3.203 \(\int \frac{a+b \log (c (d+e x)^n)}{(d+e x) (f+g x)^{5/2}} \, dx\)

Optimal. Leaf size=406 \[ -\frac{2 b e^{3/2} n \text{PolyLog}\left (2,1-\frac{2}{1-\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}}\right )}{(e f-d g)^{5/2}}-\frac{2 e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{(e f-d g)^{5/2}}+\frac{2 e \left (a+b \log \left (c (d+e x)^n\right )\right )}{\sqrt{f+g x} (e f-d g)^2}+\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 (f+g x)^{3/2} (e f-d g)}+\frac{2 b e^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )^2}{(e f-d g)^{5/2}}+\frac{16 b e^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )}{3 (e f-d g)^{5/2}}-\frac{4 b e^{3/2} n \log \left (\frac{2}{1-\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}}\right ) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )}{(e f-d g)^{5/2}}-\frac{4 b e n}{3 \sqrt{f+g x} (e f-d g)^2} \]

[Out]

(-4*b*e*n)/(3*(e*f - d*g)^2*Sqrt[f + g*x]) + (16*b*e^(3/2)*n*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])
/(3*(e*f - d*g)^(5/2)) + (2*b*e^(3/2)*n*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]]^2)/(e*f - d*g)^(5/2)
+ (2*(a + b*Log[c*(d + e*x)^n]))/(3*(e*f - d*g)*(f + g*x)^(3/2)) + (2*e*(a + b*Log[c*(d + e*x)^n]))/((e*f - d*
g)^2*Sqrt[f + g*x]) - (2*e^(3/2)*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]]*(a + b*Log[c*(d + e*x)^n]))/
(e*f - d*g)^(5/2) - (4*b*e^(3/2)*n*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]]*Log[2/(1 - (Sqrt[e]*Sqrt[f
+ g*x])/Sqrt[e*f - d*g])])/(e*f - d*g)^(5/2) - (2*b*e^(3/2)*n*PolyLog[2, 1 - 2/(1 - (Sqrt[e]*Sqrt[f + g*x])/S
qrt[e*f - d*g])])/(e*f - d*g)^(5/2)

________________________________________________________________________________________

Rubi [A]  time = 1.38772, antiderivative size = 406, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 14, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.452, Rules used = {2411, 2347, 63, 208, 2348, 12, 1587, 6741, 5984, 5918, 2402, 2315, 2319, 51} \[ -\frac{2 b e^{3/2} n \text{PolyLog}\left (2,1-\frac{2}{1-\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}}\right )}{(e f-d g)^{5/2}}-\frac{2 e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{(e f-d g)^{5/2}}+\frac{2 e \left (a+b \log \left (c (d+e x)^n\right )\right )}{\sqrt{f+g x} (e f-d g)^2}+\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 (f+g x)^{3/2} (e f-d g)}+\frac{2 b e^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )^2}{(e f-d g)^{5/2}}+\frac{16 b e^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )}{3 (e f-d g)^{5/2}}-\frac{4 b e^{3/2} n \log \left (\frac{2}{1-\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}}\right ) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )}{(e f-d g)^{5/2}}-\frac{4 b e n}{3 \sqrt{f+g x} (e f-d g)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d + e*x)^n])/((d + e*x)*(f + g*x)^(5/2)),x]

[Out]

(-4*b*e*n)/(3*(e*f - d*g)^2*Sqrt[f + g*x]) + (16*b*e^(3/2)*n*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])
/(3*(e*f - d*g)^(5/2)) + (2*b*e^(3/2)*n*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]]^2)/(e*f - d*g)^(5/2)
+ (2*(a + b*Log[c*(d + e*x)^n]))/(3*(e*f - d*g)*(f + g*x)^(3/2)) + (2*e*(a + b*Log[c*(d + e*x)^n]))/((e*f - d*
g)^2*Sqrt[f + g*x]) - (2*e^(3/2)*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]]*(a + b*Log[c*(d + e*x)^n]))/
(e*f - d*g)^(5/2) - (4*b*e^(3/2)*n*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]]*Log[2/(1 - (Sqrt[e]*Sqrt[f
+ g*x])/Sqrt[e*f - d*g])])/(e*f - d*g)^(5/2) - (2*b*e^(3/2)*n*PolyLog[2, 1 - 2/(1 - (Sqrt[e]*Sqrt[f + g*x])/S
qrt[e*f - d*g])])/(e*f - d*g)^(5/2)

