Optimal. Leaf size=167 \[ \frac {(B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(-7 B+i A) \sqrt {a+i a \tan (c+d x)}}{3 a^2 d}+\frac {(-B+i A) \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {-11 B+5 i A}{6 a d \sqrt {a+i a \tan (c+d x)}} \]
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Rubi [A] time = 0.35, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3595, 3592, 3526, 3480, 206} \[ \frac {(-7 B+i A) \sqrt {a+i a \tan (c+d x)}}{3 a^2 d}+\frac {(B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(-B+i A) \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {-11 B+5 i A}{6 a d \sqrt {a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 206
Rule 3480
Rule 3526
Rule 3592
Rule 3595
Rubi steps
\begin {align*} \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx &=\frac {(i A-B) \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {\tan (c+d x) \left (2 a (i A-B)+\frac {1}{2} a (A+7 i B) \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=\frac {(i A-B) \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(i A-7 B) \sqrt {a+i a \tan (c+d x)}}{3 a^2 d}-\frac {\int \frac {-\frac {1}{2} a (A+7 i B)+2 a (i A-B) \tan (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=\frac {(i A-B) \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {5 i A-11 B}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {(i A-7 B) \sqrt {a+i a \tan (c+d x)}}{3 a^2 d}-\frac {(A-i B) \int \sqrt {a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac {(i A-B) \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {5 i A-11 B}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {(i A-7 B) \sqrt {a+i a \tan (c+d x)}}{3 a^2 d}+\frac {(i A+B) \operatorname {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) \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {5 i A-11 B}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {(i A-7 B) \sqrt {a+i a \tan (c+d x)}}{3 a^2 d}\\ \end {align*}
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Mathematica [A] time = 3.33, size = 167, normalized size = 1.00 \[ \frac {3 (A-i B) e^{3 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )+A \left (7 e^{2 i (c+d x)}+8 e^{4 i (c+d x)}-1\right )+i B \left (13 e^{2 i (c+d x)}+38 e^{4 i (c+d x)}-1\right )}{3 a d \left (1+e^{2 i (c+d x)}\right )^2 (\tan (c+d x)-i) \sqrt {a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.64, size = 372, normalized size = 2.23 \[ \frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac {{\left (4 \, \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} + {\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-\frac {{\left (4 \, \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} - {\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) + \sqrt {2} {\left ({\left (8 i \, A - 38 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (7 i \, A - 13 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{12 \, a^{2} d} \]
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 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{2}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.23, size = 116, normalized size = 0.69 \[ -\frac {2 i \left (-i B \sqrt {a +i a \tan \left (d x +c \right )}-\frac {\sqrt {a}\, \left (-i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8}-\frac {a \left (5 i B +3 A \right )}{4 \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {a^{2} \left (i B +A \right )}{6 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}\right )}{d \,a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.63, size = 137, normalized size = 0.82 \[ -\frac {i \, {\left (3 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 48 i \, \sqrt {i \, a \tan \left (d x + c\right ) + a} B a - \frac {4 \, {\left (3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (3 \, A + 5 i \, B\right )} a^{2} - 2 \, {\left (A + i \, B\right )} a^{3}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\right )}}{24 \, a^{3} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.70, size = 186, normalized size = 1.11 \[ -\frac {\frac {A\,1{}\mathrm {i}}{3\,d}-\frac {A\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}+\frac {\frac {B\,a}{3}-\frac {5\,B\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{2}}{a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {2\,B\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^2\,d}+\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{4\,{\left (-a\right )}^{3/2}\,d}+\frac {\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{4\,a^{3/2}\,d} \]
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 \[ \int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{2}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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