3.3.35 \(\int \frac {(A+B \log (e (a+b x)^n (c+d x)^{-n}))^p}{(a f+b f x) (c g+d g x)} \, dx\) [235]

Optimal. Leaf size=55 \[ \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^{1+p}}{B (b c-a d) f g n (1+p)} \]

[Out]

(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^(1+p)/B/(-a*d+b*c)/f/g/n/(1+p)

________________________________________________________________________________________

Rubi [A]
time = 0.24, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2573, 2561, 2339, 30} \begin {gather*} \frac {\left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^{p+1}}{B f g n (p+1) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^(1 + p)/(B*(b*c - a*d)*f*g*n*(1 + p))

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2561

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m
_.)*((h_.) + (i_.)*(x_))^(q_.), x_Symbol] :> Dist[(b*c - a*d)^(m + q + 1)*(g/b)^m*(i/d)^q, Subst[Int[x^m*((A +
B*Log[e*x^n])^p/(b - d*x)^(m + q + 2)), x], x, (a + b*x)/(c + d*x)], x] /; FreeQ[{a, b, c, d, e, f, g, h, i,
A, B, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[b*f - a*g, 0] && EqQ[d*h - c*i, 0] && IntegersQ[m, q]

Rule 2573

Int[((A_.) + Log[(e_.)*(u_)^(n_.)*(v_)^(mn_)]*(B_.))^(p_.)*(w_.), x_Symbol] :> Subst[Int[w*(A + B*Log[e*(u/v)^
n])^p, x], e*(u/v)^n, e*(u^n/v^n)] /; FreeQ[{e, A, B, n, p}, x] && EqQ[n + mn, 0] && LinearQ[{u, v}, x] &&  !I
ntegerQ[n]

Rubi steps

\begin {align*} \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^p}{(a f+b f x) (c g+d g x)} \, dx &=\frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^{1+p}}{B (b c-a d) f g n (1+p)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.04, size = 51, normalized size = 0.93 \begin {gather*} \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^{1+p}}{(b B c f g n-a B d f g n) (1+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^(1 + p)/((b*B*c*f*g*n - a*B*d*f*g*n)*(1 + p))

________________________________________________________________________________________

Maple [F]
time = 0.26, size = 0, normalized size = 0.00 \[\int \frac {\left (A +B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )\right )^{p}}{\left (b f x +a f \right ) \left (d g x +c g \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^p/(b*f*x+a*f)/(d*g*x+c*g),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.36, size = 79, normalized size = 1.44 \begin {gather*} \frac {{\left (B n \log \left (b x + a\right ) - B n \log \left (d x + c\right ) + A + B\right )} {\left (B n \log \left (b x + a\right ) - B n \log \left (d x + c\right ) + A + B\right )}^{p}}{{\left (B b c - B a d\right )} f g n p + {\left (B b c - B a d\right )} f g n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^p/(b*f*x+a*f)/(d*g*x+c*g),x, algorithm="fricas")

[Out]

(B*n*log(b*x + a) - B*n*log(d*x + c) + A + B)*(B*n*log(b*x + a) - B*n*log(d*x + c) + A + B)^p/((B*b*c - B*a*d)
*f*g*n*p + (B*b*c - B*a*d)*f*g*n)

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]
time = 4.35, size = 50, normalized size = 0.91 \begin {gather*} \frac {{\left (B n \log \left (b x + a\right ) - B n \log \left (d x + c\right ) + A + B\right )}^{p + 1}}{{\left (B b c f g n - B a d f g n\right )} {\left (p + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^p/(b*f*x+a*f)/(d*g*x+c*g),x, algorithm="giac")

[Out]

(B*n*log(b*x + a) - B*n*log(d*x + c) + A + B)^(p + 1)/((B*b*c*f*g*n - B*a*d*f*g*n)*(p + 1))

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (A+B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )\right )}^p}{\left (a\,f+b\,f\,x\right )\,\left (c\,g+d\,g\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________