3.57 \(\int \frac{x (a+b \log (c x^n))}{(d+e x)^4} \, dx\)

Optimal. Leaf size=117 \[ -\frac{a+b \log \left (c x^n\right )}{2 e^2 (d+e x)^2}+\frac{d \left (a+b \log \left (c x^n\right )\right )}{3 e^2 (d+e x)^3}+\frac{b n \log (x)}{6 d^2 e^2}-\frac{b n \log (d+e x)}{6 d^2 e^2}+\frac{b n}{6 d e^2 (d+e x)}-\frac{b n}{6 e^2 (d+e x)^2} \]

[Out]

-(b*n)/(6*e^2*(d + e*x)^2) + (b*n)/(6*d*e^2*(d + e*x)) + (b*n*Log[x])/(6*d^2*e^2) + (d*(a + b*Log[c*x^n]))/(3*
e^2*(d + e*x)^3) - (a + b*Log[c*x^n])/(2*e^2*(d + e*x)^2) - (b*n*Log[d + e*x])/(6*d^2*e^2)

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Rubi [A]  time = 0.0871504, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {43, 2350, 12, 77} \[ -\frac{a+b \log \left (c x^n\right )}{2 e^2 (d+e x)^2}+\frac{d \left (a+b \log \left (c x^n\right )\right )}{3 e^2 (d+e x)^3}+\frac{b n \log (x)}{6 d^2 e^2}-\frac{b n \log (d+e x)}{6 d^2 e^2}+\frac{b n}{6 d e^2 (d+e x)}-\frac{b n}{6 e^2 (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(x*(a + b*Log[c*x^n]))/(d + e*x)^4,x]

[Out]

-(b*n)/(6*e^2*(d + e*x)^2) + (b*n)/(6*d*e^2*(d + e*x)) + (b*n*Log[x])/(6*d^2*e^2) + (d*(a + b*Log[c*x^n]))/(3*
e^2*(d + e*x)^3) - (a + b*Log[c*x^n])/(2*e^2*(d + e*x)^2) - (b*n*Log[d + e*x])/(6*d^2*e^2)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2350

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = IntHide[(f*x)^m*(d + e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[SimplifyIntegrand[u/x,
x], x], x] /; ((EqQ[r, 1] || EqQ[r, 2]) && IntegerQ[m] && IntegerQ[q - 1/2]) || InverseFunctionFreeQ[u, x]] /
; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && IntegerQ[2*q] && ((IntegerQ[m] && IntegerQ[r]) || IGtQ[q, 0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{x \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx &=\frac{d \left (a+b \log \left (c x^n\right )\right )}{3 e^2 (d+e x)^3}-\frac{a+b \log \left (c x^n\right )}{2 e^2 (d+e x)^2}-(b n) \int \frac{-d-3 e x}{6 e^2 x (d+e x)^3} \, dx\\ &=\frac{d \left (a+b \log \left (c x^n\right )\right )}{3 e^2 (d+e x)^3}-\frac{a+b \log \left (c x^n\right )}{2 e^2 (d+e x)^2}-\frac{(b n) \int \frac{-d-3 e x}{x (d+e x)^3} \, dx}{6 e^2}\\ &=\frac{d \left (a+b \log \left (c x^n\right )\right )}{3 e^2 (d+e x)^3}-\frac{a+b \log \left (c x^n\right )}{2 e^2 (d+e x)^2}-\frac{(b n) \int \left (-\frac{1}{d^2 x}-\frac{2 e}{(d+e x)^3}+\frac{e}{d (d+e x)^2}+\frac{e}{d^2 (d+e x)}\right ) \, dx}{6 e^2}\\ &=-\frac{b n}{6 e^2 (d+e x)^2}+\frac{b n}{6 d e^2 (d+e x)}+\frac{b n \log (x)}{6 d^2 e^2}+\frac{d \left (a+b \log \left (c x^n\right )\right )}{3 e^2 (d+e x)^3}-\frac{a+b \log \left (c x^n\right )}{2 e^2 (d+e x)^2}-\frac{b n \log (d+e x)}{6 d^2 e^2}\\ \end{align*}

Mathematica [A]  time = 0.0897124, size = 135, normalized size = 1.15 \[ -\frac{a+b \log \left (c x^n\right )}{2 e^2 (d+e x)^2}+\frac{d \left (a+b \log \left (c x^n\right )\right )}{3 e^2 (d+e x)^3}-\frac{b n \left (-\frac{2 \log (d+e x)}{d^2}+\frac{2 \log (x)}{d^2}+\frac{2}{d (d+e x)}+\frac{1}{(d+e x)^2}\right )}{6 e^2}+\frac{b n \left (-\frac{\log (d+e x)}{d^2}+\frac{\log (x)}{d^2}+\frac{1}{d (d+e x)}\right )}{2 e^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*(a + b*Log[c*x^n]))/(d + e*x)^4,x]

