Integrand size = 36, antiderivative size = 205 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}+\frac {(i A-B) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {4 (5 A+7 i B) \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {(5 A+7 i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {(25 A+23 i B) (a+i a \tan (c+d x))^{3/2}}{15 a^2 d} \]
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Time = 0.61 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3676, 3678, 3673, 3608, 3561, 212} \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {(25 A+23 i B) (a+i a \tan (c+d x))^{3/2}}{15 a^2 d}+\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}+\frac {(-B+i A) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {(5 A+7 i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 a d}+\frac {4 (5 A+7 i B) \sqrt {a+i a \tan (c+d x)}}{5 a d} \]
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Rule 212
Rule 3561
Rule 3608
Rule 3673
Rule 3676
Rule 3678
Rubi steps \begin{align*} \text {integral}& = \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) \text {arctanh}\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*}
Time = 2.43 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.63 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}+\frac {35 A+61 i B+(10 i A-38 B) \tan (c+d x)+2 (5 A+i B) \tan ^2(c+d x)+6 B \tan ^3(c+d x)}{15 d \sqrt {a+i a \tan (c+d x)}} \]
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Time = 0.17 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.81
method | result | size |
derivativedivides | \(\frac {\frac {2 i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {4 i B a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-\frac {2 A a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+4 i a^{2} B \sqrt {a +i a \tan \left (d x +c \right )}+2 A \,a^{2} \sqrt {a +i a \tan \left (d x +c \right )}+\frac {a^{\frac {5}{2}} \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2}+\frac {a^{3} \left (i B +A \right )}{\sqrt {a +i a \tan \left (d x +c \right )}}}{a^{3} d}\) | \(167\) |
default | \(\frac {\frac {2 i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {4 i B a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-\frac {2 A a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+4 i a^{2} B \sqrt {a +i a \tan \left (d x +c \right )}+2 A \,a^{2} \sqrt {a +i a \tan \left (d x +c \right )}+\frac {a^{\frac {5}{2}} \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2}+\frac {a^{3} \left (i B +A \right )}{\sqrt {a +i a \tan \left (d x +c \right )}}}{a^{3} d}\) | \(167\) |
parts | \(\frac {2 A \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+a \sqrt {a +i a \tan \left (d x +c \right )}+\frac {a^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}+\frac {a^{2}}{2 \sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d \,a^{2}}+\frac {2 i B \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}+\frac {a^{3}}{2 \sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d \,a^{3}}\) | \(207\) |
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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 = 0.28 (sec) , antiderivative size = 447, normalized size of antiderivative = 2.18 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {15 \, \sqrt {2} {\left (a d e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a d^{2}}} \log \left (-\frac {4 \, {\left ({\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )} + {\left (i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a d^{2}}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 15 \, \sqrt {2} {\left (a d e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a d^{2}}} \log \left (-\frac {4 \, {\left ({\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )} + {\left (-i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a d^{2}}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 2 \, \sqrt {2} {\left ({\left (35 \, A + 103 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 5 \, {\left (25 \, A + 41 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 15 \, {\left (7 \, A + 11 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 15 \, A + 15 i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{60 \, {\left (a d e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )}} \]
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\[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{3}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]
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Time = 0.31 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.77 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {15 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 24 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} B a + 40 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} {\left (A + 2 i \, B\right )} a^{2} - 120 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (A + 2 i \, B\right )} a^{3} - \frac {60 \, {\left (A + i \, B\right )} a^{4}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}}{60 \, a^{4} d} \]
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\[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{3}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]
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Time = 8.70 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.15 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {A}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}+\frac {B\,1{}\mathrm {i}}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}+\frac {2\,A\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a\,d}-\frac {2\,A\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a^2\,d}+\frac {B\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,4{}\mathrm {i}}{a\,d}-\frac {B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,4{}\mathrm {i}}{3\,a^2\,d}+\frac {B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,2{}\mathrm {i}}{5\,a^3\,d}+\frac {\sqrt {2}\,B\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{2\,\sqrt {-a}\,d}-\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,1{}\mathrm {i}}{2\,\sqrt {a}\,d} \]
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