Optimal. Leaf size=176 \[ \frac{(c-i d)^2 (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a)}-\frac{(c+i d)^2 (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (-b+i a)}+\frac{d^2 (a+b \tan (e+f x))^{m+1}}{b f (m+1)} \]
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Rubi [A] time = 0.213298, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3543, 3539, 3537, 68} \[ \frac{(c-i d)^2 (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a)}-\frac{(c+i d)^2 (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (-b+i a)}+\frac{d^2 (a+b \tan (e+f x))^{m+1}}{b f (m+1)} \]
Antiderivative was successfully verified.
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Rule 3543
Rule 3539
Rule 3537
Rule 68
Rubi steps
\begin{align*} \int (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 f (1+m)}+\int (a+b \tan (e+f x))^m \left (c^2-d^2+2 c d \tan (e+f x)\right ) \, dx\\ &=\frac{d^2 (a+b \tan (e+f x))^{1+m}}{b f (1+m)}+\frac{1}{2} (c-i d)^2 \int (1+i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx+\frac{1}{2} (c+i d)^2 \int (1-i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx\\ &=\frac{d^2 (a+b \tan (e+f x))^{1+m}}{b f (1+m)}+\frac{\left (i (c-i d)^2\right ) \operatorname{Subst}\left (\int \frac{(a-i b x)^m}{-1+x} \, dx,x,i \tan (e+f x)\right )}{2 f}-\frac{\left (i (c+i d)^2\right ) \operatorname{Subst}\left (\int \frac{(a+i b x)^m}{-1+x} \, dx,x,-i \tan (e+f x)\right )}{2 f}\\ &=\frac{d^2 (a+b \tan (e+f x))^{1+m}}{b f (1+m)}+\frac{(c-i d)^2 \, _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) f (1+m)}-\frac{(c+i d)^2 \, _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) f (1+m)}\\ \end{align*}
Mathematica [A] time = 0.202632, size = 135, normalized size = 0.77 \[ \frac{(a+b \tan (e+f x))^{m+1} \left (-\frac{i (c-i d)^2 \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{a-i b}+\frac{i (c+i d)^2 \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{a+i b}+\frac{2 d^2}{b}\right )}{2 f (m+1)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.259, size = 0, normalized size = 0. \begin{align*} \int \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{\left (d \tan \left (f x + e\right ) + c\right )}^{2}{\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 (d^{2} \tan \left (f x + e\right )^{2} + 2 \, c d \tan \left (f x + e\right ) + c^{2}\right )}{\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} \left (c + d \tan{\left (e + f x \right )}\right )^{2}\, 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 (d \tan \left (f x + e\right ) + c\right )}^{2}{\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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