Integrand size = 34, antiderivative size = 156 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {2 A \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {A+i B}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {3 A+i B}{2 a d \sqrt {a+i a \tan (c+d x)}} \]
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Time = 0.71 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {3677, 3681, 3561, 212, 3680, 65, 214} \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {2 A \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {A+i B}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {3 A+i B}{2 a d \sqrt {a+i a \tan (c+d x)}} \]
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Rule 65
Rule 212
Rule 214
Rule 3561
Rule 3677
Rule 3680
Rule 3681
Rubi steps \begin{align*} \text {integral}& = \frac {A+i B}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\cot (c+d x) \left (3 a A-\frac {3}{2} a (i A-B) \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{3 a^2} \\ & = \frac {A+i B}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {3 A+i B}{2 a d \sqrt {a+i a \tan (c+d x)}}+\frac {\int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \left (3 a^2 A-\frac {3}{4} a^2 (3 i A-B) \tan (c+d x)\right ) \, dx}{3 a^4} \\ & = \frac {A+i B}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {3 A+i B}{2 a d \sqrt {a+i a \tan (c+d x)}}+\frac {A \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx}{a^3}+\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 {3 A+i B}{2 a d \sqrt {a+i a \tan (c+d x)}}+\frac {A \text {Subst}\left (\int \frac {1}{x \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{a 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) \text {arctanh}\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 {3 A+i B}{2 a d \sqrt {a+i a \tan (c+d x)}}-\frac {(2 i A) \text {Subst}\left (\int \frac {1}{i-\frac {i x^2}{a}} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{a^2 d} \\ & = -\frac {2 A \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {(A-i B) \text {arctanh}\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 {3 A+i B}{2 a d \sqrt {a+i a \tan (c+d x)}} \\ \end{align*}
Time = 2.79 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {-6 a^3 A \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )+\frac {3 a^3 (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2}}+\frac {a^{9/2} (A+i B)}{(a+i a \tan (c+d x))^{3/2}}+\frac {3 a^{7/2} (3 A+i B)}{2 \sqrt {a+i a \tan (c+d x)}}}{3 a^{9/2} d} \]
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Time = 0.26 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.81
method | result | size |
derivativedivides | \(\frac {2 a \left (-\frac {-i B -3 A}{4 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {-i B -A}{6 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {\left (i B -A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 a^{\frac {5}{2}}}-\frac {A \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{a^{\frac {5}{2}}}\right )}{d}\) | \(126\) |
default | \(\frac {2 a \left (-\frac {-i B -3 A}{4 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {-i B -A}{6 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {\left (i B -A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 a^{\frac {5}{2}}}-\frac {A \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{a^{\frac {5}{2}}}\right )}{d}\) | \(126\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 624 vs. \(2 (117) = 234\).
Time = 0.26 (sec) , antiderivative size = 624, normalized size of antiderivative = 4.00 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\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 ) + 6 \, a^{2} d \sqrt {\frac {A^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac {16 \, {\left (3 \, A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + A a^{2} + 2 \, \sqrt {2} {\left (a^{3} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{3} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2}}{a^{3} d^{2}}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{A}\right ) - 6 \, a^{2} d \sqrt {\frac {A^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac {16 \, {\left (3 \, A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + A a^{2} - 2 \, \sqrt {2} {\left (a^{3} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{3} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2}}{a^{3} d^{2}}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{A}\right ) - \sqrt {2} {\left (2 \, {\left (5 \, A + 2 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (11 \, A + 5 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} \]
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\[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \cot {\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \]
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Time = 0.30 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.03 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {\frac {3 \, \sqrt {2} {\left (A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {3}{2}}} - \frac {24 \, A \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} - \frac {4 \, {\left (3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (3 \, A + i \, B\right )} + 2 \, {\left (A + i \, B\right )} a\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a}}{24 \, d} \]
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\[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \cot \left (d x + c\right )}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
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Time = 7.95 (sec) , antiderivative size = 563, normalized size of antiderivative = 3.61 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\frac {A+B\,1{}\mathrm {i}}{3\,d}+\frac {\left (3\,A+B\,1{}\mathrm {i}\right )\,\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\,\mathrm {atanh}\left (\frac {31\,A^3\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{\sqrt {a^3}\,\left (\frac {31\,A^3\,d}{a}+\frac {A\,B^2\,d}{a}+\frac {A^2\,B\,d\,2{}\mathrm {i}}{a}\right )}+\frac {A\,B^2\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{\sqrt {a^3}\,\left (\frac {31\,A^3\,d}{a}+\frac {A\,B^2\,d}{a}+\frac {A^2\,B\,d\,2{}\mathrm {i}}{a}\right )}+\frac {A^2\,B\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{\sqrt {a^3}\,\left (\frac {31\,A^3\,d}{a}+\frac {A\,B^2\,d}{a}+\frac {A^2\,B\,d\,2{}\mathrm {i}}{a}\right )}\right )}{d\,\sqrt {a^3}}+\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,A^3\,d\,\sqrt {-a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,31{}\mathrm {i}}{16\,\left (\frac {31\,d\,A^3\,a^2}{8}-\frac {29{}\mathrm {i}\,d\,A^2\,B\,a^2}{8}+\frac {3\,d\,A\,B^2\,a^2}{8}-\frac {1{}\mathrm {i}\,d\,B^3\,a^2}{8}\right )}+\frac {\sqrt {2}\,B^3\,d\,\sqrt {-a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{16\,\left (\frac {31\,d\,A^3\,a^2}{8}-\frac {29{}\mathrm {i}\,d\,A^2\,B\,a^2}{8}+\frac {3\,d\,A\,B^2\,a^2}{8}-\frac {1{}\mathrm {i}\,d\,B^3\,a^2}{8}\right )}+\frac {\sqrt {2}\,A\,B^2\,d\,\sqrt {-a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,3{}\mathrm {i}}{16\,\left (\frac {31\,d\,A^3\,a^2}{8}-\frac {29{}\mathrm {i}\,d\,A^2\,B\,a^2}{8}+\frac {3\,d\,A\,B^2\,a^2}{8}-\frac {1{}\mathrm {i}\,d\,B^3\,a^2}{8}\right )}+\frac {29\,\sqrt {2}\,A^2\,B\,d\,\sqrt {-a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{16\,\left (\frac {31\,d\,A^3\,a^2}{8}-\frac {29{}\mathrm {i}\,d\,A^2\,B\,a^2}{8}+\frac {3\,d\,A\,B^2\,a^2}{8}-\frac {1{}\mathrm {i}\,d\,B^3\,a^2}{8}\right )}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,\sqrt {-a^3}}{4\,a^3\,d} \]
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