Rule 2411

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

Rule 2347

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[((
d + e*x)^(q + 1)*(a + b*Log[c*x^n])^p)/x, x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 2348

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.))/(x_), x_Symbol] :> With[{u = IntHi
de[(d + e*x^r)^q/x, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[Dist[1/x, u, x], x], x]] /; FreeQ[{a, b
, c, d, e, n, r}, x] && IntegerQ[q - 1/2]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1587

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*Log[RemoveConte
nt[Qq, x]])/(q*Coeff[Qq, x, q]), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]*D[Qq, x])/(q*Coeff[Q
q, x, q])]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 5984

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*e*(p + 1)), x] + Dist[1/(c*d), Int[(a + b*ArcTanh[c*x])^p/(1 - c*x), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]

Rule 5918

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^p*
Log[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[2/(1 + (e*x)/d)])/(1 - c^2
*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2319

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1
)*(a + b*Log[c*x^n])^p)/(e*(q + 1)), x] - Dist[(b*n*p)/(e*(q + 1)), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^
(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers
Q[2*p, 2*q] &&  !IGtQ[q, 0]) || (EqQ[p, 2] && NeQ[q, 1]))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rubi steps

\begin{align*} \int \frac{a+b \log \left (c (d+e x)^n\right )}{(d+e x) (f+g x)^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a+b \log \left (c x^n\right )}{x \left (\frac{e f-d g}{e}+\frac{g x}{e}\right )^{5/2}} \, dx,x,d+e x\right )}{e}\\ &=\frac{\operatorname{Subst}\left (\int \frac{a+b \log \left (c x^n\right )}{x \left (\frac{e f-d g}{e}+\frac{g x}{e}\right )^{3/2}} \, dx,x,d+e x\right )}{e f-d g}-\frac{g \operatorname{Subst}\left (\int \frac{a+b \log \left (c x^n\right )}{\left (\frac{e f-d g}{e}+\frac{g x}{e}\right )^{5/2}} \, dx,x,d+e x\right )}{e (e f-d g)}\\ &=\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 (e f-d g) (f+g x)^{3/2}}+\frac{e \operatorname{Subst}\left (\int \frac{a+b \log \left (c x^n\right )}{x \sqrt{\frac{e f-d g}{e}+\frac{g x}{e}}} \, dx,x,d+e x\right )}{(e f-d g)^2}-\frac{g \operatorname{Subst}\left (\int \frac{a+b \log \left (c x^n\right )}{\left (\frac{e f-d g}{e}+\frac{g x}{e}\right )^{3/2}} \, dx,x,d+e x\right )}{(e f-d g)^2}-\frac{(2 b n) \operatorname{Subst}\left (\int \frac{1}{x \left (\frac{e f-d g}{e}+\frac{g x}{e}\right )^{3/2}} \, dx,x,d+e x\right )}{3 (e f-d g)}\\ &=-\frac{4 b e n}{3 (e f-d g)^2 \sqrt{f+g x}}+\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 (e f-d g) (f+g x)^{3/2}}+\frac{2 e \left (a+b \log \left (c (d+e x)^n\right )\right )}{(e f-d g)^2 \sqrt{f+g x}}-\frac{2 e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{(e f-d g)^{5/2}}-\frac{(2 b e n) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{\frac{e f-d g}{e}+\frac{g x}{e}}} \, dx,x,d+e x\right )}{3 (e f-d g)^2}-\frac{(b e n) \operatorname{Subst}\left (\int -\frac{2 \sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f-\frac{d g}{e}+\frac{g x}{e}}}{\sqrt{e f-d g}}\right )}{\sqrt{e f-d g} x} \, dx,x,d+e x\right )}{(e f-d g)^2}-\frac{(2 