Optimal. Leaf size=363 \[ \frac{(c-i d)^2 (A-i B-C) (a+b \tan (e+f x))^{m+1} \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a)}+\frac{(c+i d)^2 (i A-B-i C) (a+b \tan (e+f x))^{m+1} \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (a+i b)}+\frac{(a+b \tan (e+f x))^{m+1} \left (2 a^2 C d^2-a b d (m+3) (B d+2 c C)+b^2 (m+2) \left (d^2 (m+3) (A-C)+2 B c d (m+3)+2 c^2 C\right )\right )}{b^3 f (m+1) (m+2) (m+3)}-\frac{d \tan (e+f x) (2 a C d-b (B d (m+3)+2 c C)) (a+b \tan (e+f x))^{m+1}}{b^2 f (m+2) (m+3)}+\frac{C (c+d \tan (e+f x))^2 (a+b \tan (e+f x))^{m+1}}{b f (m+3)} \]
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Rubi [A] time = 1.15153, antiderivative size = 360, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {3647, 3637, 3630, 3539, 3537, 68} \[ \frac{(a+b \tan (e+f x))^{m+1} \left (2 a^2 C d^2-a b d (m+3) (B d+2 c C)+b^2 (m+2) \left (d^2 (m+3) (A-C)+2 B c d (m+3)+2 c^2 C\right )\right )}{b^3 f (m+1) (m+2) (m+3)}+\frac{(c-i d)^2 (A-i B-C) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a)}+\frac{(c+i d)^2 (i A-B-i C) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (a+i b)}+\frac{d \tan (e+f x) (-2 a C d+b B d (m+3)+2 b c C) (a+b \tan (e+f x))^{m+1}}{b^2 f (m+2) (m+3)}+\frac{C (c+d \tan (e+f x))^2 (a+b \tan (e+f x))^{m+1}}{b f (m+3)} \]
Antiderivative was successfully verified.
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Rule 3647
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))^2 \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx &=\frac{C (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^2}{b f (3+m)}+\frac{\int (a+b \tan (e+f x))^m (c+d \tan (e+f x)) \left (A b c (3+m)-C (2 a d+b c (1+m))+b (B c+(A-C) d) (3+m) \tan (e+f x)+(2 b c C-2 a C d+b B d (3+m)) \tan ^2(e+f x)\right ) \, dx}{b (3+m)}\\ &=\frac{d (2 b c C-2 a C d+b B d (3+m)) \tan (e+f x) (a+b \tan (e+f x))^{1+m}}{b^2 f (2+m) (3+m)}+\frac{C (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^2}{b f (3+m)}-\frac{\int (a+b \tan (e+f x))^m \left (a d (2 b c C-2 a C d+b B d (3+m))-b c (2+m) (A b c (3+m)-C (2 a d+b c (1+m)))-b^2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) (2+m) (3+m) \tan (e+f x)-\left (b c (2+m) (2 b c C-2 a C d+b B d (3+m))+d \left (b^2 (B c+(A-C) d) (2+m) (3+m)-a (2 b c C-2 a C d+b B d (3+m))\right )\right ) \tan ^2(e+f x)\right ) \, dx}{b^2 (2+m) (3+m)}\\ &=\frac{\left (2 a^2 C d^2-a b d (2 c C+B d) (3+m)+b^2 (2+m) \left (2 c^2 C+2 B c d (3+m)+(A-C) d^2 (3+m)\right )\right ) (a+b \tan (e+f x))^{1+m}}{b^3 f (1+m) (2+m) (3+m)}+\frac{d (2 b c C-2 a C d+b B d (3+m)) \tan (e+f x) (a+b \tan (e+f x))^{1+m}}{b^2 f (2+m) (3+m)}+\frac{C (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^2}{b f (3+m)}-\frac{\int (a+b \tan (e+f x))^m \left (-b^2 \left (A c^2-c^2 C-2 B c d-A d^2+C d^2\right ) (2+m) (3+m)-b^2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) (2+m) (3+m) \tan (e+f x)\right ) \, dx}{b^2 (2+m) (3+m)}\\ &=\frac{\left (2 a^2 C d^2-a b d (2 c C+B d) (3+m)+b^2 (2+m) \left (2 c^2 C+2 B c d (3+m)+(A-C) d^2 (3+m)\right )\right ) (a+b \tan (e+f x))^{1+m}}{b^3 f (1+m) (2+m) (3+m)}+\frac{d (2 b c C-2 a C d+b B d (3+m)) \tan (e+f x) (a+b \tan (e+f x))^{1+m}}{b^2 f (2+m) (3+m)}+\frac{C (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^2}{b f (3+m)}+\frac{1}{2} \left ((A-i B-C) (c-i d)^2\right ) \int (1+i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx+\frac{1}{2} \left ((A+i B-C) (c+i d)^2\right ) \int (1-i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx\\ &=\frac{\left (2 a^2 C d^2-a b d (2 c C+B d) (3+m)+b^2 (2+m) \left (2 c^2 C+2 B c d (3+m)+(A-C) d^2 (3+m)\right )\right ) (a+b \tan (e+f x))^{1+m}}{b^3 f (1+m) (2+m) (3+m)}+\frac{d (2 b c C-2 a C d+b B d (3+m)) \tan (e+f x) (a+b \tan (e+f x))^{1+m}}{b^2 f (2+m) (3+m)}+\frac{C (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^2}{b f (3+m)}+\frac{\left ((i A+B-i C) (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 (A+i B-C) (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 (2 a^2 C d^2-a b d (2 c C+B d) (3+m)+b^2 (2+m) \left (2 c^2 C+2 B c d (3+m)+(A-C) d^2 (3+m)\right )\right ) (a+b \tan (e+f x))^{1+m}}{b^3 f (1+m) (2+m) (3+m)}-\frac{(i A+B-i C) (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 (a-i b) f (1+m)}-\frac{(A+i B-C) (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{d (2 b c C-2 a C d+b B d (3+m)) \tan (e+f x) (a+b \tan (e+f x))^{1+m}}{b^2 f (2+m) (3+m)}+\frac{C (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^2}{b f (3+m)}\\ \end{align*}
Mathematica [A] time = 6.33159, size = 505, normalized size = 1.39 \[ \frac{C (c+d \tan (e+f x))^2 (a+b \tan (e+f x))^{m+1}}{b f (m+3)}+\frac{\frac{d \tan (e+f x) (-2 a C d+b B d (m+3)+2 b c C) (a+b \tan (e+f x))^{m+1}}{b f (m+2)}-\frac{\frac{i (a+b \tan (e+f x))^{m+1} \left (b^2 (m+2) (m+3) \left (-\left (A c^2-A d^2-2 B c d-c^2 C+C d^2\right )\right )-i b^2 (m+2) (m+3) \left (2 c d (A-C)+B \left (c^2-d^2\right )\right )\right ) \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{-i a-i b \tan (e+f x)}{b-i a}\right )}{2 f (m+1) (a+i b)}-\frac{i (a+b \tan (e+f x))^{m+1} \left (-b^2 (m+2) (m+3) \left (A c^2-A d^2-2 B c d-c^2 C+C d^2\right )+i b^2 (m+2) (m+3) \left (2 c d (A-C)+B \left (c^2-d^2\right )\right )\right ) \text{Hypergeometric2F1}\left (1,m+1,m+2,-\frac{i a+i b \tan (e+f x)}{-b-i a}\right )}{2 f (m+1) (a-i b)}+\frac{(a+b \tan (e+f x))^{m+1} \left (-d \left (b^2 (m+2) (m+3) (d (A-C)+B c)-a (-2 a C d+b B d (m+3)+2 b c C)\right )-b c (m+2) (-2 a C d+b B d (m+3)+2 b c C)\right )}{b f (m+1)}}{b (m+2)}}{b (m+3)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.591, 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} \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^{2} \tan \left (f x + e\right )^{4} +{\left (2 \, C c d + B d^{2}\right )} \tan \left (f x + e\right )^{3} + A c^{2} +{\left (C c^{2} + 2 \, B c d + A d^{2}\right )} \tan \left (f x + e\right )^{2} +{\left (B c^{2} + 2 \, A c 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(-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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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 )}^{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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