3.22.14 \(\int \frac {a+b x+c x^2}{(d+e x)^3} \, dx\) [2114]

Optimal. Leaf size=62 \[ -\frac {c d^2-b d e+a e^2}{2 e^3 (d+e x)^2}+\frac {2 c d-b e}{e^3 (d+e x)}+\frac {c \log (d+e x)}{e^3} \]

[Out]

1/2*(-a*e^2+b*d*e-c*d^2)/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 = 0.03, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {712} \begin {gather*} -\frac {a e^2-b d e+c d^2}{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 {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)/(d + e*x)^3,x]

[Out]

-1/2*(c*d^2 - b*d*e + a*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 712

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 {a+b x+c x^2}{(d+e x)^3} \, dx &=\int \left (\frac {c d^2-b d e+a e^2}{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 {c d^2-b d e+a e^2}{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 = 0.01, size = 57, normalized size = 0.92 \begin {gather*} \frac {c d (3 d+4 e x)-e (b d+a e+2 b e x)+2 c (d+e x)^2 \log (d+e x)}{2 e^3 (d+e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.53, size = 61, normalized size = 0.98

method	result	size
norman	\(\frac {-\frac {e^{2} a +b d e -3 c \,d^{2}}{2 e^{3}}-\frac {\left (b e -2 c d \right ) x}{e^{2}}}{\left (e x +d \right )^{2}}+\frac {c \ln \left (e x +d \right )}{e^{3}}\)	\(57\)
risch	\(\frac {-\frac {e^{2} a +b d e -3 c \,d^{2}}{2 e^{3}}-\frac {\left (b e -2 c d \right ) x}{e^{2}}}{\left (e x +d \right )^{2}}+\frac {c \ln \left (e x +d \right )}{e^{3}}\)	\(57\)
default	\(-\frac {b e -2 c d}{e^{3} \left (e x +d \right )}+\frac {c \ln \left (e x +d \right )}{e^{3}}-\frac {e^{2} a -b d e +c \,d^{2}}{2 e^{3} \left (e x +d \right )^{2}}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)/(e*x+d)^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 68, normalized size = 1.10 \begin {gather*} c e^{\left (-3\right )} \log \left (x e + d\right ) + \frac {3 \, c d^{2} - b d e + 2 \, {\left (2 \, c d e - b e^{2}\right )} x - a e^{2}}{2 \, {\left (x^{2} e^{5} + 2 \, d x e^{4} + d^{2} e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 3.79, size = 82, normalized size = 1.32 \begin {gather*} \frac {3 \, c d^{2} - {\left (2 \, b x + a\right )} e^{2} + {\left (4 \, c d x - b d\right )} e + 2 \, {\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )} \log \left (x e + d\right )}{2 \, {\left (x^{2} e^{5} + 2 \, d x e^{4} + d^{2} e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.32, size = 68, normalized size = 1.10 \begin {gather*} \frac {c \log {\left (d + e x \right )}}{e^{3}} + \frac {- a e^{2} - 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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*log(d + e*x)/e**3 + (-a*e**2 - 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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Giac [A]
time = 1.20, size = 60, normalized size = 0.97 \begin {gather*} 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 - a e^{2}\right )} e^{\left (-1\right )}\right )} e^{\left (-2\right )}}{2 \, {\left (x e + d\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)/(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 - a*e^2)*e^(-1))*e^(-2)/(x*e + d)^2

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Mupad [B]
time = 0.07, size = 67, normalized size = 1.08 \begin {gather*} \frac {c\,\ln \left (d+e\,x\right )}{e^3}-\frac {\frac {-3\,c\,d^2+b\,d\,e+a\,e^2}{2\,e^3}+\frac {x\,\left (b\,e-2\,c\,d\right )}{e^2}}{d^2+2\,d\,e\,x+e^2\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x + c*x^2)/(d + e*x)^3,x)

[Out]

(c*log(d + e*x))/e^3 - ((a*e^2 - 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)

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