\(\int \frac {g+h x}{a+b \log (c (d (e+f x)^p)^q)} \, dx\) [465]

Optimal result

Integrand size = 26, antiderivative size = 179 \[ \int \frac {g+h x}{a+b \log \left (c \left (d (e+f x)^p\right )^q\right )} \, dx=\frac {e^{-\frac {a}{b p q}} (f g-e h) (e+f x) \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac {1}{p q}} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{b p q}\right )}{b f^2 p q}+\frac {e^{-\frac {2 a}{b p q}} h (e+f x)^2 \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac {2}{p q}} \operatorname {ExpIntegralEi}\left (\frac {2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{b p q}\right )}{b f^2 p q} \] Output:

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.92 \[ \int \frac {g+h x}{a+b \log \left (c \left (d (e+f x)^p\right )^q\right )} \, dx=\frac {e^{-\frac {2 a}{b p q}} (e+f x) \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac {2}{p q}} \left (e^{\frac {a}{b p q}} (f g-e h) \left (c \left (d (e+f x)^p\right )^q\right )^{\frac {1}{p q}} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{b p q}\right )+h (e+f x) \operatorname {ExpIntegralEi}\left (\frac {2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{b p q}\right )\right )}{b f^2 p q} \] Input:

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

Output:

((e + f*x)*(E^(a/(b*p*q))*(f*g - e*h)*(c*(d*(e + f*x)^p)^q)^(1/(p*q))*ExpI 
ntegralEi[(a + b*Log[c*(d*(e + f*x)^p)^q])/(b*p*q)] + h*(e + f*x)*ExpInteg 
ralEi[(2*(a + b*Log[c*(d*(e + f*x)^p)^q]))/(b*p*q)]))/(b*E^((2*a)/(b*p*q)) 
*f^2*p*q*(c*(d*(e + f*x)^p)^q)^(2/(p*q)))
 

Rubi [A] (verified)

Time = 1.30 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {2895, 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.

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

\(\Big \downarrow \) 2895

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

\(\Big \downarrow \) 2846

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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]
 

rule 2895

Int[((a_.) + Log[(c_.)*((d_.)*((e_.) + (f_.)*(x_))^(m_.))^(n_)]*(b_.))^(p_. 
)*(u_.), x_Symbol] :> Subst[Int[u*(a + b*Log[c*d^n*(e + f*x)^(m*n)])^p, x], 
 c*d^n*(e + f*x)^(m*n), c*(d*(e + f*x)^m)^n] /; FreeQ[{a, b, c, d, e, f, m, 
 n, p}, x] &&  !IntegerQ[n] &&  !(EqQ[d, 1] && EqQ[m, 1]) && IntegralFreeQ[ 
IntHide[u*(a + b*Log[c*d^n*(e + f*x)^(m*n)])^p, x]]
 

Maple [F]

\[\int \frac {h x +g}{a +b \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right )}d x\]

Input:

int((h*x+g)/(a+b*ln(c*(d*(f*x+e)^p)^q)),x)
 

Output:

int((h*x+g)/(a+b*ln(c*(d*(f*x+e)^p)^q)),x)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.78 \[ \int \frac {g+h x}{a+b \log \left (c \left (d (e+f x)^p\right )^q\right )} \, dx=\frac {{\left ({\left (f g - e h\right )} e^{\left (\frac {b q \log \left (d\right ) + b \log \left (c\right ) + a}{b p q}\right )} \operatorname {log\_integral}\left ({\left (f x + e\right )} e^{\left (\frac {b q \log \left (d\right ) + b \log \left (c\right ) + a}{b p q}\right )}\right ) + h \operatorname {log\_integral}\left ({\left (f^{2} x^{2} + 2 \, e f x + e^{2}\right )} e^{\left (\frac {2 \, {\left (b q \log \left (d\right ) + b \log \left (c\right ) + a\right )}}{b p q}\right )}\right )\right )} e^{\left (-\frac {2 \, {\left (b q \log \left (d\right ) + b \log \left (c\right ) + a\right )}}{b p q}\right )}}{b f^{2} p q} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.39 \[ \int \frac {g+h x}{a+b \log \left (c \left (d (e+f x)^p\right )^q\right )} \, dx=\frac {g {\rm Ei}\left (\frac {\log \left (d\right )}{p} + \frac {\log \left (c\right )}{p q} + \frac {a}{b p q} + \log \left (f x + e\right )\right ) e^{\left (-\frac {a}{b p q}\right )}}{b c^{\frac {1}{p q}} d^{\left (\frac {1}{p}\right )} f p q} - \frac {e h {\rm Ei}\left (\frac {\log \left (d\right )}{p} + \frac {\log \left (c\right )}{p q} + \frac {a}{b p q} + \log \left (f x + e\right )\right ) e^{\left (-\frac {a}{b p q}\right )}}{b c^{\frac {1}{p q}} d^{\left (\frac {1}{p}\right )} f^{2} p q} + \frac {h {\rm Ei}\left (\frac {2 \, \log \left (d\right )}{p} + \frac {2 \, \log \left (c\right )}{p q} + \frac {2 \, a}{b p q} + 2 \, \log \left (f x + e\right )\right ) e^{\left (-\frac {2 \, a}{b p q}\right )}}{b c^{\frac {2}{p q}} d^{\frac {2}{p}} f^{2} p q} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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