3.3.49 \(\int \frac {x^3 (a+b \log (c (d+e x)^n))}{(f+g x)^2} \, dx\) [249]

Optimal. Leaf size=265 \[ -\frac {2 a f x}{g^3}+\frac {2 b f n x}{g^3}+\frac {b d n x}{2 e g^2}-\frac {b n x^2}{4 g^2}-\frac {b d^2 n \log (d+e x)}{2 e^2 g^2}-\frac {b e f^3 n \log (d+e x)}{g^4 (e f-d g)}-\frac {2 b f (d+e x) \log \left (c (d+e x)^n\right )}{e g^3}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac {f^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^4 (f+g x)}+\frac {b e f^3 n \log (f+g x)}{g^4 (e f-d g)}+\frac {3 f^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g^4}+\frac {3 b f^2 n \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{g^4} \]

[Out]

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

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Rubi [A]
time = 0.19, antiderivative size = 265, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {45, 2463, 2436, 2332, 2442, 36, 31, 2441, 2440, 2438} \begin {gather*} \frac {3 b f^2 n \text {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g^4}+\frac {f^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^4 (f+g x)}+\frac {3 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^4}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}-\frac {2 a f x}{g^3}-\frac {2 b f (d+e x) \log \left (c (d+e x)^n\right )}{e g^3}-\frac {b d^2 n \log (d+e x)}{2 e^2 g^2}-\frac {b e f^3 n \log (d+e x)}{g^4 (e f-d g)}+\frac {b e f^3 n \log (f+g x)}{g^4 (e f-d g)}+\frac {b d n x}{2 e g^2}+\frac {2 b f n x}{g^3}-\frac {b n x^2}{4 g^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

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^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x)^2} \, dx &=\int \left (-\frac {2 f \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3}+\frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2}-\frac {f^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3 (f+g x)^2}+\frac {3 f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3 (f+g x)}\right ) \, dx\\ &=-\frac {(2 f) \int \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx}{g^3}+\frac {\left (3 f^2\right ) \int \frac {a+b \log \left (c (d+e x)^n\right )}{f+g x} \, dx}{g^3}-\frac {f^3 \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^2} \, dx}{g^3}+\frac {\int x \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx}{g^2}\\ &=-\frac {2 a f x}{g^3}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac {f^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^4 (f+g x)}+\frac {3 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^4}-\frac {(2 b f) \int \log \left (c (d+e x)^n\right ) \, dx}{g^3}-\frac {\left (3 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^4}-\frac {\left (b e f^3 n\right ) \int \frac {1}{(d+e x) (f+g x)} \, dx}{g^4}-\frac {(b e n) \int \frac {x^2}{d+e x} \, dx}{2 g^2}\\ &=-\frac {2 a f x}{g^3}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac {f^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^4 (f+g x)}+\frac {3 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^4}-\frac {(2 b f) \text {Subst}\left (\int \log \left (c x^n\right ) \, dx,x,d+e x\right )}{e g^3}-\frac {\left (3 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^4}-\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^2}-\frac {\left (b e^2 f^3 n\right ) \int \frac {1}{d+e x} \, dx}{g^4 (e f-d g)}+\frac {\left (b e f^3 n\right ) \int \frac {1}{f+g x} \, dx}{g^3 (e f-d g)}\\ &=-\frac {2 a f x}{g^3}+\frac {2 b f n x}{g^3}+\frac {b d n x}{2 e g^2}-\frac {b n x^2}{4 g^2}-\frac {b d^2 n \log (d+e x)}{2 e^2 g^2}-\frac {b e f^3 n \log (d+e x)}{g^4 (e f-d g)}-\frac {2 b f (d+e x) \log \left (c (d+e x)^n\right )}{e g^3}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac {f^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^4 (f+g x)}+\frac {b e f^3 n \log (f+g x)}{g^4 (e f-d g)}+\frac {3 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^4}+\frac {3 b f^2 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 = 0.20, size = 220, normalized size = 0.83 \begin {gather*} \frac {-8 a f g x+8 b f g n x-\frac {b g^2 n \left (e x (-2 d+e x)+2 d^2 \log (d+e x)\right )}{e^2}-\frac {8 b f g (d+e x) \log \left (c (d+e x)^n\right )}{e}+2 g^2 x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {4 f^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x}-\frac {4 b e f^3 n (\log (d+e x)-\log (f+g x))}{e f-d g}+12 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 )+12 b f^2 n \text {Li}_2\left (\frac {g (d+e x)}{-e f+d g}\right )}{4 g^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a*f*g*x + 8*b*f*g*n*x - (b*g^2*n*(e*x*(-2*d + e*x) + 2*d^2*Log[d + e*x]))/e^2 - (8*b*f*g*(d + e*x)*Log[c*(
d + e*x)^n])/e + 2*g^2*x^2*(a + b*Log[c*(d + e*x)^n]) + (4*f^3*(a + b*Log[c*(d + e*x)^n]))/(f + g*x) - (4*b*e*
f^3*n*(Log[d + e*x] - Log[f + g*x]))/(e*f - d*g) + 12*f^2*(a + b*Log[c*(d + e*x)^n])*Log[(e*(f + g*x))/(e*f -
d*g)] + 12*b*f^2*n*PolyLog[2, (g*(d + e*x))/(-(e*f) + d*g)])/(4*g^4)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.40, size = 1063, normalized size = 4.01

