3.3.43 \(\int \frac {x^2 (a+b \log (c (d+e x)^n))}{f+g x} \, dx\) [243]

Optimal. Leaf size=181 \[ -\frac {a f x}{g^2}+\frac {b f n x}{g^2}+\frac {b d n x}{2 e g}-\frac {b n x^2}{4 g}-\frac {b d^2 n \log (d+e x)}{2 e^2 g}-\frac {b f (d+e x) \log \left (c (d+e x)^n\right )}{e g^2}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}+\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g^3}+\frac {b f^2 n \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{g^3} \]

[Out]

-a*f*x/g^2+b*f*n*x/g^2+1/2*b*d*n*x/e/g-1/4*b*n*x^2/g-1/2*b*d^2*n*ln(e*x+d)/e^2/g-b*f*(e*x+d)*ln(c*(e*x+d)^n)/e
/g^2+1/2*x^2*(a+b*ln(c*(e*x+d)^n))/g+f^2*(a+b*ln(c*(e*x+d)^n))*ln(e*(g*x+f)/(-d*g+e*f))/g^3+b*f^2*n*polylog(2,
-g*(e*x+d)/(-d*g+e*f))/g^3

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Rubi [A]
time = 0.14, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {45, 2463, 2436, 2332, 2442, 2441, 2440, 2438} \begin {gather*} \frac {b f^2 n \text {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g^3}+\frac {f^2 \log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}-\frac {a f x}{g^2}-\frac {b f (d+e x) \log \left (c (d+e x)^n\right )}{e g^2}-\frac {b d^2 n \log (d+e x)}{2 e^2 g}+\frac {b d n x}{2 e g}+\frac {b f n x}{g^2}-\frac {b n x^2}{4 g} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*f*x)/g^2) + (b*f*n*x)/g^2 + (b*d*n*x)/(2*e*g) - (b*n*x^2)/(4*g) - (b*d^2*n*Log[d + e*x])/(2*e^2*g) - (b*f
*(d + e*x)*Log[c*(d + e*x)^n])/(e*g^2) + (x^2*(a + b*Log[c*(d + e*x)^n]))/(2*g) + (f^2*(a + b*Log[c*(d + e*x)^
n])*Log[(e*(f + g*x))/(e*f - d*g)])/g^3 + (b*f^2*n*PolyLog[2, -((g*(d + e*x))/(e*f - d*g))])/g^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 2436

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

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]

Rule 2440

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + c*e*(x/g)])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2441

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

Rule 2442

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

Rule 2463

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rubi steps

\begin {align*} \int \frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x} \, dx &=\int \left (-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2}+\frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}+\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2 (f+g x)}\right ) \, dx\\ &=-\frac {f \int \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx}{g^2}+\frac {f^2 \int \frac {a+b \log \left (c (d+e x)^n\right )}{f+g x} \, dx}{g^2}+\frac {\int x \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx}{g}\\ &=-\frac {a f x}{g^2}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}+\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g^3}-\frac {(b f) \int \log \left (c (d+e x)^n\right ) \, dx}{g^2}-\frac {\left (b e f^2 n\right ) \int \frac {\log \left (\frac {e (f+g x)}{e f-d g}\right )}{d+e x} \, dx}{g^3}-\frac {(b e n) \int \frac {x^2}{d+e x} \, dx}{2 g}\\ &=-\frac {a f x}{g^2}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}+\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g^3}-\frac {(b f) \text {Subst}\left (\int \log \left (c x^n\right ) \, dx,x,d+e x\right )}{e g^2}-\frac {\left (b f^2 n\right ) \text {Subst}\left (\int \frac {\log \left (1+\frac {g x}{e f-d g}\right )}{x} \, dx,x,d+e x\right )}{g^3}-\frac {(b e n) \int \left (-\frac {d}{e^2}+\frac {x}{e}+\frac {d^2}{e^2 (d+e x)}\right ) \, dx}{2 g}\\ &=-\frac {a f x}{g^2}+\frac {b f n x}{g^2}+\frac {b d n x}{2 e g}-\frac {b n x^2}{4 g}-\frac {b d^2 n \log (d+e x)}{2 e^2 g}-\frac {b f (d+e x) \log \left (c (d+e x)^n\right )}{e g^2}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}+\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g^3}+\frac {b f^2 n \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{g^3}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 170, normalized size = 0.94 \begin {gather*} -\frac {a f x}{g^2}+\frac {b f n x}{g^2}+\frac {b n \left (\frac {2 d x}{e}-x^2-\frac {2 d^2 \log (d+e x)}{e^2}\right )}{4 g}-\frac {b f (d+e x) \log \left (c (d+e x)^n\right )}{e g^2}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}+\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g^3}+\frac {b f^2 n \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{g^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*f*x)/g^2) + (b*f*n*x)/g^2 + (b*n*((2*d*x)/e - x^2 - (2*d^2*Log[d + e*x])/e^2))/(4*g) - (b*f*(d + e*x)*Log
[c*(d + e*x)^n])/(e*g^2) + (x^2*(a + b*Log[c*(d + e*x)^n]))/(2*g) + (f^2*(a + b*Log[c*(d + e*x)^n])*Log[(e*(f
+ g*x))/(e*f - d*g)])/g^3 + (b*f^2*n*PolyLog[2, -((g*(d + e*x))/(e*f - d*g))])/g^3

