3.56 \(\int \frac {x^2 (a+b \log (c x^n))}{(d+e x)^4} \, dx\)

Optimal. Leaf size=79 \[ \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 d (d+e x)^3}-\frac {2 b n}{3 e^3 (d+e x)}+\frac {b d n}{6 e^3 (d+e x)^2}-\frac {b n \log (d+e x)}{3 d e^3} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2335, 43} \[ \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 d (d+e x)^3}-\frac {2 b n}{3 e^3 (d+e x)}+\frac {b d n}{6 e^3 (d+e x)^2}-\frac {b n \log (d+e x)}{3 d e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2335

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp
[((f*x)^(m + 1)*(d + e*x^r)^(q + 1)*(a + b*Log[c*x^n]))/(d*f*(m + 1)), x] - Dist[(b*n)/(d*(m + 1)), Int[(f*x)^
m*(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && EqQ[m + r*(q + 1) + 1, 0] && NeQ[
m, -1]

Rubi steps

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

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Mathematica [B]  time = 0.12, size = 172, normalized size = 2.18 \[ -\frac {a d^2}{3 e^3 (d+e x)^3}+\frac {a d}{e^3 (d+e x)^2}-\frac {a}{e^3 (d+e x)}-\frac {b d^2 \log \left (c x^n\right )}{3 e^3 (d+e x)^3}+\frac {b d \log \left (c x^n\right )}{e^3 (d+e x)^2}-\frac {b \log \left (c x^n\right )}{e^3 (d+e x)}+\frac {b d n}{6 e^3 (d+e x)^2}-\frac {2 b n}{3 e^3 (d+e x)}+\frac {b n \log (x)}{3 d e^3}-\frac {b n \log (d+e x)}{3 d e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.70, size = 178, normalized size = 2.25 \[ \frac {2 \, b e^{3} n x^{3} \log \relax (x) - 3 \, b d^{3} n - 2 \, a d^{3} - 2 \, {\left (2 \, b d e^{2} n + 3 \, a d e^{2}\right )} x^{2} - {\left (7 \, b d^{2} e n + 6 \, a d^{2} e\right )} x - 2 \, {\left (b e^{3} n x^{3} + 3 \, b d e^{2} n x^{2} + 3 \, b d^{2} e n x + b d^{3} n\right )} \log \left (e x + d\right ) - 2 \, {\left (3 \, b d e^{2} x^{2} + 3 \, b d^{2} e x + b d^{3}\right )} \log \relax (c)}{6 \, {\left (d e^{6} x^{3} + 3 \, d^{2} e^{5} x^{2} + 3 \, d^{3} e^{4} x + d^{4} e^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*log(c*x^n))/(e*x+d)^4,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.30, size = 193, normalized size = 2.44 \[ -\frac {2 \, b n x^{3} e^{3} \log \left (x e + d\right ) + 6 \, b d n x^{2} e^{2} \log \left (x e + d\right ) + 6 \, b d^{2} n x e \log \left (x e + d\right ) - 2 \, b n x^{3} e^{3} \log \relax (x) + 4 \, b d n x^{2} e^{2} + 7 \, b d^{2} n x e + 2 \, b d^{3} n \log \left (x e + d\right ) + 6 \, b d x^{2} e^{2} \log \relax (c) + 6 \, b d^{2} x e \log \relax (c) + 3 \, b d^{3} n + 6 \, a d x^{2} e^{2} + 6 \, a d^{2} x e + 2 \, b d^{3} \log \relax (c) + 2 \, a d^{3}}{6 \, {\left (d x^{3} e^{6} + 3 \, d^{2} x^{2} e^{5} + 3 \, d^{3} x e^{4} + d^{4} e^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*b*n*x^3*e^3*log(x*e + d) + 6*b*d*n*x^2*e^2*log(x*e + d) + 6*b*d^2*n*x*e*log(x*e + d) - 2*b*n*x^3*e^3*l
og(x) + 4*b*d*n*x^2*e^2 + 7*b*d^2*n*x*e + 2*b*d^3*n*log(x*e + d) + 6*b*d*x^2*e^2*log(c) + 6*b*d^2*x*e*log(c) +
3*b*d^3*n + 6*a*d*x^2*e^2 + 6*a*d^2*x*e + 2*b*d^3*log(c) + 2*a*d^3)/(d*x^3*e^6 + 3*d^2*x^2*e^5 + 3*d^3*x*e^4
+ d^4*e^3)