[Out]

(d*(a + b*Log[c*x^n]))/(3*e^2*(d + e*x)^3) - (a + b*Log[c*x^n])/(2*e^2*(d + e*x)^2) - (b*n*((d + e*x)^(-2) + 2
/(d*(d + e*x)) + (2*Log[x])/d^2 - (2*Log[d + e*x])/d^2))/(6*e^2) + (b*n*(1/(d*(d + e*x)) + Log[x]/d^2 - Log[d
+ e*x]/d^2))/(2*e^2)

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Maple [C]  time = 0.115, size = 403, normalized size = 3.4 \begin{align*} -{\frac{b \left ( 3\,ex+d \right ) \ln \left ({x}^{n} \right ) }{6\, \left ( ex+d \right ) ^{3}{e}^{2}}}-{\frac{3\,i\pi \,b{d}^{2}ex \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +i\pi \,b{d}^{3}{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}+i\pi \,b{d}^{3} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +3\,i\pi \,b{d}^{2}ex{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}-3\,i\pi \,b{d}^{2}ex \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}-i\pi \,b{d}^{3} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}-i\pi \,b{d}^{3}{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ) -3\,i\pi \,b{d}^{2}ex{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ) +2\,\ln \left ( ex+d \right ) b{e}^{3}n{x}^{3}-2\,\ln \left ( -x \right ) b{e}^{3}n{x}^{3}+6\,\ln \left ( ex+d \right ) bd{e}^{2}n{x}^{2}-6\,\ln \left ( -x \right ) bd{e}^{2}n{x}^{2}+6\,\ln \left ( ex+d \right ) b{d}^{2}enx-6\,\ln \left ( -x \right ) b{d}^{2}enx-2\,bd{e}^{2}n{x}^{2}+6\,\ln \left ( c \right ) b{d}^{2}ex+2\,\ln \left ( ex+d \right ) b{d}^{3}n-2\,\ln \left ( -x \right ) b{d}^{3}n-2\,b{d}^{2}enx+2\,\ln \left ( c \right ) b{d}^{3}+6\,a{d}^{2}ex+2\,a{d}^{3}}{12\,{e}^{2}{d}^{2} \left ( ex+d \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*ln(c*x^n))/(e*x+d)^4,x)

[Out]

-1/6*b*(3*e*x+d)/(e*x+d)^3/e^2*ln(x^n)-1/12*(3*I*Pi*b*d^2*e*x*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*b*d^3*csgn(I*x^n)
*csgn(I*c*x^n)^2+I*Pi*b*d^3*csgn(I*c*x^n)^2*csgn(I*c)+3*I*Pi*b*d^2*e*x*csgn(I*x^n)*csgn(I*c*x^n)^2-3*I*Pi*b*d^
2*e*x*csgn(I*c*x^n)^3-I*Pi*b*d^3*csgn(I*c*x^n)^3-I*Pi*b*d^3*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-3*I*Pi*b*d^2*e
*x*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+2*ln(e*x+d)*b*e^3*n*x^3-2*ln(-x)*b*e^3*n*x^3+6*ln(e*x+d)*b*d*e^2*n*x^2-
6*ln(-x)*b*d*e^2*n*x^2+6*ln(e*x+d)*b*d^2*e*n*x-6*ln(-x)*b*d^2*e*n*x-2*b*d*e^2*n*x^2+6*ln(c)*b*d^2*e*x+2*ln(e*x
+d)*b*d^3*n-2*ln(-x)*b*d^3*n-2*b*d^2*e*n*x+2*ln(c)*b*d^3+6*a*d^2*e*x+2*a*d^3)/e^2/d^2/(e*x+d)^3

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Maxima [A]  time = 1.15431, size = 203, normalized size = 1.74 \begin{align*} \frac{1}{6} \, b n{\left (\frac{x}{d e^{3} x^{2} + 2 \, d^{2} e^{2} x + d^{3} e} - \frac{\log \left (e x + d\right )}{d^{2} e^{2}} + \frac{\log \left (x\right )}{d^{2} e^{2}}\right )} - \frac{{\left (3 \, e x + d\right )} b \log \left (c x^{n}\right )}{6 \,{\left (e^{5} x^{3} + 3 \, d e^{4} x^{2} + 3 \, d^{2} e^{3} x + d^{3} e^{2}\right )}} - \frac{{\left (3 \, e x + d\right )} a}{6 \,{\left (e^{5} x^{3} + 3 \, d e^{4} x^{2} + 3 \, d^{2} e^{3} x + d^{3} e^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*log(c*x^n))/(e*x+d)^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.07109, size = 346, normalized size = 2.96 \begin{align*} \frac{b d e^{2} n x^{2} - a d^{3} +{\left (b d^{2} e n - 3 \, a d^{2} e\right )} x -{\left (b e^{3} n x^{3} + 3 \, b d e^{2} n x^{2} + 3 \, b d^{2} e n x + b d^{3} n\right )} \log \left (e x + d\right ) -{\left (3 \, b d^{2} e x + b d^{3}\right )} \log \left (c\right ) +{\left (b e^{3} n x^{3} + 3 \, b d e^{2} n x^{2}\right )} \log \left (x\right )}{6 \,{\left (d^{2} e^{5} x^{3} + 3 \, d^{3} e^{4} x^{2} + 3 \, d^{4} e^{3} x + d^{5} e^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*log(c*x^n))/(e*x+d)^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 11.6869, size = 799, normalized size = 6.83 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*ln(c*x**n))/(e*x+d)**4,x)

