\(\int \frac {(f+g x)^2}{a+b \log (c (d+e x)^n)} \, dx\) [89]

Optimal result

Integrand size = 24, antiderivative size = 219 \[ \int \frac {(f+g x)^2}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {e^{-\frac {a}{b n}} (e f-d g)^2 (d+e x) \left (c (d+e x)^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b e^3 n}+\frac {2 e^{-\frac {2 a}{b n}} g (e f-d g) (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \operatorname {ExpIntegralEi}\left (\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )}{b e^3 n}+\frac {e^{-\frac {3 a}{b n}} g^2 (d+e x)^3 \left (c (d+e x)^n\right )^{-3/n} \operatorname {ExpIntegralEi}\left (\frac {3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )}{b e^3 n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2446, 2436, 2337, 2209, 2437, 2347} \[ \int \frac {(f+g x)^2}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {2 g e^{-\frac {2 a}{b n}} (d+e x)^2 (e f-d g) \left (c (d+e x)^n\right )^{-2/n} \operatorname {ExpIntegralEi}\left (\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )}{b e^3 n}+\frac {e^{-\frac {a}{b n}} (d+e x) (e f-d g)^2 \left (c (d+e x)^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b e^3 n}+\frac {g^2 e^{-\frac {3 a}{b n}} (d+e x)^3 \left (c (d+e x)^n\right )^{-3/n} \operatorname {ExpIntegralEi}\left (\frac {3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )}{b e^3 n} \]

[In]

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

[Out]

((e*f - d*g)^2*(d + e*x)*ExpIntegralEi[(a + b*Log[c*(d + e*x)^n])/(b*n)])/(b*e^3*E^(a/(b*n))*n*(c*(d + e*x)^n)
^n^(-1)) + (2*g*(e*f - d*g)*(d + e*x)^2*ExpIntegralEi[(2*(a + b*Log[c*(d + e*x)^n]))/(b*n)])/(b*e^3*E^((2*a)/(
b*n))*n*(c*(d + e*x)^n)^(2/n)) + (g^2*(d + e*x)^3*ExpIntegralEi[(3*(a + b*Log[c*(d + e*x)^n]))/(b*n)])/(b*e^3*
E^((3*a)/(b*n))*n*(c*(d + e*x)^n)^(3/n))

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2337

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

Rule 2347

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*n*(c*x^n
)^((m + 1)/n)), Subst[Int[E^(((m + 1)/n)*x)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}
, 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 2437

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

Rule 2446

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

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.90 \[ \int \frac {(f+g x)^2}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {e^{-\frac {3 a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-3/n} \left (e^{\frac {2 a}{b n}} (e f-d g)^2 \left (c (d+e x)^n\right )^{2/n} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )-g (d+e x) \left (-2 e^{\frac {a}{b n}} (e f-d g) \left (c (d+e x)^n\right )^{\frac {1}{n}} \operatorname {ExpIntegralEi}\left (\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )-g (d+e x) \operatorname {ExpIntegralEi}\left (\frac {3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )\right )\right )}{b e^3 n} \]

[In]

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

[Out]

((d + e*x)*(E^((2*a)/(b*n))*(e*f - d*g)^2*(c*(d + e*x)^n)^(2/n)*ExpIntegralEi[(a + b*Log[c*(d + e*x)^n])/(b*n)
] - g*(d + e*x)*(-2*E^(a/(b*n))*(e*f - d*g)*(c*(d + e*x)^n)^n^(-1)*ExpIntegralEi[(2*(a + b*Log[c*(d + e*x)^n])
)/(b*n)] - g*(d + e*x)*ExpIntegralEi[(3*(a + b*Log[c*(d + e*x)^n]))/(b*n)])))/(b*e^3*E^((3*a)/(b*n))*n*(c*(d +
e*x)^n)^(3/n))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 1.34 (sec) , antiderivative size = 1889, normalized size of antiderivative = 8.63

method	result	size
risch	\(\text {Expression too large to display}\)	\(1889\)

[In]

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

[Out]

