\(\int \frac {(c+d x)^2}{(a+b x)^4 \log (e (\frac {a+b x}{c+d x})^n)} \, dx\) [248]

Optimal result

Integrand size = 35, antiderivative size = 75 \[ \int \frac {(c+d x)^2}{(a+b x)^4 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\frac {\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{3/n} (c+d x)^3 \operatorname {ExpIntegralEi}\left (-\frac {3 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{n}\right )}{(b c-a d) n (a+b x)^3} \]

[Out]

(e*((b*x+a)/(d*x+c))^n)^(3/n)*(d*x+c)^3*Ei(-3*ln(e*((b*x+a)/(d*x+c))^n)/n)/(-a*d+b*c)/n/(b*x+a)^3

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {2561, 2347, 2209} \[ \int \frac {(c+d x)^2}{(a+b x)^4 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\frac {(c+d x)^3 \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{3/n} \operatorname {ExpIntegralEi}\left (-\frac {3 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{n}\right )}{n (a+b x)^3 (b c-a d)} \]

[In]

Int[(c + d*x)^2/((a + b*x)^4*Log[e*((a + b*x)/(c + d*x))^n]),x]

[Out]

((e*((a + b*x)/(c + d*x))^n)^(3/n)*(c + d*x)^3*ExpIntegralEi[(-3*Log[e*((a + b*x)/(c + d*x))^n])/n])/((b*c - a
*d)*n*(a + b*x)^3)

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 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 2561

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m
_.)*((h_.) + (i_.)*(x_))^(q_.), x_Symbol] :> Dist[(b*c - a*d)^(m + q + 1)*(g/b)^m*(i/d)^q, Subst[Int[x^m*((A +
B*Log[e*x^n])^p/(b - d*x)^(m + q + 2)), x], x, (a + b*x)/(c + d*x)], x] /; FreeQ[{a, b, c, d, e, f, g, h, i,
A, B, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[b*f - a*g, 0] && EqQ[d*h - c*i, 0] && IntegersQ[m, q]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{x^4 \log \left (e x^n\right )} \, dx,x,\frac {a+b x}{c+d x}\right )}{b c-a d} \\ & = \frac {\left (\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{3/n} (c+d x)^3\right ) \text {Subst}\left (\int \frac {e^{-\frac {3 x}{n}}}{x} \, dx,x,\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(b c-a d) n (a+b x)^3} \\ & = \frac {\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{3/n} (c+d x)^3 \text {Ei}\left (-\frac {3 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{n}\right )}{(b c-a d) n (a+b x)^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00 \[ \int \frac {(c+d x)^2}{(a+b x)^4 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\frac {\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{3/n} (c+d x)^3 \operatorname {ExpIntegralEi}\left (-\frac {3 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{n}\right )}{(b c-a d) n (a+b x)^3} \]

[In]

Integrate[(c + d*x)^2/((a + b*x)^4*Log[e*((a + b*x)/(c + d*x))^n]),x]

[Out]

((e*((a + b*x)/(c + d*x))^n)^(3/n)*(c + d*x)^3*ExpIntegralEi[(-3*Log[e*((a + b*x)/(c + d*x))^n])/n])/((b*c - a
*d)*n*(a + b*x)^3)

Maple [F]

\[\int \frac {\left (d x +c \right )^{2}}{\left (b x +a \right )^{4} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )}d x\]

[In]

int((d*x+c)^2/(b*x+a)^4/ln(e*((b*x+a)/(d*x+c))^n),x)

[Out]

int((d*x+c)^2/(b*x+a)^4/ln(e*((b*x+a)/(d*x+c))^n),x)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.17 \[ \int \frac {(c+d x)^2}{(a+b x)^4 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\frac {e^{\frac {3}{n}} \operatorname {log\_integral}\left (\frac {d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}{{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )} e^{\frac {3}{n}}}\right )}{{\left (b c - a d\right )} n} \]

[In]

integrate((d*x+c)^2/(b*x+a)^4/log(e*((b*x+a)/(d*x+c))^n),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(c+d x)^2}{(a+b x)^4 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\text {Timed out} \]

[In]

integrate((d*x+c)**2/(b*x+a)**4/ln(e*((b*x+a)/(d*x+c))**n),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(c+d x)^2}{(a+b x)^4 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\int { \frac {{\left (d x + c\right )}^{2}}{{\left (b x + a\right )}^{4} \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right )} \,d x } \]

[In]

integrate((d*x+c)^2/(b*x+a)^4/log(e*((b*x+a)/(d*x+c))^n),x, algorithm="maxima")

[Out]

integrate((d*x + c)^2/((b*x + a)^4*log(e*((b*x + a)/(d*x + c))^n)), x)

Giac [F]

\[ \int \frac {(c+d x)^2}{(a+b x)^4 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\int { \frac {{\left (d x + c\right )}^{2}}{{\left (b x + a\right )}^{4} \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right )} \,d x } \]

[In]

integrate((d*x+c)^2/(b*x+a)^4/log(e*((b*x+a)/(d*x+c))^n),x, algorithm="giac")

[Out]

integrate((d*x + c)^2/((b*x + a)^4*log(e*((b*x + a)/(d*x + c))^n)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^2}{(a+b x)^4 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\int \frac {{\left (c+d\,x\right )}^2}{\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\,{\left (a+b\,x\right )}^4} \,d x \]

[In]

int((c + d*x)^2/(log(e*((a + b*x)/(c + d*x))^n)*(a + b*x)^4),x)

[Out]

int((c + d*x)^2/(log(e*((a + b*x)/(c + d*x))^n)*(a + b*x)^4), x)