Optimal. Leaf size=104 \[ \frac {64 i a^3 \sec (c+d x)}{15 d \sqrt {a+i a \tan (c+d x)}}+\frac {16 i a^2 \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d}+\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d} \]
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Rubi [A] time = 0.10, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3494, 3493} \[ \frac {64 i a^3 \sec (c+d x)}{15 d \sqrt {a+i a \tan (c+d x)}}+\frac {16 i a^2 \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d}+\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d} \]
Antiderivative was successfully verified.
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Rule 3493
Rule 3494
Rubi steps
\begin {align*} \int \sec (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx &=\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}+\frac {1}{5} (8 a) \int \sec (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\\ &=\frac {16 i a^2 \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d}+\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}+\frac {1}{15} \left (32 a^2\right ) \int \sec (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\\ &=\frac {64 i a^3 \sec (c+d x)}{15 d \sqrt {a+i a \tan (c+d x)}}+\frac {16 i a^2 \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d}+\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\\ \end {align*}
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Mathematica [A] time = 0.39, size = 93, normalized size = 0.89 \[ \frac {2 a^2 \sec ^2(c+d x) \sqrt {a+i a \tan (c+d x)} (\sin (c-d x)+i \cos (c-d x)) (7 i \sin (2 (c+d x))+23 \cos (2 (c+d x))+20)}{15 d (\cos (d x)+i \sin (d x))^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.62, size = 83, normalized size = 0.80 \[ \frac {\sqrt {2} {\left (120 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 160 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 64 i \, a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
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 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.90, size = 90, normalized size = 0.87 \[ \frac {2 \left (32 i \left (\cos ^{3}\left (d x +c \right )\right )+32 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+11 i \cos \left (d x +c \right )-3 \sin \left (d x +c \right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{2}}{15 d \cos \left (d x +c \right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 6.08, size = 105, normalized size = 1.01 \[ \frac {8\,a^2\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,20{}\mathrm {i}+{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,15{}\mathrm {i}+8{}\mathrm {i}\right )}{15\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2} \]
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 \[ \int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}} \sec {\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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