3.212 \(\int (f+g x)^3 \log (c (d+e x^n)^p) \, dx\)

Optimal. Leaf size=234 \[ \frac {(f+g x)^4 \log \left (c \left (d+e x^n\right )^p\right )}{4 g}-\frac {f^4 p \log \left (d+e x^n\right )}{4 g}-\frac {e f^3 n p x^{n+1} \, _2F_1\left (1,1+\frac {1}{n};2+\frac {1}{n};-\frac {e x^n}{d}\right )}{d (n+1)}-\frac {3 e f^2 g n p x^{n+2} \, _2F_1\left (1,\frac {n+2}{n};2 \left (1+\frac {1}{n}\right );-\frac {e x^n}{d}\right )}{2 d (n+2)}-\frac {e f g^2 n p x^{n+3} \, _2F_1\left (1,\frac {n+3}{n};2+\frac {3}{n};-\frac {e x^n}{d}\right )}{d (n+3)}-\frac {e g^3 n p x^{n+4} \, _2F_1\left (1,\frac {n+4}{n};2 \left (1+\frac {2}{n}\right );-\frac {e x^n}{d}\right )}{4 d (n+4)} \]

[Out]

-e*f^3*n*p*x^(1+n)*hypergeom([1, 1+1/n],[2+1/n],-e*x^n/d)/d/(1+n)-3/2*e*f^2*g*n*p*x^(2+n)*hypergeom([1, (2+n)/
n],[2+2/n],-e*x^n/d)/d/(2+n)-e*f*g^2*n*p*x^(3+n)*hypergeom([1, (3+n)/n],[2+3/n],-e*x^n/d)/d/(3+n)-1/4*e*g^3*n*
p*x^(4+n)*hypergeom([1, (4+n)/n],[2+4/n],-e*x^n/d)/d/(4+n)-1/4*f^4*p*ln(d+e*x^n)/g+1/4*(g*x+f)^4*ln(c*(d+e*x^n
)^p)/g

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Rubi [A]  time = 0.23, antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2463, 1844, 260, 364} \[ \frac {(f+g x)^4 \log \left (c \left (d+e x^n\right )^p\right )}{4 g}-\frac {3 e f^2 g n p x^{n+2} \, _2F_1\left (1,\frac {n+2}{n};2 \left (1+\frac {1}{n}\right );-\frac {e x^n}{d}\right )}{2 d (n+2)}-\frac {f^4 p \log \left (d+e x^n\right )}{4 g}-\frac {e f^3 n p x^{n+1} \, _2F_1\left (1,1+\frac {1}{n};2+\frac {1}{n};-\frac {e x^n}{d}\right )}{d (n+1)}-\frac {e f g^2 n p x^{n+3} \, _2F_1\left (1,\frac {n+3}{n};2+\frac {3}{n};-\frac {e x^n}{d}\right )}{d (n+3)}-\frac {e g^3 n p x^{n+4} \, _2F_1\left (1,\frac {n+4}{n};2 \left (1+\frac {2}{n}\right );-\frac {e x^n}{d}\right )}{4 d (n+4)} \]

Antiderivative was successfully verified.

[In]

Int[(f + g*x)^3*Log[c*(d + e*x^n)^p],x]

[Out]

-((e*f^3*n*p*x^(1 + n)*Hypergeometric2F1[1, 1 + n^(-1), 2 + n^(-1), -((e*x^n)/d)])/(d*(1 + n))) - (3*e*f^2*g*n
*p*x^(2 + n)*Hypergeometric2F1[1, (2 + n)/n, 2*(1 + n^(-1)), -((e*x^n)/d)])/(2*d*(2 + n)) - (e*f*g^2*n*p*x^(3
+ n)*Hypergeometric2F1[1, (3 + n)/n, 2 + 3/n, -((e*x^n)/d)])/(d*(3 + n)) - (e*g^3*n*p*x^(4 + n)*Hypergeometric
2F1[1, (4 + n)/n, 2*(1 + 2/n), -((e*x^n)/d)])/(4*d*(4 + n)) - (f^4*p*Log[d + e*x^n])/(4*g) + ((f + g*x)^4*Log[
c*(d + e*x^n)^p])/(4*g)

