3.242 \(\int \frac {x^3 (a+b \log (c (d+e x)^n))}{f+g x} \, dx\)

Optimal. Leaf size=281 \[ -\frac {f^3 \log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^4}-\frac {f x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g}+\frac {a f^2 x}{g^3}+\frac {b f^2 (d+e x) \log \left (c (d+e x)^n\right )}{e g^3}+\frac {b d^3 n \log (d+e x)}{3 e^3 g}+\frac {b d^2 f n \log (d+e x)}{2 e^2 g^2}-\frac {b d^2 n x}{3 e^2 g}-\frac {b f^3 n \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{g^4}-\frac {b d f n x}{2 e g^2}+\frac {b d n x^2}{6 e g}-\frac {b f^2 n x}{g^3}+\frac {b f n x^2}{4 g^2}-\frac {b n x^3}{9 g} \]

[Out]

a*f^2*x/g^3-b*f^2*n*x/g^3-1/2*b*d*f*n*x/e/g^2-1/3*b*d^2*n*x/e^2/g+1/4*b*f*n*x^2/g^2+1/6*b*d*n*x^2/e/g-1/9*b*n*
x^3/g+1/2*b*d^2*f*n*ln(e*x+d)/e^2/g^2+1/3*b*d^3*n*ln(e*x+d)/e^3/g+b*f^2*(e*x+d)*ln(c*(e*x+d)^n)/e/g^3-1/2*f*x^
2*(a+b*ln(c*(e*x+d)^n))/g^2+1/3*x^3*(a+b*ln(c*(e*x+d)^n))/g-f^3*(a+b*ln(c*(e*x+d)^n))*ln(e*(g*x+f)/(-d*g+e*f))
/g^4-b*f^3*n*polylog(2,-g*(e*x+d)/(-d*g+e*f))/g^4

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Rubi [A]  time = 0.28, antiderivative size = 281, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {43, 2416, 2389, 2295, 2395, 2394, 2393, 2391} \[ -\frac {b f^3 n \text {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g^4}-\frac {f^3 \log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^4}-\frac {f x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g}+\frac {a f^2 x}{g^3}+\frac {b f^2 (d+e x) \log \left (c (d+e x)^n\right )}{e g^3}+\frac {b d^2 f n \log (d+e x)}{2 e^2 g^2}-\frac {b d^2 n x}{3 e^2 g}+\frac {b d^3 n \log (d+e x)}{3 e^3 g}-\frac {b d f n x}{2 e g^2}+\frac {b d n x^2}{6 e g}-\frac {b f^2 n x}{g^3}+\frac {b f n x^2}{4 g^2}-\frac {b n x^3}{9 g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*f^2*x)/g^3 - (b*f^2*n*x)/g^3 - (b*d*f*n*x)/(2*e*g^2) - (b*d^2*n*x)/(3*e^2*g) + (b*f*n*x^2)/(4*g^2) + (b*d*n
*x^2)/(6*e*g) - (b*n*x^3)/(9*g) + (b*d^2*f*n*Log[d + e*x])/(2*e^2*g^2) + (b*d^3*n*Log[d + e*x])/(3*e^3*g) + (b
*f^2*(d + e*x)*Log[c*(d + e*x)^n])/(e*g^3) - (f*x^2*(a + b*Log[c*(d + e*x)^n]))/(2*g^2) + (x^3*(a + b*Log[c*(d
+ e*x)^n]))/(3*g) - (f^3*(a + b*Log[c*(d + e*x)^n])*Log[(e*(f + g*x))/(e*f - d*g)])/g^4 - (b*f^3*n*PolyLog[2,
-((g*(d + e*x))/(e*f - d*g))])/g^4

