3.1.40 \(\int \frac {x^2 (a+b \log (c x^n))}{(d+e x)^2} \, dx\) [40]

Optimal. Leaf size=98 \[ -\frac {b n x}{e^2}+\frac {2 x \left (a+b \log \left (c x^n\right )\right )}{e^2}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{e (d+e x)}-\frac {d \left (2 a+b n+2 b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^3}-\frac {2 b d n \text {Li}_2\left (-\frac {e x}{d}\right )}{e^3} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 107, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2384, 45, 2393, 2332, 2354, 2438} \begin {gather*} -\frac {2 b d n \text {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^3}-\frac {d \log \left (\frac {e x}{d}+1\right ) \left (2 a+2 b \log \left (c x^n\right )+b n\right )}{e^3}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{e (d+e x)}+\frac {x (2 a+b n)}{e^2}+\frac {2 b x \log \left (c x^n\right )}{e^2}-\frac {2 b n x}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 45

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 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

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 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 2393

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

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*} \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx &=\int \left (\frac {a+b \log \left (c x^n\right )}{e^2}+\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{e^2 (d+e x)^2}-\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^2 (d+e x)}\right ) \, dx\\ &=\frac {\int \left (a+b \log \left (c x^n\right )\right ) \, dx}{e^2}-\frac {(2 d) \int \frac {a+b \log \left (c x^n\right )}{d+e x} \, dx}{e^2}+\frac {d^2 \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^2} \, dx}{e^2}\\ &=\frac {a x}{e^2}+\frac {d x \left (a+b \log \left (c x^n\right )\right )}{e^2 (d+e x)}-\frac {2 d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^3}+\frac {b \int \log \left (c x^n\right ) \, dx}{e^2}+\frac {(2 b d n) \int \frac {\log \left (1+\frac {e x}{d}\right )}{x} \, dx}{e^3}-\frac {(b d n) \int \frac {1}{d+e x} \, dx}{e^2}\\ &=\frac {a x}{e^2}-\frac {b n x}{e^2}+\frac {b x \log \left (c x^n\right )}{e^2}+\frac {d x \left (a+b \log \left (c x^n\right )\right )}{e^2 (d+e x)}-\frac {b d n \log (d+e x)}{e^3}-\frac {2 d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^3}-\frac {2 b d n \text {Li}_2\left (-\frac {e x}{d}\right )}{e^3}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 98, normalized size = 1.00 \begin {gather*} \frac {a e x-b e n x+b e x \log \left (c x^n\right )-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{d+e x}+b d n (\log (x)-\log (d+e x))-2 d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )-2 b d n \text {Li}_2\left (-\frac {e x}{d}\right )}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 558, normalized size = 5.69

method	result	size
risch	\(\frac {2 b n d \dilog \left (-\frac {e x}{d}\right )}{e^{3}}-\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} x}{2 e^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} d^{2}}{2 e^{3} \left (e x +d \right )}-\frac {b n d}{e^{3}}+\frac {b \ln \left (x^{n}\right ) x}{e^{2}}-\frac {a \,d^{2}}{e^{3} \left (e x +d \right )}-\frac {2 a d \ln \left (e x +d \right )}{e^{3}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} x}{2 e^{2}}+\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} x}{2 e^{2}}+\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} d^{2}}{2 e^{3} \left (e x +d \right )}+\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} d \ln \left (e x +d \right )}{e^{3}}+\frac {b \ln \left (c \right ) x}{e^{2}}+\frac {2 b n d \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{e^{3}}-\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} d \ln \left (e x +d \right )}{e^{3}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) x}{2 e^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} d \ln \left (e x +d \right )}{e^{3}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} d^{2}}{2 e^{3} \left (e x +d \right )}-\frac {b \ln \left (x^{n}\right ) d^{2}}{e^{3} \left (e x +d \right )}-\frac {2 b \ln \left (x^{n}\right ) d \ln \left (e x +d \right )}{e^{3}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) d^{2}}{2 e^{3} \left (e x +d \right )}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) d \ln \left (e x +d \right )}{e^{3}}-\frac {b \ln \left (c \right ) d^{2}}{e^{3} \left (e x +d \right )}-\frac {2 b \ln \left (c \right ) d \ln \left (e x +d \right )}{e^{3}}-\frac {b n d \ln \left (e x +d \right )}{e^{3}}+\frac {b n d \ln \left (e x \right )}{e^{3}}-\frac {b n x}{e^{2}}+\frac {a x}{e^{2}}\)	\(558\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 13.63, size = 269, normalized size = 2.74 \begin {gather*} \frac {a d^{2} \left (\begin {cases} \frac {x}{d^{2}} & \text {for}\: e = 0 \\- \frac {1}{d e + e^{2} x} & \text {otherwise} \end {cases}\right )}{e^{2}} - \frac {2 a d \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right )}{e^{2}} + \frac {a x}{e^{2}} - \frac {b d^{2} n \left (\begin {cases} \frac {x}{d^{2}} & \text {for}\: e = 0 \\- \frac {\log {\left (x \right )}}{d e} + \frac {\log {\left (\frac {d}{e} + x \right )}}{d e} & \text {otherwise} \end {cases}\right )}{e^{2}} + \frac {b d^{2} \left (\begin {cases} \frac {x}{d^{2}} & \text {for}\: e = 0 \\- \frac {1}{d e + e^{2} x} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{e^{2}} + \frac {2 b d n \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\begin {cases} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (d \right )} \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (d \right )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (d \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (d \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {otherwise} \end {cases}}{e} & \text {otherwise} \end {cases}\right )}{e^{2}} - \frac {2 b d \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{e^{2}} - \frac {b n x}{e^{2}} + \frac {b x \log {\left (c x^{n} \right )}}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*d**2*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), True))/e**2 - 2*a*d*Piecewise((x/d, Eq(e, 0)), (log(d
+ e*x)/e, True))/e**2 + a*x/e**2 - b*d**2*n*Piecewise((x/d**2, Eq(e, 0)), (-log(x)/(d*e) + log(d/e + x)/(d*e)
, True))/e**2 + b*d**2*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), True))*log(c*x**n)/e**2 + 2*b*d*n*Pie
cewise((x/d, Eq(e, 0)), (Piecewise((-polylog(2, e*x*exp_polar(I*pi)/d), (Abs(x) < 1) & (1/Abs(x) < 1)), (log(d
)*log(x) - polylog(2, e*x*exp_polar(I*pi)/d), Abs(x) < 1), (-log(d)*log(1/x) - polylog(2, e*x*exp_polar(I*pi)/
d), 1/Abs(x) < 1), (-meijerg(((), (1, 1)), ((0, 0), ()), x)*log(d) + meijerg(((1, 1), ()), ((), (0, 0)), x)*lo
g(d) - polylog(2, e*x*exp_polar(I*pi)/d), True))/e, True))/e**2 - 2*b*d*Piecewise((x/d, Eq(e, 0)), (log(d + e*
x)/e, True))*log(c*x**n)/e**2 - b*n*x/e**2 + b*x*log(c*x**n)/e**2

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (d+e\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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