3.171 \(\int (f+g x)^2 (a+b \log (c (d+e x)^n))^n \, dx\)

Optimal. Leaf size=348 \[ \frac {g 2^{-n} e^{-\frac {2 a}{b n}} (d+e x)^2 (e f-d g) \left (c (d+e x)^n\right )^{-2/n} \left (a+b \log \left (c (d+e x)^n\right )\right )^n \left (-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )^{-n} \Gamma \left (n+1,-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )}{e^3}+\frac {e^{-\frac {a}{b n}} (d+e x) (e f-d g)^2 \left (c (d+e x)^n\right )^{-1/n} \left (a+b \log \left (c (d+e x)^n\right )\right )^n \left (-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )^{-n} \Gamma \left (n+1,-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{e^3}+\frac {g^2 3^{-n-1} e^{-\frac {3 a}{b n}} (d+e x)^3 \left (c (d+e x)^n\right )^{-3/n} \left (a+b \log \left (c (d+e x)^n\right )\right )^n \left (-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )^{-n} \Gamma \left (n+1,-\frac {3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )}{e^3} \]

[Out]

3^(-1-n)*g^2*(e*x+d)^3*GAMMA(1+n,-3*(a+b*ln(c*(e*x+d)^n))/b/n)*(a+b*ln(c*(e*x+d)^n))^n/e^3/exp(3*a/b/n)/((c*(e
*x+d)^n)^(3/n))/(((-a-b*ln(c*(e*x+d)^n))/b/n)^n)+g*(-d*g+e*f)*(e*x+d)^2*GAMMA(1+n,-2*(a+b*ln(c*(e*x+d)^n))/b/n
)*(a+b*ln(c*(e*x+d)^n))^n/(2^n)/e^3/exp(2*a/b/n)/((c*(e*x+d)^n)^(2/n))/(((-a-b*ln(c*(e*x+d)^n))/b/n)^n)+(-d*g+
e*f)^2*(e*x+d)*GAMMA(1+n,(-a-b*ln(c*(e*x+d)^n))/b/n)*(a+b*ln(c*(e*x+d)^n))^n/e^3/exp(a/b/n)/((c*(e*x+d)^n)^(1/
n))/(((-a-b*ln(c*(e*x+d)^n))/b/n)^n)

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Rubi [A]  time = 0.36, antiderivative size = 348, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2401, 2389, 2300, 2181, 2390, 2310} \[ \frac {g 2^{-n} e^{-\frac {2 a}{b n}} (d+e x)^2 (e f-d g) \left (c (d+e x)^n\right )^{-2/n} \left (a+b \log \left (c (d+e x)^n\right )\right )^n \left (-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )^{-n} \text {Gamma}\left (n+1,-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )}{e^3}+\frac {e^{-\frac {a}{b n}} (d+e x) (e f-d g)^2 \left (c (d+e x)^n\right )^{-1/n} \left (a+b \log \left (c (d+e x)^n\right )\right )^n \left (-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )^{-n} \text {Gamma}\left (n+1,-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{e^3}+\frac {g^2 3^{-n-1} e^{-\frac {3 a}{b n}} (d+e x)^3 \left (c (d+e x)^n\right )^{-3/n} \left (a+b \log \left (c (d+e x)^n\right )\right )^n \left (-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )^{-n} \text {Gamma}\left (n+1,-\frac {3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )}{e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3^(-1 - n)*g^2*(d + e*x)^3*Gamma[1 + n, (-3*(a + b*Log[c*(d + e*x)^n]))/(b*n)]*(a + b*Log[c*(d + e*x)^n])^n)/
(e^3*E^((3*a)/(b*n))*(c*(d + e*x)^n)^(3/n)*(-((a + b*Log[c*(d + e*x)^n])/(b*n)))^n) + (g*(e*f - d*g)*(d + e*x)
^2*Gamma[1 + n, (-2*(a + b*Log[c*(d + e*x)^n]))/(b*n)]*(a + b*Log[c*(d + e*x)^n])^n)/(2^n*e^3*E^((2*a)/(b*n))*
(c*(d + e*x)^n)^(2/n)*(-((a + b*Log[c*(d + e*x)^n])/(b*n)))^n) + ((e*f - d*g)^2*(d + e*x)*Gamma[1 + n, -((a +
b*Log[c*(d + e*x)^n])/(b*n))]*(a + b*Log[c*(d + e*x)^n])^n)/(e^3*E^(a/(b*n))*(c*(d + e*x)^n)^n^(-1)*(-((a + b*
Log[c*(d + e*x)^n])/(b*n)))^n)

Rule 2181

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(F^(g*(e - (c*f)/d))*(c +
d*x)^FracPart[m]*Gamma[m + 1, (-((f*g*Log[F])/d))*(c + d*x)])/(d*(-((f*g*Log[F])/d))^(IntPart[m] + 1)*(-((f*g*
Log[F]*(c + d*x))/d))^FracPart[m]), x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 2300

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[E^(x/n)*(a +
b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2310