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maple [C]  time = 0.23, size = 553, normalized size = 7.00 \[ -\frac {\left (3 e^{2} x^{2}+3 d e x +d^{2}\right ) b \ln \left (x^{n}\right )}{3 \left (e x +d \right )^{3} e^{3}}-\frac {6 b d \,e^{2} x^{2} \ln \relax (c )+6 b \,d^{2} e x \ln \relax (c )+6 a d \,e^{2} x^{2}+6 a \,d^{2} e x +2 a \,d^{3}+3 b \,d^{3} n +2 b \,d^{3} n \ln \left (e x +d \right )-2 b \,d^{3} n \ln \left (-x \right )+2 b \,d^{3} \ln \relax (c )-3 i \pi b \,d^{2} e x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )-3 i \pi b d \,e^{2} x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )-i \pi b \,d^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+3 i \pi b d \,e^{2} x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+3 i \pi b d \,e^{2} x^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+2 b \,e^{3} n \,x^{3} \ln \left (e x +d \right )-2 b \,e^{3} n \,x^{3} \ln \left (-x \right )-i \pi b \,d^{3} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+6 b d \,e^{2} n \,x^{2} \ln \left (e x +d \right )+6 b \,d^{2} e n x \ln \left (e x +d \right )-6 b d \,e^{2} n \,x^{2} \ln \left (-x \right )-6 b \,d^{2} e n x \ln \left (-x \right )+3 i \pi b \,d^{2} e x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+3 i \pi b \,d^{2} e x \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-3 i \pi b d \,e^{2} x^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}-3 i \pi b \,d^{2} e x \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+i \pi b \,d^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+i \pi b \,d^{3} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+7 b \,d^{2} e n x +4 b d \,e^{2} n \,x^{2}}{6 \left (e x +d \right )^{3} d \,e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*ln(c*x^n)+a)/(e*x+d)^4,x)

[Out]

-1/3*b*(3*e^2*x^2+3*d*e*x+d^2)/(e*x+d)^3/e^3*ln(x^n)-1/6*(6*b*d*e^2*x^2*ln(c)+6*b*d^2*e*x*ln(c)+3*I*Pi*b*d*e^2
*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2+3*I*Pi*b*d^2*e*x*csgn(I*c*x^n)^2*csgn(I*c)+6*a*d*e^2*x^2+6*a*d^2*e*x+2*a*d^3+
3*I*Pi*b*d^2*csgn(I*x^n)*csgn(I*c*x^n)^2*e*x+3*I*Pi*b*d*e^2*x^2*csgn(I*c*x^n)^2*csgn(I*c)+3*b*d^3*n+2*ln(e*x+d
)*b*d^3*n-2*ln(-x)*b*d^3*n+2*b*d^3*ln(c)-3*I*Pi*b*d*e^2*x^2*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)-I*Pi*b*d^3*csgn(I*c*x^n)^3+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(e*x+d)*b*d^2*e*n*x-6*ln(-x)*b*d*e^2*n*x^2-6*ln(-x)*b*d^2*e*n*x+I*Pi*b*d^3*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*x^n)*csgn(I*c*x^n)*csgn(I*c)-3*
I*Pi*b*d*e^2*x^2*csgn(I*c*x^n)^3-3*I*Pi*b*d^2*e*x*csgn(I*c*x^n)^3+7*b*d^2*e*n*x+4*b*d*e^2*n*x^2)/d/e^3/(e*x+d)
^3

