Optimal. Leaf size=264 \[ -\frac{d \left (c^2 \left (m^2-5 m+6\right )+2 i c d (3-m) m-d^2 \left (m^2-m+2\right )\right ) (a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;-\frac{d (i \tan (e+f x)+1)}{i c-d}\right )}{2 f m \left (c^2+d^2\right )^3}-\frac{d (c (4-m)+i d m) (a+i a \tan (e+f x))^m}{2 f \left (c^2+d^2\right )^2 (c+d \tan (e+f x))}-\frac{d (a+i a \tan (e+f x))^m}{2 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^2}-\frac{(a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;\frac{1}{2} (i \tan (e+f x)+1)\right )}{2 f m (d+i c)^3} \]
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Rubi [A] time = 0.883234, antiderivative size = 264, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3561, 3598, 3600, 3481, 68, 3599} \[ -\frac{d \left (c^2 \left (m^2-5 m+6\right )+2 i c d (3-m) m-d^2 \left (m^2-m+2\right )\right ) (a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;-\frac{d (i \tan (e+f x)+1)}{i c-d}\right )}{2 f m \left (c^2+d^2\right )^3}-\frac{d (c (4-m)+i d m) (a+i a \tan (e+f x))^m}{2 f \left (c^2+d^2\right )^2 (c+d \tan (e+f x))}-\frac{d (a+i a \tan (e+f x))^m}{2 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^2}-\frac{(a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;\frac{1}{2} (i \tan (e+f x)+1)\right )}{2 f m (d+i c)^3} \]
Antiderivative was successfully verified.
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Rule 3561
Rule 3598
Rule 3600
Rule 3481
Rule 68
Rule 3599
Rubi steps
\begin{align*} \int \frac{(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^3} \, dx &=-\frac{d (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}+\frac{\int \frac{(a+i a \tan (e+f x))^m (a (2 c+i d m)-a d (2-m) \tan (e+f x))}{(c+d \tan (e+f x))^2} \, dx}{2 a \left (c^2+d^2\right )}\\ &=-\frac{d (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}-\frac{d (c (4-m)+i d m) (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right )^2 f (c+d \tan (e+f x))}+\frac{\int \frac{(a+i a \tan (e+f x))^m \left (a^2 \left (2 c^2+i c d (5-m) m-d^2 \left (2-m+m^2\right )\right )-a^2 d (1-m) (c (4-m)+i d m) \tan (e+f x)\right )}{c+d \tan (e+f x)} \, dx}{2 a^2 \left (c^2+d^2\right )^2}\\ &=-\frac{d (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}-\frac{d (c (4-m)+i d m) (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right )^2 f (c+d \tan (e+f x))}+\frac{\int (a+i a \tan (e+f x))^m \, dx}{(c-i d)^3}-\frac{\left (d \left (2 i c d (3-m) m+c^2 \left (6-5 m+m^2\right )-d^2 \left (2-m+m^2\right )\right )\right ) \int \frac{(a-i a \tan (e+f x)) (a+i a \tan (e+f x))^m}{c+d \tan (e+f x)} \, dx}{2 a (c+i d)^2 (i c+d)^3}\\ &=-\frac{d (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}-\frac{d (c (4-m)+i d m) (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right )^2 f (c+d \tan (e+f x))}-\frac{a \operatorname{Subst}\left (\int \frac{(a+x)^{-1+m}}{a-x} \, dx,x,i a \tan (e+f x)\right )}{(i c+d)^3 f}-\frac{\left (a d \left (2 i c d (3-m) m+c^2 \left (6-5 m+m^2\right )-d^2 \left (2-m+m^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{(a+i a x)^{-1+m}}{c+d x} \, dx,x,\tan (e+f x)\right )}{2 (c+i d)^2 (i c+d)^3 f}\\ &=-\frac{\, _2F_1\left (1,m;1+m;\frac{1}{2} (1+i \tan (e+f x))\right ) (a+i a \tan (e+f x))^m}{2 (i c+d)^3 f m}-\frac{d \left (2 i c d (3-m) m+c^2 \left (6-5 m+m^2\right )-d^2 \left (2-m+m^2\right )\right ) \, _2F_1\left (1,m;1+m;-\frac{d (1+i \tan (e+f x))}{i c-d}\right ) (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right )^3 f m}-\frac{d (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}-\frac{d (c (4-m)+i d m) (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right )^2 f (c+d \tan (e+f x))}\\ \end{align*}
Mathematica [F] time = 40.6961, size = 0, normalized size = 0. \[ \int \frac{(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^3} \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.876, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{m}}{ \left ( c+d\tan \left ( fx+e \right ) \right ) ^{3}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (\frac{2 \, a e^{\left (2 i \, f x + 2 i \, e\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{m}{\left (-i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 3 i \, e^{\left (4 i \, f x + 4 i \, e\right )} - 3 i \, e^{\left (2 i \, f x + 2 i \, e\right )} - i\right )}}{-i \, c^{3} + 3 \, c^{2} d + 3 i \, c d^{2} - d^{3} +{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-3 i \, c^{3} - 3 \, c^{2} d - 3 i \, c d^{2} - 3 \, d^{3}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-3 i \, c^{3} + 3 \, c^{2} d - 3 i \, c d^{2} + 3 \, d^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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