Optimal. Leaf size=702 \[ \frac{(a+b \tan (e+f x))^{m+1} \left (2 a^2 d^3 \left (d (A-C) \left (3 c^2-d^2\right )-B \left (c^3-3 c d^2\right )\right )-2 a b d^2 \left (2 c d (A-C) \left (c^2 (3-m)-d^2 (m+1)\right )+B \left (6 c^2 d^2+c^4 (-(2-m))-d^4 m\right )\right )-b^2 \left (A d^2 \left (2 c^2 d^2 \left (-m^2+3 m+1\right )+c^4 \left (-\left (m^2-5 m+6\right )\right )+d^4 (1-m) m\right )+B c d \left (-2 c^2 d^2 \left (-m^2+m+3\right )+c^4 \left (m^2-3 m+2\right )+d^4 m (m+1)\right )+c^2 C \left (2 c^2 d^2 \left (-m^2-m+3\right )+c^4 (1-m) m-d^4 \left (m^2+3 m+2\right )\right )\right )\right ) \text{Hypergeometric2F1}\left (1,m+1,m+2,-\frac{d (a+b \tan (e+f x))}{b c-a d}\right )}{2 f (m+1) \left (c^2+d^2\right )^3 (b c-a d)^3}+\frac{(A-i B-C) (a+b \tan (e+f x))^{m+1} \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a) (c-i d)^3}+\frac{(A+i B-C) (a+b \tan (e+f x))^{m+1} \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (a+i b) (-d+i c)^3}-\frac{(a+b \tan (e+f x))^{m+1} \left (2 a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )-b \left (c^2 d^2 (A (5-m)-C (m+3))+A d^4 (1-m)-B c^3 d (3-m)+B c d^3 (m+1)+c^4 C (1-m)\right )\right )}{2 f \left (c^2+d^2\right )^2 (b c-a d)^2 (c+d \tan (e+f x))}+\frac{\left (A d^2-B c d+c^2 C\right ) (a+b \tan (e+f x))^{m+1}}{2 f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))^2} \]
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Rubi [A] time = 2.93764, antiderivative size = 702, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {3649, 3653, 3539, 3537, 68, 3634} \[ \frac{(a+b \tan (e+f x))^{m+1} \left (2 a^2 d^3 \left (d (A-C) \left (3 c^2-d^2\right )-B \left (c^3-3 c d^2\right )\right )-2 a b d^2 \left (2 c d (A-C) \left (c^2 (3-m)-d^2 (m+1)\right )+B \left (6 c^2 d^2+c^4 (-(2-m))-d^4 m\right )\right )-b^2 \left (A d^2 \left (2 c^2 d^2 \left (-m^2+3 m+1\right )+c^4 \left (-\left (m^2-5 m+6\right )\right )+d^4 (1-m) m\right )+B \left (-2 c^3 d^3 \left (-m^2+m+3\right )+c^5 d \left (m^2-3 m+2\right )+c d^5 m (m+1)\right )+c^2 C \left (2 c^2 d^2 \left (-m^2-m+3\right )+c^4 (1-m) m-d^4 \left (m^2+3 m+2\right )\right )\right )\right ) \, _2F_1\left (1,m+1;m+2;-\frac{d (a+b \tan (e+f x))}{b c-a d}\right )}{2 f (m+1) \left (c^2+d^2\right )^3 (b c-a d)^3}-\frac{(a+b \tan (e+f x))^{m+1} \left (2 a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )-b \left (c^2 d^2 (A (5-m)-C (m+3))+A d^4 (1-m)-B c^3 d (3-m)+B c d^3 (m+1)+c^4 C (1-m)\right )\right )}{2 f \left (c^2+d^2\right )^2 (b c-a d)^2 (c+d \tan (e+f x))}+\frac{\left (A d^2-B c d+c^2 C\right ) (a+b \tan (e+f x))^{m+1}}{2 f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))^2}+\frac{(A-i B-C) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a) (c-i d)^3}+\frac{(A+i B-C) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (a+i b) (-d+i c)^3} \]
Antiderivative was successfully verified.
