Optimal. Leaf size=139 \[ \frac {256 i a^4 \sec (c+d x)}{35 d \sqrt {a+i a \tan (c+d x)}}+\frac {64 i a^3 \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {24 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{35 d}+\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d} \]
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Rubi [A]
time = 0.11, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3575, 3574}
\begin {gather*} \frac {256 i a^4 \sec (c+d x)}{35 d \sqrt {a+i a \tan (c+d x)}}+\frac {64 i a^3 \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {24 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{35 d}+\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d} \end {gather*}
Antiderivative was successfully verified.
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Rule 3574
Rule 3575
Rubi steps
\begin {align*} \int \sec (c+d x) (a+i a \tan (c+d x))^{7/2} \, dx &=\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}+\frac {1}{7} (12 a) \int \sec (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\\ &=\frac {24 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{35 d}+\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}+\frac {1}{35} \left (96 a^2\right ) \int \sec (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\\ &=\frac {64 i a^3 \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {24 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{35 d}+\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}+\frac {1}{35} \left (128 a^3\right ) \int \sec (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\\ &=\frac {256 i a^4 \sec (c+d x)}{35 d \sqrt {a+i a \tan (c+d x)}}+\frac {64 i a^3 \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {24 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{35 d}+\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}\\ \end {align*}
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Mathematica [A]
time = 0.87, size = 109, normalized size = 0.78 \begin {gather*} \frac {2 a^3 \sec ^2(c+d x) (i \cos (c-2 d x)+\sin (c-2 d x)) (75+102 \cos (2 (c+d x))+19 i \sec (c+d x) \sin (3 (c+d x))+14 i \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{35 d (\cos (d x)+i \sin (d x))^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.60, size = 100, normalized size = 0.72
method | result | size |
default | \(\frac {2 \left (128 i \left (\cos ^{4}\left (d x +c \right )\right )+128 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+54 i \left (\cos ^{2}\left (d x +c \right )\right )-22 \sin \left (d x +c \right ) \cos \left (d x +c \right )-5 i\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{3}}{35 d \cos \left (d x +c \right )^{3}}\) | \(100\) |
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.39, size = 109, normalized size = 0.78 \begin {gather*} -\frac {16 \, \sqrt {2} {\left (-35 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 70 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 56 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 16 i \, a^{3}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{35 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\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 [B]
time = 6.00, size = 286, normalized size = 2.06 \begin {gather*} \frac {a^3\,{\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}}\,16{}\mathrm {i}}{d}-\frac {a^3\,{\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}}\,16{}\mathrm {i}}{d\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}+\frac {a^3\,{\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}}\,48{}\mathrm {i}}{5\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {a^3\,{\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}}\,16{}\mathrm {i}}{7\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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