Optimal. Leaf size=538 \[ \frac{c (d+e x)^{m+1} \left (e m (-2 a B e+A b e-2 A c d+b B d)-\frac{2 b \left (a B e^2+2 A c d e+B c d^2\right )-4 c \left (A \left (a e^2 (1-m)+c d^2\right )+a B d e m\right )+b^2 (-e) (A e m+B d (2-m))}{\sqrt{b^2-4 a c}}\right ) \, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )}{(m+1) \left (b^2-4 a c\right ) \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right ) \left (a e^2-b d e+c d^2\right )}+\frac{c (d+e x)^{m+1} \left (\frac{2 b \left (a B e^2+2 A c d e+B c d^2\right )-4 c \left (A \left (a e^2 (1-m)+c d^2\right )+a B d e m\right )+b^2 (-e) (A e m+B d (2-m))}{\sqrt{b^2-4 a c}}+e m (-2 a B e+A b e-2 A c d+b B d)\right ) \, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{(m+1) \left (b^2-4 a c\right ) \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right ) \left (a e^2-b d e+c d^2\right )}+\frac{(d+e x)^{m+1} \left (-A \left (2 a c e+b^2 (-e)+b c d\right )+c x (-2 a B e+A b e-2 A c d+b B d)+a B (2 c d-b e)\right )}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 3.49374, antiderivative size = 536, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {822, 830, 68} \[ \frac{c (d+e x)^{m+1} \left (e m (-2 a B e+A b e-2 A c d+b B d)-\frac{2 b \left (a B e^2+2 A c d e+B c d^2\right )-4 c \left (a A e^2 (1-m)+a B d e m+A c d^2\right )+b^2 (-e) (A e m+B d (2-m))}{\sqrt{b^2-4 a c}}\right ) \, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )}{(m+1) \left (b^2-4 a c\right ) \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right ) \left (a e^2-b d e+c d^2\right )}+\frac{c (d+e x)^{m+1} \left (\frac{2 b \left (a B e^2+2 A c d e+B c d^2\right )-4 c \left (a A e^2 (1-m)+a B d e m+A c d^2\right )+b^2 (-e) (A e m+B d (2-m))}{\sqrt{b^2-4 a c}}+e m (-2 a B e+A b e-2 A c d+b B d)\right ) \, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{(m+1) \left (b^2-4 a c\right ) \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right ) \left (a e^2-b d e+c d^2\right )}+\frac{(d+e x)^{m+1} \left (-A \left (2 a c e+b^2 (-e)+b c d\right )+c x (-2 a B e+A b e-2 A c d+b B d)+a B (2 c d-b e)\right )}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 822
Rule 830
Rule 68
Rubi steps
\begin{align*} \int \frac{(A+B x) (d+e x)^m}{\left (a+b x+c x^2\right )^2} \, dx &=\frac{(d+e x)^{1+m} \left (a B (2 c d-b e)-A \left (b c d-b^2 e+2 a c e\right )+c (b B d-2 A c d+A b e-2 a B e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}-\frac{\int \frac{(d+e x)^m \left (b^2 e (B d+A e m)+2 c \left (A c d^2+a A e^2 (1-m)+a B d e m\right )-b \left (B c d^2+a B e^2 (1+m)+A c d e (2+m)\right )+c e (b B d-2 A c d+A b e-2 a B e) m x\right )}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )}\\ &=\frac{(d+e x)^{1+m} \left (a B (2 c d-b e)-A \left (b c d-b^2 e+2 a c e\right )+c (b B d-2 A c d+A b e-2 a B e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}-\frac{\int \left (\frac{\left (c e (b B d-2 A c d+A b e-2 a B e) m-\frac{c \left (2 b B c d^2-4 A c^2 d^2-2 b^2 B d e+4 A b c d e+2 a b B e^2-4 a A c e^2+b^2 B d e m-4 a B c d e m-A b^2 e^2 m+4 a A c e^2 m\right )}{\sqrt{b^2-4 a c}}\right ) (d+e x)^m}{b-\sqrt{b^2-4 a c}+2 c x}+\frac{\left (c e (b B d-2 A c d+A b e-2 a B e) m+\frac{c \left (2 b B c d^2-4 A c^2 d^2-2 b^2 B d e+4 A b c d e+2 a b B e^2-4 a A c e^2+b^2 B d e m-4 a B c d e m-A b^2 e^2 m+4 a A c e^2 m\right )}{\sqrt{b^2-4 a c}}\right ) (d+e x)^m}{b+\sqrt{b^2-4 a c}+2 c x}\right ) \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )}\\ &=\frac{(d+e x)^{1+m} \left (a B (2 c d-b e)-A \left (b c d-b^2 e+2 a c e\right )+c (b B d-2 A c d+A b e-2 a B e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}-\frac{\left (c \left (e (b B d-2 A c d+A b e-2 a B e) m-\frac{2 b \left (B c d^2+2 A c d e+a B e^2\right )-b^2 e (B d (2-m)+A e m)-4 c \left (A c d^2+a A e^2 (1-m)+a B d e m\right )}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{(d+e x)^m}{b-\sqrt{b^2-4 a c}+2 c x} \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )}-\frac{\left (c \left (e (b B d-2 A c d+A b e-2 a B e) m+\frac{2 b \left (B c d^2+2 A c d e+a B e^2\right )-b^2 e (B d (2-m)+A e m)-4 c \left (A c d^2+a A e^2 (1-m)+a B d e m\right )}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{(d+e x)^m}{b+\sqrt{b^2-4 a c}+2 c x} \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )}\\ &=\frac{(d+e x)^{1+m} \left (a B (2 c d-b e)-A \left (b c d-b^2 e+2 a c e\right )+c (b B d-2 A c d+A b e-2 a B e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}+\frac{c \left (e (b B d-2 A c d+A b e-2 a B e) m-\frac{2 b \left (B c d^2+2 A c d e+a B e^2\right )-b^2 e (B d (2-m)+A e m)-4 c \left (A c d^2+a A e^2 (1-m)+a B d e m\right )}{\sqrt{b^2-4 a c}}\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )}{\left (b^2-4 a c\right ) \left (2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e\right ) \left (c d^2-b d e+a e^2\right ) (1+m)}+\frac{c \left (e (b B d-2 A c d+A b e-2 a B e) m+\frac{2 b \left (B c d^2+2 A c d e+a B e^2\right )-b^2 e (B d (2-m)+A e m)-4 c \left (A c d^2+a A e^2 (1-m)+a B d e m\right )}{\sqrt{b^2-4 a c}}\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{\left (b^2-4 a c\right ) \left (2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e\right ) \left (c d^2-b d e+a e^2\right ) (1+m)}\\ \end{align*}
Mathematica [A] time = 6.30184, size = 452, normalized size = 0.84 \[ \frac{(d+e x)^{m+1} \left (\frac{c \left (\frac{-2 b \left (a B e^2+2 A c d e+B c d^2\right )+4 c \left (-a A e^2 (m-1)+a B d e m+A c d^2\right )+b^2 e (A e m-B d (m-2))}{\sqrt{b^2-4 a c}}+e m (-2 a B e+A b e-2 A c d+b B d)\right ) \, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d+\left (\sqrt{b^2-4 a c}-b\right ) e}\right )}{(m+1) \left (e \left (\sqrt{b^2-4 a c}-b\right )+2 c d\right )}+\frac{c \left (\frac{2 b \left (a B e^2+2 A c d e+B c d^2\right )-4 c \left (-a A e^2 (m-1)+a B d e m+A c d^2\right )+b^2 e (B d (m-2)-A e m)}{\sqrt{b^2-4 a c}}+e m (-2 a B e+A b e-2 A c d+b B d)\right ) \, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{(m+1) \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )}+\frac{-2 A c (a e+c d x)+a B (2 c (d-e x)-b e)+A b^2 e+A b c (e x-d)+b B c d x}{a+x (b+c x)}\right )}{\left (b^2-4 a c\right ) \left (e (a e-b d)+c d^2\right )} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [F] time = 0.112, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( Bx+A \right ) \left ( ex+d \right ) ^{m}}{ \left ( c{x}^{2}+bx+a \right ) ^{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x + A\right )}{\left (e x + d\right )}^{m}}{{\left (c x^{2} + b x + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B x + A\right )}{\left (e x + d\right )}^{m}}{c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x +{\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x + A\right )}{\left (e x + d\right )}^{m}}{{\left (c x^{2} + b x + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]