3.1292 \(\int \frac{(A+B x) (a+c x^2)}{(d+e x)^3} \, dx\)

Optimal. Leaf size=94 \[ \frac{\left (a e^2+c d^2\right ) (B d-A e)}{2 e^4 (d+e x)^2}-\frac{a B e^2-2 A c d e+3 B c d^2}{e^4 (d+e x)}-\frac{c (3 B d-A e) \log (d+e x)}{e^4}+\frac{B c x}{e^3} \]

[Out]

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

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Rubi [A]  time = 0.0720507, antiderivative size = 94, 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} \[ \frac{\left (a e^2+c d^2\right ) (B d-A e)}{2 e^4 (d+e x)^2}-\frac{a B e^2-2 A c d e+3 B c d^2}{e^4 (d+e x)}-\frac{c (3 B d-A e) \log (d+e x)}{e^4}+\frac{B c x}{e^3} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2))/(d + e*x)^3,x]

[Out]

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

Mathematica [A]  time = 0.0857045, size = 88, normalized size = 0.94 \[ \frac{\frac{\left (a e^2+c d^2\right ) (B d-A e)}{(d+e x)^2}-\frac{2 \left (a B e^2-2 A c d e+3 B c d^2\right )}{d+e x}+2 \log (d+e x) (A c e-3 B c d)+2 B c e x}{2 e^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + c*x^2))/(d + e*x)^3,x]

[Out]

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

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Maple [A]  time = 0.007, size = 144, normalized size = 1.5 \begin{align*}{\frac{Bcx}{{e}^{3}}}+2\,{\frac{Acd}{{e}^{3} \left ( ex+d \right ) }}-{\frac{aB}{{e}^{2} \left ( ex+d \right ) }}-3\,{\frac{Bc{d}^{2}}{{e}^{4} \left ( ex+d \right ) }}-{\frac{aA}{2\,e \left ( ex+d \right ) ^{2}}}-{\frac{Ac{d}^{2}}{2\,{e}^{3} \left ( ex+d \right ) ^{2}}}+{\frac{aBd}{2\,{e}^{2} \left ( ex+d \right ) ^{2}}}+{\frac{Bc{d}^{3}}{2\,{e}^{4} \left ( ex+d \right ) ^{2}}}+{\frac{\ln \left ( ex+d \right ) Ac}{{e}^{3}}}-3\,{\frac{\ln \left ( ex+d \right ) Bcd}{{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)^3,x)

[Out]

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

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Maxima [A]  time = 1.03202, size = 150, normalized size = 1.6 \begin{align*} -\frac{5 \, B c d^{3} - 3 \, A c d^{2} e + B a d e^{2} + A a e^{3} + 2 \,{\left (3 \, B c d^{2} e - 2 \, A c d e^{2} + B a e^{3}\right )} x}{2 \,{\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} + \frac{B c x}{e^{3}} - \frac{{\left (3 \, B c d - A c 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+a)/(e*x+d)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.8462, size = 359, normalized size = 3.82 \begin{align*} \frac{2 \, B c e^{3} x^{3} + 4 \, B c d e^{2} x^{2} - 5 \, B c d^{3} + 3 \, A c d^{2} e - B a d e^{2} - A a e^{3} - 2 \,{\left (2 \, B c d^{2} e - 2 \, A c d e^{2} + B a e^{3}\right )} x - 2 \,{\left (3 \, B c d^{3} - A c d^{2} e +{\left (3 \, B c d e^{2} - A c e^{3}\right )} x^{2} + 2 \,{\left (3 \, B c d^{2} e - A c d e^{2}\right )} x\right )} \log \left (e x + d\right )}{2 \,{\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} 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)^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 3.4963, size = 117, normalized size = 1.24 \begin{align*} \frac{B c x}{e^{3}} - \frac{c \left (- A e + 3 B d\right ) \log{\left (d + e x \right )}}{e^{4}} - \frac{A a e^{3} - 3 A c d^{2} e + B a d e^{2} + 5 B c d^{3} + x \left (- 4 A c d e^{2} + 2 B a e^{3} + 6 B c d^{2} e\right )}{2 d^{2} e^{4} + 4 d e^{5} x + 2 e^{6} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)/(e*x+d)**3,x)

[Out]

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

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Giac [A]  time = 1.23724, size = 130, normalized size = 1.38 \begin{align*} B c x e^{\left (-3\right )} -{\left (3 \, B c d - A c e\right )} e^{\left (-4\right )} \log \left ({\left | x e + d \right |}\right ) - \frac{{\left (5 \, B c d^{3} - 3 \, A c d^{2} e + B a d e^{2} + A a e^{3} + 2 \,{\left (3 \, B c d^{2} e - 2 \, A c d e^{2} + B a e^{3}\right )} x\right )} e^{\left (-4\right )}}{2 \,{\left (x e + d\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)/(e*x+d)^3,x, algorithm="giac")

[Out]

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