3.228 \(\int \frac {b x+c x^2}{(d+e x)^3} \, dx\)

Optimal. Leaf size=55 \[ -\frac {d (c d-b e)}{2 e^3 (d+e x)^2}+\frac {2 c d-b e}{e^3 (d+e x)}+\frac {c \log (d+e x)}{e^3} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {698} \[ -\frac {d (c d-b e)}{2 e^3 (d+e x)^2}+\frac {2 c d-b e}{e^3 (d+e x)}+\frac {c \log (d+e x)}{e^3} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)/(d + e*x)^3,x]

[Out]

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

Rule 698

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 52, normalized size = 0.95 \[ \frac {-b e (d+2 e x)+c d (3 d+4 e x)+2 c (d+e x)^2 \log (d+e x)}{2 e^3 (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.98, size = 81, normalized size = 1.47 \[ \frac {3 \, c d^{2} - b d e + 2 \, {\left (2 \, c d e - b e^{2}\right )} x + 2 \, {\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )} \log \left (e x + d\right )}{2 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)/(e*x+d)^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 55, normalized size = 1.00 \[ c e^{\left (-3\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {{\left (2 \, {\left (2 \, c d - b e\right )} x + {\left (3 \, c d^{2} - b d e\right )} e^{\left (-1\right )}\right )} e^{\left (-2\right )}}{2 \, {\left (x e + d\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 70, normalized size = 1.27 \[ \frac {b d}{2 \left (e x +d \right )^{2} e^{2}}-\frac {c \,d^{2}}{2 \left (e x +d \right )^{2} e^{3}}-\frac {b}{\left (e x +d \right ) e^{2}}+\frac {2 c d}{\left (e x +d \right ) e^{3}}+\frac {c \ln \left (e x +d \right )}{e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x)/(e*x+d)^3,x)

[Out]

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

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maxima [A]  time = 1.41, size = 65, normalized size = 1.18 \[ \frac {3 \, c d^{2} - b d e + 2 \, {\left (2 \, c d e - b e^{2}\right )} x}{2 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} + \frac {c \log \left (e x + d\right )}{e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)/(e*x+d)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.19, size = 63, normalized size = 1.15 \[ \frac {\frac {3\,c\,d^2-b\,d\,e}{2\,e^3}-\frac {x\,\left (b\,e-2\,c\,d\right )}{e^2}}{d^2+2\,d\,e\,x+e^2\,x^2}+\frac {c\,\ln \left (d+e\,x\right )}{e^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)/(d + e*x)^3,x)

[Out]

((3*c*d^2 - b*d*e)/(2*e^3) - (x*(b*e - 2*c*d))/e^2)/(d^2 + e^2*x^2 + 2*d*e*x) + (c*log(d + e*x))/e^3

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sympy [A]  time = 0.36, size = 63, normalized size = 1.15 \[ \frac {c \log {\left (d + e x \right )}}{e^{3}} + \frac {- b d e + 3 c d^{2} + x \left (- 2 b e^{2} + 4 c d e\right )}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)/(e*x+d)**3,x)

[Out]

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

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