3.6.47 \(\int \frac {1}{(g+h x) (i+j x)^2 (a+b \log (c (d (e+f x)^p)^q))^2} \, dx\) [547]
Optimal. Leaf size=38 \[ \text {Int}\left (\frac {1}{(g+h x) (i+j x)^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2},x\right ) \]
[Out]
Unintegrable(1/(h*x+g)/(j*x+i)^2/(a+b*ln(c*(d*(f*x+e)^p)^q))^2,x)
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Rubi [A]
time = 0.22, 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)^2 \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)^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*}
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Mathematica [A]
time = 23.59, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}
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 = 0.35, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (h x +g \right ) \left (j x +i \right )^{2} \left (a +b \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right )\right )^{2}}\, dx\]
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 = 0.00, 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)^2/(a+b*log(c*(d*(f*x+e)^p)^q))^2,x, algorithm="maxima")
[Out]
(f*x + e)/(a*b*f*g*p*q - (a*b*f*h*j^2*p*q + (f*h*j^2*p*q^2*log(d) + f*h*j^2*p*q*log(c))*b^2)*x^3 + (f*g*p*q^2*
log(d) + f*g*p*q*log(c))*b^2 - ((g*j^2*p*q + 2*I*h*j*p*q)*a*b*f + ((g*j^2*p*q + 2*I*h*j*p*q)*f*log(c) + (g*j^2
*p*q^2 + 2*I*h*j*p*q^2)*f*log(d))*b^2)*x^2 - ((2*I*g*j*p*q - h*p*q)*a*b*f + ((2*I*g*j*p*q - h*p*q)*f*log(c) +
(2*I*g*j*p*q^2 - h*p*q^2)*f*log(d))*b^2)*x - (b^2*f*h*j^2*p*q*x^3 - b^2*f*g*p*q + (g*j^2*p*q + 2*I*h*j*p*q)*b^
2*f*x^2 + (2*I*g*j*p*q - h*p*q)*b^2*f*x)*log(((f*x + e)^p)^q)) - integrate((2*f*h*j*x^2 - I*f*g + (f*g*j + 3*h
*j*e)*x + (2*g*j + I*h)*e)/(-I*a*b*f*g^2*p*q + (a*b*f*h^2*j^3*p*q + (f*h^2*j^3*p*q^2*log(d) + f*h^2*j^3*p*q*lo
g(c))*b^2)*x^5 + ((2*g*h*j^3*p*q + 3*I*h^2*j^2*p*q)*a*b*f + ((2*g*h*j^3*p*q + 3*I*h^2*j^2*p*q)*f*log(c) + (2*g
*h*j^3*p*q^2 + 3*I*h^2*j^2*p*q^2)*f*log(d))*b^2)*x^4 + ((g^2*j^3*p*q + 6*I*g*h*j^2*p*q - 3*h^2*j*p*q)*a*b*f +
((g^2*j^3*p*q + 6*I*g*h*j^2*p*q - 3*h^2*j*p*q)*f*log(c) + (g^2*j^3*p*q^2 + 6*I*g*h*j^2*p*q^2 - 3*h^2*j*p*q^2)*
f*log(d))*b^2)*x^3 + (-I*f*g^2*p*q^2*log(d) - I*f*g^2*p*q*log(c))*b^2 + ((3*I*g^2*j^2*p*q - 6*g*h*j*p*q - I*h^
2*p*q)*a*b*f + ((3*I*g^2*j^2*p*q - 6*g*h*j*p*q - I*h^2*p*q)*f*log(c) + (3*I*g^2*j^2*p*q^2 - 6*g*h*j*p*q^2 - I*
h^2*p*q^2)*f*log(d))*b^2)*x^2 - ((3*g^2*j*p*q + 2*I*g*h*p*q)*a*b*f + ((3*g^2*j*p*q + 2*I*g*h*p*q)*f*log(c) + (
3*g^2*j*p*q^2 + 2*I*g*h*p*q^2)*f*log(d))*b^2)*x + (b^2*f*h^2*j^3*p*q*x^5 + (2*g*h*j^3*p*q + 3*I*h^2*j^2*p*q)*b
^2*f*x^4 - I*b^2*f*g^2*p*q + (g^2*j^3*p*q + 6*I*g*h*j^2*p*q - 3*h^2*j*p*q)*b^2*f*x^3 + (3*I*g^2*j^2*p*q - 6*g*
h*j*p*q - I*h^2*p*q)*b^2*f*x^2 - (3*g^2*j*p*q + 2*I*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)^2/(a+b*log(c*(d*(f*x+e)^p)^q))^2,x, algorithm="fricas")