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maxima [B]  time = 0.67, size = 179, normalized size = 2.27 \[ -\frac {1}{6} \, b n {\left (\frac {4 \, e x + 3 \, d}{e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}} + \frac {2 \, \log \left (e x + d\right )}{d e^{3}} - \frac {2 \, \log \relax (x)}{d e^{3}}\right )} - \frac {{\left (3 \, e^{2} x^{2} + 3 \, d e x + d^{2}\right )} b \log \left (c x^{n}\right )}{3 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} - \frac {{\left (3 \, e^{2} x^{2} + 3 \, d e x + d^{2}\right )} a}{3 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*log(c*x^n))/(e*x+d)^4,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.02, size = 167, normalized size = 2.11 \[ -\frac {x^2\,\left (3\,a\,e^2+2\,b\,e^2\,n\right )+a\,d^2+x\,\left (3\,a\,d\,e+\frac {7\,b\,d\,e\,n}{2}\right )+\frac {3\,b\,d^2\,n}{2}}{3\,d^3\,e^3+9\,d^2\,e^4\,x+9\,d\,e^5\,x^2+3\,e^6\,x^3}-\frac {\ln \left (c\,x^n\right )\,\left (\frac {b\,d^2}{3\,e^3}+\frac {b\,x^2}{e}+\frac {b\,d\,x}{e^2}\right )}{d^3+3\,d^2\,e\,x+3\,d\,e^2\,x^2+e^3\,x^3}-\frac {2\,b\,n\,\mathrm {atanh}\left (\frac {2\,e\,x}{d}+1\right )}{3\,d\,e^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(a + b*log(c*x^n)))/(d + e*x)^4,x)

[Out]

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

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sympy [A]  time = 15.67, size = 748, normalized size = 9.47 \[ \begin {cases} \tilde {\infty } \left (- \frac {a}{x} - \frac {b n \log {\relax (x )}}{x} - \frac {b n}{x} - \frac {b \log {\relax (c )}}{x}\right ) & \text {for}\: d = 0 \wedge e = 0 \\\frac {\frac {a x^{3}}{3} + \frac {b n x^{3} \log {\relax (x )}}{3} - \frac {b n x^{3}}{9} + \frac {b x^{3} \log {\relax (c )}}{3}}{d^{4}} & \text {for}\: e = 0 \\\frac {- \frac {a}{x} - \frac {b n \log {\relax (x )}}{x} - \frac {b n}{x} - \frac {b \log {\relax (c )}}{x}}{e^{4}} & \text {for}\: d = 0 \\- \frac {2 a d^{3}}{6 d^{4} e^{3} + 18 d^{3} e^{4} x + 18 d^{2} e^{5} x^{2} + 6 d e^{6} x^{3}} - \frac {6 a d^{2} e x}{6 d^{4} e^{3} + 18 d^{3} e^{4} x + 18 d^{2} e^{5} x^{2} + 6 d e^{6} x^{3}} - \frac {6 a d e^{2} x^{2}}{6 d^{4} e^{3} + 18 d^{3} e^{4} x + 18 d^{2} e^{5} x^{2} + 6 d e^{6} x^{3}} - \frac {2 b d^{3} n \log {\left (\frac {d}{e} + x \right )}}{6 d^{4} e^{3} + 18 d^{3} e^{4} x + 18 d^{2} e^{5} x^{2} + 6 d e^{6} x^{3}} - \frac {3 b d^{3} n}{6 d^{4} e^{3} + 18 d^{3} e^{4} x + 18 d^{2} e^{5} x^{2} + 6 d e^{6} x^{3}} - \frac {6 b d^{2} e n x \log {\left (\frac {d}{e} + x \right )}}{6 d^{4} e^{3} + 18 d^{3} e^{4} x + 18 d^{2} e^{5} x^{2} + 6 d e^{6} x^{3}} - \frac {7 b d^{2} e n x}{6 d^{4} e^{3} + 18 d^{3} e^{4} x + 18 d^{2} e^{5} x^{2} + 6 d e^{6} x^{3}} - \frac {6 b d e^{2} n x^{2} \log {\left (\frac {d}{e} + x \right )}}{6 d^{4} e^{3} + 18 d^{3} e^{4} x + 18 d^{2} e^{5} x^{2} + 6 d e^{6} x^{3}} - \frac {4 b d e^{2} n x^{2}}{6 d^{4} e^{3} + 18 d^{3} e^{4} x + 18 d^{2} e^{5} x^{2} + 6 d e^{6} x^{3}} + \frac {2 b e^{3} n x^{3} \log {\relax (x )}}{6 d^{4} e^{3} + 18 d^{3} e^{4} x + 18 d^{2} e^{5} x^{2} + 6 d e^{6} x^{3}} - \frac {2 b e^{3} n x^{3} \log {\left (\frac {d}{e} + x \right )}}{6 d^{4} e^{3} + 18 d^{3} e^{4} x + 18 d^{2} e^{5} x^{2} + 6 d e^{6} x^{3}} + \frac {2 b e^{3} x^{3} \log {\relax (c )}}{6 d^{4} e^{3} + 18 d^{3} e^{4} x + 18 d^{2} e^{5} x^{2} + 6 d e^{6} x^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*ln(c*x**n))/(e*x+d)**4,x)