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.33, size = 724, normalized size = 4.00

method	result	size
risch	\(-\frac {b n \,f^{2} \ln \left (g x +f \right ) \ln \left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{g^{3}}+\frac {b \ln \left (c \right ) f^{2} \ln \left (g x +f \right )}{g^{3}}-\frac {b \ln \left (c \right ) x f}{g^{2}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right ) x f}{2 g^{2}}+\frac {b \ln \left (\left (e x +d \right )^{n}\right ) x^{2}}{2 g}+\frac {a \,x^{2}}{2 g}+\frac {a \,f^{2} \ln \left (g x +f \right )}{g^{3}}-\frac {b n \,f^{2} \dilog \left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{g^{3}}-\frac {i b \pi \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} f^{2} \ln \left (g x +f \right )}{2 g^{3}}+\frac {i b \pi \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} x^{2}}{4 g}+\frac {i b \pi \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} x f}{2 g^{2}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} f^{2} \ln \left (g x +f \right )}{2 g^{3}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right ) x^{2}}{4 g}+\frac {5 b n \,f^{2}}{4 g^{3}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} x f}{2 g^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} x f}{2 g^{2}}+\frac {i b \pi \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} f^{2} \ln \left (g x +f \right )}{2 g^{3}}-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) x f}{g^{2}}-\frac {b n \,d^{2} \ln \left (\left (g x +f \right ) e +d g -e f \right )}{2 e^{2} g}-\frac {i b \pi \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} x^{2}}{4 g}+\frac {b \ln \left (c \right ) x^{2}}{2 g}+\frac {b \ln \left (\left (e x +d \right )^{n}\right ) f^{2} \ln \left (g x +f \right )}{g^{3}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} x^{2}}{4 g}-\frac {b n d \ln \left (\left (g x +f \right ) e +d g -e f \right ) f}{e \,g^{2}}+\frac {b n d f}{2 e \,g^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right ) f^{2} \ln \left (g x +f \right )}{2 g^{3}}-\frac {a f x}{g^{2}}+\frac {b f n x}{g^{2}}+\frac {b d n x}{2 e g}-\frac {b n \,x^{2}}{4 g}\)	\(724\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*n/g^3*f^2*ln(g*x+f)*ln(((g*x+f)*e+d*g-e*f)/(d*g-e*f))-1/2*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+
d)^n)*f^2/g^3*ln(g*x+f)+1/2*I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2*f^2/g^3*ln(g*x+f)+b*ln(c)*f^2/g^3*ln(g*x+f)
-b*ln(c)/g^2*x*f-b*n/g^3*f^2*dilog(((g*x+f)*e+d*g-e*f)/(d*g-e*f))+1/2*b*ln((e*x+d)^n)/g*x^2+1/2*a/g*x^2+1/2*I*
b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)/g^2*x*f+a*f^2/g^3*ln(g*x+f)-1/4*I*b*Pi*csgn(I*c)*csgn(I*(
e*x+d)^n)*csgn(I*c*(e*x+d)^n)/g*x^2-1/2*I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2/g^2*x*f-1/2*I*b*Pi*csgn(I*c*(e*
x+d)^n)^3*f^2/g^3*ln(g*x+f)-1/2*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2/g^2*x*f+5/4*b*n/g^3*f^2+1/2*I*b
*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2*f^2/g^3*ln(g*x+f)-b*ln((e*x+d)^n)/g^2*x*f-1/2*b/e^2*n/g*d^2*ln((g*
x+f)*e+d*g-e*f)+1/2*b*ln(c)/g*x^2+b*ln((e*x+d)^n)*f^2/g^3*ln(g*x+f)-1/4*I*b*Pi*csgn(I*c*(e*x+d)^n)^3/g*x^2+1/4
*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2/g*x^2-b/e*n/g^2*d*ln((g*x+f)*e+d*g-e*f)*f+1/2*I*b*Pi*csgn(I*c*
(e*x+d)^n)^3/g^2*x*f+1/2*b/e*n/g^2*d*f+1/4*I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2/g*x^2-a*f*x/g^2+b*f*n*x/g^2+
1/2*b*d*n*x/e/g-1/4*b*n*x^2/g

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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*(e*x+d)^n))/(g*x+f),x, algorithm="maxima")

[Out]

1/2*a*(2*f^2*log(g*x + f)/g^3 + (g*x^2 - 2*f*x)/g^2) + b*integrate((x^2*log((x*e + d)^n) + x^2*log(c))/(g*x +
f), 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*(e*x+d)^n))/(g*x+f),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \left (a + b \log {\left (c \left (d + e x\right )^{n} \right )}\right )}{f + g x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*(e*x+d)^n))/(g*x+f),x, algorithm="giac")

[Out]

integrate((b*log((x*e + d)^n*c) + a)*x^2/(g*x + f), 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\,{\left (d+e\,x\right )}^n\right )\right )}{f+g\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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