\(\int \frac {a+b x}{(c+d x)^3 \log (e (\frac {a+b x}{c+d x})^n)} \, dx\) [243]

Optimal result

Integrand size = 33, antiderivative size = 75 \[ \int \frac {a+b x}{(c+d x)^3 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\frac {(a+b x)^2 \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-2/n} \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{n}\right )}{(b c-a d) n (c+d x)^2} \] Output:

(b*x+a)^2*Ei(2*ln(e*((b*x+a)/(d*x+c))^n)/n)/(-a*d+b*c)/n/((e*((b*x+a)/(d*x 
+c))^n)^(2/n))/(d*x+c)^2
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00 \[ \int \frac {a+b x}{(c+d x)^3 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\frac {(a+b x)^2 \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-2/n} \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{n}\right )}{(b c-a d) n (c+d x)^2} \] Input:

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

Output:

((a + b*x)^2*ExpIntegralEi[(2*Log[e*((a + b*x)/(c + d*x))^n])/n])/((b*c - 
a*d)*n*(e*((a + b*x)/(c + d*x))^n)^(2/n)*(c + d*x)^2)
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2961, 2747, 2609}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {a+b x}{(c+d x)^3 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx\)

\(\Big \downarrow \) 2961

\(\displaystyle \frac {\int \frac {a+b x}{(c+d x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}d\frac {a+b x}{c+d x}}{b c-a d}\)

\(\Big \downarrow \) 2747

\(\displaystyle \frac {(a+b x)^2 \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-2/n} \int \frac {\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{2/n} (c+d x)}{a+b x}d\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{n (c+d x)^2 (b c-a d)}\)

\(\Big \downarrow \) 2609

\(\displaystyle \frac {(a+b x)^2 \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{-2/n} \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{n}\right )}{n (c+d x)^2 (b c-a d)}\)

Input:

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

Output:

((a + b*x)^2*ExpIntegralEi[(2*Log[e*((a + b*x)/(c + d*x))^n])/n])/((b*c - 
a*d)*n*(e*((a + b*x)/(c + d*x))^n)^(2/n)*(c + d*x)^2)
 

Defintions of rubi rules used

rule 2609

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

rule 2747

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol 
] :> Simp[(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 2961

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*( 
B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m_.)*((h_.) + (i_.)*(x_))^(q_.), x_Symbol 
] :> Simp[(b*c - a*d)^(m + q + 1)*(g/b)^m*(i/d)^q   Subst[Int[x^m*((A + B*L 
og[e*x^n])^p/(b - d*x)^(m + q + 2)), x], x, (a + b*x)/(c + d*x)], x] /; Fre 
eQ[{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]
 

Maple [F]

\[\int \frac {b x +a}{\left (d x +c \right )^{3} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )}d x\]

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.88 \[ \int \frac {a+b x}{(c+d x)^3 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\frac {\operatorname {log\_integral}\left (\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} e^{\frac {2}{n}}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{{\left (b c - a d\right )} e^{\frac {2}{n}} n} \] Input:

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

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b x}{(c+d x)^3 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

\[ \int \frac {a+b x}{(c+d x)^3 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\int { \frac {b x + a}{{\left (d x + c\right )}^{3} \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right )} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {a+b x}{(c+d x)^3 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\int { \frac {b x + a}{{\left (d x + c\right )}^{3} \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right )} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b x}{(c+d x)^3 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\int \frac {a+b\,x}{\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\,{\left (c+d\,x\right )}^3} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {a+b x}{(c+d x)^3 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )} \, dx=\text {too large to display} \] Input:

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

Output:

(2*int(x/(log(((a + b*x)**n*e)/(c + d*x)**n)*a*c**3 + 3*log(((a + b*x)**n* 
e)/(c + d*x)**n)*a*c**2*d*x + 3*log(((a + b*x)**n*e)/(c + d*x)**n)*a*c*d** 
2*x**2 + log(((a + b*x)**n*e)/(c + d*x)**n)*a*d**3*x**3 + log(((a + b*x)** 
n*e)/(c + d*x)**n)*b*c**3*x + 3*log(((a + b*x)**n*e)/(c + d*x)**n)*b*c**2* 
d*x**2 + 3*log(((a + b*x)**n*e)/(c + d*x)**n)*b*c*d**2*x**3 + log(((a + b* 
x)**n*e)/(c + d*x)**n)*b*d**3*x**4),x)*a**2*b*d**3*n - 4*int(x/(log(((a + 
b*x)**n*e)/(c + d*x)**n)*a*c**3 + 3*log(((a + b*x)**n*e)/(c + d*x)**n)*a*c 
**2*d*x + 3*log(((a + b*x)**n*e)/(c + d*x)**n)*a*c*d**2*x**2 + log(((a + b 
*x)**n*e)/(c + d*x)**n)*a*d**3*x**3 + log(((a + b*x)**n*e)/(c + d*x)**n)*b 
*c**3*x + 3*log(((a + b*x)**n*e)/(c + d*x)**n)*b*c**2*d*x**2 + 3*log(((a + 
 b*x)**n*e)/(c + d*x)**n)*b*c*d**2*x**3 + log(((a + b*x)**n*e)/(c + d*x)** 
n)*b*d**3*x**4),x)*a*b**2*c*d**2*n + 2*int(x/(log(((a + b*x)**n*e)/(c + d* 
x)**n)*a*c**3 + 3*log(((a + b*x)**n*e)/(c + d*x)**n)*a*c**2*d*x + 3*log((( 
a + b*x)**n*e)/(c + d*x)**n)*a*c*d**2*x**2 + log(((a + b*x)**n*e)/(c + d*x 
)**n)*a*d**3*x**3 + log(((a + b*x)**n*e)/(c + d*x)**n)*b*c**3*x + 3*log((( 
a + b*x)**n*e)/(c + d*x)**n)*b*c**2*d*x**2 + 3*log(((a + b*x)**n*e)/(c + d 
*x)**n)*b*c*d**2*x**3 + log(((a + b*x)**n*e)/(c + d*x)**n)*b*d**3*x**4),x) 
*b**3*c**2*d*n + int(1/(log(((a + b*x)**n*e)/(c + d*x)**n)*a*c**3 + 3*log( 
((a + b*x)**n*e)/(c + d*x)**n)*a*c**2*d*x + 3*log(((a + b*x)**n*e)/(c + d* 
x)**n)*a*c*d**2*x**2 + log(((a + b*x)**n*e)/(c + d*x)**n)*a*d**3*x**3 +...