3.609 \(\int F^{f (a+b \log (c (d+e x)^n))^2} (g+h x)^m \, dx\)

Optimal. Leaf size=31 \[ \text {Int}\left ((g+h x)^m F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2},x\right ) \]

[Out]

Unintegrable(F^(f*(a+b*ln(c*(e*x+d)^n))^2)*(h*x+g)^m,x)

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Rubi [A]  time = 0.09, 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 = {} \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (g+h x)^m \, dx \]

Verification is Not applicable to the result.

[In]

Int[F^(f*(a + b*Log[c*(d + e*x)^n])^2)*(g + h*x)^m,x]

[Out]

Defer[Int][F^(f*(a + b*Log[c*(d + e*x)^n])^2)*(g + h*x)^m, x]

Rubi steps

\begin {align*} \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (g+h x)^m \, dx &=\int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (g+h x)^m \, dx\\ \end {align*}

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Mathematica [A]  time = 1.42, size = 0, normalized size = 0.00 \[ \int F^{f \left (a+b \log \left (c (d+e x)^n\right )\right )^2} (g+h x)^m \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[F^(f*(a + b*Log[c*(d + e*x)^n])^2)*(g + h*x)^m,x]

[Out]

Integrate[F^(f*(a + b*Log[c*(d + e*x)^n])^2)*(g + h*x)^m, x]

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fricas [A]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (h x + g\right )}^{m} F^{b^{2} f \log \left ({\left (e x + d\right )}^{n} c\right )^{2} + 2 \, a b f \log \left ({\left (e x + d\right )}^{n} c\right ) + a^{2} f}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(f*(a+b*log(c*(e*x+d)^n))^2)*(h*x+g)^m,x, algorithm="fricas")

[Out]

integral((h*x + g)^m*F^(b^2*f*log((e*x + d)^n*c)^2 + 2*a*b*f*log((e*x + d)^n*c) + a^2*f), x)

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giac [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (h x + g\right )}^{m} F^{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2} f}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(f*(a+b*log(c*(e*x+d)^n))^2)*(h*x+g)^m,x, algorithm="giac")

[Out]

integrate((h*x + g)^m*F^((b*log((e*x + d)^n*c) + a)^2*f), x)

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maple [A]  time = 1.02, size = 0, normalized size = 0.00 \[ \int F^{\left (b \ln \left (c \left (e x +d \right )^{n}\right )+a \right )^{2} f} \left (h x +g \right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^((b*ln(c*(e*x+d)^n)+a)^2*f)*(h*x+g)^m,x)

[Out]

int(F^((b*ln(c*(e*x+d)^n)+a)^2*f)*(h*x+g)^m,x)

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maxima [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (h x + g\right )}^{m} F^{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2} f}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(f*(a+b*log(c*(e*x+d)^n))^2)*(h*x+g)^m,x, algorithm="maxima")

[Out]

integrate((h*x + g)^m*F^((b*log((e*x + d)^n*c) + a)^2*f), x)

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mupad [A]  time = 0.00, size = -1, normalized size = -0.03 \[ \int {\mathrm {e}}^{f\,\ln \relax (F)\,{\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}^2}\,{\left (g+h\,x\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(f*(a + b*log(c*(d + e*x)^n))^2)*(g + h*x)^m,x)

[Out]

int(exp(f*log(F)*(a + b*log(c*(d + e*x)^n))^2)*(g + h*x)^m, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(f*(a+b*ln(c*(e*x+d)**n))**2)*(h*x+g)**m,x)

[Out]

Timed out

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