Optimal. Leaf size=209 \[ \frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(i A-B) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(3 A+5 i B) \tan ^2(c+d x)}{2 a d \sqrt {a+i a \tan (c+d x)}}-\frac {2 (3 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{a^2 d}+\frac {(11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{6 a^3 d} \]
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Rubi [A]
time = 0.36, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3676, 3673,
3608, 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 {(11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{6 a^3 d}-\frac {2 (3 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{a^2 d}+\frac {(-B+i A) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(3 A+5 i B) \tan ^2(c+d x)}{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 3608
Rule 3673
Rule 3676
Rubi steps
\begin {align*} \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx &=\frac {(i A-B) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {\tan ^2(c+d x) \left (3 a (i A-B)+\frac {3}{2} a (A+3 i B) \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=\frac {(i A-B) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(3 A+5 i B) \tan ^2(c+d x)}{2 a d \sqrt {a+i a \tan (c+d x)}}+\frac {\int \tan (c+d x) \sqrt {a+i a \tan (c+d x)} \left (-3 a^2 (3 A+5 i B)+\frac {3}{4} a^2 (11 i A-21 B) \tan (c+d x)\right ) \, dx}{3 a^4}\\ &=\frac {(i A-B) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(3 A+5 i B) \tan ^2(c+d x)}{2 a d \sqrt {a+i a \tan (c+d x)}}+\frac {(11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{6 a^3 d}+\frac {\int \sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{4} a^2 (11 i A-21 B)-3 a^2 (3 A+5 i B) \tan (c+d x)\right ) \, dx}{3 a^4}\\ &=\frac {(i A-B) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(3 A+5 i B) \tan ^2(c+d x)}{2 a d \sqrt {a+i a \tan (c+d x)}}-\frac {2 (3 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{a^2 d}+\frac {(11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{6 a^3 d}+\frac {(i A+B) \int \sqrt {a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac {(i A-B) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(3 A+5 i B) \tan ^2(c+d x)}{2 a d \sqrt {a+i a \tan (c+d x)}}-\frac {2 (3 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{a^2 d}+\frac {(11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{6 a^3 d}+\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 {(i A-B) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(3 A+5 i B) \tan ^2(c+d x)}{2 a d \sqrt {a+i a \tan (c+d x)}}-\frac {2 (3 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{a^2 d}+\frac {(11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{6 a^3 d}\\ \end {align*}
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Mathematica [A]
time = 4.03, size = 176, normalized size = 0.84 \begin {gather*} \frac {-\frac {24 i (A-i B) e^{3 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )}{\left (1+e^{2 i (c+d x)}\right )^{3/2}}+i \sec ^3(c+d x) (21 (3 A+5 i B) \cos (c+d x)+(37 A+51 i B) \cos (3 (c+d x))+2 i (39 A+61 i B+(39 A+53 i B) \cos (2 (c+d x))) \sin (c+d x))}{24 a d (-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 = 0.11, size = 153, normalized size = 0.73
method | result | size |
derivativedivides | \(-\frac {2 \left (-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 i a B \sqrt {a +i a \tan \left (d x +c \right )}+a A \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a^{\frac {3}{2}} \left (-i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8}+\frac {a^{2} \left (7 i B +5 A \right )}{4 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {a^{3} \left (i B +A \right )}{6 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}\right )}{d \,a^{3}}\) | \(153\) |
default | \(-\frac {2 \left (-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 i a B \sqrt {a +i a \tan \left (d x +c \right )}+a A \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a^{\frac {3}{2}} \left (-i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8}+\frac {a^{2} \left (7 i B +5 A \right )}{4 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {a^{3} \left (i B +A \right )}{6 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}\right )}{d \,a^{3}}\) | \(153\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.52, size = 160, normalized size = 0.77 \begin {gather*} -\frac {3 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {5}{2}} \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 ) - 16 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} B a + 48 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (A + 2 i \, B\right )} a^{2} + \frac {4 \, {\left (3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (5 \, A + 7 i \, B\right )} a^{3} - 2 \, {\left (A + i \, B\right )} a^{4}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}}{24 \, a^{4} 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. 440 vs. \(2 (162) = 324\).
time = 1.79, size = 440, normalized size = 2.11 \begin {gather*} -\frac {3 \, \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} \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}} {\left (a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} \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 (19 \, A + 26 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, {\left (17 \, A + 29 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, {\left (2 \, A + 3 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}}}{12 \, {\left (a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )}} \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 ^{3}{\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 = 7.44, size = 233, normalized size = 1.11 \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 )\,7{}\mathrm {i}}{2\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}+\frac {\frac {A\,a}{3}-\frac {5\,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 {2\,A\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^2\,d}-\frac {B\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,4{}\mathrm {i}}{a^2\,d}+\frac {B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,2{}\mathrm {i}}{3\,a^3\,d}-\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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