3.3.44 \(\int \frac {x (a+b \log (c (d+e x)^n))}{f+g x} \, dx\) [244]

Optimal. Leaf size=104 \[ \frac {a x}{g}-\frac {b n x}{g}+\frac {b (d+e x) \log \left (c (d+e x)^n\right )}{e g}-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g^2}-\frac {b f n \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{g^2} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {45, 2463, 2436, 2332, 2441, 2440, 2438} \begin {gather*} -\frac {b f n \text {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g^2}-\frac {f \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^2}+\frac {a x}{g}+\frac {b (d+e x) \log \left (c (d+e x)^n\right )}{e g}-\frac {b n x}{g} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 0.05, size = 95, normalized size = 0.91 \begin {gather*} \frac {a g x-b g n x+\frac {b g (d+e x) \log \left (c (d+e x)^n\right )}{e}-f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )-b f n \text {Li}_2\left (\frac {g (d+e x)}{-e f+d g}\right )}{g^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.37, size = 463, normalized size = 4.45

method	result	size
risch	\(\frac {b \ln \left (\left (e x +d \right )^{n}\right ) x}{g}-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) f \ln \left (g x +f \right )}{g^{2}}-\frac {b n x}{g}-\frac {b n f}{g^{2}}+\frac {b n d \ln \left (\left (g x +f \right ) e +d g -e f \right )}{e g}+\frac {b n f \dilog \left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{g^{2}}+\frac {b n f \ln \left (g x +f \right ) \ln \left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{g^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} f \ln \left (g x +f \right )}{2 g^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} f \ln \left (g x +f \right )}{2 g^{2}}+\frac {i b \pi \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} x}{2 g}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right ) x}{2 g}+\frac {i b \pi \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} f \ln \left (g x +f \right )}{2 g^{2}}-\frac {i b \pi \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} x}{2 g}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} x}{2 g}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right ) f \ln \left (g x +f \right )}{2 g^{2}}+\frac {b \ln \left (c \right ) x}{g}-\frac {b \ln \left (c \right ) f \ln \left (g x +f \right )}{g^{2}}+\frac {a x}{g}-\frac {a f \ln \left (g x +f \right )}{g^{2}}\)	\(463\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \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*(a+b*ln(c*(e*x+d)**n))/(g*x+f),x)

[Out]

Integral(x*(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*(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/(g*x + f), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x\,\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*(a + b*log(c*(d + e*x)^n)))/(f + g*x),x)

[Out]

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

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