[Out]

Piecewise((zoo*(-a/(2*x**2) - b*n*log(x)/(2*x**2) - b*n/(4*x**2) - b*log(c)/(2*x**2)), Eq(d, 0) & Eq(e, 0)), (
(a*x**2/2 + b*n*x**2*log(x)/2 - b*n*x**2/4 + b*x**2*log(c)/2)/d**4, Eq(e, 0)), ((-a/(2*x**2) - b*n*log(x)/(2*x
**2) - b*n/(4*x**2) - b*log(c)/(2*x**2))/e**4, Eq(d, 0)), (-3*a*d**3/(18*d**5*e**2 + 54*d**4*e**3*x + 54*d**3*
e**4*x**2 + 18*d**2*e**5*x**3) - 9*a*d**2*e*x/(18*d**5*e**2 + 54*d**4*e**3*x + 54*d**3*e**4*x**2 + 18*d**2*e**
5*x**3) - 3*b*d**3*n*log(d/e + x)/(18*d**5*e**2 + 54*d**4*e**3*x + 54*d**3*e**4*x**2 + 18*d**2*e**5*x**3) - b*
d**3*n/(18*d**5*e**2 + 54*d**4*e**3*x + 54*d**3*e**4*x**2 + 18*d**2*e**5*x**3) - 9*b*d**2*e*n*x*log(d/e + x)/(
18*d**5*e**2 + 54*d**4*e**3*x + 54*d**3*e**4*x**2 + 18*d**2*e**5*x**3) + 9*b*d*e**2*n*x**2*log(x)/(18*d**5*e**
2 + 54*d**4*e**3*x + 54*d**3*e**4*x**2 + 18*d**2*e**5*x**3) - 9*b*d*e**2*n*x**2*log(d/e + x)/(18*d**5*e**2 + 5
4*d**4*e**3*x + 54*d**3*e**4*x**2 + 18*d**2*e**5*x**3) + 9*b*d*e**2*x**2*log(c)/(18*d**5*e**2 + 54*d**4*e**3*x
+ 54*d**3*e**4*x**2 + 18*d**2*e**5*x**3) + 3*b*e**3*n*x**3*log(x)/(18*d**5*e**2 + 54*d**4*e**3*x + 54*d**3*e*
*4*x**2 + 18*d**2*e**5*x**3) - 3*b*e**3*n*x**3*log(d/e + x)/(18*d**5*e**2 + 54*d**4*e**3*x + 54*d**3*e**4*x**2
+ 18*d**2*e**5*x**3) - b*e**3*n*x**3/(18*d**5*e**2 + 54*d**4*e**3*x + 54*d**3*e**4*x**2 + 18*d**2*e**5*x**3)
+ 3*b*e**3*x**3*log(c)/(18*d**5*e**2 + 54*d**4*e**3*x + 54*d**3*e**4*x**2 + 18*d**2*e**5*x**3), True))

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Giac [A]  time = 1.27931, size = 238, normalized size = 2.03 \begin{align*} -\frac{b n x^{3} e^{3} \log \left (x e + d\right ) + 3 \, b d n x^{2} e^{2} \log \left (x e + d\right ) + 3 \, b d^{2} n x e \log \left (x e + d\right ) - b n x^{3} e^{3} \log \left (x\right ) - 3 \, b d n x^{2} e^{2} \log \left (x\right ) - b d n x^{2} e^{2} - b d^{2} n x e + b d^{3} n \log \left (x e + d\right ) + 3 \, b d^{2} x e \log \left (c\right ) + 3 \, a d^{2} x e + b d^{3} \log \left (c\right ) + a d^{3}}{6 \,{\left (d^{2} x^{3} e^{5} + 3 \, d^{3} x^{2} e^{4} + 3 \, d^{4} x e^{3} + d^{5} e^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*log(c*x^n))/(e*x+d)^4,x, algorithm="giac")

[Out]

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