Optimal. Leaf size=155 \[ -\frac {(B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {B+i A}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {-B+i A}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {B+i A}{6 a d (a+i a \tan (c+d x))^{3/2}} \]
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Rubi [A] time = 0.13, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3526, 3479, 3480, 206} \[ \frac {B+i A}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {(B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {-B+i A}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {B+i A}{6 a d (a+i a \tan (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 206
Rule 3479
Rule 3480
Rule 3526
Rubi steps
\begin {align*} \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\frac {i A-B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(A-i B) \int \frac {1}{(a+i a \tan (c+d x))^{3/2}} \, dx}{2 a}\\ &=\frac {i A-B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i A+B}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {(A-i B) \int \frac {1}{\sqrt {a+i a \tan (c+d x)}} \, dx}{4 a^2}\\ &=\frac {i A-B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i A+B}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {i A+B}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {(A-i B) \int \sqrt {a+i a \tan (c+d x)} \, dx}{8 a^3}\\ &=\frac {i A-B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i A+B}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {i A+B}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}-\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 )}{4 a^2 d}\\ &=-\frac {(i A+B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {i A-B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i A+B}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {i A+B}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 2.93, size = 176, normalized size = 1.14 \[ -\frac {e^{-6 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{3/2} \sec ^2(c+d x) \left (\sqrt {1+e^{2 i (c+d x)}} \left (B \left (e^{2 i (c+d x)}-17 e^{4 i (c+d x)}+3\right )-i A \left (11 e^{2 i (c+d x)}+23 e^{4 i (c+d x)}+3\right )\right )+15 (B+i A) e^{5 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{240 a^2 d \sqrt {a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.63, size = 392, normalized size = 2.53 \[ -\frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {{\left (4 \, \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} + {\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-\frac {{\left (4 \, \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} - {\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - \sqrt {2} {\left ({\left (23 i \, A + 17 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (34 i \, A + 16 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (14 i \, A - 4 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, A - 3 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \]
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 \[ \int \frac {B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.18, size = 123, normalized size = 0.79 \[ \frac {2 i \left (-\frac {\left (-i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {5}{2}}}-\frac {-\frac {A}{2}-\frac {i B}{2}}{5 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {i B -A}{12 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {i B -A}{8 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.70, size = 139, normalized size = 0.90 \[ \frac {i \, {\left (\frac {15 \, \sqrt {2} {\left (A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {3}{2}}} + \frac {4 \, {\left (15 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} {\left (A - i \, B\right )} + 10 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (A - i \, B\right )} a + 12 \, {\left (A + i \, B\right )} a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a}\right )}}{240 \, a d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 6.58, size = 205, normalized size = 1.32 \[ \frac {\frac {A\,1{}\mathrm {i}}{5\,d}+\frac {A\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{6\,a\,d}+\frac {A\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2\,1{}\mathrm {i}}{4\,a^2\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {-\frac {B}{5}+\frac {B\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{6\,a}+\frac {B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{4\,a^2}}{d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{8\,{\left (-a\right )}^{5/2}\,d}-\frac {\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{8\,a^{5/2}\,d} \]
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 \[ \int \frac {A + B \tan {\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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