Optimal. Leaf size=167 \[ \frac{b (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-\sqrt{-b^2}}\right )}{2 \sqrt{-b^2} f (m+1) \left (a-\sqrt{-b^2}\right )}-\frac{b (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+\sqrt{-b^2}}\right )}{2 \sqrt{-b^2} f (m+1) \left (a+\sqrt{-b^2}\right )} \]
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Rubi [A] time = 0.251154, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3485, 712, 68} \[ \frac{b (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-\sqrt{-b^2}}\right )}{2 \sqrt{-b^2} f (m+1) \left (a-\sqrt{-b^2}\right )}-\frac{b (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+\sqrt{-b^2}}\right )}{2 \sqrt{-b^2} f (m+1) \left (a+\sqrt{-b^2}\right )} \]
Antiderivative was successfully verified.
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Rule 3485
Rule 712
Rule 68
Rubi steps
\begin{align*} \int (a+b \tan (e+f x))^m \, dx &=\frac{b \operatorname{Subst}\left (\int \frac{(a+x)^m}{b^2+x^2} \, dx,x,b \tan (e+f x)\right )}{f}\\ &=\frac{b \operatorname{Subst}\left (\int \left (\frac{\sqrt{-b^2} (a+x)^m}{2 b^2 \left (\sqrt{-b^2}-x\right )}+\frac{\sqrt{-b^2} (a+x)^m}{2 b^2 \left (\sqrt{-b^2}+x\right )}\right ) \, dx,x,b \tan (e+f x)\right )}{f}\\ &=-\frac{b \operatorname{Subst}\left (\int \frac{(a+x)^m}{\sqrt{-b^2}-x} \, dx,x,b \tan (e+f x)\right )}{2 \sqrt{-b^2} f}-\frac{b \operatorname{Subst}\left (\int \frac{(a+x)^m}{\sqrt{-b^2}+x} \, dx,x,b \tan (e+f x)\right )}{2 \sqrt{-b^2} f}\\ &=\frac{b \, _2F_1\left (1,1+m;2+m;\frac{a+b \tan (e+f x)}{a-\sqrt{-b^2}}\right ) (a+b \tan (e+f x))^{1+m}}{2 \sqrt{-b^2} \left (a-\sqrt{-b^2}\right ) f (1+m)}-\frac{b \, _2F_1\left (1,1+m;2+m;\frac{a+b \tan (e+f x)}{a+\sqrt{-b^2}}\right ) (a+b \tan (e+f x))^{1+m}}{2 \sqrt{-b^2} \left (a+\sqrt{-b^2}\right ) f (1+m)}\\ \end{align*}
Mathematica [C] time = 0.141003, size = 118, normalized size = 0.71 \[ \frac{(a+b \tan (e+f x))^{m+1} \left ((a+i b) \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )-(a-i b) \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )\right )}{2 f (m+1) (a+i b) (b+i a)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.185, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\tan \left ( fx+e \right ) \right ) ^{m}\, 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{\left (b \tan \left (f x + e\right ) + a\right )}^{m}\,{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 ({\left (b \tan \left (f x + e\right ) + a\right )}^{m}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan{\left (e + f x \right )}\right )^{m}\, dx \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{\left (b \tan \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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