Optimal. Leaf size=301 \[ -\frac{d^2 \left (2 a c d-b \left (c^2 (2-m)-d^2 m\right )\right ) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{d (a+b \tan (e+f x))}{b c-a d}\right )}{f (m+1) \left (c^2+d^2\right )^2 (b c-a d)^2}+\frac{d^2 (a+b \tan (e+f x))^{m+1}}{f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))}+\frac{(a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a) (c-i d)^2}-\frac{(a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (-b+i a) (c+i d)^2} \]
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Rubi [A] time = 0.804799, antiderivative size = 299, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3569, 3653, 3539, 3537, 68, 3634} \[ -\frac{d^2 \left (2 a c d-b c^2 (2-m)+b d^2 m\right ) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{d (a+b \tan (e+f x))}{b c-a d}\right )}{f (m+1) \left (c^2+d^2\right )^2 (b c-a d)^2}+\frac{d^2 (a+b \tan (e+f x))^{m+1}}{f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))}+\frac{(a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a) (c-i d)^2}-\frac{(a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (-b+i a) (c+i d)^2} \]
Antiderivative was successfully verified.
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Rule 3569
Rule 3653
Rule 3539
Rule 3537
Rule 68
Rule 3634
Rubi steps
\begin{align*} \int \frac{(a+b \tan (e+f x))^m}{(c+d \tan (e+f x))^2} \, dx &=\frac{d^2 (a+b \tan (e+f x))^{1+m}}{(b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))}+\frac{\int \frac{(a+b \tan (e+f x))^m \left (-a c d+b \left (c^2-d^2 m\right )-d (b c-a d) \tan (e+f x)-b d^2 m \tan ^2(e+f x)\right )}{c+d \tan (e+f x)} \, dx}{(b c-a d) \left (c^2+d^2\right )}\\ &=\frac{d^2 (a+b \tan (e+f x))^{1+m}}{(b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))}+\frac{\int (a+b \tan (e+f x))^m \left ((b c-a d) \left (c^2-d^2\right )-2 c d (b c-a d) \tan (e+f x)\right ) \, dx}{(b c-a d) \left (c^2+d^2\right )^2}-\frac{\left (d^2 \left (2 a c d-b c^2 (2-m)+b d^2 m\right )\right ) \int \frac{(a+b \tan (e+f x))^m \left (1+\tan ^2(e+f x)\right )}{c+d \tan (e+f x)} \, dx}{(b c-a d) \left (c^2+d^2\right )^2}\\ &=\frac{d^2 (a+b \tan (e+f x))^{1+m}}{(b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))}+\frac{\int (1+i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx}{2 (c-i d)^2}+\frac{\int (1-i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx}{2 (c+i d)^2}-\frac{\left (d^2 \left (2 a c d-b c^2 (2-m)+b d^2 m\right )\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^m}{c+d x} \, dx,x,\tan (e+f x)\right )}{(b c-a d) \left (c^2+d^2\right )^2 f}\\ &=-\frac{d^2 \left (2 a c d-b c^2 (2-m)+b d^2 m\right ) \, _2F_1\left (1,1+m;2+m;-\frac{d (a+b \tan (e+f x))}{b c-a d}\right ) (a+b \tan (e+f x))^{1+m}}{(b c-a d)^2 \left (c^2+d^2\right )^2 f (1+m)}+\frac{d^2 (a+b \tan (e+f x))^{1+m}}{(b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))}+\frac{i \operatorname{Subst}\left (\int \frac{(a-i b x)^m}{-1+x} \, dx,x,i \tan (e+f x)\right )}{2 (c-i d)^2 f}-\frac{i \operatorname{Subst}\left (\int \frac{(a+i b x)^m}{-1+x} \, dx,x,-i \tan (e+f x)\right )}{2 (c+i d)^2 f}\\ &=\frac{\, _2F_1\left (1,1+m;2+m;\frac{a+b \tan (e+f x)}{a-i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (i a+b) (c-i d)^2 f (1+m)}-\frac{\, _2F_1\left (1,1+m;2+m;\frac{a+b \tan (e+f x)}{a+i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (i a-b) (c+i d)^2 f (1+m)}-\frac{d^2 \left (2 a c d-b c^2 (2-m)+b d^2 m\right ) \, _2F_1\left (1,1+m;2+m;-\frac{d (a+b \tan (e+f x))}{b c-a d}\right ) (a+b \tan (e+f x))^{1+m}}{(b c-a d)^2 \left (c^2+d^2\right )^2 f (1+m)}+\frac{d^2 (a+b \tan (e+f x))^{1+m}}{(b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))}\\ \end{align*}
Mathematica [A] time = 4.01135, size = 266, normalized size = 0.88 \[ \frac{(a+b \tan (e+f x))^{m+1} \left (-\frac{2 d^2 \left (2 a c d+b c^2 (m-2)+b d^2 m\right ) \, _2F_1\left (1,m+1;m+2;\frac{d (a+b \tan (e+f x))}{a d-b c}\right )}{(m+1) \left (c^2+d^2\right ) (a d-b c)}-\frac{i \left (\frac{(c-i d)^2 (b c-a d) \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{a+i b}+\frac{(c+i d)^2 (a d-b c) \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{a-i b}\right )}{(m+1) \left (c^2+d^2\right )}-\frac{2 d^2}{c+d \tan (e+f x)}\right )}{2 f \left (c^2+d^2\right ) (a d-b c)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.355, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\tan \left ( fx+e \right ) \right ) ^{m}}{ \left ( c+d\tan \left ( fx+e \right ) \right ) ^{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \tan \left (f x + e\right ) + a\right )}^{m}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{2}}\,{d x} \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 (b \tan \left (f x + e\right ) + a\right )}^{m}}{d^{2} \tan \left (f x + e\right )^{2} + 2 \, c d \tan \left (f x + e\right ) + c^{2}}, 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 (b \tan \left (f x + e\right ) + a\right )}^{m}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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