3.2.72 \(\int (f+g x) (a+b \log (c (d+e x)^n))^n \, dx\) [172]

Optimal. Leaf size=225 \[ \frac {2^{-1-n} e^{-\frac {2 a}{b n}} g (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \Gamma \left (1+n,-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^n \left (-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )^{-n}}{e^2}+\frac {e^{-\frac {a}{b n}} (e f-d g) (d+e x) \left (c (d+e x)^n\right )^{-1/n} \Gamma \left (1+n,-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^n \left (-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )^{-n}}{e^2} \]

[Out]

2^(-1-n)*g*(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/e^2/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)*(e*x+d)*GAMMA(1+n,(-a-b*ln(c*(e*x+d)^n))/b/n)*(a+b*l
n(c*(e*x+d)^n))^n/e^2/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 = 0.15, antiderivative size = 225, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2448, 2436, 2337, 2212, 2437, 2347} \begin {gather*} \frac {e^{-\frac {a}{b n}} (d+e x) (e f-d g) \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^2}+\frac {g 2^{-n-1} e^{-\frac {2 a}{b n}} (d+e x)^2 \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^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(-1 - n)*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)/(e
^2*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)*(d + e*x)*Gam
ma[1 + n, -((a + b*Log[c*(d + e*x)^n])/(b*n))]*(a + b*Log[c*(d + e*x)^n])^n)/(e^2*E^(a/(b*n))*(c*(d + e*x)^n)^
n^(-1)*(-((a + b*Log[c*(d + e*x)^n])/(b*n)))^n)

Rule 2212

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

Rule 2337

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 2347

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)/n)*x)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}
, x]

Rule 2436

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 2437

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 2448

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

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Mathematica [A]
time = 0.14, size = 181, normalized size = 0.80 \begin {gather*} \frac {2^{-1-n} e^{-\frac {2 a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-2/n} \left (g (d+e x) \Gamma \left (1+n,-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )+2^{1+n} e^{\frac {a}{b n}} (e f-d g) \left (c (d+e x)^n\right )^{\frac {1}{n}} \Gamma \left (1+n,-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )\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^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \left (g x +f \right ) \left (a +b \ln \left (c \left (e x +d \right )^{n}\right )\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
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((g*x+f)*(a+b*log(c*(e*x+d)^n))^n,x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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