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

Optimal. Leaf size=112 \[ \frac{(f x)^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 f n}-\frac{d^2 p x^{-2 n} (f x)^{2 n} \log \left (d+e x^n\right )}{2 e^2 f n}+\frac{d p x^{-n} (f x)^{2 n}}{2 e f n}-\frac{p (f x)^{2 n}}{4 f n} \]

[Out]

-(p*(f*x)^(2*n))/(4*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)

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Rubi [A]  time = 0.0571477, antiderivative size = 112, normalized size of antiderivative = 1., 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} \[ \frac{(f x)^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 f n}-\frac{d^2 p x^{-2 n} (f x)^{2 n} \log \left (d+e x^n\right )}{2 e^2 f n}+\frac{d p x^{-n} (f x)^{2 n}}{2 e f n}-\frac{p (f x)^{2 n}}{4 f n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(p*(f*x)^(2*n))/(4*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 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]

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 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 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])

Rubi steps

\begin{align*} \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{(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 ) \operatorname{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 ) \operatorname{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]  time = 0.0407131, size = 74, normalized size = 0.66 \[ -\frac{x^{-2 n} (f x)^{2 n} \left (e x^n \left (-2 e x^n \log \left (c \left (d+e x^n\right )^p\right )-2 d p+e p x^n\right )+2 d^2 p \log \left (d+e x^n\right )\right )}{4 e^2 f n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((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])))/(4*e^2*f*n*x
^(2*n))

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Maple [F]  time = 1.887, size = 0, normalized size = 0. \begin{align*} \int \left ( fx \right ) ^{-1+2\,n}\ln \left ( c \left ( d+e{x}^{n} \right ) ^{p} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.09319, size = 204, normalized size = 1.82 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (f x\right )^{2 \, n - 1} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)