[Out]
-(f*x - (a*b*f*h*j^2*p*q*x^3 - a*b*f*g*p*q + (a*b*f*g*j^2 + 2*I*a*b*f*h*j)*p*q*x^2 + (2*I*a*b*f*g*j - a*b*f*h)
*p*q*x + (b^2*f*h*j^2*p^2*q^2*x^3 - b^2*f*g*p^2*q^2 + (b^2*f*g*j^2 + 2*I*b^2*f*h*j)*p^2*q^2*x^2 + (2*I*b^2*f*g
*j - b^2*f*h)*p^2*q^2*x)*log(f*x + e) + (b^2*f*h*j^2*p*q*x^3 - b^2*f*g*p*q + (b^2*f*g*j^2 + 2*I*b^2*f*h*j)*p*q
*x^2 + (2*I*b^2*f*g*j - b^2*f*h)*p*q*x)*log(c) + (b^2*f*h*j^2*p*q^2*x^3 - b^2*f*g*p*q^2 + (b^2*f*g*j^2 + 2*I*b
^2*f*h*j)*p*q^2*x^2 + (2*I*b^2*f*g*j - b^2*f*h)*p*q^2*x)*log(d))*integral(-(2*f*h*j*x^2 + f*g*j*x - I*f*g + (3
*h*j*x + 2*g*j + I*h)*e)/(a*b*f*h^2*j^3*p*q*x^5 - I*a*b*f*g^2*p*q + (2*a*b*f*g*h*j^3 + 3*I*a*b*f*h^2*j^2)*p*q*
x^4 + (a*b*f*g^2*j^3 + 6*I*a*b*f*g*h*j^2 - 3*a*b*f*h^2*j)*p*q*x^3 + (3*I*a*b*f*g^2*j^2 - 6*a*b*f*g*h*j - I*a*b
*f*h^2)*p*q*x^2 - (3*a*b*f*g^2*j + 2*I*a*b*f*g*h)*p*q*x + (b^2*f*h^2*j^3*p^2*q^2*x^5 - I*b^2*f*g^2*p^2*q^2 + (
2*b^2*f*g*h*j^3 + 3*I*b^2*f*h^2*j^2)*p^2*q^2*x^4 + (b^2*f*g^2*j^3 + 6*I*b^2*f*g*h*j^2 - 3*b^2*f*h^2*j)*p^2*q^2
*x^3 + (3*I*b^2*f*g^2*j^2 - 6*b^2*f*g*h*j - I*b^2*f*h^2)*p^2*q^2*x^2 - (3*b^2*f*g^2*j + 2*I*b^2*f*g*h)*p^2*q^2
*x)*log(f*x + e) + (b^2*f*h^2*j^3*p*q*x^5 - I*b^2*f*g^2*p*q + (2*b^2*f*g*h*j^3 + 3*I*b^2*f*h^2*j^2)*p*q*x^4 +
(b^2*f*g^2*j^3 + 6*I*b^2*f*g*h*j^2 - 3*b^2*f*h^2*j)*p*q*x^3 + (3*I*b^2*f*g^2*j^2 - 6*b^2*f*g*h*j - I*b^2*f*h^2
)*p*q*x^2 - (3*b^2*f*g^2*j + 2*I*b^2*f*g*h)*p*q*x)*log(c) + (b^2*f*h^2*j^3*p*q^2*x^5 - I*b^2*f*g^2*p*q^2 + (2*
b^2*f*g*h*j^3 + 3*I*b^2*f*h^2*j^2)*p*q^2*x^4 + (b^2*f*g^2*j^3 + 6*I*b^2*f*g*h*j^2 - 3*b^2*f*h^2*j)*p*q^2*x^3 +
(3*I*b^2*f*g^2*j^2 - 6*b^2*f*g*h*j - I*b^2*f*h^2)*p*q^2*x^2 - (3*b^2*f*g^2*j + 2*I*b^2*f*g*h)*p*q^2*x)*log(d)
), x) + e)/(a*b*f*h*j^2*p*q*x^3 - a*b*f*g*p*q + (a*b*f*g*j^2 + 2*I*a*b*f*h*j)*p*q*x^2 + (2*I*a*b*f*g*j - a*b*f
*h)*p*q*x + (b^2*f*h*j^2*p^2*q^2*x^3 - b^2*f*g*p^2*q^2 + (b^2*f*g*j^2 + 2*I*b^2*f*h*j)*p^2*q^2*x^2 + (2*I*b^2*
f*g*j - b^2*f*h)*p^2*q^2*x)*log(f*x + e) + (b^2*f*h*j^2*p*q*x^3 - b^2*f*g*p*q + (b^2*f*g*j^2 + 2*I*b^2*f*h*j)*
p*q*x^2 + (2*I*b^2*f*g*j - b^2*f*h)*p*q*x)*log(c) + (b^2*f*h*j^2*p*q^2*x^3 - b^2*f*g*p*q^2 + (b^2*f*g*j^2 + 2*
I*b^2*f*h*j)*p*q^2*x^2 + (2*I*b^2*f*g*j - 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 )^{2}}\, dx \end {gather*}
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]
Integral(1/((a + b*log(c*(d*(e + f*x)**p)**q))**2*(g + h*x)*(i + j*x)**2), 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)^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)
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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 )}^2\,{\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)^2*(a + b*log(c*(d*(e + f*x)^p)^q))^2),x)
[Out]
int(1/((g + h*x)*(i + j*x)^2*(a + b*log(c*(d*(e + f*x)^p)^q))^2), x)
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