3.64 \(\int (f x)^{-1+3 n} \log (c (d+e x^n)^p) \, dx\)

Optimal. Leaf size=141 \[ \frac {(f x)^{3 n} \log \left (c \left (d+e x^n\right )^p\right )}{3 f n}+\frac {d^3 p x^{-3 n} (f x)^{3 n} \log \left (d+e x^n\right )}{3 e^3 f n}-\frac {d^2 p x^{-2 n} (f x)^{3 n}}{3 e^2 f n}+\frac {d p x^{-n} (f x)^{3 n}}{6 e f n}-\frac {p (f x)^{3 n}}{9 f n} \]

[Out]

-1/9*p*(f*x)^(3*n)/f/n-1/3*d^2*p*(f*x)^(3*n)/e^2/f/n/(x^(2*n))+1/6*d*p*(f*x)^(3*n)/e/f/n/(x^n)+1/3*d^3*p*(f*x)
^(3*n)*ln(d+e*x^n)/e^3/f/n/(x^(3*n))+1/3*(f*x)^(3*n)*ln(c*(d+e*x^n)^p)/f/n

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Rubi [A]  time = 0.08, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2455, 20, 266, 43} \[ \frac {(f x)^{3 n} \log \left (c \left (d+e x^n\right )^p\right )}{3 f n}-\frac {d^2 p x^{-2 n} (f x)^{3 n}}{3 e^2 f n}+\frac {d^3 p x^{-3 n} (f x)^{3 n} \log \left (d+e x^n\right )}{3 e^3 f n}+\frac {d p x^{-n} (f x)^{3 n}}{6 e f n}-\frac {p (f x)^{3 n}}{9 f n} \]

Antiderivative was successfully verified.

[In]

Int[(f*x)^(-1 + 3*n)*Log[c*(d + e*x^n)^p],x]

[Out]

-(p*(f*x)^(3*n))/(9*f*n) - (d^2*p*(f*x)^(3*n))/(3*e^2*f*n*x^(2*n)) + (d*p*(f*x)^(3*n))/(6*e*f*n*x^n) + (d^3*p*
(f*x)^(3*n)*Log[d + e*x^n])/(3*e^3*f*n*x^(3*n)) + ((f*x)^(3*n)*Log[c*(d + e*x^n)^p])/(3*f*n)

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(b^IntPart[n]*(b*v)^FracPart[n])/(a^IntPart[n
]*(a*v)^FracPart[n]), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2455

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

Rubi steps

\begin {align*} \int (f x)^{-1+3 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx &=\frac {(f x)^{3 n} \log \left (c \left (d+e x^n\right )^p\right )}{3 f n}-\frac {(e p) \int \frac {x^{-1+n} (f x)^{3 n}}{d+e x^n} \, dx}{3 f}\\ &=\frac {(f x)^{3 n} \log \left (c \left (d+e x^n\right )^p\right )}{3 f n}-\frac {\left (e p x^{-3 n} (f x)^{3 n}\right ) \int \frac {x^{-1+4 n}}{d+e x^n} \, dx}{3 f}\\ &=\frac {(f x)^{3 n} \log \left (c \left (d+e x^n\right )^p\right )}{3 f n}-\frac {\left (e p x^{-3 n} (f x)^{3 n}\right ) \operatorname {Subst}\left (\int \frac {x^3}{d+e x} \, dx,x,x^n\right )}{3 f n}\\ &=\frac {(f x)^{3 n} \log \left (c \left (d+e x^n\right )^p\right )}{3 f n}-\frac {\left (e p x^{-3 n} (f x)^{3 n}\right ) \operatorname {Subst}\left (\int \left (\frac {d^2}{e^3}-\frac {d x}{e^2}+\frac {x^2}{e}-\frac {d^3}{e^3 (d+e x)}\right ) \, dx,x,x^n\right )}{3 f n}\\ &=-\frac {p (f x)^{3 n}}{9 f n}-\frac {d^2 p x^{-2 n} (f x)^{3 n}}{3 e^2 f n}+\frac {d p x^{-n} (f x)^{3 n}}{6 e f n}+\frac {d^3 p x^{-3 n} (f x)^{3 n} \log \left (d+e x^n\right )}{3 e^3 f n}+\frac {(f x)^{3 n} \log \left (c \left (d+e x^n\right )^p\right )}{3 f n}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 92, normalized size = 0.65 \[ \frac {x^{-3 n} (f x)^{3 n} \left (6 e^3 x^{3 n} \log \left (c \left (d+e x^n\right )^p\right )+6 d^3 p \log \left (d+e x^n\right )-e p x^n \left (6 d^2-3 d e x^n+2 e^2 x^{2 n}\right )\right )}{18 e^3 f n} \]

Antiderivative was successfully verified.

