Optimal. Leaf size=161 \[ \frac {F_1(1+n;-m,1;2+n;-\tan (e+f x),-i \tan (e+f x)) (d \tan (e+f x))^{1+n} (1+\tan (e+f x))^{-m} (a+a \tan (e+f x))^m}{2 d f (1+n)}+\frac {F_1(1+n;-m,1;2+n;-\tan (e+f x),i \tan (e+f x)) (d \tan (e+f x))^{1+n} (1+\tan (e+f x))^{-m} (a+a \tan (e+f x))^m}{2 d f (1+n)} \]
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Rubi [A]
time = 0.12, antiderivative size = 161, normalized size of antiderivative = 1.00, 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 = {3656, 926, 140,
138} \begin {gather*} \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)} \end {gather*}
Antiderivative was successfully verified.
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Rule 138
Rule 140
Rule 926
Rule 3656
Rubi steps
\begin {align*} \int (d \tan (e+f x))^n (a+a \tan (e+f x))^m \, dx &=\frac {\text {Subst}\left (\int \frac {(d x)^n (a+a x)^m}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\text {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 \text {Subst}\left (\int \frac {(d x)^n (a+a x)^m}{i-x} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac {i \text {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 ) \text {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 ) \text {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*}
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Mathematica [F]
time = 0.82, size = 0, normalized size = 0.00 \begin {gather*} \int (d \tan (e+f x))^n (a+a \tan (e+f x))^m \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [F]
time = 0.40, size = 0, normalized size = 0.00 \[\int \left (d \tan \left (f x +e \right )\right )^{n} \left (a +a \tan \left (f x +e \right )\right )^{m}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \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 {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\right )}^m \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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