method	result	size
risch	\(\text {Expression too large to display}\)	\(1063\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*b*n/g^4*f^2*ln(g*x+f)*ln(((g*x+f)*e+d*g-e*f)/(d*g-e*f))+3/2*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2/
g^4*f^2*ln(g*x+f)+1/2*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2*f^3/g^4/(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^2*x^2-1/2*I*b*Pi*csgn(I*c*(e*x+d)^n)^3*f^3/g^4/(g*x+f)+1/2*a/g^2*x^2+3*b*l
n(c)/g^4*f^2*ln(g*x+f)-3/2*I*b*Pi*csgn(I*c*(e*x+d)^n)^3/g^4*f^2*ln(g*x+f)+3/2*I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d
)^n)^2/g^4*f^2*ln(g*x+f)-I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2/g^3*x*f-I*b*Pi*csgn(I*c)*csgn(I*c*(e*x
+d)^n)^2/g^3*x*f+1/4*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2/g^2*x^2+1/2*b*ln((e*x+d)^n)/g^2*x^2+1/2*b*
ln(c)/g^2*x^2+a*f^3/g^4/(g*x+f)+3*a/g^4*f^2*ln(g*x+f)+1/2*I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2*f^3/g^4/(g*x+
f)-3/2*b/e*n/g^2/(d*g-e*f)*ln((g*x+f)*e+d*g-e*f)*d^2*f-2*b*ln((e*x+d)^n)/g^3*x*f+3*b*ln((e*x+d)^n)/g^4*f^2*ln(
g*x+f)+b*ln((e*x+d)^n)*f^3/g^4/(g*x+f)+9/4*b*n/g^4*f^2-3*b*n/g^4*f^2*dilog(((g*x+f)*e+d*g-e*f)/(d*g-e*f))+I*b*
Pi*csgn(I*c*(e*x+d)^n)^3/g^3*x*f-1/2*b/e^2*n/g/(d*g-e*f)*ln((g*x+f)*e+d*g-e*f)*d^3+2*b*n/g^3/(d*g-e*f)*ln((g*x
+f)*e+d*g-e*f)*d*f^2+b*e*n/g^4/(d*g-e*f)*ln((g*x+f)*e+d*g-e*f)*f^3-b*e*n/g^4*f^3/(d*g-e*f)*ln(g*x+f)+I*b*Pi*cs
gn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)/g^3*x*f-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/(g*x+f)-3/2*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)/g^4*f^2*ln(g*x+f)+b*ln(c)*f^3/g^
4/(g*x+f)-2*b*ln(c)/g^3*x*f+1/2*b/e*n/g^3*d*f-1/4*I*b*Pi*csgn(I*c*(e*x+d)^n)^3/g^2*x^2+1/4*I*b*Pi*csgn(I*c)*cs
gn(I*c*(e*x+d)^n)^2/g^2*x^2+2*b*f*n*x/g^3+1/2*b*d*n*x/e/g^2-2*a*f*x/g^3-1/4*b*n*x^2/g^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^3*(a+b*log(c*(e*x+d)^n))/(g*x+f)^2,x, algorithm="maxima")

[Out]

1/2*(2*f^3/(g^5*x + f*g^4) + 6*f^2*log(g*x + f)/g^4 + (g*x^2 - 4*f*x)/g^3)*a + b*integrate((x^3*log((x*e + d)^
n) + x^3*log(c))/(g^2*x^2 + 2*f*g*x + f^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^3*(a+b*log(c*(e*x+d)^n))/(g*x+f)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \left (a + b \log {\left (c \left (d + e x\right )^{n} \right )}\right )}{\left (f + g x\right )^{2}}\, dx \end {gather*}

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)**2,x)

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^3\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}{{\left (f+g\,x\right )}^2} \,d x \end {gather*}

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)^2,x)

[Out]

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

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