3.1291 \(\int \frac{(A+B x) (a+c x^2)}{(d+e x)^2} \, dx\)

Optimal. Leaf size=89 \[ \frac{\left (a e^2+c d^2\right ) (B d-A e)}{e^4 (d+e x)}+\frac{\log (d+e x) \left (a B e^2-2 A c d e+3 B c d^2\right )}{e^4}-\frac{c x (2 B d-A e)}{e^3}+\frac{B c x^2}{2 e^2} \]

[Out]

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

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Rubi [A]  time = 0.0777694, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {772} \[ \frac{\left (a e^2+c d^2\right ) (B d-A e)}{e^4 (d+e x)}+\frac{\log (d+e x) \left (a B e^2-2 A c d e+3 B c d^2\right )}{e^4}-\frac{c x (2 B d-A e)}{e^3}+\frac{B c x^2}{2 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 772

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 0.0663175, size = 86, normalized size = 0.97 \[ \frac{\frac{2 \left (a e^2+c d^2\right ) (B d-A e)}{d+e x}+2 \log (d+e x) \left (a B e^2-2 A c d e+3 B c d^2\right )+2 c e x (A e-2 B d)+B c e^2 x^2}{2 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 131, normalized size = 1.5 \begin{align*}{\frac{Bc{x}^{2}}{2\,{e}^{2}}}+{\frac{Acx}{{e}^{2}}}-2\,{\frac{Bcdx}{{e}^{3}}}-{\frac{aA}{e \left ( ex+d \right ) }}-{\frac{Ac{d}^{2}}{{e}^{3} \left ( ex+d \right ) }}+{\frac{aBd}{{e}^{2} \left ( ex+d \right ) }}+{\frac{Bc{d}^{3}}{{e}^{4} \left ( ex+d \right ) }}-2\,{\frac{\ln \left ( ex+d \right ) Acd}{{e}^{3}}}+{\frac{\ln \left ( ex+d \right ) aB}{{e}^{2}}}+3\,{\frac{\ln \left ( ex+d \right ) Bc{d}^{2}}{{e}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.07845, size = 136, normalized size = 1.53 \begin{align*} \frac{B c d^{3} - A c d^{2} e + B a d e^{2} - A a e^{3}}{e^{5} x + d e^{4}} + \frac{B c e x^{2} - 2 \,{\left (2 \, B c d - A c e\right )} x}{2 \, e^{3}} + \frac{{\left (3 \, B c d^{2} - 2 \, A c d e + B a e^{2}\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+a)/(e*x+d)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.78224, size = 335, normalized size = 3.76 \begin{align*} \frac{B c e^{3} x^{3} + 2 \, B c d^{3} - 2 \, A c d^{2} e + 2 \, B a d e^{2} - 2 \, A a e^{3} -{\left (3 \, B c d e^{2} - 2 \, A c e^{3}\right )} x^{2} - 2 \,{\left (2 \, B c d^{2} e - A c d e^{2}\right )} x + 2 \,{\left (3 \, B c d^{3} - 2 \, A c d^{2} e + B a d e^{2} +{\left (3 \, B c d^{2} e - 2 \, A c d e^{2} + B a e^{3}\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+a)/(e*x+d)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.33886, size = 102, normalized size = 1.15 \begin{align*} \frac{B c x^{2}}{2 e^{2}} + \frac{- A a e^{3} - A c d^{2} e + B a d e^{2} + B c d^{3}}{d e^{4} + e^{5} x} - \frac{x \left (- A c e + 2 B c d\right )}{e^{3}} + \frac{\left (- 2 A c d e + B a e^{2} + 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+a)/(e*x+d)**2,x)

[Out]

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

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Giac [A]  time = 1.19729, size = 204, normalized size = 2.29 \begin{align*} \frac{1}{2} \,{\left (B c - \frac{2 \,{\left (3 \, B c d e - 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 \, A c d e + B a 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{A c d^{2} e^{3}}{x e + d} + \frac{B a d e^{4}}{x e + d} - \frac{A a e^{5}}{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+a)/(e*x+d)^2,x, algorithm="giac")

[Out]

1/2*(B*c - 2*(3*B*c*d*e - A*c*e^2)*e^(-1)/(x*e + d))*(x*e + d)^2*e^(-4) - (3*B*c*d^2 - 2*A*c*d*e + B*a*e^2)*e^
(-4)*log(abs(x*e + d)*e^(-1)/(x*e + d)^2) + (B*c*d^3*e^2/(x*e + d) - A*c*d^2*e^3/(x*e + d) + B*a*d*e^4/(x*e +
d) - A*a*e^5/(x*e + d))*e^(-6)