3.2314 \(\int \frac{(A+B x) (a+b x+c x^2)}{(d+e x)^4} \, dx\)

Optimal. Leaf size=127 \[ \frac{A e (2 c d-b e)-B \left (3 c d^2-e (2 b d-a e)\right )}{2 e^4 (d+e x)^2}+\frac{(B d-A e) \left (a e^2-b d e+c d^2\right )}{3 e^4 (d+e x)^3}+\frac{-A c e-b B e+3 B c d}{e^4 (d+e x)}+\frac{B c \log (d+e x)}{e^4} \]

[Out]

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

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Rubi [A]  time = 0.12017, antiderivative size = 126, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 1, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.043, Rules used = {771} \[ -\frac{-B e (2 b d-a e)-A e (2 c d-b e)+3 B c d^2}{2 e^4 (d+e x)^2}+\frac{(B d-A e) \left (a e^2-b d e+c d^2\right )}{3 e^4 (d+e x)^3}+\frac{-A c e-b B e+3 B c d}{e^4 (d+e x)}+\frac{B c \log (d+e x)}{e^4} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + b*x + c*x^2))/(d + e*x)^4,x]

[Out]

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

Mathematica [A]  time = 0.090788, size = 130, normalized size = 1.02 \[ \frac{-A e \left (e (2 a e+b d+3 b e x)+2 c \left (d^2+3 d e x+3 e^2 x^2\right )\right )+B \left (c d \left (11 d^2+27 d e x+18 e^2 x^2\right )-e \left (a e (d+3 e x)+2 b \left (d^2+3 d e x+3 e^2 x^2\right )\right )\right )+6 B c (d+e x)^3 \log (d+e x)}{6 e^4 (d+e x)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + b*x + c*x^2))/(d + e*x)^4,x]

[Out]

(-(A*e*(e*(b*d + 2*a*e + 3*b*e*x) + 2*c*(d^2 + 3*d*e*x + 3*e^2*x^2))) + B*(c*d*(11*d^2 + 27*d*e*x + 18*e^2*x^2
) - e*(a*e*(d + 3*e*x) + 2*b*(d^2 + 3*d*e*x + 3*e^2*x^2))) + 6*B*c*(d + e*x)^3*Log[d + e*x])/(6*e^4*(d + e*x)^
3)

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Maple [A]  time = 0.005, size = 225, normalized size = 1.8 \begin{align*}{\frac{Bc\ln \left ( ex+d \right ) }{{e}^{4}}}-{\frac{aA}{3\,e \left ( ex+d \right ) ^{3}}}+{\frac{Abd}{3\,{e}^{2} \left ( ex+d \right ) ^{3}}}-{\frac{Ac{d}^{2}}{3\,{e}^{3} \left ( ex+d \right ) ^{3}}}+{\frac{aBd}{3\,{e}^{2} \left ( ex+d \right ) ^{3}}}-{\frac{B{d}^{2}b}{3\,{e}^{3} \left ( ex+d \right ) ^{3}}}+{\frac{Bc{d}^{3}}{3\,{e}^{4} \left ( ex+d \right ) ^{3}}}-{\frac{Ab}{2\,{e}^{2} \left ( ex+d \right ) ^{2}}}+{\frac{Acd}{{e}^{3} \left ( ex+d \right ) ^{2}}}-{\frac{aB}{2\,{e}^{2} \left ( ex+d \right ) ^{2}}}+{\frac{Bbd}{{e}^{3} \left ( ex+d \right ) ^{2}}}-{\frac{3\,Bc{d}^{2}}{2\,{e}^{4} \left ( ex+d \right ) ^{2}}}-{\frac{Ac}{{e}^{3} \left ( ex+d \right ) }}-{\frac{bB}{{e}^{3} \left ( ex+d \right ) }}+3\,{\frac{Bcd}{{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+a)/(e*x+d)^4,x)

[Out]

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

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Maxima [A]  time = 1.04028, size = 208, normalized size = 1.64 \begin{align*} \frac{11 \, B c d^{3} - 2 \, A a e^{3} - 2 \,{\left (B b + A c\right )} d^{2} e -{\left (B a + A b\right )} d e^{2} + 6 \,{\left (3 \, B c d e^{2} -{\left (B b + A c\right )} e^{3}\right )} x^{2} + 3 \,{\left (9 \, B c d^{2} e - 2 \,{\left (B b + A c\right )} d e^{2} -{\left (B a + A b\right )} e^{3}\right )} x}{6 \,{\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} + \frac{B c \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+a)/(e*x+d)^4,x, algorithm="maxima")

[Out]