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*n*(c*x^n
)^((m + 1)/n)), Subst[Int[E^(((m + 1)*x)/n)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}
, x]

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 2401

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Int[Exp
andIntegrand[(f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[
e*f - d*g, 0] && IGtQ[q, 0]

Rubi steps

\begin {align*} \int (f+g x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^n \, dx &=\int \left (\frac {(e f-d g)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^n}{e^2}+\frac {2 g (e f-d g) (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^n}{e^2}+\frac {g^2 (d+e x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^n}{e^2}\right ) \, dx\\ &=\frac {g^2 \int (d+e x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^n \, dx}{e^2}+\frac {(2 g (e f-d g)) \int (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^n \, dx}{e^2}+\frac {(e f-d g)^2 \int \left (a+b \log \left (c (d+e x)^n\right )\right )^n \, dx}{e^2}\\ &=\frac {g^2 \operatorname {Subst}\left (\int x^2 \left (a+b \log \left (c x^n\right )\right )^n \, dx,x,d+e x\right )}{e^3}+\frac {(2 g (e f-d g)) \operatorname {Subst}\left (\int x \left (a+b \log \left (c x^n\right )\right )^n \, dx,x,d+e x\right )}{e^3}+\frac {(e f-d g)^2 \operatorname {Subst}\left (\int \left (a+b \log \left (c x^n\right )\right )^n \, dx,x,d+e x\right )}{e^3}\\ &=\frac {\left (g^2 (d+e x)^3 \left (c (d+e x)^n\right )^{-3/n}\right ) \operatorname {Subst}\left (\int e^{\frac {3 x}{n}} (a+b x)^n \, dx,x,\log \left (c (d+e x)^n\right )\right )}{e^3 n}+\frac {\left (2 g (e f-d g) (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n}\right ) \operatorname {Subst}\left (\int e^{\frac {2 x}{n}} (a+b x)^n \, dx,x,\log \left (c (d+e x)^n\right )\right )}{e^3 n}+\frac {\left ((e f-d g)^2 (d+e x) \left (c (d+e x)^n\right )^{-1/n}\right ) \operatorname {Subst}\left (\int e^{\frac {x}{n}} (a+b x)^n \, dx,x,\log \left (c (d+e x)^n\right )\right )}{e^3 n}\\ &=\frac {3^{-1-n} e^{-\frac {3 a}{b n}} g^2 (d+e x)^3 \left (c (d+e x)^n\right )^{-3/n} \Gamma \left (1+n,-\frac {3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^n \left (-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )^{-n}}{e^3}+\frac {2^{-n} e^{-\frac {2 a}{b n}} g (e f-d g) (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \Gamma \left (1+n,-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^n \left (-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )^{-n}}{e^3}+\frac {e^{-\frac {a}{b n}} (e f-d g)^2 (d+e x) \left (c (d+e x)^n\right )^{-1/n} \Gamma \left (1+n,-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^n \left (-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )^{-n}}{e^3}\\ \end {align*}

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Mathematica [A]  time = 0.62, size = 262, normalized size = 0.75 \[ \frac {2^{-n} 3^{-n-1} e^{-\frac {3 a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-3/n} \left (a+b \log \left (c (d+e x)^n\right )\right )^n \left (-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )^{-n} \left (3^{n+1} e^{\frac {a}{b n}} (e f-d g) \left (c (d+e x)^n\right )^{\frac {1}{n}} \left (2^n e^{\frac {a}{b n}} (e f-d g) \left (c (d+e x)^n\right )^{\frac {1}{n}} \Gamma \left (n+1,-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )+g (d+e x) \Gamma \left (n+1,-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )\right )+g^2 2^n (d+e x)^2 \Gamma \left (n+1,-\frac {3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )\right )}{e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3^(-1 - n)*(d + e*x)*(2^n*g^2*(d + e*x)^2*Gamma[1 + n, (-3*(a + b*Log[c*(d + e*x)^n]))/(b*n)] + 3^(1 + n)*E^(
a/(b*n))*(e*f - d*g)*(c*(d + e*x)^n)^n^(-1)*(g*(d + e*x)*Gamma[1 + n, (-2*(a + b*Log[c*(d + e*x)^n]))/(b*n)] +
2^n*E^(a/(b*n))*(e*f - d*g)*(c*(d + e*x)^n)^n^(-1)*Gamma[1 + n, -((a + b*Log[c*(d + e*x)^n])/(b*n))]))*(a + b
*Log[c*(d + e*x)^n])^n)/(2^n*e^3*E^((3*a)/(b*n))*(c*(d + e*x)^n)^(3/n)*(-((a + b*Log[c*(d + e*x)^n])/(b*n)))^n
)

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (g^{2} x^{2} + 2 \, f g x + f^{2}\right )} {\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 2.25, size = 0, normalized size = 0.00 \[ \int \left (g x +f \right )^{2} \left (b \ln \left (c \left (e x +d \right )^{n}\right )+a \right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*log(c*(d + e*x)**n))**n*(f + g*x)**2, x)

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