sqrt {e}}+2 x}{-\frac {a}{e}-\frac {\sqrt {2} \sqrt {a} x}{\sqrt {e}}-x^2} \, dx,x,\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{2 \sqrt {2} \sqrt {a} d}-\frac {\left (i e^{5/2}\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {a}}{\sqrt {e}}-2 x}{-\frac {a}{e}+\frac {\sqrt {2} \sqrt {a} x}{\sqrt {e}}-x^2} \, dx,x,\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{2 \sqrt {2} \sqrt {a} d}\\ &=-\frac {i e^{5/2} \log \left (a-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{2 \sqrt {2} \sqrt {a} d}+\frac {i e^{5/2} \log \left (a+\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{2 \sqrt {2} \sqrt {a} d}-\frac {i e^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}{a d}-\frac {\left (i e^{5/2}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} d}+\frac {\left (i e^{5/2}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} d}\\ &=\frac {i e^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} d}-\frac {i e^{5/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} d}-\frac {i e^{5/2} \log \left (a-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{2 \sqrt {2} \sqrt {a} d}+\frac {i e^{5/2} \log \left (a+\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{2 \sqrt {2} \sqrt {a} d}-\frac {i e^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}{a d}\\ \end {align*}
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Mathematica [A]
time = 3.93, size = 350, normalized size = 0.95 \begin {gather*} \frac {e^3 \left (\sec (c+d x) \sqrt {1+\cos (2 c)+i \sin (2 c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}-i \tanh ^{-1}\left (\frac {\sqrt {1-i \cos (c)+\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1-i \cos (c)-\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}}\right ) \sqrt {-1-i \cos (c)-\sin (c)} \sqrt {1+i \cos (c)-\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}+i \tanh ^{-1}\left (\frac {\sqrt {1+i \cos (c)-\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1+i \cos (c)+\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}}\right ) \sqrt {1-i \cos (c)+\sin (c)} \sqrt {-1+i \cos (c)+\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}\right ) (-i+\tan (c+d x))}{d \sqrt {e \sec (c+d x)} \sqrt {1+\cos (2 c)+i \sin (2 c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )} \sqrt {a+i a \tan (c+d x)}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 3.92, size = 316, normalized size = 0.86
method | result | size |
default | \(-\frac {\left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (\cos ^{2}\left (d x +c \right )\right ) \left (-1+\cos \left (d x +c \right )\right )^{3} \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (i \cos \left (d x +c \right ) \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right )}{2}\right )-i \cos \left (d x +c \right ) \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right )}{2}\right )+2 i \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}-\cos \left (d x +c \right ) \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right )}{2}\right )-\cos \left (d x +c \right ) \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right )}{2}\right )+2 \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}+2 \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\right )}{2 d \sin \left (d x +c \right )^{5} \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )-1\right ) \left (\frac {1}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} a}\) | \(316\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than
twice the leaf count of optimal. 2137 vs. \(2 (246) = 492\).
time = 0.63, size = 2137, normalized size = 5.79 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.42, size = 411, normalized size = 1.11 \begin {gather*} \frac {a d \sqrt {\frac {i \, e^{5}}{a d^{2}}} \log \left (2 \, {\left (a d \sqrt {\frac {i \, e^{5}}{a d^{2}}} + \frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\frac {5}{2}} + e^{\left (2 i \, d x + 2 i \, c + \frac {5}{2}\right )}\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{\sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-\frac {5}{2}\right )}\right ) - a d \sqrt {\frac {i \, e^{5}}{a d^{2}}} \log \left (-2 \, {\left (a d \sqrt {\frac {i \, e^{5}}{a d^{2}}} - \frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\frac {5}{2}} + e^{\left (2 i \, d x + 2 i \, c + \frac {5}{2}\right )}\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{\sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-\frac {5}{2}\right )}\right ) - a d \sqrt {-\frac {i \, e^{5}}{a d^{2}}} \log \left (2 \, {\left (a d \sqrt {-\frac {i \, e^{5}}{a d^{2}}} + \frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\frac {5}{2}} + e^{\left (2 i \, d x + 2 i \, c + \frac {5}{2}\right )}\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{\sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-\frac {5}{2}\right )}\right ) + a d \sqrt {-\frac {i \, e^{5}}{a d^{2}}} \log \left (-2 \, {\left (a d \sqrt {-\frac {i \, e^{5}}{a d^{2}}} - \frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\frac {5}{2}} + e^{\left (2 i \, d x + 2 i \, c + \frac {5}{2}\right )}\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{\sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-\frac {5}{2}\right )}\right ) - \frac {4 i \, \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {3}{2} i \, d x + \frac {3}{2} i \, c + \frac {5}{2}\right )}}{\sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{2 \, a d} \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 (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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