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

Optimal. Leaf size=38 \[ \text {Int}\left (\frac {1}{(g+h x) (i+j x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2},x\right ) \]

[Out]

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

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Rubi [A]
time = 0.19, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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*} \int \frac {1}{(g+h x) (546+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) (546+j x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2} \, dx\\ \end {align*}

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Mathematica [A]
time = 16.55, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [A]
time = 0.43, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (h x +g \right ) \left (j x +i \right ) \left (a +b \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right )\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(f*x + e)/(I*a*b*f*g*p*q + (I*f*g*p*q^2*log(d) + I*f*g*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 + ((g*j*p*q + I*h*p*q)*a*b*f + ((g*j*p*q + I*h*p*q)*f*log(c) + (g*j*p*q^2 + I*h*
p*q^2)*f*log(d))*b^2)*x + (b^2*f*h*j*p*q*x^2 + I*b^2*f*g*p*q + (g*j*p*q + I*h*p*q)*b^2*f*x)*log(((f*x + e)^p)^
q)) - integrate(-(f*h*j*x^2 + 2*h*j*x*e - I*f*g + (g*j + I*h)*e)/(a*b*f*g^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*((g*h*j^2*p*q + I*h^2*j*p*q)*a*b*f + ((g*h*j^2*p*q + I*h
^2*j*p*q)*f*log(c) + (g*h*j^2*p*q^2 + I*h^2*j*p*q^2)*f*log(d))*b^2)*x^3 + (f*g^2*p*q^2*log(d) + f*g^2*p*q*log(
c))*b^2 - ((g^2*j^2*p*q + 4*I*g*h*j*p*q - h^2*p*q)*a*b*f + ((g^2*j^2*p*q + 4*I*g*h*j*p*q - h^2*p*q)*f*log(c) +
(g^2*j^2*p*q^2 + 4*I*g*h*j*p*q^2 - h^2*p*q^2)*f*log(d))*b^2)*x^2 + 2*((-I*g^2*j*p*q + g*h*p*q)*a*b*f + ((-I*g
^2*j*p*q + g*h*p*q)*f*log(c) + (-I*g^2*j*p*q^2 + g*h*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*p*q + 2*(g*h*j^2*p*q + I*h^2*j*p*q)*b^2*f*x^3 + (g^2*j^2*p*q + 4*I*g*h*j*p*q - h^2*p*q)*b^2*f*x^2 - 2*(-I*
g^2*j*p*q + g*h*p*q)*b^2*f*x)*log(((f*x + e)^p)^q)), x)

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Fricas [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(f*x - (a*b*f*h*j*p*q*x^2 + I*a*b*f*g*p*q + (a*b*f*g*j + I*a*b*f*h)*p*q*x + (b^2*f*h*j*p^2*q^2*x^2 + I*b^2*f*
g*p^2*q^2 + (b^2*f*g*j + I*b^2*f*h)*p^2*q^2*x)*log(f*x + e) + (b^2*f*h*j*p*q*x^2 + I*b^2*f*g*p*q + (b^2*f*g*j
+ I*b^2*f*h)*p*q*x)*log(c) + (b^2*f*h*j*p*q^2*x^2 + I*b^2*f*g*p*q^2 + (b^2*f*g*j + I*b^2*f*h)*p*q^2*x)*log(d))
*integral(-(f*h*j*x^2 - I*f*g + (2*h*j*x + g*j + I*h)*e)/(a*b*f*h^2*j^2*p*q*x^4 - a*b*f*g^2*p*q + 2*(a*b*f*g*h
*j^2 + I*a*b*f*h^2*j)*p*q*x^3 + (a*b*f*g^2*j^2 + 4*I*a*b*f*g*h*j - a*b*f*h^2)*p*q*x^2 - 2*(-I*a*b*f*g^2*j + a*
b*f*g*h)*p*q*x + (b^2*f*h^2*j^2*p^2*q^2*x^4 - b^2*f*g^2*p^2*q^2 + 2*(b^2*f*g*h*j^2 + I*b^2*f*h^2*j)*p^2*q^2*x^
3 + (b^2*f*g^2*j^2 + 4*I*b^2*f*g*h*j - b^2*f*h^2)*p^2*q^2*x^2 - 2*(-I*b^2*f*g^2*j + b^2*f*g*h)*p^2*q^2*x)*log(
f*x + e) + (b^2*f*h^2*j^2*p*q*x^4 - b^2*f*g^2*p*q + 2*(b^2*f*g*h*j^2 + I*b^2*f*h^2*j)*p*q*x^3 + (b^2*f*g^2*j^2
+ 4*I*b^2*f*g*h*j - b^2*f*h^2)*p*q*x^2 - 2*(-I*b^2*f*g^2*j + b^2*f*g*h)*p*q*x)*log(c) + (b^2*f*h^2*j^2*p*q^2*
x^4 - b^2*f*g^2*p*q^2 + 2*(b^2*f*g*h*j^2 + I*b^2*f*h^2*j)*p*q^2*x^3 + (b^2*f*g^2*j^2 + 4*I*b^2*f*g*h*j - b^2*f
*h^2)*p*q^2*x^2 - 2*(-I*b^2*f*g^2*j + b^2*f*g*h)*p*q^2*x)*log(d)), x) + e)/(a*b*f*h*j*p*q*x^2 + I*a*b*f*g*p*q
+ (a*b*f*g*j + I*a*b*f*h)*p*q*x + (b^2*f*h*j*p^2*q^2*x^2 + I*b^2*f*g*p^2*q^2 + (b^2*f*g*j + I*b^2*f*h)*p^2*q^2
*x)*log(f*x + e) + (b^2*f*h*j*p*q*x^2 + I*b^2*f*g*p*q + (b^2*f*g*j + I*b^2*f*h)*p*q*x)*log(c) + (b^2*f*h*j*p*q
^2*x^2 + I*b^2*f*g*p*q^2 + (b^2*f*g*j + I*b^2*f*h)*p*q^2*x)*log(d))

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Sympy [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [A]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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