-1/e^3*g^2/b/n*(e*x+d)^3*((e*x+d)^n)^(-3/n)*c^(-3/n)*exp(-3/2*(-I*b*Pi*csgn(I*c*(e*x+d)^n)*csgn(I*c)*csgn(I*(e
*x+d)^n)+I*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2*b+I*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2*b-I*Pi*csgn(I*c*(
e*x+d)^n)^3*b+2*a)/b/n)*Ei(1,-3*ln(e*x+d)-3/2*(-I*b*Pi*csgn(I*c*(e*x+d)^n)*csgn(I*c)*csgn(I*(e*x+d)^n)+I*Pi*cs
gn(I*c)*csgn(I*c*(e*x+d)^n)^2*b+I*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2*b-I*Pi*csgn(I*c*(e*x+d)^n)^3*b+2*
b*ln(c)+2*b*(ln((e*x+d)^n)-n*ln(e*x+d))+2*a)/b/n)-1/e*f^2/b/n*(e*x+d)*((e*x+d)^n)^(-1/n)*c^(-1/n)*exp(-1/2*(-I
*b*Pi*csgn(I*c*(e*x+d)^n)*csgn(I*c)*csgn(I*(e*x+d)^n)+I*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2*b+I*Pi*csgn(I*(e*x+
d)^n)*csgn(I*c*(e*x+d)^n)^2*b-I*Pi*csgn(I*c*(e*x+d)^n)^3*b+2*a)/b/n)*Ei(1,-ln(e*x+d)-1/2*(-I*b*Pi*csgn(I*c*(e*
x+d)^n)*csgn(I*c)*csgn(I*(e*x+d)^n)+I*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2*b+I*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*
x+d)^n)^2*b-I*Pi*csgn(I*c*(e*x+d)^n)^3*b+2*b*ln(c)+2*b*(ln((e*x+d)^n)-n*ln(e*x+d))+2*a)/b/n)-1/e^3*g^2*d^2/b/n
*(e*x+d)*((e*x+d)^n)^(-1/n)*c^(-1/n)*exp(-1/2*(-I*b*Pi*csgn(I*c*(e*x+d)^n)*csgn(I*c)*csgn(I*(e*x+d)^n)+I*Pi*cs
gn(I*c)*csgn(I*c*(e*x+d)^n)^2*b+I*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2*b-I*Pi*csgn(I*c*(e*x+d)^n)^3*b+2*
a)/b/n)*Ei(1,-ln(e*x+d)-1/2*(-I*b*Pi*csgn(I*c*(e*x+d)^n)*csgn(I*c)*csgn(I*(e*x+d)^n)+I*Pi*csgn(I*c)*csgn(I*c*(
e*x+d)^n)^2*b+I*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2*b-I*Pi*csgn(I*c*(e*x+d)^n)^3*b+2*b*ln(c)+2*b*(ln((e
*x+d)^n)-n*ln(e*x+d))+2*a)/b/n)+2/e^3*d*g^2/b/n*(e*x+d)^2*((e*x+d)^n)^(-2/n)*c^(-2/n)*exp(-(-I*b*Pi*csgn(I*c*(
e*x+d)^n)*csgn(I*c)*csgn(I*(e*x+d)^n)+I*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2*b+I*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(
e*x+d)^n)^2*b-I*Pi*csgn(I*c*(e*x+d)^n)^3*b+2*a)/b/n)*Ei(1,-2*ln(e*x+d)-(-I*b*Pi*csgn(I*c*(e*x+d)^n)*csgn(I*c)*
csgn(I*(e*x+d)^n)+I*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2*b+I*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2*b-I*Pi*c
sgn(I*c*(e*x+d)^n)^3*b+2*b*ln(c)+2*b*(ln((e*x+d)^n)-n*ln(e*x+d))+2*a)/b/n)-2/e^2*f*g/b/n*(e*x+d)^2*((e*x+d)^n)
^(-2/n)*c^(-2/n)*exp(-(-I*b*Pi*csgn(I*c*(e*x+d)^n)*csgn(I*c)*csgn(I*(e*x+d)^n)+I*Pi*csgn(I*c)*csgn(I*c*(e*x+d)
^n)^2*b+I*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2*b-I*Pi*csgn(I*c*(e*x+d)^n)^3*b+2*a)/b/n)*Ei(1,-2*ln(e*x+d
)-(-I*b*Pi*csgn(I*c*(e*x+d)^n)*csgn(I*c)*csgn(I*(e*x+d)^n)+I*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2*b+I*Pi*csgn(I*
(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2*b-I*Pi*csgn(I*c*(e*x+d)^n)^3*b+2*b*ln(c)+2*b*(ln((e*x+d)^n)-n*ln(e*x+d))+2*a)
/b/n)+2/e^2*d*f*g/b/n*(e*x+d)*((e*x+d)^n)^(-1/n)*c^(-1/n)*exp(-1/2*(-I*b*Pi*csgn(I*c*(e*x+d)^n)*csgn(I*c)*csgn
(I*(e*x+d)^n)+I*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2*b+I*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2*b-I*Pi*csgn(
I*c*(e*x+d)^n)^3*b+2*a)/b/n)*Ei(1,-ln(e*x+d)-1/2*(-I*b*Pi*csgn(I*c*(e*x+d)^n)*csgn(I*c)*csgn(I*(e*x+d)^n)+I*Pi
*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2*b+I*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2*b-I*Pi*csgn(I*c*(e*x+d)^n)^3*b
+2*b*ln(c)+2*b*(ln((e*x+d)^n)-n*ln(e*x+d))+2*a)/b/n)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.88 \[ \int \frac {(f+g x)^2}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {{\left (g^{2} \operatorname {log\_integral}\left ({\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} e^{\left (\frac {3 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )}\right ) + 2 \, {\left (e f g - d g^{2}\right )} e^{\left (\frac {b \log \left (c\right ) + a}{b n}\right )} \operatorname {log\_integral}\left ({\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )} e^{\left (\frac {2 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )}\right ) + {\left (e^{2} f^{2} - 2 \, d e f g + d^{2} g^{2}\right )} e^{\left (\frac {2 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )} \operatorname {log\_integral}\left ({\left (e x + d\right )} e^{\left (\frac {b \log \left (c\right ) + a}{b n}\right )}\right )\right )} e^{\left (-\frac {3 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )}}{b e^{3} n} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.54 \[ \int \frac {(f+g x)^2}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {f^{2} {\rm Ei}\left (\frac {\log \left (c\right )}{n} + \frac {a}{b n} + \log \left (e x + d\right )\right ) e^{\left (-\frac {a}{b n}\right )}}{b c^{\left (\frac {1}{n}\right )} e n} - \frac {2 \, d f g {\rm Ei}\left (\frac {\log \left (c\right )}{n} + \frac {a}{b n} + \log \left (e x + d\right )\right ) e^{\left (-\frac {a}{b n}\right )}}{b c^{\left (\frac {1}{n}\right )} e^{2} n} + \frac {d^{2} g^{2} {\rm Ei}\left (\frac {\log \left (c\right )}{n} + \frac {a}{b n} + \log \left (e x + d\right )\right ) e^{\left (-\frac {a}{b n}\right )}}{b c^{\left (\frac {1}{n}\right )} e^{3} n} + \frac {2 \, f g {\rm Ei}\left (\frac {2 \, \log \left (c\right )}{n} + \frac {2 \, a}{b n} + 2 \, \log \left (e x + d\right )\right ) e^{\left (-\frac {2 \, a}{b n}\right )}}{b c^{\frac {2}{n}} e^{2} n} - \frac {2 \, d g^{2} {\rm Ei}\left (\frac {2 \, \log \left (c\right )}{n} + \frac {2 \, a}{b n} + 2 \, \log \left (e x + d\right )\right ) e^{\left (-\frac {2 \, a}{b n}\right )}}{b c^{\frac {2}{n}} e^{3} n} + \frac {g^{2} {\rm Ei}\left (\frac {3 \, \log \left (c\right )}{n} + \frac {3 \, a}{b n} + 3 \, \log \left (e x + d\right )\right ) e^{\left (-\frac {3 \, a}{b n}\right )}}{b c^{\frac {3}{n}} e^{3} n} \]

