Optimal. Leaf size=247 \[ \frac{(c-i d) (A-i B-C) (a+b \tan (e+f x))^{m+1} \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a)}-\frac{(c+i d) (A+i B-C) (a+b \tan (e+f x))^{m+1} \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (-b+i a)}-\frac{(a C d-b (m+2) (B d+c C)) (a+b \tan (e+f x))^{m+1}}{b^2 f (m+1) (m+2)}+\frac{C d \tan (e+f x) (a+b \tan (e+f x))^{m+1}}{b f (m+2)} \]
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Rubi [A] time = 0.528326, antiderivative size = 247, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.116, Rules used = {3637, 3630, 3539, 3537, 68} \[ \frac{(c-i d) (A-i B-C) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a)}-\frac{(c+i d) (A+i B-C) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (-b+i a)}-\frac{(a C d-b (m+2) (B d+c C)) (a+b \tan (e+f x))^{m+1}}{b^2 f (m+1) (m+2)}+\frac{C d \tan (e+f x) (a+b \tan (e+f x))^{m+1}}{b f (m+2)} \]
Antiderivative was successfully verified.
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Rule 3637
Rule 3630
Rule 3539
Rule 3537
Rule 68
Rubi steps
\begin{align*} \int (a+b \tan (e+f x))^m (c+d \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx &=\frac{C d \tan (e+f x) (a+b \tan (e+f x))^{1+m}}{b f (2+m)}-\frac{\int (a+b \tan (e+f x))^m \left (a C d-A b c (2+m)-b (B c+(A-C) d) (2+m) \tan (e+f x)+(a C d-b (c C+B d) (2+m)) \tan ^2(e+f x)\right ) \, dx}{b (2+m)}\\ &=-\frac{(a C d-b (c C+B d) (2+m)) (a+b \tan (e+f x))^{1+m}}{b^2 f (1+m) (2+m)}+\frac{C d \tan (e+f x) (a+b \tan (e+f x))^{1+m}}{b f (2+m)}-\frac{\int (a+b \tan (e+f x))^m (-b (A c-c C-B d) (2+m)-b (B c+(A-C) d) (2+m) \tan (e+f x)) \, dx}{b (2+m)}\\ &=-\frac{(a C d-b (c C+B d) (2+m)) (a+b \tan (e+f x))^{1+m}}{b^2 f (1+m) (2+m)}+\frac{C d \tan (e+f x) (a+b \tan (e+f x))^{1+m}}{b f (2+m)}+\frac{1}{2} ((A-i B-C) (c-i d)) \int (1+i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx+\frac{1}{2} ((A+i B-C) (c+i d)) \int (1-i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx\\ &=-\frac{(a C d-b (c C+B d) (2+m)) (a+b \tan (e+f x))^{1+m}}{b^2 f (1+m) (2+m)}+\frac{C d \tan (e+f x) (a+b \tan (e+f x))^{1+m}}{b f (2+m)}-\frac{(i (A+i B-C) (c+i d)) \operatorname{Subst}\left (\int \frac{(a+i b x)^m}{-1+x} \, dx,x,-i \tan (e+f x)\right )}{2 f}+\frac{((A-i B-C) (i c+d)) \operatorname{Subst}\left (\int \frac{(a-i b x)^m}{-1+x} \, dx,x,i \tan (e+f x)\right )}{2 f}\\ &=-\frac{(a C d-b (c C+B d) (2+m)) (a+b \tan (e+f x))^{1+m}}{b^2 f (1+m) (2+m)}-\frac{(A-i B-C) (i c+d) \, _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 (a-i b) f (1+m)}-\frac{(A+i B-C) (c+i d) \, _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 d \tan (e+f x) (a+b \tan (e+f x))^{1+m}}{b f (2+m)}\\ \end{align*}
Mathematica [A] time = 2.71089, size = 202, normalized size = 0.82 \[ \frac{(a+b \tan (e+f x))^{m+1} \left (-\frac{i b (m+2) (c-i d) (A-i B-C) \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{a+b \tan (e+f x)}{a-i b}\right )}{(m+1) (a-i b)}+\frac{i b (m+2) (c+i d) (A+i B-C) \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{a+b \tan (e+f x)}{a+i b}\right )}{(m+1) (a+i b)}+\frac{2 b (m+2) (B d+c C)-2 a C d}{b (m+1)}+2 C d \tan (e+f x)\right )}{2 b f (m+2)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.492, 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 ) \left ( A+B\tan \left ( fx+e \right ) +C \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) \, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [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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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C d \tan \left (f x + e\right )^{3} +{\left (C c + B d\right )} \tan \left (f x + e\right )^{2} + A c +{\left (B c + A d\right )} \tan \left (f x + e\right )\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 ) \left (A + B \tan{\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )\, 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 (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )}{\left (d \tan \left (f x + e\right ) + c\right )}{\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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