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 2295

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

Rule 2389

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 2391

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 2393

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 2394

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 2395

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 2416

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^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x} \, dx &=\int \left (\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3}-\frac {f x \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}-\frac {f^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3 (f+g x)}\right ) \, dx\\ &=\frac {f^2 \int \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx}{g^3}-\frac {f^3 \int \frac {a+b \log \left (c (d+e x)^n\right )}{f+g x} \, dx}{g^3}-\frac {f \int x \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx}{g^2}+\frac {\int x^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx}{g}\\ &=\frac {a f^2 x}{g^3}-\frac {f x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g}-\frac {f^3 \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^4}+\frac {\left (b f^2\right ) \int \log \left (c (d+e x)^n\right ) \, dx}{g^3}+\frac {\left (b e f^3 n\right ) \int \frac {\log \left (\frac {e (f+g x)}{e f-d g}\right )}{d+e x} \, dx}{g^4}+\frac {(b e f n) \int \frac {x^2}{d+e x} \, dx}{2 g^2}-\frac {(b e n) \int \frac {x^3}{d+e x} \, dx}{3 g}\\ &=\frac {a f^2 x}{g^3}-\frac {f x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g}-\frac {f^3 \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^4}+\frac {\left (b f^2\right ) \operatorname {Subst}\left (\int \log \left (c x^n\right ) \, dx,x,d+e x\right )}{e g^3}+\frac {\left (b f^3 n\right ) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {g x}{e f-d g}\right )}{x} \, dx,x,d+e x\right )}{g^4}+\frac {(b e f n) \int \left (-\frac {d}{e^2}+\frac {x}{e}+\frac {d^2}{e^2 (d+e x)}\right ) \, dx}{2 g^2}-\frac {(b e n) \int \left (\frac {d^2}{e^3}-\frac {d x}{e^2}+\frac {x^2}{e}-\frac {d^3}{e^3 (d+e x)}\right ) \, dx}{3 g}\\ &=\frac {a f^2 x}{g^3}-\frac {b f^2 n x}{g^3}-\frac {b d f n x}{2 e g^2}-\frac {b d^2 n x}{3 e^2 g}+\frac {b f n x^2}{4 g^2}+\frac {b d n x^2}{6 e g}-\frac {b n x^3}{9 g}+\frac {b d^2 f n \log (d+e x)}{2 e^2 g^2}+\frac {b d^3 n \log (d+e x)}{3 e^3 g}+\frac {b f^2 (d+e x) \log \left (c (d+e x)^n\right )}{e g^3}-\frac {f x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g}-\frac {f^3 \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^4}-\frac {b f^3 n \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{g^4}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 241, normalized size = 0.86 \[ \frac {e \left (g x \left (6 a e^2 \left (6 f^2-3 f g x+2 g^2 x^2\right )-b n \left (12 d^2 g^2-6 d e g (g x-3 f)+e^2 \left (36 f^2-9 f g x+4 g^2 x^2\right )\right )\right )-36 a e^2 f^3 \log \left (\frac {e (f+g x)}{e f-d g}\right )+6 b e \log \left (c (d+e x)^n\right ) \left (-6 e f^3 \log \left (\frac {e (f+g x)}{e f-d g}\right )+6 d f^2 g+e g x \left (6 f^2-3 f g x+2 g^2 x^2\right )\right )\right )+6 b d^2 g^2 n (2 d g+3 e f) \log (d+e x)-36 b e^3 f^3 n \text {Li}_2\left (\frac {g (d+e x)}{d g-e f}\right )}{36 e^3 g^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*b*d^2*g^2*(3*e*f + 2*d*g)*n*Log[d + e*x] + e*(g*x*(6*a*e^2*(6*f^2 - 3*f*g*x + 2*g^2*x^2) - b*n*(12*d^2*g^2
- 6*d*e*g*(-3*f + g*x) + e^2*(36*f^2 - 9*f*g*x + 4*g^2*x^2))) - 36*a*e^2*f^3*Log[(e*(f + g*x))/(e*f - d*g)] +
6*b*e*Log[c*(d + e*x)^n]*(6*d*f^2*g + e*g*x*(6*f^2 - 3*f*g*x + 2*g^2*x^2) - 6*e*f^3*Log[(e*(f + g*x))/(e*f - d
*g)])) - 36*b*e^3*f^3*n*PolyLog[2, (g*(d + e*x))/(-(e*f) + d*g)])/(36*e^3*g^4)

