3.1109 \(\int \frac{(A+B x) (b x+c x^2)}{(d+e x)^2} \, dx\)

Optimal. Leaf size=99 \[ -\frac{x (-A c e-b B e+2 B c d)}{e^3}+\frac{d (B d-A e) (c d-b e)}{e^4 (d+e x)}+\frac{\log (d+e x) (B d (3 c d-2 b e)-A e (2 c d-b e))}{e^4}+\frac{B c x^2}{2 e^2} \]

[Out]

-(((2*B*c*d - b*B*e - A*c*e)*x)/e^3) + (B*c*x^2)/(2*e^2) + (d*(B*d - A*e)*(c*d - b*e))/(e^4*(d + e*x)) + ((B*d
*(3*c*d - 2*b*e) - A*e*(2*c*d - b*e))*Log[d + e*x])/e^4

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Rubi [A]  time = 0.114505, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.045, Rules used = {771} \[ -\frac{x (-A c e-b B e+2 B c d)}{e^3}+\frac{d (B d-A e) (c d-b e)}{e^4 (d+e x)}+\frac{\log (d+e x) (B d (3 c d-2 b e)-A e (2 c d-b e))}{e^4}+\frac{B c x^2}{2 e^2} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2))/(d + e*x)^2,x]

[Out]

-(((2*B*c*d - b*B*e - A*c*e)*x)/e^3) + (B*c*x^2)/(2*e^2) + (d*(B*d - A*e)*(c*d - b*e))/(e^4*(d + e*x)) + ((B*d
*(3*c*d - 2*b*e) - A*e*(2*c*d - b*e))*Log[d + e*x])/e^4

Rule 771

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && N
eQ[b^2 - 4*a*c, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin{align*} \int \frac{(A+B x) \left (b x+c x^2\right )}{(d+e x)^2} \, dx &=\int \left (\frac{-2 B c d+b B e+A c e}{e^3}+\frac{B c x}{e^2}-\frac{d (B d-A e) (c d-b e)}{e^3 (d+e x)^2}+\frac{B d (3 c d-2 b e)-A e (2 c d-b e)}{e^3 (d+e x)}\right ) \, dx\\ &=-\frac{(2 B c d-b B e-A c e) x}{e^3}+\frac{B c x^2}{2 e^2}+\frac{d (B d-A e) (c d-b e)}{e^4 (d+e x)}+\frac{(B d (3 c d-2 b e)-A e (2 c d-b e)) \log (d+e x)}{e^4}\\ \end{align*}

Mathematica [A]  time = 0.084098, size = 93, normalized size = 0.94 \[ \frac{2 e x (A c e+b B e-2 B c d)+\frac{2 d (B d-A e) (c d-b e)}{d+e x}+2 \log (d+e x) (A e (b e-2 c d)+B d (3 c d-2 b e))+B c e^2 x^2}{2 e^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2))/(d + e*x)^2,x]

[Out]

(2*e*(-2*B*c*d + b*B*e + A*c*e)*x + B*c*e^2*x^2 + (2*d*(B*d - A*e)*(c*d - b*e))/(d + e*x) + 2*(B*d*(3*c*d - 2*
b*e) + A*e*(-2*c*d + b*e))*Log[d + e*x])/(2*e^4)

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Maple [A]  time = 0.009, size = 155, normalized size = 1.6 \begin{align*}{\frac{Bc{x}^{2}}{2\,{e}^{2}}}+{\frac{Acx}{{e}^{2}}}+{\frac{bBx}{{e}^{2}}}-2\,{\frac{Bcdx}{{e}^{3}}}+{\frac{\ln \left ( ex+d \right ) Ab}{{e}^{2}}}-2\,{\frac{\ln \left ( ex+d \right ) Acd}{{e}^{3}}}-2\,{\frac{\ln \left ( ex+d \right ) Bbd}{{e}^{3}}}+3\,{\frac{\ln \left ( ex+d \right ) Bc{d}^{2}}{{e}^{4}}}+{\frac{dAb}{{e}^{2} \left ( ex+d \right ) }}-{\frac{Ac{d}^{2}}{{e}^{3} \left ( ex+d \right ) }}-{\frac{B{d}^{2}b}{{e}^{3} \left ( ex+d \right ) }}+{\frac{Bc{d}^{3}}{{e}^{4} \left ( ex+d \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)/(e*x+d)^2,x)

[Out]

1/2*B*c*x^2/e^2+1/e^2*A*c*x+1/e^2*b*B*x-2/e^3*B*c*d*x+1/e^2*ln(e*x+d)*A*b-2/e^3*ln(e*x+d)*A*c*d-2/e^3*ln(e*x+d
)*B*b*d+3/e^4*ln(e*x+d)*B*c*d^2+d/e^2/(e*x+d)*A*b-d^2/e^3/(e*x+d)*A*c-d^2/e^3/(e*x+d)*B*b+d^3/e^4/(e*x+d)*B*c