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 1844

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a +
b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !IGtQ[m, 0]

Rule 2463

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

Rubi steps

\begin {align*} \int (f+g x)^3 \log \left (c \left (d+e x^n\right )^p\right ) \, dx &=\frac {(f+g x)^4 \log \left (c \left (d+e x^n\right )^p\right )}{4 g}-\frac {(e n p) \int \frac {x^{-1+n} (f+g x)^4}{d+e x^n} \, dx}{4 g}\\ &=\frac {(f+g x)^4 \log \left (c \left (d+e x^n\right )^p\right )}{4 g}-\frac {(e n p) \int \left (\frac {f^4 x^{-1+n}}{d+e x^n}+\frac {4 f^3 g x^n}{d+e x^n}+\frac {6 f^2 g^2 x^{1+n}}{d+e x^n}+\frac {4 f g^3 x^{2+n}}{d+e x^n}+\frac {g^4 x^{3+n}}{d+e x^n}\right ) \, dx}{4 g}\\ &=\frac {(f+g x)^4 \log \left (c \left (d+e x^n\right )^p\right )}{4 g}-\left (e f^3 n p\right ) \int \frac {x^n}{d+e x^n} \, dx-\frac {\left (e f^4 n p\right ) \int \frac {x^{-1+n}}{d+e x^n} \, dx}{4 g}-\frac {1}{2} \left (3 e f^2 g n p\right ) \int \frac {x^{1+n}}{d+e x^n} \, dx-\left (e f g^2 n p\right ) \int \frac {x^{2+n}}{d+e x^n} \, dx-\frac {1}{4} \left (e g^3 n p\right ) \int \frac {x^{3+n}}{d+e x^n} \, dx\\ &=-\frac {e f^3 n p x^{1+n} \, _2F_1\left (1,1+\frac {1}{n};2+\frac {1}{n};-\frac {e x^n}{d}\right )}{d (1+n)}-\frac {3 e f^2 g n p x^{2+n} \, _2F_1\left (1,\frac {2+n}{n};2 \left (1+\frac {1}{n}\right );-\frac {e x^n}{d}\right )}{2 d (2+n)}-\frac {e f g^2 n p x^{3+n} \, _2F_1\left (1,\frac {3+n}{n};2+\frac {3}{n};-\frac {e x^n}{d}\right )}{d (3+n)}-\frac {e g^3 n p x^{4+n} \, _2F_1\left (1,\frac {4+n}{n};2 \left (1+\frac {2}{n}\right );-\frac {e x^n}{d}\right )}{4 d (4+n)}-\frac {f^4 p \log \left (d+e x^n\right )}{4 g}+\frac {(f+g x)^4 \log \left (c \left (d+e x^n\right )^p\right )}{4 g}\\ \end {align*}

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Mathematica [A]  time = 0.49, size = 224, normalized size = 0.96 \[ \frac {(f+g x)^4 \log \left (c \left (d+e x^n\right )^p\right )-e n p \left (\frac {f^4 \log \left (d+e x^n\right )}{e n}+\frac {4 f^3 g x^{n+1} \, _2F_1\left (1,1+\frac {1}{n};2+\frac {1}{n};-\frac {e x^n}{d}\right )}{d (n+1)}+\frac {6 f^2 g^2 x^{n+2} \, _2F_1\left (1,\frac {n+2}{n};2 \left (1+\frac {1}{n}\right );-\frac {e x^n}{d}\right )}{d (n+2)}+\frac {4 f g^3 x^{n+3} \, _2F_1\left (1,\frac {n+3}{n};2+\frac {3}{n};-\frac {e x^n}{d}\right )}{d (n+3)}+\frac {g^4 x^{n+4} \, _2F_1\left (1,\frac {n+4}{n};2+\frac {4}{n};-\frac {e x^n}{d}\right )}{d (n+4)}\right )}{4 g} \]

Antiderivative was successfully verified.