1/6*(11*B*c*d^3 - 2*A*a*e^3 - 2*(B*b + A*c)*d^2*e - (B*a + A*b)*d*e^2 + 6*(3*B*c*d*e^2 - (B*b + A*c)*e^3)*x^2
+ 3*(9*B*c*d^2*e - 2*(B*b + A*c)*d*e^2 - (B*a + A*b)*e^3)*x)/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4) +
B*c*log(e*x + d)/e^4

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Fricas [A]  time = 1.18765, size = 398, normalized size = 3.13 \begin{align*} \frac{11 \, B c d^{3} - 2 \, A a e^{3} - 2 \,{\left (B b + A c\right )} d^{2} e -{\left (B a + A b\right )} d e^{2} + 6 \,{\left (3 \, B c d e^{2} -{\left (B b + A c\right )} e^{3}\right )} x^{2} + 3 \,{\left (9 \, B c d^{2} e - 2 \,{\left (B b + A c\right )} d e^{2} -{\left (B a + A b\right )} e^{3}\right )} x + 6 \,{\left (B c e^{3} x^{3} + 3 \, B c d e^{2} x^{2} + 3 \, B c d^{2} e x + B c d^{3}\right )} \log \left (e x + d\right )}{6 \,{\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} 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+a)/(e*x+d)^4,x, algorithm="fricas")

[Out]

1/6*(11*B*c*d^3 - 2*A*a*e^3 - 2*(B*b + A*c)*d^2*e - (B*a + A*b)*d*e^2 + 6*(3*B*c*d*e^2 - (B*b + A*c)*e^3)*x^2
+ 3*(9*B*c*d^2*e - 2*(B*b + A*c)*d*e^2 - (B*a + A*b)*e^3)*x + 6*(B*c*e^3*x^3 + 3*B*c*d*e^2*x^2 + 3*B*c*d^2*e*x
+ B*c*d^3)*log(e*x + d))/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4)

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Sympy [A]  time = 16.7764, size = 184, normalized size = 1.45 \begin{align*} \frac{B c \log{\left (d + e x \right )}}{e^{4}} - \frac{2 A a e^{3} + A b d e^{2} + 2 A c d^{2} e + B a d e^{2} + 2 B b d^{2} e - 11 B c d^{3} + x^{2} \left (6 A c e^{3} + 6 B b e^{3} - 18 B c d e^{2}\right ) + x \left (3 A b e^{3} + 6 A c d e^{2} + 3 B a e^{3} + 6 B b d e^{2} - 27 B c d^{2} e\right )}{6 d^{3} e^{4} + 18 d^{2} e^{5} x + 18 d e^{6} x^{2} + 6 e^{7} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x+a)/(e*x+d)**4,x)

[Out]

B*c*log(d + e*x)/e**4 - (2*A*a*e**3 + A*b*d*e**2 + 2*A*c*d**2*e + B*a*d*e**2 + 2*B*b*d**2*e - 11*B*c*d**3 + x*
*2*(6*A*c*e**3 + 6*B*b*e**3 - 18*B*c*d*e**2) + x*(3*A*b*e**3 + 6*A*c*d*e**2 + 3*B*a*e**3 + 6*B*b*d*e**2 - 27*B
*c*d**2*e))/(6*d**3*e**4 + 18*d**2*e**5*x + 18*d*e**6*x**2 + 6*e**7*x**3)

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Giac [A]  time = 1.11572, size = 186, normalized size = 1.46 \begin{align*} B c e^{\left (-4\right )} \log \left ({\left | x e + d \right |}\right ) + \frac{{\left (6 \,{\left (3 \, B c d e - B b e^{2} - A c e^{2}\right )} x^{2} + 3 \,{\left (9 \, B c d^{2} - 2 \, B b d e - 2 \, A c d e - B a e^{2} - A b e^{2}\right )} x +{\left (11 \, B c d^{3} - 2 \, B b d^{2} e - 2 \, A c d^{2} e - B a d e^{2} - A b d e^{2} - 2 \, A a e^{3}\right )} e^{\left (-1\right )}\right )} e^{\left (-3\right )}}{6 \,{\left (x e + d\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)/(e*x+d)^4,x, algorithm="giac")

[Out]

B*c*e^(-4)*log(abs(x*e + d)) + 1/6*(6*(3*B*c*d*e - B*b*e^2 - A*c*e^2)*x^2 + 3*(9*B*c*d^2 - 2*B*b*d*e - 2*A*c*d
*e - B*a*e^2 - A*b*e^2)*x + (11*B*c*d^3 - 2*B*b*d^2*e - 2*A*c*d^2*e - B*a*d*e^2 - A*b*d*e^2 - 2*A*a*e^3)*e^(-1
))*e^(-3)/(x*e + d)^3