Optimal. Leaf size=205 \[ \frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}+\frac {(i A-B) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {4 (5 A+7 i B) \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {(5 A+7 i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {(25 A+23 i B) (a+i a \tan (c+d x))^{3/2}}{15 a^2 d} \]
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Rubi [A]
time = 0.34, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3676, 3678,
3673, 3608, 3561, 212} \begin {gather*} -\frac {(25 A+23 i B) (a+i a \tan (c+d x))^{3/2}}{15 a^2 d}+\frac {(-B+i A) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {(5 A+7 i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 a d}+\frac {4 (5 A+7 i B) \sqrt {a+i a \tan (c+d x)}}{5 a d}+\frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d} \end {gather*}
Antiderivative was successfully verified.
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Rule 212
Rule 3561
Rule 3608
Rule 3673
Rule 3676
Rule 3678
Rubi steps
\begin {align*} \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx &=\frac {(i A-B) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \left (3 a (i A-B)+\frac {1}{2} a (5 A+7 i B) \tan (c+d x)\right ) \, dx}{a^2}\\ &=\frac {(i A-B) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {(5 A+7 i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {2 \int \tan (c+d x) \sqrt {a+i a \tan (c+d x)} \left (-a^2 (5 A+7 i B)+\frac {1}{4} a^2 (25 i A-23 B) \tan (c+d x)\right ) \, dx}{5 a^3}\\ &=\frac {(i A-B) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {(5 A+7 i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {(25 A+23 i B) (a+i a \tan (c+d x))^{3/2}}{15 a^2 d}-\frac {2 \int \sqrt {a+i a \tan (c+d x)} \left (-\frac {1}{4} a^2 (25 i A-23 B)-a^2 (5 A+7 i B) \tan (c+d x)\right ) \, dx}{5 a^3}\\ &=\frac {(i A-B) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {4 (5 A+7 i B) \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {(5 A+7 i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {(25 A+23 i B) (a+i a \tan (c+d x))^{3/2}}{15 a^2 d}+\frac {(i A+B) \int \sqrt {a+i a \tan (c+d x)} \, dx}{2 a}\\ &=\frac {(i A-B) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {4 (5 A+7 i B) \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {(5 A+7 i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {(25 A+23 i B) (a+i a \tan (c+d x))^{3/2}}{15 a^2 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 )}{d}\\ &=\frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}+\frac {(i A-B) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {4 (5 A+7 i B) \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {(5 A+7 i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {(25 A+23 i B) (a+i a \tan (c+d x))^{3/2}}{15 a^2 d}\\ \end {align*}
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Mathematica [A]
time = 3.17, size = 176, normalized size = 0.86 \begin {gather*} \frac {\left ((A-i B) \sqrt {1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )+\frac {1}{30} \sec ^2(c+d x) (5 (23 A+37 i B) \cos (c+d x)+(25 A+59 i B) \cos (3 (c+d x))+4 i (5 A+16 i B+(5 A+22 i B) \cos (2 (c+d x))) \sin (c+d x))\right ) (A+B \tan (c+d x))}{2 d (A \cos (c+d x)+B \sin (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 = 168, normalized size = 0.82
method | result | size |
derivativedivides | \(-\frac {2 \left (-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 i B a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+\frac {A a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 i a^{2} B \sqrt {a +i a \tan \left (d x +c \right )}-a^{2} A \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a^{\frac {5}{2}} \left (-i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}-\frac {a^{3} \left (i B +A \right )}{2 \sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d \,a^{3}}\) | \(168\) |
default | \(-\frac {2 \left (-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 i B a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+\frac {A a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 i a^{2} B \sqrt {a +i a \tan \left (d x +c \right )}-a^{2} A \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a^{\frac {5}{2}} \left (-i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}-\frac {a^{3} \left (i B +A \right )}{2 \sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d \,a^{3}}\) | \(168\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.50, size = 157, normalized size = 0.77 \begin {gather*} -\frac {15 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {7}{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 ) - 24 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} B a + 40 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} {\left (A + 2 i \, B\right )} a^{2} - 120 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (A + 2 i \, B\right )} a^{3} - \frac {60 \, {\left (A + i \, B\right )} a^{4}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}}{60 \, 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. 447 vs. \(2 (163) = 326\).
time = 2.27, size = 447, normalized size = 2.18 \begin {gather*} -\frac {15 \, \sqrt {2} {\left (a d e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a d^{2}}} \log \left (-\frac {4 \, {\left ({\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )} + {\left (i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a 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 d^{2}}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 15 \, \sqrt {2} {\left (a d e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a d^{2}}} \log \left (-\frac {4 \, {\left ({\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )} + {\left (-i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a 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 d^{2}}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 2 \, \sqrt {2} {\left ({\left (35 \, A + 103 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 5 \, {\left (25 \, A + 41 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 15 \, {\left (7 \, A + 11 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 15 \, A + 15 i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{60 \, {\left (a d e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + 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 )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, 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 = 7.95, size = 236, normalized size = 1.15 \begin {gather*} \frac {A}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}+\frac {B\,1{}\mathrm {i}}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}+\frac {2\,A\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a\,d}-\frac {2\,A\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a^2\,d}+\frac {B\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,4{}\mathrm {i}}{a\,d}-\frac {B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,4{}\mathrm {i}}{3\,a^2\,d}+\frac {B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,2{}\mathrm {i}}{5\,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}}{2\,\sqrt {-a}\,d}-\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,1{}\mathrm {i}}{2\,\sqrt {a}\,d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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