[In]

Integrate[(f*x)^(-1 + 3*n)*Log[c*(d + e*x^n)^p],x]

[Out]

((f*x)^(3*n)*(-(e*p*x^n*(6*d^2 - 3*d*e*x^n + 2*e^2*x^(2*n))) + 6*d^3*p*Log[d + e*x^n] + 6*e^3*x^(3*n)*Log[c*(d
+ e*x^n)^p]))/(18*e^3*f*n*x^(3*n))

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fricas [A]  time = 0.48, size = 112, normalized size = 0.79 \[ \frac {3 \, d e^{2} f^{3 \, n - 1} p x^{2 \, n} - 6 \, d^{2} e f^{3 \, n - 1} p x^{n} - 2 \, {\left (e^{3} p - 3 \, e^{3} \log \relax (c)\right )} f^{3 \, n - 1} x^{3 \, n} + 6 \, {\left (e^{3} f^{3 \, n - 1} p x^{3 \, n} + d^{3} f^{3 \, n - 1} p\right )} \log \left (e x^{n} + d\right )}{18 \, e^{3} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(-1+3*n)*log(c*(d+e*x^n)^p),x, algorithm="fricas")

[Out]

1/18*(3*d*e^2*f^(3*n - 1)*p*x^(2*n) - 6*d^2*e*f^(3*n - 1)*p*x^n - 2*(e^3*p - 3*e^3*log(c))*f^(3*n - 1)*x^(3*n)
+ 6*(e^3*f^(3*n - 1)*p*x^(3*n) + d^3*f^(3*n - 1)*p)*log(e*x^n + d))/(e^3*n)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (f x\right )^{3 \, n - 1} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(-1+3*n)*log(c*(d+e*x^n)^p),x, algorithm="giac")

[Out]

integrate((f*x)^(3*n - 1)*log((e*x^n + d)^p*c), x)

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maple [F]  time = 1.57, size = 0, normalized size = 0.00 \[ \int \left (f x \right )^{3 n -1} \ln \left (c \left (e \,x^{n}+d \right )^{p}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^(-1+3*n)*ln(c*(e*x^n+d)^p),x)

[Out]

int((f*x)^(-1+3*n)*ln(c*(e*x^n+d)^p),x)

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maxima [A]  time = 0.77, size = 115, normalized size = 0.82 \[ \frac {e p {\left (\frac {6 \, d^{3} f^{3 \, n} \log \left (\frac {e x^{n} + d}{e}\right )}{e^{4} n} - \frac {2 \, e^{2} f^{3 \, n} x^{3 \, n} - 3 \, d e f^{3 \, n} x^{2 \, n} + 6 \, d^{2} f^{3 \, n} x^{n}}{e^{3} n}\right )}}{18 \, f} + \frac {\left (f x\right )^{3 \, n} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{3 \, f n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(-1+3*n)*log(c*(d+e*x^n)^p),x, algorithm="maxima")

[Out]

1/18*e*p*(6*d^3*f^(3*n)*log((e*x^n + d)/e)/(e^4*n) - (2*e^2*f^(3*n)*x^(3*n) - 3*d*e*f^(3*n)*x^(2*n) + 6*d^2*f^
(3*n)*x^n)/(e^3*n))/f + 1/3*(f*x)^(3*n)*log((e*x^n + d)^p*c)/(f*n)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \ln \left (c\,{\left (d+e\,x^n\right )}^p\right )\,{\left (f\,x\right )}^{3\,n-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(d + e*x^n)^p)*(f*x)^(3*n - 1),x)

[Out]

int(log(c*(d + e*x^n)^p)*(f*x)^(3*n - 1), 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*x)**(-1+3*n)*ln(c*(d+e*x**n)**p),x)

[Out]

Timed out

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