[Out]

Piecewise((zoo*(-a/x - b*n*log(x)/x - b*n/x - b*log(c)/x), Eq(d, 0) & Eq(e, 0)), ((a*x**3/3 + b*n*x**3*log(x)/
3 - b*n*x**3/9 + b*x**3*log(c)/3)/d**4, Eq(e, 0)), ((-a/x - b*n*log(x)/x - b*n/x - b*log(c)/x)/e**4, Eq(d, 0))
, (-2*a*d**3/(6*d**4*e**3 + 18*d**3*e**4*x + 18*d**2*e**5*x**2 + 6*d*e**6*x**3) - 6*a*d**2*e*x/(6*d**4*e**3 +
18*d**3*e**4*x + 18*d**2*e**5*x**2 + 6*d*e**6*x**3) - 6*a*d*e**2*x**2/(6*d**4*e**3 + 18*d**3*e**4*x + 18*d**2*
e**5*x**2 + 6*d*e**6*x**3) - 2*b*d**3*n*log(d/e + x)/(6*d**4*e**3 + 18*d**3*e**4*x + 18*d**2*e**5*x**2 + 6*d*e
**6*x**3) - 3*b*d**3*n/(6*d**4*e**3 + 18*d**3*e**4*x + 18*d**2*e**5*x**2 + 6*d*e**6*x**3) - 6*b*d**2*e*n*x*log
(d/e + x)/(6*d**4*e**3 + 18*d**3*e**4*x + 18*d**2*e**5*x**2 + 6*d*e**6*x**3) - 7*b*d**2*e*n*x/(6*d**4*e**3 + 1
8*d**3*e**4*x + 18*d**2*e**5*x**2 + 6*d*e**6*x**3) - 6*b*d*e**2*n*x**2*log(d/e + x)/(6*d**4*e**3 + 18*d**3*e**
4*x + 18*d**2*e**5*x**2 + 6*d*e**6*x**3) - 4*b*d*e**2*n*x**2/(6*d**4*e**3 + 18*d**3*e**4*x + 18*d**2*e**5*x**2
+ 6*d*e**6*x**3) + 2*b*e**3*n*x**3*log(x)/(6*d**4*e**3 + 18*d**3*e**4*x + 18*d**2*e**5*x**2 + 6*d*e**6*x**3)
- 2*b*e**3*n*x**3*log(d/e + x)/(6*d**4*e**3 + 18*d**3*e**4*x + 18*d**2*e**5*x**2 + 6*d*e**6*x**3) + 2*b*e**3*x
**3*log(c)/(6*d**4*e**3 + 18*d**3*e**4*x + 18*d**2*e**5*x**2 + 6*d*e**6*x**3), True))

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