Optimal. Leaf size=119 \[ -\frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {A+i B}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {A+3 i B}{2 a d \sqrt {a+i a \tan (c+d x)}} \]
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Rubi [A]
time = 0.13, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3671, 3607,
3561, 212} \begin {gather*} -\frac {(A-i 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 {A+i B}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {A+3 i B}{2 a d \sqrt {a+i a \tan (c+d x)}} \end {gather*}
Antiderivative was successfully verified.
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Rule 212
Rule 3561
Rule 3607
Rule 3671
Rubi steps
\begin {align*} \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx &=-\frac {A+i B}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {i \int \frac {a (A+i B)+2 a B \tan (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {A+i B}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {A+3 i B}{2 a d \sqrt {a+i a \tan (c+d x)}}-\frac {(i A+B) \int \sqrt {a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=-\frac {A+i B}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {A+3 i B}{2 a d \sqrt {a+i a \tan (c+d x)}}-\frac {(A-i 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 {(A-i 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 {A+i B}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {A+3 i B}{2 a d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}
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Mathematica [A]
time = 2.35, size = 145, normalized size = 1.22 \begin {gather*} \frac {\sqrt {1+e^{2 i (c+d x)}} \left (-i A \left (-1+2 e^{2 i (c+d x)}\right )+B \left (-1+8 e^{2 i (c+d x)}\right )\right )+3 (i A+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 = 0.81
method | result | size |
derivativedivides | \(\frac {-\frac {\left (\frac {A}{4}-\frac {i B}{4}\right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{\sqrt {a}}-\frac {2 \left (-\frac {A}{4}-\frac {3 i B}{4}\right )}{\sqrt {a +i a \tan \left (d x +c \right )}}-\frac {a \left (i B +A \right )}{3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}}{a d}\) | \(96\) |
default | \(\frac {-\frac {\left (\frac {A}{4}-\frac {i B}{4}\right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{\sqrt {a}}-\frac {2 \left (-\frac {A}{4}-\frac {3 i B}{4}\right )}{\sqrt {a +i a \tan \left (d x +c \right )}}-\frac {a \left (i B +A \right )}{3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}}{a d}\) | \(96\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.50, size = 116, normalized size = 0.97 \begin {gather*} \frac {3 \, \sqrt {2} {\left (A - i \, B\right )} \sqrt {a} \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 ) + \frac {4 \, {\left (3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (A + 3 i \, B\right )} a - 2 \, {\left (A + i \, B\right )} a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}}{24 \, a^{2} 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. 369 vs. \(2 (88) = 176\).
time = 2.07, size = 369, normalized size = 3.10 \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 (i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, 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 (-i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, 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 (A + 4 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (A + 7 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - A - i \, 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 {\left (A + B \tan {\left (c + d x \right )}\right ) \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 = 7.07, size = 163, normalized size = 1.37 \begin {gather*} -\frac {\frac {B\,1{}\mathrm {i}}{3\,d}-\frac {B\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {\frac {A}{3}-\frac {A\,\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}\,B\,\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}\,A\,\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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