\(\int \frac {f+g x}{a+b \log (c (d+e x)^n)} \, dx\) [90]

Optimal result

Integrand size = 22, antiderivative size = 139 \[ \int \frac {f+g x}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {e^{-\frac {a}{b n}} (e f-d g) (d+e x) \left (c (d+e x)^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b e^2 n}+\frac {e^{-\frac {2 a}{b n}} g (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \operatorname {ExpIntegralEi}\left (\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )}{b e^2 n} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.91 \[ \int \frac {f+g x}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {e^{-\frac {2 a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-2/n} \left (e^{\frac {a}{b n}} (e f-d g) \left (c (d+e x)^n\right )^{\frac {1}{n}} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )+g (d+e x) \operatorname {ExpIntegralEi}\left (\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )\right )}{b e^2 n} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2846, 2009}

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.

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

\(\Big \downarrow \) 2846

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {e^{-\frac {a}{b n}} (d+e x) (e f-d g) \left (c (d+e x)^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b e^2 n}+\frac {g e^{-\frac {2 a}{b n}} (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \operatorname {ExpIntegralEi}\left (\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )}{b e^2 n}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2846

Int[((f_.) + (g_.)*(x_))^(q_.)/((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.) 
]*(b_.)), x_Symbol] :> Int[ExpandIntegrand[(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]
 

Maple [C] (warning: unable to verify)

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

Time = 1.97 (sec) , antiderivative size = 937, normalized size of antiderivative = 6.74

method	result	size
risch	\(\text {Expression too large to display}\)	\(937\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.76 \[ \int \frac {f+g x}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {{\left ({\left (e f - d g\right )} e^{\left (\frac {b \log \left (c\right ) + a}{b n}\right )} \operatorname {log\_integral}\left ({\left (e x + d\right )} e^{\left (\frac {b \log \left (c\right ) + a}{b n}\right )}\right ) + g \operatorname {log\_integral}\left ({\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )} e^{\left (\frac {2 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )}\right )\right )} e^{\left (-\frac {2 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )}}{b e^{2} n} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.14 \[ \int \frac {f+g x}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {f {\rm Ei}\left (\frac {\log \left (c\right )}{n} + \frac {a}{b n} + \log \left (e x + d\right )\right ) e^{\left (-\frac {a}{b n}\right )}}{b c^{\left (\frac {1}{n}\right )} e n} - \frac {d g {\rm Ei}\left (\frac {\log \left (c\right )}{n} + \frac {a}{b n} + \log \left (e x + d\right )\right ) e^{\left (-\frac {a}{b n}\right )}}{b c^{\left (\frac {1}{n}\right )} e^{2} n} + \frac {g {\rm Ei}\left (\frac {2 \, \log \left (c\right )}{n} + \frac {2 \, a}{b n} + 2 \, \log \left (e x + d\right )\right ) e^{\left (-\frac {2 \, a}{b n}\right )}}{b c^{\frac {2}{n}} e^{2} n} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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