Optimal. Leaf size=161 \[ \frac{(\tan (e+f x)+1)^{-m} (a \tan (e+f x)+a)^m (d \tan (e+f x))^{n+1} F_1(n+1;-m,1;n+2;-\tan (e+f x),-i \tan (e+f x))}{2 d f (n+1)}+\frac{(\tan (e+f x)+1)^{-m} (a \tan (e+f x)+a)^m (d \tan (e+f x))^{n+1} F_1(n+1;-m,1;n+2;-\tan (e+f x),i \tan (e+f x))}{2 d f (n+1)} \]
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Rubi [A] time = 0.14935, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3575, 912, 135, 133} \[ \frac{(\tan (e+f x)+1)^{-m} (a \tan (e+f x)+a)^m (d \tan (e+f x))^{n+1} F_1(n+1;-m,1;n+2;-\tan (e+f x),-i \tan (e+f x))}{2 d f (n+1)}+\frac{(\tan (e+f x)+1)^{-m} (a \tan (e+f x)+a)^m (d \tan (e+f x))^{n+1} F_1(n+1;-m,1;n+2;-\tan (e+f x),i \tan (e+f x))}{2 d f (n+1)} \]
Antiderivative was successfully verified.
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Rule 3575
Rule 912
Rule 135
Rule 133
Rubi steps
\begin{align*} \int (d \tan (e+f x))^n (a+a \tan (e+f x))^m \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(d x)^n (a+a x)^m}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{i (d x)^n (a+a x)^m}{2 (i-x)}+\frac{i (d x)^n (a+a x)^m}{2 (i+x)}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{i \operatorname{Subst}\left (\int \frac{(d x)^n (a+a x)^m}{i-x} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac{i \operatorname{Subst}\left (\int \frac{(d x)^n (a+a x)^m}{i+x} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=\frac{\left (i (1+\tan (e+f x))^{-m} (a+a \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \frac{(d x)^n (1+x)^m}{i-x} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac{\left (i (1+\tan (e+f x))^{-m} (a+a \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \frac{(d x)^n (1+x)^m}{i+x} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=\frac{F_1(1+n;-m,1;2+n;-\tan (e+f x),-i \tan (e+f x)) (d \tan (e+f x))^{1+n} (1+\tan (e+f x))^{-m} (a+a \tan (e+f x))^m}{2 d f (1+n)}+\frac{F_1(1+n;-m,1;2+n;-\tan (e+f x),i \tan (e+f x)) (d \tan (e+f x))^{1+n} (1+\tan (e+f x))^{-m} (a+a \tan (e+f x))^m}{2 d f (1+n)}\\ \end{align*}
Mathematica [F] time = 0.677661, size = 0, normalized size = 0. \[ \int (d \tan (e+f x))^n (a+a \tan (e+f x))^m \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.615, size = 0, normalized size = 0. \begin{align*} \int \left ( d\tan \left ( fx+e \right ) \right ) ^{n} \left ( a+a\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 (a \tan \left (f x + e\right ) + a\right )}^{m} \left (d \tan \left (f x + e\right )\right )^{n}\,{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 (a \tan \left (f x + e\right ) + a\right )}^{m} \left (d \tan \left (f x + e\right )\right )^{n}, 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 \left (\tan{\left (e + f x \right )} + 1\right )\right )^{m} \left (d \tan{\left (e + f x \right )}\right )^{n}\, 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 (a \tan \left (f x + e\right ) + a\right )}^{m} \left (d \tan \left (f x + e\right )\right )^{n}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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