Optimal. Leaf size=512 \[ \frac {3 i \sqrt {a} e^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{4 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac {3 i \sqrt {a} e^{5/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{4 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac {3 i \sqrt {a} e^{5/2} \log \left (-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{8 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\frac {3 i \sqrt {a} e^{5/2} \log \left (\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{8 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac {3 i \cos ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d (e \cos (c+d x))^{5/2}}+\frac {i a}{2 d \sqrt {a+i a \tan (c+d x)} (e \cos (c+d x))^{5/2}} \]
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Rubi [A] time = 0.56, 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 = {3515, 3498, 3501, 3495, 297, 1162, 617, 204, 1165, 628} \[ \frac {3 i \sqrt {a} e^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{4 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac {3 i \sqrt {a} e^{5/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{4 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac {3 i \sqrt {a} e^{5/2} \log \left (-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{8 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\frac {3 i \sqrt {a} e^{5/2} \log \left (\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{8 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac {3 i \cos ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d (e \cos (c+d x))^{5/2}}+\frac {i a}{2 d \sqrt {a+i a \tan (c+d x)} (e \cos (c+d x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 204
Rule 297
Rule 617
Rule 628
Rule 1162
Rule 1165
Rule 3495
Rule 3498
Rule 3501
Rule 3515
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 ) \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 )}{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 ) \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 )}{4 d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac {\left (3 i a e^3\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 )}{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 ) \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 )}{8 d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac {\left (3 i a e^2\right ) \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 )}{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 ) \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 )}{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 ) \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 )}{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 ) \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 )}{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 ) \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 )}{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 = 2.42, size = 227, normalized size = 0.44 \[ \frac {\sqrt {\cos (c+d x)} \sqrt {a+i a \tan (c+d x)} \left (-3 i \cos ^{\frac {3}{2}}(c+d x)+2 \sqrt {\cos (c+d x)} (\sin (c+d x)+i \cos (c+d x))+\frac {3 i \left (-e^{-2 i c}\right )^{3/4} e^{-\frac {1}{2} i (2 c+5 d x)} \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (1+e^{2 i (c+d x)}\right )^2 \left (\tan ^{-1}\left (\frac {e^{\frac {i d x}{2}}}{\sqrt [4]{-e^{-2 i c}}}\right )-\tanh ^{-1}\left (\frac {e^{\frac {i d x}{2}}}{\sqrt [4]{-e^{-2 i c}}}\right )\right )}{4 \sqrt {2}}\right )}{4 d (e \cos (c+d x))^{5/2}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.71, size = 569, normalized size = 1.11 \[ \frac {\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \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 )} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} - {\left (d e^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{3}\right )} \sqrt {\frac {9 i \, a}{16 \, d^{2} e^{5}}} \log \left (\frac {4}{3} i \, d e^{3} \sqrt {\frac {9 i \, a}{16 \, d^{2} e^{5}}} + \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{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 )}\right ) + {\left (d e^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{3}\right )} \sqrt {\frac {9 i \, a}{16 \, d^{2} e^{5}}} \log \left (-\frac {4}{3} i \, d e^{3} \sqrt {\frac {9 i \, a}{16 \, d^{2} e^{5}}} + \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{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 )}\right ) - {\left (d e^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{3}\right )} \sqrt {-\frac {9 i \, a}{16 \, d^{2} e^{5}}} \log \left (\frac {4}{3} i \, d e^{3} \sqrt {-\frac {9 i \, a}{16 \, d^{2} e^{5}}} + \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{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 )}\right ) + {\left (d e^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{3}\right )} \sqrt {-\frac {9 i \, a}{16 \, d^{2} e^{5}}} \log \left (-\frac {4}{3} i \, d e^{3} \sqrt {-\frac {9 i \, a}{16 \, d^{2} e^{5}}} + \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{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 )}\right )}{2 \, {\left (d e^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{3}\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 \[ \int \frac {\sqrt {i \, a \tan \left (d x + c\right ) + a}}{\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.55, size = 366, normalized size = 0.71 \[ -\frac {\sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \left (-1+\cos \left (d x +c \right )\right )^{3} \left (3 i \left (\cos ^{2}\left (d x +c \right )\right ) \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right )}{2}\right )+3 i \left (\cos ^{2}\left (d x +c \right )\right ) \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (-\cos \left (d x +c \right )-1+\sin \left (d x +c \right )\right )}{2}\right )+6 i \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+6 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}-3 \left (\cos ^{2}\left (d x +c \right )\right ) \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right )}{2}\right )+3 \left (\cos ^{2}\left (d x +c \right )\right ) \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (-\cos \left (d x +c \right )-1+\sin \left (d x +c \right )\right )}{2}\right )+4 i \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}+2 \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}-4 \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\right )}{8 d \sin \left (d x +c \right )^{5} \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )-1\right ) \left (\frac {1}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (e \cos \left (d x +c \right )\right )^{\frac {5}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 1.27, size = 2265, normalized size = 4.42 \[ \text {result too large to display} \]
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 \[ \int \frac {\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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