Optimal. Leaf size=512 \[ \frac {3 i \sqrt {a} e^{5/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{4 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac {3 i \sqrt {a} e^{5/2} \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{4 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac {3 i \sqrt {a} e^{5/2} \log \left (a-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{8 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\frac {3 i \sqrt {a} e^{5/2} \log \left (a+\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{8 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\frac {i a}{2 d (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}-\frac {3 i \cos ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d (e \cos (c+d x))^{5/2}} \]
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Rubi [A]
time = 0.38, antiderivative size = 512, normalized size of antiderivative = 1.00, number of steps
used = 13, number of rules used = 10, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3596, 3579,
3582, 3576, 303, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {3 i \sqrt {a} 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 )}{4 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac {3 i \sqrt {a} 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 )}{4 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac {3 i \sqrt {a} 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 )}{8 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\frac {3 i \sqrt {a} 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 )}{8 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac {3 i \cos ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d (e \cos (c+d x))^{5/2}}+\frac {i a}{2 d \sqrt {a+i a \tan (c+d x)} (e \cos (c+d x))^{5/2}} \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 3579
Rule 3582
Rule 3596
Rubi steps
\begin {align*} \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \cos (c+d x))^{5/2}} \, dx &=\frac {\int (e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)} \, dx}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\\ &=\frac {i a}{2 d (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}+\frac {(3 a) \int \frac {(e \sec (c+d x))^{5/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx}{4 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\\ &=\frac {i a}{2 d (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}-\frac {3 i \cos ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d (e \cos (c+d x))^{5/2}}+\frac {\left (3 e^2\right ) \int \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx}{8 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\\ &=\frac {i a}{2 d (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}-\frac {3 i \cos ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d (e \cos (c+d x))^{5/2}}-\frac {\left (3 i a 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 )}{2 d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\\ &=\frac {i a}{2 d (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}-\frac {3 i \cos ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d (e \cos (c+d x))^{5/2}}+\frac {\left (3 i a 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 )}{4 d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac {\left (3 i a 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 )}{4 d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\\ &=\frac {i a}{2 d (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}-\frac {3 i \cos ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d (e \cos (c+d x))^{5/2}}-\frac {\left (3 i a 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 )}{8 d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac {\left (3 i a 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 )}{8 d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac {\left (3 i \sqrt {a} 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 )}{8 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac {\left (3 i \sqrt {a} 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 )}{8 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\\ &=-\frac {3 i \sqrt {a} 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 )}{8 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\frac {3 i \sqrt {a} 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 )}{8 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\frac {i a}{2 d (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}-\frac {3 i \cos ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d (e \cos (c+d x))^{5/2}}-\frac {\left (3 i \sqrt {a} 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 )}{4 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\frac {\left (3 i \sqrt {a} 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 )}{4 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\\ &=\frac {3 i \sqrt {a} 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 )}{4 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac {3 i \sqrt {a} 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 )}{4 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac {3 i \sqrt {a} 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 )}{8 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\frac {3 i \sqrt {a} 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 )}{8 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\frac {i a}{2 d (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}-\frac {3 i \cos ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d (e \cos (c+d x))^{5/2}}\\ \end {align*}
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Mathematica [A]
time = 2.59, size = 227, normalized size = 0.44 \begin {gather*} \frac {\sqrt {\cos (c+d x)} \left (\frac {3 i e^{-\frac {1}{2} i (2 c+5 d x)} \left (-e^{-2 i c}\right )^{3/4} \left (1+e^{2 i (c+d x)}\right )^2 \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (\text {ArcTan}\left (\frac {e^{\frac {i d x}{2}}}{\sqrt [4]{-e^{-2 i c}}}\right )-\tanh ^{-1}\left (\frac {e^{\frac {i d x}{2}}}{\sqrt [4]{-e^{-2 i c}}}\right )\right )}{4 \sqrt {2}}-3 i \cos ^{\frac {3}{2}}(c+d x)+2 \sqrt {\cos (c+d x)} (i \cos (c+d x)+\sin (c+d x))\right ) \sqrt {a+i a \tan (c+d x)}}{4 d (e \cos (c+d x))^{5/2}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 1.14, size = 366, normalized size = 0.71
method | result | size |
default | \(-\frac {\sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \left (-1+\cos \left (d x +c \right )\right )^{3} \left (3 i \left (\cos ^{2}\left (d x +c \right )\right ) \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right )}{2}\right )-3 i \left (\cos ^{2}\left (d x +c \right )\right ) \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right )}{2}\right )+6 i \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}-3 \left (\cos ^{2}\left (d x +c \right )\right ) \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right )}{2}\right )-3 \left (\cos ^{2}\left (d x +c \right )\right ) \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right )}{2}\right )+6 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}+4 i \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}+2 \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )-4 \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\right )}{8 d \sin \left (d x +c \right )^{5} \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )-1\right ) \left (\frac {1}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (e \cos \left (d x +c \right )\right )^{\frac {5}{2}}}\) | \(366\) |
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. 2232 vs. \(2 (336) = 672\).
time = 0.73, size = 2232, normalized size = 4.36 \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 = 0.35, size = 522, normalized size = 1.02 \begin {gather*} \frac {\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-3 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} - {\left (d e^{\frac {5}{2}} + d e^{\left (4 i \, d x + 4 i \, c + \frac {5}{2}\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c + \frac {5}{2}\right )}\right )} \sqrt {\frac {9 i \, a e^{\left (-5\right )}}{16 \, d^{2}}} \log \left (\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} + \frac {4}{3} i \, d \sqrt {\frac {9 i \, a e^{\left (-5\right )}}{16 \, d^{2}}} e^{\frac {5}{2}}\right ) + {\left (d e^{\frac {5}{2}} + d e^{\left (4 i \, d x + 4 i \, c + \frac {5}{2}\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c + \frac {5}{2}\right )}\right )} \sqrt {\frac {9 i \, a e^{\left (-5\right )}}{16 \, d^{2}}} \log \left (\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} - \frac {4}{3} i \, d \sqrt {\frac {9 i \, a e^{\left (-5\right )}}{16 \, d^{2}}} e^{\frac {5}{2}}\right ) - {\left (d e^{\frac {5}{2}} + d e^{\left (4 i \, d x + 4 i \, c + \frac {5}{2}\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c + \frac {5}{2}\right )}\right )} \sqrt {-\frac {9 i \, a e^{\left (-5\right )}}{16 \, d^{2}}} \log \left (\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} + \frac {4}{3} i \, d \sqrt {-\frac {9 i \, a e^{\left (-5\right )}}{16 \, d^{2}}} e^{\frac {5}{2}}\right ) + {\left (d e^{\frac {5}{2}} + d e^{\left (4 i \, d x + 4 i \, c + \frac {5}{2}\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c + \frac {5}{2}\right )}\right )} \sqrt {-\frac {9 i \, a e^{\left (-5\right )}}{16 \, d^{2}}} \log \left (\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} - \frac {4}{3} i \, d \sqrt {-\frac {9 i \, a e^{\left (-5\right )}}{16 \, d^{2}}} e^{\frac {5}{2}}\right )}{2 \, {\left (d e^{\frac {5}{2}} + d e^{\left (4 i \, d x + 4 i \, c + \frac {5}{2}\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c + \frac {5}{2}\right )}\right )}} \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 {\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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