\(\int (f x)^{-1+2 n} \log (c (d+e x^n)^p) \, dx\) [65]

Optimal result

Integrand size = 22, antiderivative size = 112 \[ \int (f x)^{-1+2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {p (f x)^{2 n}}{4 f n}+\frac {d p x^{-n} (f x)^{2 n}}{2 e f n}-\frac {d^2 p x^{-2 n} (f x)^{2 n} \log \left (d+e x^n\right )}{2 e^2 f n}+\frac {(f x)^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 f n} \]

[Out]

-1/4*p*(f*x)^(2*n)/f/n+1/2*d*p*(f*x)^(2*n)/e/f/n/(x^n)-1/2*d^2*p*(f*x)^(2*n)*ln(d+e*x^n)/e^2/f/n/(x^(2*n))+1/2
*(f*x)^(2*n)*ln(c*(d+e*x^n)^p)/f/n

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2505, 20, 272, 45} \[ \int (f x)^{-1+2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {(f x)^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 f n}-\frac {d^2 p x^{-2 n} (f x)^{2 n} \log \left (d+e x^n\right )}{2 e^2 f n}+\frac {d p x^{-n} (f x)^{2 n}}{2 e f n}-\frac {p (f x)^{2 n}}{4 f n} \]

[In]

Int[(f*x)^(-1 + 2*n)*Log[c*(d + e*x^n)^p],x]

[Out]

-1/4*(p*(f*x)^(2*n))/(f*n) + (d*p*(f*x)^(2*n))/(2*e*f*n*x^n) - (d^2*p*(f*x)^(2*n)*Log[d + e*x^n])/(2*e^2*f*n*x
^(2*n)) + ((f*x)^(2*n)*Log[c*(d + e*x^n)^p])/(2*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 45

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 272

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 2505

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.66 \[ \int (f x)^{-1+2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {x^{-2 n} (f x)^{2 n} \left (2 d^2 p \log \left (d+e x^n\right )+e x^n \left (-2 d p+e p x^n-2 e x^n \log \left (c \left (d+e x^n\right )^p\right )\right )\right )}{4 e^2 f n} \]

[In]

Integrate[(f*x)^(-1 + 2*n)*Log[c*(d + e*x^n)^p],x]

[Out]

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

Maple [F]

\[\int \left (f x \right )^{-1+2 n} \ln \left (c \left (d +e \,x^{n}\right )^{p}\right )d x\]

[In]

int((f*x)^(-1+2*n)*ln(c*(d+e*x^n)^p),x)

[Out]

int((f*x)^(-1+2*n)*ln(c*(d+e*x^n)^p),x)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.82 \[ \int (f x)^{-1+2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\frac {2 \, d e f^{2 \, n - 1} p x^{n} - {\left (e^{2} p - 2 \, e^{2} \log \left (c\right )\right )} f^{2 \, n - 1} x^{2 \, n} + 2 \, {\left (e^{2} f^{2 \, n - 1} p x^{2 \, n} - d^{2} f^{2 \, n - 1} p\right )} \log \left (e x^{n} + d\right )}{4 \, e^{2} n} \]

[In]

integrate((f*x)^(-1+2*n)*log(c*(d+e*x^n)^p),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int (f x)^{-1+2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\int \left (f x\right )^{2 n - 1} \log {\left (c \left (d + e x^{n}\right )^{p} \right )}\, dx \]

[In]

integrate((f*x)**(-1+2*n)*ln(c*(d+e*x**n)**p),x)

[Out]

Integral((f*x)**(2*n - 1)*log(c*(d + e*x**n)**p), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.85 \[ \int (f x)^{-1+2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=-\frac {e p {\left (\frac {2 \, d^{2} f^{2 \, n} \log \left (\frac {e x^{n} + d}{e}\right )}{e^{3} n} + \frac {e f^{2 \, n} x^{2 \, n} - 2 \, d f^{2 \, n} x^{n}}{e^{2} n}\right )}}{4 \, f} + \frac {\left (f x\right )^{2 \, n} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{2 \, f n} \]

[In]

integrate((f*x)^(-1+2*n)*log(c*(d+e*x^n)^p),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int (f x)^{-1+2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\int { \left (f x\right )^{2 \, n - 1} \log \left ({\left (e x^{n} + d\right )}^{p} c\right ) \,d x } \]

[In]

integrate((f*x)^(-1+2*n)*log(c*(d+e*x^n)^p),x, algorithm="giac")

[Out]

integrate((f*x)^(2*n - 1)*log((e*x^n + d)^p*c), x)

Mupad [F(-1)]

Timed out. \[ \int (f x)^{-1+2 n} \log \left (c \left (d+e x^n\right )^p\right ) \, dx=\int \ln \left (c\,{\left (d+e\,x^n\right )}^p\right )\,{\left (f\,x\right )}^{2\,n-1} \,d x \]

[In]

int(log(c*(d + e*x^n)^p)*(f*x)^(2*n - 1),x)

[Out]

int(log(c*(d + e*x^n)^p)*(f*x)^(2*n - 1), x)