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Rule 3649
Rule 3653
Rule 3539
Rule 3537
Rule 68
Rule 3634
Rubi steps
\begin{align*} \int \frac{(a+b \tan (e+f x))^m \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^3} \, dx &=\frac{\left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^{1+m}}{2 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}+\frac{\int \frac{(a+b \tan (e+f x))^m \left (A \left (2 c (b c-a d)+b d^2 (1-m)\right )+(c C-B d) (2 a d-b c (1+m))+2 (b c-a d) (B c-(A-C) d) \tan (e+f x)+b \left (c^2 C-B c d+A d^2\right ) (1-m) \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^2} \, dx}{2 (b c-a d) \left (c^2+d^2\right )}\\ &=\frac{\left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^{1+m}}{2 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}-\frac{\left (2 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )-b \left (c^4 C (1-m)+A d^4 (1-m)-B c^3 d (3-m)+B c d^3 (1+m)+c^2 d^2 (A (5-m)-C (3+m))\right )\right ) (a+b \tan (e+f x))^{1+m}}{2 (b c-a d)^2 \left (c^2+d^2\right )^2 f (c+d \tan (e+f x))}+\frac{\int \frac{(a+b \tan (e+f x))^m \left (-\left (2 d (b c-a d) (B c-(A-C) d)-b c \left (c^2 C-B c d+A d^2\right ) (1-m)\right ) (a d-b c (1+m))-\left (a c d-b \left (c^2-d^2 m\right )\right ) \left (A \left (2 c (b c-a d)+b d^2 (1-m)\right )+(c C-B d) (2 a d-b c (1+m))\right )-2 (b c-a d)^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right ) \tan (e+f x)+b m \left (2 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )-b \left (A d^4 (1-m)-B c^3 d (3-m)+B c d^3 (1+m)+c^4 (C-C m)+c^2 d^2 (A (5-m)-C (3+m))\right )\right ) \tan ^2(e+f x)\right )}{c+d \tan (e+f x)} \, dx}{2 (b c-a d)^2 \left (c^2+d^2\right )^2}\\ &=\frac{\left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^{1+m}}{2 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}-\frac{\left (2 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )-b \left (c^4 C (1-m)+A d^4 (1-m)-B c^3 d (3-m)+B c d^3 (1+m)+c^2 d^2 (A (5-m)-C (3+m))\right )\right ) (a+b \tan (e+f x))^{1+m}}{2 (b c-a d)^2 \left (c^2+d^2\right )^2 f (c+d \tan (e+f x))}+\frac{\int (a+b \tan (e+f x))^m \left (2 (b c-a d)^2 \left (A c^3-c^3 C+3 B c^2 d-3 A c d^2+3 c C d^2-B d^3\right )-2 (b c-a d)^2 \left ((A-C) d \left (3 c^2-d^2\right )-B \left (c^3-3 c d^2\right )\right ) \tan (e+f x)\right ) \, dx}{2 (b c-a d)^2 \left (c^2+d^2\right )^3}+\frac{\left (2 a^2 d^3 \left ((A-C) d \left (3 c^2-d^2\right )-B \left (c^3-3 c d^2\right )\right )-2 a b d^2 \left (B \left (6 c^2 d^2-c^4 (2-m)-d^4 m\right )+2 c (A-C) d \left (c^2 (3-m)-d^2 (1+m)\right )\right )-b^2 \left (A d^2 \left (d^4 (1-m) m+2 c^2 d^2 \left (1+3 m-m^2\right )-c^4 \left (6-5 m+m^2\right )\right )+B \left (c d^5 m (1+m)-2 c^3 d^3 \left (3+m-m^2\right )+c^5 d \left (2-3 m+m^2\right )\right )+c^2 C \left (c^4 (1-m) m+2 c^2 d^2 \left (3-m-m^2\right )-d^4 \left (2+3 m+m^2\right )\right )\right )\right ) \int \frac{(a+b \tan (e+f x))^m \left (1+\tan ^2(e+f x)\right )}{c+d \tan (e+f x)} \, dx}{2 (b c-a d)^2 \left (c^2+d^2\right )^3}\\ &=\frac{\left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^{1+m}}{2 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}-\frac{\left (2 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )-b \left (c^4 C (1-m)+A d^4 (1-m)-B c^3 d (3-m)+B c d^3 (1+m)+c^2 d^2 (A (5-m)-C (3+m))\right )\right ) (a+b \tan (e+f x))^{1+m}}{2 (b c-a d)^2 \left (c^2+d^2\right )^2 f (c+d \tan (e+f x))}+\frac{(A-i B-C) \int (1+i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx}{2 (c-i d)^3}+\frac{(A+i B-C) \int (1-i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx}{2 (c+i d)^3}+\frac{\left (2 a^2 d^3 \left ((A-C) d \left (3 c^2-d^2\right )-B \left (c^3-3 c d^2\right )\right )-2 a b d^2 \left (B \left (6 c^2 d^2-c^4 (2-m)-d^4 m\right )+2 c (A-C) d \left (c^2 (3-m)-d^2 (1+m)\right )\right )-b^2 \left (A d^2 \left (d^4 (1-m) m+2 c^2 d^2 \left (1+3 m-m^2\right )-c^4 \left (6-5 m+m^2\right )\right )+B \left (c d^5 m (1+m)-2 c^3 d^3 \left (3+m-m^2\right )+c^5 d \left (2-3 m+m^2\right )\right )+c^2 C \left (c^4 (1-m) m+2 c^2 d^2 \left (3-m-m^2\right )-d^4 \left (2+3 m+m^2\right )\right )\right )\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^m}{c+d x} \, dx,x,\tan (e+f x)\right )}{2 (b c-a d)^2 \left (c^2+d^2\right )^3 f}\\ &=\frac{\left (2 a^2 d^3 \left ((A-C) d \left (3 c^2-d^2\right )-B \left (c^3-3 c d^2\right )\right )-2 a b d^2 \left (B \left (6 c^2 d^2-c^4 (2-m)-d^4 m\right )+2 c (A-C) d \left (c^2 (3-m)-d^2 (1+m)\right )\right )-b^2 \left (A d^2 \left (d^4 (1-m) m+2 c^2 d^2 \left (1+3 m-m^2\right )-c^4 \left (6-5 m+m^2\right )\right )+B \left (c d^5 m (1+m)-2 c^3 d^3 \left (3+m-m^2\right )+c^5 d \left (2-3 m+m^2\right )\right )+c^2 C \left (c^4 (1-m) m+2 c^2 d^2 \left (3-m-m^2\right )-d^4 \left (2+3 m+m^2\right )\right )\right )\right ) \, _2F_1\left (1,1+m;2+m;-\frac{d (a+b \tan (e+f x))}{b c-a d}\right ) (a+b \tan (e+f x))^{1+m}}{2 (b c-a d)^3 \left (c^2+d^2\right )^3 f (1+m)}+\frac{\left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^{1+m}}{2 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}-\frac{\left (2 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )-b \left (c^4 C (1-m)+A d^4 (1-m)-B c^3 d (3-m)+B c d^3 (1+m)+c^2 d^2 (A (5-m)-C (3+m))\right )\right ) (a+b \tan (e+f x))^{1+m}}{2 (b c-a d)^2 \left (c^2+d^2\right )^2 f (c+d \tan (e+f x))}-\frac{(i (A+i B-C)) \operatorname{Subst}\left (\int \frac{(a+i b x)^m}{-1+x} \, dx,x,-i \tan (e+f x)\right )}{2 (c+i d)^3 f}+\frac{(A-i B-C) \operatorname{Subst}\left (\int \frac{(a-i b x)^m}{-1+x} \, dx,x,i \tan (e+f x)\right )}{2 (i c+d)^3 f}\\ &=-\frac{(A-i B-C) \, _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) (i c+d)^3 f (1+m)}-\frac{(A+i B-C) \, _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) (c+i d)^3 f (1+m)}+\frac{\left (2 a^2 d^3 \left ((A-C) d \left (3 c^2-d^2\right )-B \left (c^3-3 c d^2\right )\right )-2 a b d^2 \left (B \left (6 c^2 d^2-c^4 (2-m)-d^4 m\right )+2 c (A-C) d \left (c^2 (3-m)-d^2 (1+m)\right )\right )-b^2 \left (A d^2 \left (d^4 (1-m) m+2 c^2 d^2 \left (1+3 m-m^2\right )-c^4 \left (6-5 m+m^2\right )\right )+B \left (c d^5 m (1+m)-2 c^3 d^3 \left (3+m-m^2\right )+c^5 d \left (2-3 m+m^2\right )\right )+c^2 C \left (c^4 (1-m) m+2 c^2 d^2 \left (3-m-m^2\right )-d^4 \left (2+3 m+m^2\right )\right )\right )\right ) \, _2F_1\left (1,1+m;2+m;-\frac{d (a+b \tan (e+f x))}{b c-a d}\right ) (a+b \tan (e+f x))^{1+m}}{2 (b c-a d)^3 \left (c^2+d^2\right )^3 f (1+m)}+\frac{\left (c^2 C-B c d+A d^2\right ) (a+b \tan (e+f x))^{1+m}}{2 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}-\frac{\left (2 a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )-b \left (c^4 C (1-m)+A d^4 (1-m)-B c^3 d (3-m)+B c d^3 (1+m)+c^2 d^2 (A (5-m)-C (3+m))\right )\right ) (a+b \tan (e+f x))^{1+m}}{2 (b c-a d)^2 \left (c^2+d^2\right )^2 f (c+d \tan (e+f x))}\\ \end{align*}
Mathematica [B] time = 6.23456, size = 2238, normalized size = 3.19 \[ \text{Result too large to show} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.805, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\tan \left ( fx+e \right ) \right ) ^{m} \left ( A+B\tan \left ( fx+e \right ) +C \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }{ \left ( c+d\tan \left ( fx+e \right ) \right ) ^{3}}}\, 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 (\frac{{\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )}{\left (b \tan \left (f x + e\right ) + a\right )}^{m}}{d^{3} \tan \left (f x + e\right )^{3} + 3 \, c d^{2} \tan \left (f x + e\right )^{2} + 3 \, c^{2} d \tan \left (f x + e\right ) + c^{3}}, 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 \frac{{\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )}{\left (b \tan \left (f x + e\right ) + a\right )}^{m}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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