b e n) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{\frac{e f-d g}{e}+\frac{g x}{e}}} \, dx,x,d+e x\right )}{(e f-d g)^2}\\ &=-\frac{4 b e n}{3 (e f-d g)^2 \sqrt{f+g x}}+\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 (e f-d g) (f+g x)^{3/2}}+\frac{2 e \left (a+b \log \left (c (d+e x)^n\right )\right )}{(e f-d g)^2 \sqrt{f+g x}}-\frac{2 e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{(e f-d g)^{5/2}}+\frac{\left (2 b e^{3/2} n\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f-\frac{d g}{e}+\frac{g x}{e}}}{\sqrt{e f-d g}}\right )}{x} \, dx,x,d+e x\right )}{(e f-d g)^{5/2}}-\frac{\left (4 b e^2 n\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{e f-d g}{g}+\frac{e x^2}{g}} \, dx,x,\sqrt{f+g x}\right )}{3 g (e f-d g)^2}-\frac{\left (4 b e^2 n\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{e f-d g}{g}+\frac{e x^2}{g}} \, dx,x,\sqrt{f+g x}\right )}{g (e f-d g)^2}\\ &=-\frac{4 b e n}{3 (e f-d g)^2 \sqrt{f+g x}}+\frac{16 b e^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )}{3 (e f-d g)^{5/2}}+\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 (e f-d g) (f+g x)^{3/2}}+\frac{2 e \left (a+b \log \left (c (d+e x)^n\right )\right )}{(e f-d g)^2 \sqrt{f+g x}}-\frac{2 e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{(e f-d g)^{5/2}}+\frac{\left (4 b e^{5/2} n\right ) \operatorname{Subst}\left (\int \frac{x \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{e f-d g}}\right )}{d g+e \left (-f+x^2\right )} \, dx,x,\sqrt{f+g x}\right )}{(e f-d g)^{5/2}}\\ &=-\frac{4 b e n}{3 (e f-d g)^2 \sqrt{f+g x}}+\frac{16 b e^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )}{3 (e f-d g)^{5/2}}+\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 (e f-d g) (f+g x)^{3/2}}+\frac{2 e \left (a+b \log \left (c (d+e x)^n\right )\right )}{(e f-d g)^2 \sqrt{f+g x}}-\frac{2 e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{(e f-d g)^{5/2}}+\frac{\left (4 b e^{5/2} n\right ) \operatorname{Subst}\left (\int \frac{x \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{e f-d g}}\right )}{-e f+d g+e x^2} \, dx,x,\sqrt{f+g x}\right )}{(e f-d g)^{5/2}}\\ &=-\frac{4 b e n}{3 (e f-d g)^2 \sqrt{f+g x}}+\frac{16 b e^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )}{3 (e f-d g)^{5/2}}+\frac{2 b e^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )^2}{(e f-d g)^{5/2}}+\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 (e f-d g) (f+g x)^{3/2}}+\frac{2 e \left (a+b \log \left (c (d+e x)^n\right )\right )}{(e f-d g)^2 \sqrt{f+g x}}-\frac{2 e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{(e f-d g)^{5/2}}-\frac{\left (4 b e^2 n\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{e f-d g}}\right )}{1-\frac{\sqrt{e} x}{\sqrt{e f-d g}}} \, dx,x,\sqrt{f+g x}\right )}{(e f-d g)^3}\\ &=-\frac{4 b e n}{3 (e f-d g)^2 \sqrt{f+g x}}+\frac{16 b e^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )}{3 (e f-d g)^{5/2}}+\frac{2 b e^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )^2}{(e f-d g)^{5/2}}+\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 (e f-d g) (f+g x)^{3/2}}+\frac{2 e \left (a+b \log \left (c (d+e x)^n\right )\right )}{(e f-d g)^2 \sqrt{f+g x}}-\frac{2 e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{(e f-d g)^{5/2}}-\frac{4 b e^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \log \left (\frac{2}{1-\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}}\right )}{(e f-d g)^{5/2}}+\frac{\left (4 b e^2 n\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{2}{1-\frac{\sqrt{e} x}{\sqrt{e f-d g}}}\right )}{1-\frac{e x^2}{e f-d g}} \, dx,x,\sqrt{f+g x}\right )}{(e f-d g)^3}\\ &=-\frac{4 b e n}{3 (e f-d g)^2 \sqrt{f+g x}}+\frac{16 b e^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )}{3 (e f-d g)^{5/2}}+\frac{2 b e^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )^2}{(e f-d g)^{5/2}}+\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 (e f-d g) (f+g x)^{3/2}}+\frac{2 e \left (a+b \log \left (c (d+e x)^n\right )\right )}{(e f-d g)^2 \sqrt{f+g x}}-\frac{2 e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{(e f-d g)^{5/2}}-\frac{4 b e^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \log \left (\frac{2}{1-\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}}\right )}{(e f-d g)^{5/2}}-\frac{\left (4 b e^{3/2} n\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}}\right )}{(e f-d g)^{5/2}}\\ &=-\frac{4 b e n}{3 (e f-d g)^2 \sqrt{f+g x}}+\frac{16 b e^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )}{3 (e f-d g)^{5/2}}+\frac{2 b e^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )^2}{(e f-d g)^{5/2}}+\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 (e f-d g) (f+g x)^{3/2}}+\frac{2 e \left (a+b \log \left (c (d+e x)^n\right )\right )}{(e f-d g)^2 \sqrt{f+g x}}-\frac{2 e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{(e f-d g)^{5/2}}-\frac{4 b e^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \log \left (\frac{2}{1-\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}}\right )}{(e f-d g)^{5/2}}-\frac{2 b e^{3/2} n \text{Li}_2\left (1-\frac{2}{1-\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}}\right )}{(e f-d g)^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.781404, size = 608, normalized size = 1.5 \[ \frac{-3 b e^{3/2} n (f+g x)^{3/2} \left (2 \text{PolyLog}\left (2,\frac{1}{2}-\frac{\sqrt{e} \sqrt{f+g x}}{2 \sqrt{e f-d g}}\right )+\log \left (\sqrt{e f-d g}-\sqrt{e} \sqrt{f+g x}\right ) \left (\log \left (\sqrt{e f-d g}-\sqrt{e} \sqrt{f+g x}\right )+2 \log \left (\frac{1}{2} \left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}+1\right )\right )\right )\right )+3 b e^{3/2} n (f+g x)^{3/2} \left (2 \text{PolyLog}\left (2,\frac{1}{2} \left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}+1\right )\right )+\log \left (\sqrt{e f-d g}+\sqrt{e} \sqrt{f+g x}\right ) \left (\log \left (\sqrt{e f-d g}+\sqrt{e} \sqrt{f+g x}\right )+2 \log \left (\frac{1}{2}-\frac{\sqrt{e} \sqrt{f+g x}}{2 \sqrt{e f-d g}}\right )\right )\right )+6 e^{3/2} (f+g x)^{3/2} \log \left (\sqrt{e f-d g}-\sqrt{e} \sqrt{f+g x}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-6 e^{3/2} (f+g x)^{3/2} \log \left (\sqrt{e f-d g}+\sqrt{e} \sqrt{f+g x}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+4 (e f-d g)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )+12 e (f+g x) \sqrt{e f-d g} \left (a+b \log \left (c (d+e x)^n\right )\right )+24 b e^{3/2} n (f+g x)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )-8 b e n (f+g x) \sqrt{e f-d g} \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{e (f+g x)}{e f-d g}\right )}{6 (f+g x)^{3/2} (e f-d g)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d + e*x)^n])/((d + e*x)*(f + g*x)^(5/2)),x]

