3.1.63 \(\int \frac {x^2 (d+e x)}{(b x+c x^2)^3} \, dx\) [63]

Optimal. Leaf size=57 \[ \frac {c d-b e}{2 b c (b+c x)^2}+\frac {d}{b^2 (b+c x)}+\frac {d \log (x)}{b^3}-\frac {d \log (b+c x)}{b^3} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {779} \begin {gather*} -\frac {d \log (b+c x)}{b^3}+\frac {d \log (x)}{b^3}+\frac {d}{b^2 (b+c x)}+\frac {c d-b e}{2 b c (b+c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 779

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 53, normalized size = 0.93 \begin {gather*} \frac {\frac {b \left (3 b c d-b^2 e+2 c^2 d x\right )}{c (b+c x)^2}+2 d \log (x)-2 d \log (b+c x)}{2 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.51, size = 56, normalized size = 0.98

method	result	size
risch	\(\frac {\frac {c d x}{b^{2}}-\frac {b e -3 c d}{2 b c}}{\left (c x +b \right )^{2}}+\frac {d \ln \left (-x \right )}{b^{3}}-\frac {d \ln \left (c x +b \right )}{b^{3}}\)	\(55\)
default	\(-\frac {b e -c d}{2 b c \left (c x +b \right )^{2}}-\frac {d \ln \left (c x +b \right )}{b^{3}}+\frac {d}{b^{2} \left (c x +b \right )}+\frac {d \ln \left (x \right )}{b^{3}}\)	\(56\)
norman	\(\frac {\frac {\left (b e -2 c d \right ) x^{3}}{b^{2}}+\frac {c \left (b e -3 c d \right ) x^{4}}{2 b^{3}}}{x^{2} \left (c x +b \right )^{2}}+\frac {d \ln \left (x \right )}{b^{3}}-\frac {d \ln \left (c x +b \right )}{b^{3}}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 69, normalized size = 1.21 \begin {gather*} \frac {2 \, c^{2} d x + 3 \, b c d - b^{2} e}{2 \, {\left (b^{2} c^{3} x^{2} + 2 \, b^{3} c^{2} x + b^{4} c\right )}} - \frac {d \log \left (c x + b\right )}{b^{3}} + \frac {d \log \left (x\right )}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.05, size = 110, normalized size = 1.93 \begin {gather*} \frac {2 \, b c^{2} d x + 3 \, b^{2} c d - b^{3} e - 2 \, {\left (c^{3} d x^{2} + 2 \, b c^{2} d x + b^{2} c d\right )} \log \left (c x + b\right ) + 2 \, {\left (c^{3} d x^{2} + 2 \, b c^{2} d x + b^{2} c d\right )} \log \left (x\right )}{2 \, {\left (b^{3} c^{3} x^{2} + 2 \, b^{4} c^{2} x + b^{5} c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.21, size = 63, normalized size = 1.11 \begin {gather*} \frac {- b^{2} e + 3 b c d + 2 c^{2} d x}{2 b^{4} c + 4 b^{3} c^{2} x + 2 b^{2} c^{3} x^{2}} + \frac {d \left (\log {\left (x \right )} - \log {\left (\frac {b}{c} + x \right )}\right )}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.35, size = 60, normalized size = 1.05 \begin {gather*} -\frac {d \log \left ({\left | c x + b \right |}\right )}{b^{3}} + \frac {d \log \left ({\left | x \right |}\right )}{b^{3}} + \frac {2 \, b c^{2} d x + 3 \, b^{2} c d - b^{3} e}{2 \, {\left (c x + b\right )}^{2} b^{3} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.07, size = 62, normalized size = 1.09 \begin {gather*} -\frac {\frac {b\,e-3\,c\,d}{2\,b\,c}-\frac {c\,d\,x}{b^2}}{b^2+2\,b\,c\,x+c^2\,x^2}-\frac {2\,d\,\mathrm {atanh}\left (\frac {2\,c\,x}{b}+1\right )}{b^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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