3.5.58 \(\int \frac {1}{(g+h x) (a+b \log (c (d (e+f x)^p)^q))^3} \, dx\) [458]

Optimal. Leaf size=31 \[ \text {Int}\left (\frac {1}{(g+h x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^3},x\right ) \]

[Out]

Unintegrable(1/(h*x+g)/(a+b*ln(c*(d*(f*x+e)^p)^q))^3,x)

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Rubi [A]
time = 0.04, 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) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[1/((g + h*x)*(a + b*Log[c*(d*(e + f*x)^p)^q])^3),x]

[Out]

Defer[Int][1/((g + h*x)*(a + b*Log[c*(d*(e + f*x)^p)^q])^3), x]

Rubi steps

\begin {align*} \int \frac {1}{(g+h x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^3} \, dx &=\int \frac {1}{(g+h x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^3} \, dx\\ \end {align*}

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Mathematica [A]
time = 0.76, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{(g+h x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[1/((g + h*x)*(a + b*Log[c*(d*(e + f*x)^p)^q])^3),x]

[Out]

Integrate[1/((g + h*x)*(a + b*Log[c*(d*(e + f*x)^p)^q])^3), x]

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Maple [A]
time = 0.20, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (h x +g \right ) \left (a +b \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right )\right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(h*x+g)/(a+b*ln(c*(d*(f*x+e)^p)^q))^3,x)

[Out]

int(1/(h*x+g)/(a+b*ln(c*(d*(f*x+e)^p)^q))^3,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)/(a+b*log(c*(d*(f*x+e)^p)^q))^3,x, algorithm="maxima")

[Out]

-1/2*(b*f^2*h*p*q*x^2 + (a*f^2*g + (f^2*g*p*q + f^2*g*q*log(d) + f^2*g*log(c))*b - (a*f*h - (f*h*p*q - f*h*q*l
og(d) - f*h*log(c))*b)*e)*x - ((h*q*log(d) + h*log(c))*b + a*h)*e^2 + (a*f*g + (f*g*p*q + f*g*q*log(d) + f*g*l
og(c))*b)*e + (b*f*g*e - b*h*e^2 + (b*f^2*g - b*f*h*e)*x)*log(((f*x + e)^p)^q))/(a^2*b^2*f^2*g^2*p^2*q^2 + 2*(
f^2*g^2*p^2*q^3*log(d) + f^2*g^2*p^2*q^2*log(c))*a*b^3 + (f^2*g^2*p^2*q^4*log(d)^2 + 2*f^2*g^2*p^2*q^3*log(c)*
log(d) + f^2*g^2*p^2*q^2*log(c)^2)*b^4 + (a^2*b^2*f^2*h^2*p^2*q^2 + 2*(f^2*h^2*p^2*q^3*log(d) + f^2*h^2*p^2*q^
2*log(c))*a*b^3 + (f^2*h^2*p^2*q^4*log(d)^2 + 2*f^2*h^2*p^2*q^3*log(c)*log(d) + f^2*h^2*p^2*q^2*log(c)^2)*b^4)
*x^2 + (b^4*f^2*h^2*p^2*q^2*x^2 + 2*b^4*f^2*g*h*p^2*q^2*x + b^4*f^2*g^2*p^2*q^2)*log(((f*x + e)^p)^q)^2 + 2*(a
^2*b^2*f^2*g*h*p^2*q^2 + 2*(f^2*g*h*p^2*q^3*log(d) + f^2*g*h*p^2*q^2*log(c))*a*b^3 + (f^2*g*h*p^2*q^4*log(d)^2
+ 2*f^2*g*h*p^2*q^3*log(c)*log(d) + f^2*g*h*p^2*q^2*log(c)^2)*b^4)*x + 2*(a*b^3*f^2*g^2*p^2*q^2 + (f^2*g^2*p^
2*q^3*log(d) + f^2*g^2*p^2*q^2*log(c))*b^4 + (a*b^3*f^2*h^2*p^2*q^2 + (f^2*h^2*p^2*q^3*log(d) + f^2*h^2*p^2*q^
2*log(c))*b^4)*x^2 + 2*(a*b^3*f^2*g*h*p^2*q^2 + (f^2*g*h*p^2*q^3*log(d) + f^2*g*h*p^2*q^2*log(c))*b^4)*x)*log(
((f*x + e)^p)^q)) + integrate(1/2*(f^2*g^2 - 3*f*g*h*e + 2*h^2*e^2 - (f^2*g*h - f*h^2*e)*x)/(a*b^2*f^2*g^3*p^2
*q^2 + (f^2*g^3*p^2*q^3*log(d) + f^2*g^3*p^2*q^2*log(c))*b^3 + (a*b^2*f^2*h^3*p^2*q^2 + (f^2*h^3*p^2*q^3*log(d
) + f^2*h^3*p^2*q^2*log(c))*b^3)*x^3 + 3*(a*b^2*f^2*g*h^2*p^2*q^2 + (f^2*g*h^2*p^2*q^3*log(d) + f^2*g*h^2*p^2*
q^2*log(c))*b^3)*x^2 + 3*(a*b^2*f^2*g^2*h*p^2*q^2 + (f^2*g^2*h*p^2*q^3*log(d) + f^2*g^2*h*p^2*q^2*log(c))*b^3)
*x + (b^3*f^2*h^3*p^2*q^2*x^3 + 3*b^3*f^2*g*h^2*p^2*q^2*x^2 + 3*b^3*f^2*g^2*h*p^2*q^2*x + b^3*f^2*g^3*p^2*q^2)
*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)/(a+b*log(c*(d*(f*x+e)^p)^q))^3,x, algorithm="fricas")

[Out]

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

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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 )^{3} \left (g + h x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(h*x+g)/(a+b*ln(c*(d*(f*x+e)**p)**q))**3,x)

[Out]

Integral(1/((a + b*log(c*(d*(e + f*x)**p)**q))**3*(g + h*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)/(a+b*log(c*(d*(f*x+e)^p)^q))^3,x, algorithm="giac")

[Out]

integrate(1/((h*x + g)*(b*log(((f*x + e)^p*d)^q*c) + a)^3), 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 (a+b\,\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^p\right )}^q\right )\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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