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

Optimal. Leaf size=37 \[ \text{Unintegrable}\left (\frac{1}{(g+h x) (i+j x)^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2},x\right ) \]

[Out]

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

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Rubi [A]  time = 0.334712, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{1}{(g+h x) (i+j x)^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin{align*} \int \frac{1}{(g+h x) (547+j x)^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2} \, dx &=\int \frac{1}{(g+h x) (547+j x)^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2} \, dx\\ \end{align*}

Mathematica [A]  time = 48.1609, size = 0, normalized size = 0. \[ \int \frac{1}{(g+h x) (i+j x)^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [A]  time = 1.008, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( hx+g \right ) \left ( jx+i \right ) ^{2} \left ( a+b\ln \left ( c \left ( d \left ( fx+e \right ) ^{p} \right ) ^{q} \right ) \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(f*x + e)/(a*b*f*g*i^2*p*q + (a*b*f*h*j^2*p*q + (f*h*j^2*p*q*log(c) + f*h*j^2*p*q*log(d^q))*b^2)*x^3 + (f*g*i
^2*p*q*log(c) + f*g*i^2*p*q*log(d^q))*b^2 + ((2*h*i*j*q + g*j^2*q)*a*b*f*p + ((2*h*i*j*q + g*j^2*q)*f*p*log(c)
+ (2*h*i*j*q + g*j^2*q)*f*p*log(d^q))*b^2)*x^2 + ((h*i^2*q + 2*g*i*j*q)*a*b*f*p + ((h*i^2*q + 2*g*i*j*q)*f*p*
log(c) + (h*i^2*q + 2*g*i*j*q)*f*p*log(d^q))*b^2)*x + (b^2*f*h*j^2*p*q*x^3 + b^2*f*g*i^2*p*q + (2*h*i*j*q + g*
j^2*q)*b^2*f*p*x^2 + (h*i^2*q + 2*g*i*j*q)*b^2*f*p*x)*log(((f*x + e)^p)^q)) - integrate((2*f*h*j*x^2 - f*g*i +
(h*i + 2*g*j)*e + (f*g*j + 3*e*h*j)*x)/(a*b*f*g^2*i^3*p*q + (a*b*f*h^2*j^3*p*q + (f*h^2*j^3*p*q*log(c) + f*h^
2*j^3*p*q*log(d^q))*b^2)*x^5 + ((3*h^2*i*j^2*q + 2*g*h*j^3*q)*a*b*f*p + ((3*h^2*i*j^2*q + 2*g*h*j^3*q)*f*p*log
(c) + (3*h^2*i*j^2*q + 2*g*h*j^3*q)*f*p*log(d^q))*b^2)*x^4 + ((3*h^2*i^2*j*q + 6*g*h*i*j^2*q + g^2*j^3*q)*a*b*
f*p + ((3*h^2*i^2*j*q + 6*g*h*i*j^2*q + g^2*j^3*q)*f*p*log(c) + (3*h^2*i^2*j*q + 6*g*h*i*j^2*q + g^2*j^3*q)*f*
p*log(d^q))*b^2)*x^3 + (f*g^2*i^3*p*q*log(c) + f*g^2*i^3*p*q*log(d^q))*b^2 + ((h^2*i^3*q + 6*g*h*i^2*j*q + 3*g
^2*i*j^2*q)*a*b*f*p + ((h^2*i^3*q + 6*g*h*i^2*j*q + 3*g^2*i*j^2*q)*f*p*log(c) + (h^2*i^3*q + 6*g*h*i^2*j*q + 3
*g^2*i*j^2*q)*f*p*log(d^q))*b^2)*x^2 + ((2*g*h*i^3*q + 3*g^2*i^2*j*q)*a*b*f*p + ((2*g*h*i^3*q + 3*g^2*i^2*j*q)
*f*p*log(c) + (2*g*h*i^3*q + 3*g^2*i^2*j*q)*f*p*log(d^q))*b^2)*x + (b^2*f*h^2*j^3*p*q*x^5 + b^2*f*g^2*i^3*p*q
+ (3*h^2*i*j^2*q + 2*g*h*j^3*q)*b^2*f*p*x^4 + (3*h^2*i^2*j*q + 6*g*h*i*j^2*q + g^2*j^3*q)*b^2*f*p*x^3 + (h^2*i
^3*q + 6*g*h*i^2*j*q + 3*g^2*i*j^2*q)*b^2*f*p*x^2 + (2*g*h*i^3*q + 3*g^2*i^2*j*q)*b^2*f*p*x)*log(((f*x + e)^p)
^q)), x)

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Fricas [A]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{a^{2} h j^{2} x^{3} + a^{2} g i^{2} +{\left (2 \, a^{2} h i j + a^{2} g j^{2}\right )} x^{2} +{\left (b^{2} h j^{2} x^{3} + b^{2} g i^{2} +{\left (2 \, b^{2} h i j + b^{2} g j^{2}\right )} x^{2} +{\left (b^{2} h i^{2} + 2 \, b^{2} g i j\right )} x\right )} \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right )^{2} +{\left (a^{2} h i^{2} + 2 \, a^{2} g i j\right )} x + 2 \,{\left (a b h j^{2} x^{3} + a b g i^{2} +{\left (2 \, a b h i j + a b g j^{2}\right )} x^{2} +{\left (a b h i^{2} + 2 \, a b g i j\right )} x\right )} \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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