[In]

Integrate[(f + g*x)^3*Log[c*(d + e*x^n)^p],x]

[Out]

(-(e*n*p*((4*f^3*g*x^(1 + n)*Hypergeometric2F1[1, 1 + n^(-1), 2 + n^(-1), -((e*x^n)/d)])/(d*(1 + n)) + (6*f^2*
g^2*x^(2 + n)*Hypergeometric2F1[1, (2 + n)/n, 2*(1 + n^(-1)), -((e*x^n)/d)])/(d*(2 + n)) + (4*f*g^3*x^(3 + n)*
Hypergeometric2F1[1, (3 + n)/n, 2 + 3/n, -((e*x^n)/d)])/(d*(3 + n)) + (g^4*x^(4 + n)*Hypergeometric2F1[1, (4 +
n)/n, 2 + 4/n, -((e*x^n)/d)])/(d*(4 + n)) + (f^4*Log[d + e*x^n])/(e*n))) + (f + g*x)^4*Log[c*(d + e*x^n)^p])/
(4*g)

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (g^{3} x^{3} + 3 \, f g^{2} x^{2} + 3 \, f^{2} g x + f^{3}\right )} \log \left ({\left (e x^{n} + d\right )}^{p} c\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^3*log(c*(d+e*x^n)^p),x, algorithm="fricas")

[Out]

integral((g^3*x^3 + 3*f*g^2*x^2 + 3*f^2*g*x + f^3)*log((e*x^n + d)^p*c), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (g x + f\right )}^{3} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^3*log(c*(d+e*x^n)^p),x, algorithm="giac")

[Out]

integrate((g*x + f)^3*log((e*x^n + d)^p*c), x)

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maple [F]  time = 1.80, size = 0, normalized size = 0.00 \[ \int \left (g x +f \right )^{3} \ln \left (c \left (e \,x^{n}+d \right )^{p}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)^3*ln(c*(e*x^n+d)^p),x)

[Out]

int((g*x+f)^3*ln(c*(e*x^n+d)^p),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{16} \, {\left (g^{3} n p - 4 \, g^{3} \log \relax (c)\right )} x^{4} - \frac {1}{3} \, {\left (f g^{2} n p - 3 \, f g^{2} \log \relax (c)\right )} x^{3} - \frac {3}{4} \, {\left (f^{2} g n p - 2 \, f^{2} g \log \relax (c)\right )} x^{2} - {\left (f^{3} n p - f^{3} \log \relax (c)\right )} x + \frac {1}{4} \, {\left (g^{3} x^{4} + 4 \, f g^{2} x^{3} + 6 \, f^{2} g x^{2} + 4 \, f^{3} x\right )} \log \left ({\left (e x^{n} + d\right )}^{p}\right ) + \int \frac {d g^{3} n p x^{3} + 4 \, d f g^{2} n p x^{2} + 6 \, d f^{2} g n p x + 4 \, d f^{3} n p}{4 \, {\left (e x^{n} + d\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^3*log(c*(d+e*x^n)^p),x, algorithm="maxima")

[Out]

-1/16*(g^3*n*p - 4*g^3*log(c))*x^4 - 1/3*(f*g^2*n*p - 3*f*g^2*log(c))*x^3 - 3/4*(f^2*g*n*p - 2*f^2*g*log(c))*x
^2 - (f^3*n*p - f^3*log(c))*x + 1/4*(g^3*x^4 + 4*f*g^2*x^3 + 6*f^2*g*x^2 + 4*f^3*x)*log((e*x^n + d)^p) + integ
rate(1/4*(d*g^3*n*p*x^3 + 4*d*f*g^2*n*p*x^2 + 6*d*f^2*g*n*p*x + 4*d*f^3*n*p)/(e*x^n + d), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \ln \left (c\,{\left (d+e\,x^n\right )}^p\right )\,{\left (f+g\,x\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(d + e*x^n)^p)*(f + g*x)^3,x)

[Out]

int(log(c*(d + e*x^n)^p)*(f + g*x)^3, x)

