\(\int \frac {x (a+b \log (c x^n))^2}{(d+e x)^3} \, dx\) [109]

Optimal result

Integrand size = 21, antiderivative size = 112 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^3} \, dx=\frac {b n x \left (a+b \log \left (c x^n\right )\right )}{d e (d+e x)}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 d (d+e x)^2}-\frac {b n \left (a+b n+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{d e^2}-\frac {b^2 n^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{d e^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2381, 2384, 2354, 2438} \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^3} \, dx=-\frac {b n \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )+b n\right )}{d e^2}+\frac {b n x \left (a+b \log \left (c x^n\right )\right )}{d e (d+e x)}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 d (d+e x)^2}-\frac {b^2 n^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{d e^2} \]

[In]

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

[Out]

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

Rule 2354

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[Log[1 + e*(x/d)]*((a +
b*Log[c*x^n])^p/e), x] - Dist[b*n*(p/e), Int[Log[1 + e*(x/d)]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[
{a, b, c, d, e, n}, x] && IGtQ[p, 0]

Rule 2381

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_), x_Symbol] :> Simp
[(-(f*x)^(m + 1))*(d + e*x)^(q + 1)*((a + b*Log[c*x^n])^p/(d*f*(q + 1))), x] + Dist[b*n*(p/(d*(q + 1))), Int[(
f*x)^m*(d + e*x)^(q + 1)*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q}, x] && EqQ[m
+ q + 2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 2384

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

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps \begin{align*} \text {integral}& = \frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 d (d+e x)^2}-\frac {(b n) \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx}{d} \\ & = \frac {b n x \left (a+b \log \left (c x^n\right )\right )}{d e (d+e x)}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 d (d+e x)^2}-\frac {(b n) \int \frac {a+b n+b \log \left (c x^n\right )}{d+e x} \, dx}{d e} \\ & = \frac {b n x \left (a+b \log \left (c x^n\right )\right )}{d e (d+e x)}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 d (d+e x)^2}-\frac {b n \left (a+b n+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{d e^2}+\frac {\left (b^2 n^2\right ) \int \frac {\log \left (1+\frac {e x}{d}\right )}{x} \, dx}{d e^2} \\ & = \frac {b n x \left (a+b \log \left (c x^n\right )\right )}{d e (d+e x)}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 d (d+e x)^2}-\frac {b n \left (a+b n+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{d e^2}-\frac {b^2 n^2 \text {Li}_2\left (-\frac {e x}{d}\right )}{d e^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.38 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^3} \, dx=\frac {-\frac {2 b n \left (a+b \log \left (c x^n\right )\right )}{d+e x}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{d}+\frac {d \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^2}-\frac {2 \left (a+b \log \left (c x^n\right )\right )^2}{d+e x}+\frac {2 b^2 n^2 (\log (x)-\log (d+e x))}{d}-\frac {2 b n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{d}-\frac {2 b^2 n^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{d}}{2 e^2} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.52 (sec) , antiderivative size = 484, normalized size of antiderivative = 4.32

method	result	size
risch	\(-\frac {b^{2} \ln \left (x^{n}\right )^{2}}{e^{2} \left (e x +d \right )}+\frac {b^{2} \ln \left (x^{n}\right )^{2} d}{2 e^{2} \left (e x +d \right )^{2}}-\frac {b^{2} n \ln \left (x^{n}\right )}{e^{2} \left (e x +d \right )}-\frac {b^{2} n \ln \left (x^{n}\right ) \ln \left (e x +d \right )}{e^{2} d}+\frac {b^{2} n \ln \left (x^{n}\right ) \ln \left (x \right )}{e^{2} d}-\frac {b^{2} n^{2} \ln \left (x \right )^{2}}{2 e^{2} d}-\frac {b^{2} n^{2} \ln \left (e x +d \right )}{e^{2} d}+\frac {b^{2} n^{2} \ln \left (x \right )}{e^{2} d}+\frac {b^{2} n^{2} \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{e^{2} d}+\frac {b^{2} n^{2} \operatorname {dilog}\left (-\frac {e x}{d}\right )}{e^{2} d}+\left (-i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+2 b \ln \left (c \right )+2 a \right ) b \left (-\frac {\ln \left (x^{n}\right )}{e^{2} \left (e x +d \right )}+\frac {\ln \left (x^{n}\right ) d}{2 e^{2} \left (e x +d \right )^{2}}-\frac {n \left (\frac {\ln \left (e x +d \right )}{d}+\frac {1}{e x +d}-\frac {\ln \left (x \right )}{d}\right )}{2 e^{2}}\right )+\frac {{\left (-i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+2 b \ln \left (c \right )+2 a \right )}^{2} \left (-\frac {1}{e^{2} \left (e x +d \right )}+\frac {d}{2 e^{2} \left (e x +d \right )^{2}}\right )}{4}\)	\(484\)

[In]

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

[Out]

-b^2*ln(x^n)^2/e^2/(e*x+d)+1/2*b^2*ln(x^n)^2/e^2*d/(e*x+d)^2-b^2*n*ln(x^n)/e^2/(e*x+d)-b^2*n*ln(x^n)/e^2/d*ln(
e*x+d)+b^2*n*ln(x^n)/e^2/d*ln(x)-1/2*b^2*n^2/e^2/d*ln(x)^2-b^2*n^2/e^2/d*ln(e*x+d)+b^2*n^2/e^2/d*ln(x)+b^2*n^2
/e^2/d*ln(e*x+d)*ln(-e*x/d)+b^2*n^2/e^2/d*dilog(-e*x/d)+(-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+I*b*Pi*cs
gn(I*c)*csgn(I*c*x^n)^2+I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-I*b*Pi*csgn(I*c*x^n)^3+2*b*ln(c)+2*a)*b*(-ln(x^n)/e
^2/(e*x+d)+1/2*ln(x^n)/e^2*d/(e*x+d)^2-1/2*n/e^2*(1/d*ln(e*x+d)+1/(e*x+d)-1/d*ln(x)))+1/4*(-I*b*Pi*csgn(I*c)*c
sgn(I*x^n)*csgn(I*c*x^n)+I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-I*b*Pi*csgn(I*c*x
^n)^3+2*b*ln(c)+2*a)^2*(-1/e^2/(e*x+d)+1/2/e^2*d/(e*x+d)^2)

Fricas [F]

\[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x}{{\left (e x + d\right )}^{3}} \,d x } \]

[In]

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

[Out]

integral((b^2*x*log(c*x^n)^2 + 2*a*b*x*log(c*x^n) + a^2*x)/(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3), x)

Sympy [F]

\[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^3} \, dx=\int \frac {x \left (a + b \log {\left (c x^{n} \right )}\right )^{2}}{\left (d + e x\right )^{3}}\, dx \]

[In]

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

[Out]

Integral(x*(a + b*log(c*x**n))**2/(d + e*x)**3, x)

Maxima [F]

\[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x}{{\left (e x + d\right )}^{3}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x}{{\left (e x + d\right )}^{3}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^3} \, dx=\int \frac {x\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2}{{\left (d+e\,x\right )}^3} \,d x \]

[In]

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

[Out]

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