[Out]

(24*b*e^(3/2)*n*(f + g*x)^(3/2)*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]] - 8*b*e*Sqrt[e*f - d*g]*n*(f
+ g*x)*Hypergeometric2F1[-1/2, 1, 1/2, (e*(f + g*x))/(e*f - d*g)] + 4*(e*f - d*g)^(3/2)*(a + b*Log[c*(d + e*x)
^n]) + 12*e*Sqrt[e*f - d*g]*(f + g*x)*(a + b*Log[c*(d + e*x)^n]) + 6*e^(3/2)*(f + g*x)^(3/2)*(a + b*Log[c*(d +
e*x)^n])*Log[Sqrt[e*f - d*g] - Sqrt[e]*Sqrt[f + g*x]] - 6*e^(3/2)*(f + g*x)^(3/2)*(a + b*Log[c*(d + e*x)^n])*
Log[Sqrt[e*f - d*g] + Sqrt[e]*Sqrt[f + g*x]] - 3*b*e^(3/2)*n*(f + g*x)^(3/2)*(Log[Sqrt[e*f - d*g] - Sqrt[e]*Sq
rt[f + g*x]]*(Log[Sqrt[e*f - d*g] - Sqrt[e]*Sqrt[f + g*x]] + 2*Log[(1 + (Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g
])/2]) + 2*PolyLog[2, 1/2 - (Sqrt[e]*Sqrt[f + g*x])/(2*Sqrt[e*f - d*g])]) + 3*b*e^(3/2)*n*(f + g*x)^(3/2)*(Log
[Sqrt[e*f - d*g] + Sqrt[e]*Sqrt[f + g*x]]*(Log[Sqrt[e*f - d*g] + Sqrt[e]*Sqrt[f + g*x]] + 2*Log[1/2 - (Sqrt[e]
*Sqrt[f + g*x])/(2*Sqrt[e*f - d*g])]) + 2*PolyLog[2, (1 + (Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g])/2]))/(6*(e*
f - d*g)^(5/2)*(f + g*x)^(3/2))

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Maple [F]  time = 1.162, size = 0, normalized size = 0. \begin{align*} \int{\frac{a+b\ln \left ( c \left ( ex+d \right ) ^{n} \right ) }{ex+d} \left ( gx+f \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*(e*x+d)^n))/(e*x+d)/(g*x+f)^(5/2),x)

[Out]

int((a+b*ln(c*(e*x+d)^n))/(e*x+d)/(g*x+f)^(5/2),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((a+b*log(c*(e*x+d)^n))/(e*x+d)/(g*x+f)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{g x + f} b \log \left ({\left (e x + d\right )}^{n} c\right ) + \sqrt{g x + f} a}{e g^{3} x^{4} + d f^{3} +{\left (3 \, e f g^{2} + d g^{3}\right )} x^{3} + 3 \,{\left (e f^{2} g + d f g^{2}\right )} x^{2} +{\left (e f^{3} + 3 \, d f^{2} g\right )} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))/(e*x+d)/(g*x+f)^(5/2),x, algorithm="fricas")

[Out]

integral((sqrt(g*x + f)*b*log((e*x + d)^n*c) + sqrt(g*x + f)*a)/(e*g^3*x^4 + d*f^3 + (3*e*f*g^2 + d*g^3)*x^3 +
3*(e*f^2*g + d*f*g^2)*x^2 + (e*f^3 + 3*d*f^2*g)*x), x)

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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((a+b*ln(c*(e*x+d)**n))/(e*x+d)/(g*x+f)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (e x + d\right )}{\left (g x + f\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))/(e*x+d)/(g*x+f)^(5/2),x, algorithm="giac")

[Out]

integrate((b*log((e*x + d)^n*c) + a)/((e*x + d)*(g*x + f)^(5/2)), x)