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Maxima [A]  time = 1.00269, size = 147, normalized size = 1.48 \begin{align*} \frac{B c d^{3} + A b d e^{2} -{\left (B b + A c\right )} d^{2} e}{e^{5} x + d e^{4}} + \frac{B c e x^{2} - 2 \,{\left (2 \, B c d -{\left (B b + A c\right )} e\right )} x}{2 \, e^{3}} + \frac{{\left (3 \, B c d^{2} + A b e^{2} - 2 \,{\left (B b + A c\right )} d e\right )} \log \left (e x + d\right )}{e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)/(e*x+d)^2,x, algorithm="maxima")

[Out]

(B*c*d^3 + A*b*d*e^2 - (B*b + A*c)*d^2*e)/(e^5*x + d*e^4) + 1/2*(B*c*e*x^2 - 2*(2*B*c*d - (B*b + A*c)*e)*x)/e^
3 + (3*B*c*d^2 + A*b*e^2 - 2*(B*b + A*c)*d*e)*log(e*x + d)/e^4

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Fricas [A]  time = 1.51546, size = 373, normalized size = 3.77 \begin{align*} \frac{B c e^{3} x^{3} + 2 \, B c d^{3} + 2 \, A b d e^{2} - 2 \,{\left (B b + A c\right )} d^{2} e -{\left (3 \, B c d e^{2} - 2 \,{\left (B b + A c\right )} e^{3}\right )} x^{2} - 2 \,{\left (2 \, B c d^{2} e -{\left (B b + A c\right )} d e^{2}\right )} x + 2 \,{\left (3 \, B c d^{3} + A b d e^{2} - 2 \,{\left (B b + A c\right )} d^{2} e +{\left (3 \, B c d^{2} e + A b e^{3} - 2 \,{\left (B b + A c\right )} d e^{2}\right )} x\right )} \log \left (e x + d\right )}{2 \,{\left (e^{5} x + d e^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)/(e*x+d)^2,x, algorithm="fricas")

[Out]

1/2*(B*c*e^3*x^3 + 2*B*c*d^3 + 2*A*b*d*e^2 - 2*(B*b + A*c)*d^2*e - (3*B*c*d*e^2 - 2*(B*b + A*c)*e^3)*x^2 - 2*(
2*B*c*d^2*e - (B*b + A*c)*d*e^2)*x + 2*(3*B*c*d^3 + A*b*d*e^2 - 2*(B*b + A*c)*d^2*e + (3*B*c*d^2*e + A*b*e^3 -
2*(B*b + A*c)*d*e^2)*x)*log(e*x + d))/(e^5*x + d*e^4)

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Sympy [A]  time = 1.03381, size = 117, normalized size = 1.18 \begin{align*} \frac{B c x^{2}}{2 e^{2}} - \frac{- A b d e^{2} + A c d^{2} e + B b d^{2} e - B c d^{3}}{d e^{4} + e^{5} x} + \frac{x \left (A c e + B b e - 2 B c d\right )}{e^{3}} - \frac{\left (- A b e^{2} + 2 A c d e + 2 B b d e - 3 B c d^{2}\right ) \log{\left (d + e x \right )}}{e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)/(e*x+d)**2,x)

[Out]

B*c*x**2/(2*e**2) - (-A*b*d*e**2 + A*c*d**2*e + B*b*d**2*e - B*c*d**3)/(d*e**4 + e**5*x) + x*(A*c*e + B*b*e -
2*B*c*d)/e**3 - (-A*b*e**2 + 2*A*c*d*e + 2*B*b*d*e - 3*B*c*d**2)*log(d + e*x)/e**4

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Giac [A]  time = 1.35426, size = 225, normalized size = 2.27 \begin{align*} \frac{1}{2} \,{\left (B c - \frac{2 \,{\left (3 \, B c d e - B b e^{2} - A c e^{2}\right )} e^{\left (-1\right )}}{x e + d}\right )}{\left (x e + d\right )}^{2} e^{\left (-4\right )} -{\left (3 \, B c d^{2} - 2 \, B b d e - 2 \, A c d e + A b e^{2}\right )} e^{\left (-4\right )} \log \left (\frac{{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) +{\left (\frac{B c d^{3} e^{2}}{x e + d} - \frac{B b d^{2} e^{3}}{x e + d} - \frac{A c d^{2} e^{3}}{x e + d} + \frac{A b d e^{4}}{x e + d}\right )} e^{\left (-6\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)/(e*x+d)^2,x, algorithm="giac")

[Out]

1/2*(B*c - 2*(3*B*c*d*e - B*b*e^2 - A*c*e^2)*e^(-1)/(x*e + d))*(x*e + d)^2*e^(-4) - (3*B*c*d^2 - 2*B*b*d*e - 2
*A*c*d*e + A*b*e^2)*e^(-4)*log(abs(x*e + d)*e^(-1)/(x*e + d)^2) + (B*c*d^3*e^2/(x*e + d) - B*b*d^2*e^3/(x*e +
d) - A*c*d^2*e^3/(x*e + d) + A*b*d*e^4/(x*e + d))*e^(-6)