3.6.0 ∫ 1 _________________________ (g+hx)(i+jx)(a+b log(c(d(e+fx)p)q))2 dx [546]

Integrand size = 35, antiderivative size = 35 \[ \int \frac {1}{(g+h x) (i+j x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2} \, dx=\text {Int}\left (\frac {1}{(g+h x) (i+j x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2},x\right ) \]

Unintegrable(1/(h*x+g)/(j*x+i)/(a+b*ln(c*(d*(f*x+e)^p)^q))^2,x)

Rubi [N/A]

Time = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 0, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {1}{(g+h x) (i+j x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2} \, dx=\int \frac {1}{(g+h x) (i+j x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2} \, dx \]

Int[1/((g + h*x)*(i + j*x)*(a + b*Log[c*(d*(e + f*x)^p)^q])^2),x]

Defer[Int][1/((g + h*x)*(i + j*x)*(a + b*Log[c*(d*(e + f*x)^p)^q])^2), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{(g+h x) (i+j x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2} \, dx \\ \end{align*}

Mathematica [N/A]

Time = 13.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(g+h x) (i+j x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2} \, dx=\int \frac {1}{(g+h x) (i+j x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2} \, dx \]

Integrate[1/((g + h*x)*(i + j*x)*(a + b*Log[c*(d*(e + f*x)^p)^q])^2),x]

Integrate[1/((g + h*x)*(i + j*x)*(a + b*Log[c*(d*(e + f*x)^p)^q])^2), x]

Maple [N/A]

Time = 0.32 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00

\[\int \frac {1}{\left (h x +g \right ) \left (j x +i \right ) {\left (a +b \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right )\right )}^{2}}d x\]

int(1/(h*x+g)/(j*x+i)/(a+b*ln(c*(d*(f*x+e)^p)^q))^2,x)

Fricas [N/A]

Time = 0.33 (sec) , antiderivative size = 126, normalized size of antiderivative = 3.60 \[ \int \frac {1}{(g+h x) (i+j x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2} \, dx=\int { \frac {1}{{\left (h x + g\right )} {\left (j x + i\right )} {\left (b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + a\right )}^{2}} \,d x } \]

integrate(1/(h*x+g)/(j*x+i)/(a+b*log(c*(d*(f*x+e)^p)^q))^2,x, algorithm="fricas")

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

+ (a^2*h*i + a^2*g*j)*x + 2*(a*b*h*j*x^2 + a*b*g*i + (a*b*h*i + a*b*g*j)*x)*log(((f*x + e)^p*d)^q*c)), x)

Sympy [N/A]

Time = 16.73 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {1}{(g+h x) (i+j x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2} \, dx=\int \frac {1}{\left (a + b \log {\left (c \left (d \left (e + f x\right )^{p}\right )^{q} \right )}\right )^{2} \left (g + h x\right ) \left (i + j x\right )}\, dx \]

integrate(1/(h*x+g)/(j*x+i)/(a+b*ln(c*(d*(f*x+e)**p)**q))**2,x)

Integral(1/((a + b*log(c*(d*(e + f*x)**p)**q))**2*(g + h*x)*(i + j*x)), x)

Maxima [N/A]

Time = 1.27 (sec) , antiderivative size = 701, normalized size of antiderivative = 20.03 \[ \int \frac {1}{(g+h x) (i+j x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2} \, dx=\int { \frac {1}{{\left (h x + g\right )} {\left (j x + i\right )} {\left (b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + a\right )}^{2}} \,d x } \]

integrate(1/(h*x+g)/(j*x+i)/(a+b*log(c*(d*(f*x+e)^p)^q))^2,x, algorithm="maxima")

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

+ f*h*j*p*q*log(c))*b^2)*x^2 + ((h*i*p*q + g*j*p*q)*a*b*f + ((h*i*p*q + g*j*p*q)*f*log(c) + (h*i*p*q^2 + g*j*

p*q^2)*f*log(d))*b^2)*x + (b^2*f*h*j*p*q*x^2 + b^2*f*g*i*p*q + (h*i*p*q + g*j*p*q)*b^2*f*x)*log(((f*x + e)^p)^

q)) - integrate((f*h*j*x^2 + 2*e*h*j*x - f*g*i + (h*i + g*j)*e)/(a*b*f*g^2*i^2*p*q + (a*b*f*h^2*j^2*p*q + (f*h

^2*j^2*p*q^2*log(d) + f*h^2*j^2*p*q*log(c))*b^2)*x^4 + 2*((h^2*i*j*p*q + g*h*j^2*p*q)*a*b*f + ((h^2*i*j*p*q +

g*h*j^2*p*q)*f*log(c) + (h^2*i*j*p*q^2 + g*h*j^2*p*q^2)*f*log(d))*b^2)*x^3 + (f*g^2*i^2*p*q^2*log(d) + f*g^2*i

^2*p*q*log(c))*b^2 + ((h^2*i^2*p*q + 4*g*h*i*j*p*q + g^2*j^2*p*q)*a*b*f + ((h^2*i^2*p*q + 4*g*h*i*j*p*q + g^2*

j^2*p*q)*f*log(c) + (h^2*i^2*p*q^2 + 4*g*h*i*j*p*q^2 + g^2*j^2*p*q^2)*f*log(d))*b^2)*x^2 + 2*((g*h*i^2*p*q + g

^2*i*j*p*q)*a*b*f + ((g*h*i^2*p*q + g^2*i*j*p*q)*f*log(c) + (g*h*i^2*p*q^2 + g^2*i*j*p*q^2)*f*log(d))*b^2)*x +

(b^2*f*h^2*j^2*p*q*x^4 + b^2*f*g^2*i^2*p*q + 2*(h^2*i*j*p*q + g*h*j^2*p*q)*b^2*f*x^3 + (h^2*i^2*p*q + 4*g*h*i

*j*p*q + g^2*j^2*p*q)*b^2*f*x^2 + 2*(g*h*i^2*p*q + g^2*i*j*p*q)*b^2*f*x)*log(((f*x + e)^p)^q)), x)

Giac [N/A]

Time = 0.38 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(g+h x) (i+j x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2} \, dx=\int { \frac {1}{{\left (h x + g\right )} {\left (j x + i\right )} {\left (b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + a\right )}^{2}} \,d x } \]

integrate(1/(h*x+g)/(j*x+i)/(a+b*log(c*(d*(f*x+e)^p)^q))^2,x, algorithm="giac")

integrate(1/((h*x + g)*(j*x + i)*(b*log(((f*x + e)^p*d)^q*c) + a)^2), x)

Mupad [N/A]

Time = 1.35 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(g+h x) (i+j x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2} \, dx=\int \frac {1}{\left (g+h\,x\right )\,\left (i+j\,x\right )\,{\left (a+b\,\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^p\right )}^q\right )\right )}^2} \,d x \]

int(1/((g + h*x)*(i + j*x)*(a + b*log(c*(d*(e + f*x)^p)^q))^2),x)

int(1/((g + h*x)*(i + j*x)*(a + b*log(c*(d*(e + f*x)^p)^q))^2), x)

\(\int \frac {1}{(g+h x) (i+j x) (a+b \log (c (d (e+f x)^p)^q))^2} \, dx\) [546]

Optimal result