Optimal. Leaf size=121 \[ -\frac {(i A+B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {i A-B}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {i A+B}{2 a d \sqrt {a+i a \tan (c+d x)}} \]
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Rubi [A]
time = 0.07, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3607, 3560,
3561, 212} \begin {gather*} -\frac {(B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {-B+i A}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {B+i A}{2 a d \sqrt {a+i a \tan (c+d x)}} \end {gather*}
Antiderivative was successfully verified.
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Rule 212
Rule 3560
Rule 3561
Rule 3607
Rubi steps
\begin {align*} \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx &=\frac {i A-B}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(A-i B) \int \frac {1}{\sqrt {a+i a \tan (c+d x)}} \, dx}{2 a}\\ &=\frac {i A-B}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {i A+B}{2 a d \sqrt {a+i a \tan (c+d x)}}+\frac {(A-i B) \int \sqrt {a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac {i A-B}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {i A+B}{2 a d \sqrt {a+i a \tan (c+d x)}}-\frac {(i A+B) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{2 a d}\\ &=-\frac {(i A+B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {i A-B}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {i A+B}{2 a d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}
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Mathematica [A]
time = 2.31, size = 143, normalized size = 1.18 \begin {gather*} \frac {\sqrt {1+e^{2 i (c+d x)}} \left (A+4 A e^{2 i (c+d x)}-i B \left (-1+2 e^{2 i (c+d x)}\right )\right )-3 (A-i B) e^{3 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )}{3 a d \left (1+e^{2 i (c+d x)}\right )^{3/2} (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.10, size = 96, normalized size = 0.79
method | result | size |
derivativedivides | \(\frac {2 i \left (-\frac {\left (-i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 a^{\frac {3}{2}}}-\frac {-\frac {A}{2}-\frac {i B}{2}}{3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {i B -A}{4 a \sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d}\) | \(96\) |
default | \(\frac {2 i \left (-\frac {\left (-i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 a^{\frac {3}{2}}}-\frac {-\frac {A}{2}-\frac {i B}{2}}{3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {i B -A}{4 a \sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d}\) | \(96\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.51, size = 113, normalized size = 0.93 \begin {gather*} \frac {i \, {\left (\frac {3 \, \sqrt {2} {\left (A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{\sqrt {a}} + \frac {4 \, {\left (3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (A - i \, B\right )} + 2 \, {\left (A + i \, B\right )} a\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\right )}}{24 \, a d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 372 vs. \(2 (90) = 180\).
time = 1.72, size = 372, normalized size = 3.07 \begin {gather*} \frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} + {\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} - {\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - \sqrt {2} {\left (2 \, {\left (-2 i \, A - B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (5 i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{12 \, a^{2} d} \end {gather*}
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 \begin {gather*} \int \frac {A + B \tan {\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \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 = 6.99, size = 162, normalized size = 1.34 \begin {gather*} \frac {\frac {A\,1{}\mathrm {i}}{3\,d}+\frac {A\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {\frac {B}{3}-\frac {B\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{2\,a}}{d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{4\,{\left (-a\right )}^{3/2}\,d}-\frac {\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{4\,a^{3/2}\,d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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