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sympy [C]  time = 32.59, size = 415, normalized size = 1.77 \[ f^{3} x \log {\left (c \left (d + e x^{n}\right )^{p} \right )} + \frac {f^{3} p x \Phi \left (\frac {d x^{- n} e^{i \pi }}{e}, 1, \frac {e^{i \pi }}{n}\right ) \Gamma \left (\frac {1}{n}\right )}{n \Gamma \left (1 + \frac {1}{n}\right )} + \frac {3 f^{2} g x^{2} \log {\left (c \left (d + e x^{n}\right )^{p} \right )}}{2} + f g^{2} x^{3} \log {\left (c \left (d + e x^{n}\right )^{p} \right )} + \frac {g^{3} x^{4} \log {\left (c \left (d + e x^{n}\right )^{p} \right )}}{4} - \frac {3 e f^{2} g p x^{2} x^{n} \Phi \left (\frac {e x^{n} e^{i \pi }}{d}, 1, 1 + \frac {2}{n}\right ) \Gamma \left (1 + \frac {2}{n}\right )}{2 d \Gamma \left (2 + \frac {2}{n}\right )} - \frac {3 e f^{2} g p x^{2} x^{n} \Phi \left (\frac {e x^{n} e^{i \pi }}{d}, 1, 1 + \frac {2}{n}\right ) \Gamma \left (1 + \frac {2}{n}\right )}{d n \Gamma \left (2 + \frac {2}{n}\right )} - \frac {e f g^{2} p x^{3} x^{n} \Phi \left (\frac {e x^{n} e^{i \pi }}{d}, 1, 1 + \frac {3}{n}\right ) \Gamma \left (1 + \frac {3}{n}\right )}{d \Gamma \left (2 + \frac {3}{n}\right )} - \frac {3 e f g^{2} p x^{3} x^{n} \Phi \left (\frac {e x^{n} e^{i \pi }}{d}, 1, 1 + \frac {3}{n}\right ) \Gamma \left (1 + \frac {3}{n}\right )}{d n \Gamma \left (2 + \frac {3}{n}\right )} - \frac {e g^{3} p x^{4} x^{n} \Phi \left (\frac {e x^{n} e^{i \pi }}{d}, 1, 1 + \frac {4}{n}\right ) \Gamma \left (1 + \frac {4}{n}\right )}{4 d \Gamma \left (2 + \frac {4}{n}\right )} - \frac {e g^{3} p x^{4} x^{n} \Phi \left (\frac {e x^{n} e^{i \pi }}{d}, 1, 1 + \frac {4}{n}\right ) \Gamma \left (1 + \frac {4}{n}\right )}{d n \Gamma \left (2 + \frac {4}{n}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)**3*ln(c*(d+e*x**n)**p),x)

[Out]

f**3*x*log(c*(d + e*x**n)**p) + f**3*p*x*lerchphi(d*x**(-n)*exp_polar(I*pi)/e, 1, exp_polar(I*pi)/n)*gamma(1/n
)/(n*gamma(1 + 1/n)) + 3*f**2*g*x**2*log(c*(d + e*x**n)**p)/2 + f*g**2*x**3*log(c*(d + e*x**n)**p) + g**3*x**4
*log(c*(d + e*x**n)**p)/4 - 3*e*f**2*g*p*x**2*x**n*lerchphi(e*x**n*exp_polar(I*pi)/d, 1, 1 + 2/n)*gamma(1 + 2/
n)/(2*d*gamma(2 + 2/n)) - 3*e*f**2*g*p*x**2*x**n*lerchphi(e*x**n*exp_polar(I*pi)/d, 1, 1 + 2/n)*gamma(1 + 2/n)
/(d*n*gamma(2 + 2/n)) - e*f*g**2*p*x**3*x**n*lerchphi(e*x**n*exp_polar(I*pi)/d, 1, 1 + 3/n)*gamma(1 + 3/n)/(d*
gamma(2 + 3/n)) - 3*e*f*g**2*p*x**3*x**n*lerchphi(e*x**n*exp_polar(I*pi)/d, 1, 1 + 3/n)*gamma(1 + 3/n)/(d*n*ga
mma(2 + 3/n)) - e*g**3*p*x**4*x**n*lerchphi(e*x**n*exp_polar(I*pi)/d, 1, 1 + 4/n)*gamma(1 + 4/n)/(4*d*gamma(2
+ 4/n)) - e*g**3*p*x**4*x**n*lerchphi(e*x**n*exp_polar(I*pi)/d, 1, 1 + 4/n)*gamma(1 + 4/n)/(d*n*gamma(2 + 4/n)
)

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