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{3} \log \left ({\left (e x + d\right )}^{n} c\right ) + a x^{3}}{g x + f}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{3}}{g x + f}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.28, size = 1000, normalized size = 3.56 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*ln(c*(e*x+d)^n)+a)/(g*x+f),x)

[Out]

b*ln((e*x+d)^n)/g^3*x*f^2-b*ln((e*x+d)^n)*f^3/g^4*ln(g*x+f)-1/2*b*ln((e*x+d)^n)/g^2*f*x^2+1/6*I*b*Pi*csgn(I*(e
*x+d)^n)*csgn(I*c*(e*x+d)^n)^2/g*x^3+b*ln(c)/g^3*x*f^2-b*ln(c)*f^3/g^4*ln(g*x+f)-1/2*b*ln(c)/g^2*f*x^2+1/2*b/e
^2*n/g^2*d^2*ln(d*g-e*f+(g*x+f)*e)*f+b/e*n/g^3*d*ln(d*g-e*f+(g*x+f)*e)*f^2+b*n/g^4*f^3*dilog((d*g-e*f+(g*x+f)*
e)/(d*g-e*f))-1/6*I*b*Pi*csgn(I*c*(e*x+d)^n)^3/g*x^3+1/6*I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2/g*x^3+1/2*I*b*
Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2/g^3*x*f^2-1/4*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2/g^2*f*
x^2+1/2*I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2/g^3*x*f^2-1/4*I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2/g^2*f*x^2+
1/3*b*ln((e*x+d)^n)/g*x^3+1/4*I*b*Pi*csgn(I*c*(e*x+d)^n)^3/g^2*f*x^2+1/3*b/e^3*n/g*d^3*ln(d*g-e*f+(g*x+f)*e)-1
/2*I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2*f^3/g^4*ln(g*x+f)+1/2*I*b*Pi*csgn(I*c*(e*x+d)^n)^3*f^3/g^4*ln(g*x+f)
-49/36*b*n/g^4*f^3-1/2*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)/g^3*x*f^2-1/2*I*b*Pi*csgn(I*(e*x
+d)^n)*csgn(I*c*(e*x+d)^n)^2*f^3/g^4*ln(g*x+f)-1/2*I*b*Pi*csgn(I*c*(e*x+d)^n)^3/g^3*x*f^2-1/2*a/g^2*f*x^2-a*f^
3/g^4*ln(g*x+f)+1/3*b*ln(c)/g*x^3+1/4*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)/g^2*f*x^2+1/2*I*b
*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)*f^3/g^4*ln(g*x+f)-1/6*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*c
sgn(I*c*(e*x+d)^n)/g*x^3+b*n/g^4*f^3*ln(g*x+f)*ln((d*g-e*f+(g*x+f)*e)/(d*g-e*f))-2/3*b/e*n/g^3*d*f^2-1/3*b/e^2
*n/g^2*d^2*f+1/3*a/g*x^3+1/4*b*f*n*x^2/g^2-1/2*b*d*f*n*x/e/g^2-b*f^2*n*x/g^3-1/9*b*n*x^3/g-1/3*b*d^2*n*x/e^2/g
+1/6*b*d*n*x^2/e/g+a*f^2*x/g^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{6} \, a {\left (\frac {6 \, f^{3} \log \left (g x + f\right )}{g^{4}} - \frac {2 \, g^{2} x^{3} - 3 \, f g x^{2} + 6 \, f^{2} x}{g^{3}}\right )} + b \int \frac {x^{3} \log \left ({\left (e x + d\right )}^{n}\right ) + x^{3} \log \relax (c)}{g x + f}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^3\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}{f+g\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \left (a + b \log {\left (c \left (d + e x\right )^{n} \right )}\right )}{f + g x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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