[In]

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

[Out]

f^2*Ei(log(c)/n + a/(b*n) + log(e*x + d))*e^(-a/(b*n))/(b*c^(1/n)*e*n) - 2*d*f*g*Ei(log(c)/n + a/(b*n) + log(e
*x + d))*e^(-a/(b*n))/(b*c^(1/n)*e^2*n) + d^2*g^2*Ei(log(c)/n + a/(b*n) + log(e*x + d))*e^(-a/(b*n))/(b*c^(1/n
)*e^3*n) + 2*f*g*Ei(2*log(c)/n + 2*a/(b*n) + 2*log(e*x + d))*e^(-2*a/(b*n))/(b*c^(2/n)*e^2*n) - 2*d*g^2*Ei(2*l
og(c)/n + 2*a/(b*n) + 2*log(e*x + d))*e^(-2*a/(b*n))/(b*c^(2/n)*e^3*n) + g^2*Ei(3*log(c)/n + 3*a/(b*n) + 3*log
(e*x + d))*e^(-3*a/(b*n))/(b*c^(3/n)*e^3*n)

Mupad [F(-1)]

Timed out. \[ \int \frac {(f+g x)^2}{a+b \log \left (c (d+e x)^n\right )} \, dx=\int \frac {{\left (f+g\,x\right )}^2}{a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )} \,d x \]

[In]

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

[Out]

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