\(\int \frac {1}{(g+h x) (i+j x)^2 (a+b \log (c (d (e+f x)^p)^q))} \, dx\) [543]

Optimal result

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

[Out]

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

Rubi [N/A]

Not integrable

Time = 0.22 (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 = {} \[ \int \frac {1}{(g+h x) (i+j x)^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )} \, dx=\int \frac {1}{(g+h x) (i+j x)^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )} \, dx \]

[In]

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

[Out]

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

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

Mathematica [N/A]

Not integrable

Time = 1.97 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(g+h x) (i+j x)^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )} \, dx=\int \frac {1}{(g+h x) (i+j x)^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )} \, dx \]

[In]

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

[Out]

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

Maple [N/A]

Not integrable

Time = 0.31 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00

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

[In]

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

[Out]

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

Fricas [N/A]

Not integrable

Time = 0.29 (sec) , antiderivative size = 115, normalized size of antiderivative = 3.29 \[ \int \frac {1}{(g+h x) (i+j x)^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )} \, dx=\int { \frac {1}{{\left (h x + g\right )} {\left (j x + i\right )}^{2} {\left (b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + a\right )}} \,d x } \]

[In]

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

[Out]

integral(1/(a*h*j^2*x^3 + a*g*i^2 + (2*a*h*i*j + a*g*j^2)*x^2 + (a*h*i^2 + 2*a*g*i*j)*x + (b*h*j^2*x^3 + b*g*i
^2 + (2*b*h*i*j + b*g*j^2)*x^2 + (b*h*i^2 + 2*b*g*i*j)*x)*log(((f*x + e)^p*d)^q*c)), x)

Sympy [N/A]

Not integrable

Time = 12.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {1}{(g+h x) (i+j x)^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )} \, dx=\int \frac {1}{\left (a + b \log {\left (c \left (d \left (e + f x\right )^{p}\right )^{q} \right )}\right ) \left (g + h x\right ) \left (i + j x\right )^{2}}\, dx \]

[In]

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

[Out]

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

Maxima [N/A]

Not integrable

Time = 0.91 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(g+h x) (i+j x)^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )} \, dx=\int { \frac {1}{{\left (h x + g\right )} {\left (j x + i\right )}^{2} {\left (b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + a\right )}} \,d x } \]

[In]

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

[Out]

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

Giac [N/A]

Not integrable

Time = 0.31 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(g+h x) (i+j x)^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )} \, dx=\int { \frac {1}{{\left (h x + g\right )} {\left (j x + i\right )}^{2} {\left (b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + a\right )}} \,d x } \]

[In]

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

[Out]

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

Mupad [N/A]

Not integrable

Time = 1.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(g+h x) (i+j x)^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )} \, dx=\int \frac {1}{\left (g+h\,x\right )\,{\left (i+j\,x\right )}^2\,\left (a+b\,\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^p\right )}^q\right )\right )} \,d x \]

[In]

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

[Out]

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