Optimal. Leaf size=483 \[ -\frac {i \sqrt {2} \sqrt {a} e^{3/2} \sec (c+d x) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {i \sqrt {2} \sqrt {a} e^{3/2} \sec (c+d x) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {i \sqrt {a} e^{3/2} \sec (c+d x) \log \left (-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{\sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {i \sqrt {a} e^{3/2} \sec (c+d x) \log \left (\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{\sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]
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Rubi [A] time = 0.35, antiderivative size = 483, 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 = {3499, 3495, 297, 1162, 617, 204, 1165, 628} \[ -\frac {i \sqrt {2} \sqrt {a} e^{3/2} \sec (c+d x) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {i \sqrt {2} \sqrt {a} e^{3/2} \sec (c+d x) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {i \sqrt {a} e^{3/2} \sec (c+d x) \log \left (-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{\sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {i \sqrt {a} e^{3/2} \sec (c+d x) \log \left (\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{\sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 204
Rule 297
Rule 617
Rule 628
Rule 1162
Rule 1165
Rule 3495
Rule 3499
Rubi steps
\begin {align*} \int \frac {(e \sec (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx &=\frac {(e \sec (c+d x)) \int \sqrt {e \sec (c+d x)} \sqrt {a-i a \tan (c+d x)} \, dx}{\sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}\\ &=\frac {\left (4 i a e^3 \sec (c+d x)\right ) \operatorname {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 \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}\\ &=-\frac {\left (2 i a e^2 \sec (c+d x)\right ) \operatorname {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 \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (2 i a e^2 \sec (c+d x)\right ) \operatorname {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 \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}\\ &=\frac {(i a e \sec (c+d x)) \operatorname {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 )}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {(i a e \sec (c+d x)) \operatorname {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 )}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (i \sqrt {a} e^{3/2} \sec (c+d x)\right ) \operatorname {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 )}{\sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (i \sqrt {a} e^{3/2} \sec (c+d x)\right ) \operatorname {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 )}{\sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}\\ &=\frac {i \sqrt {a} e^{3/2} \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 ) \sec (c+d x)}{\sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {i \sqrt {a} e^{3/2} \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 ) \sec (c+d x)}{\sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (i \sqrt {2} \sqrt {a} e^{3/2} \sec (c+d x)\right ) \operatorname {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 )}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (i \sqrt {2} \sqrt {a} e^{3/2} \sec (c+d x)\right ) \operatorname {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 )}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}\\ &=-\frac {i \sqrt {2} \sqrt {a} e^{3/2} \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 ) \sec (c+d x)}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {i \sqrt {2} \sqrt {a} e^{3/2} \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 ) \sec (c+d x)}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {i \sqrt {a} e^{3/2} \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 ) \sec (c+d x)}{\sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {i \sqrt {a} e^{3/2} \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 ) \sec (c+d x)}{\sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 1.30, size = 302, normalized size = 0.63 \[ \frac {2 e \sqrt {\tan \left (\frac {d x}{2}\right )+i} (\cos (d x)+i \sin (d x)) \sqrt {e \sec (c+d x)} \left (\sqrt {-\sin (c)-i \cos (c)-1} \sqrt {-\sin (c)+i \cos (c)+1} \tanh ^{-1}\left (\frac {\sqrt {-\sin (c)+i \cos (c)+1} \sqrt {-\tan \left (\frac {d x}{2}\right )+i}}{\sqrt {\sin (c)+i \cos (c)-1} \sqrt {\tan \left (\frac {d x}{2}\right )+i}}\right )-\sqrt {\sin (c)-i \cos (c)+1} \sqrt {\sin (c)+i \cos (c)-1} \tanh ^{-1}\left (\frac {\sqrt {\sin (c)-i \cos (c)+1} \sqrt {-\tan \left (\frac {d x}{2}\right )+i}}{\sqrt {-\sin (c)-i \cos (c)-1} \sqrt {\tan \left (\frac {d x}{2}\right )+i}}\right )\right )}{d \sqrt {-\sin (c)-i \cos (c)-1} \sqrt {\sin (c)+i \cos (c)-1} \sqrt {-\tan \left (\frac {d x}{2}\right )+i} \sqrt {a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
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fricas [A] time = 1.24, size = 385, normalized size = 0.80 \[ \frac {1}{2} \, \sqrt {\frac {4 i \, e^{3}}{a d^{2}}} \log \left (\frac {2 \, {\left (e e^{\left (2 i \, d x + 2 i \, c\right )} + e\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + i \, a d \sqrt {\frac {4 i \, e^{3}}{a d^{2}}}}{e}\right ) - \frac {1}{2} \, \sqrt {\frac {4 i \, e^{3}}{a d^{2}}} \log \left (\frac {2 \, {\left (e e^{\left (2 i \, d x + 2 i \, c\right )} + e\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} - i \, a d \sqrt {\frac {4 i \, e^{3}}{a d^{2}}}}{e}\right ) + \frac {1}{2} \, \sqrt {-\frac {4 i \, e^{3}}{a d^{2}}} \log \left (\frac {2 \, {\left (e e^{\left (2 i \, d x + 2 i \, c\right )} + e\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + i \, a d \sqrt {-\frac {4 i \, e^{3}}{a d^{2}}}}{e}\right ) - \frac {1}{2} \, \sqrt {-\frac {4 i \, e^{3}}{a d^{2}}} \log \left (\frac {2 \, {\left (e e^{\left (2 i \, d x + 2 i \, c\right )} + e\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} - i \, a d \sqrt {-\frac {4 i \, e^{3}}{a d^{2}}}}{e}\right ) \]
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 \[ \int \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.25, size = 232, normalized size = 0.48 \[ \frac {\left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (\cos ^{2}\left (d x +c \right )\right ) \left (-1+\cos \left (d x +c \right )\right )^{2} \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (i \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right )}{2}\right )-i \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right )}{2}\right )+\arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right )}{2}\right )+\arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right )}{2}\right )\right )}{d \sin \left (d x +c \right )^{3} \left (\frac {1}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )-1\right ) a} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.05, size = 726, normalized size = 1.50 \[ \text {result too large to display} \]
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 \[ \int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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