3.208 \(\int \frac{\log (a+b x)}{(a+b x) (d+e x)^{5/2}} \, dx\)

Optimal. Leaf size=372 \[ -\frac{2 b^{3/2} \text{PolyLog}\left (2,1-\frac{2}{1-\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}}\right )}{(b d-a e)^{5/2}}+\frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )^2}{(b d-a e)^{5/2}}+\frac{16 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{3 (b d-a e)^{5/2}}-\frac{2 b^{3/2} \log (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{(b d-a e)^{5/2}}-\frac{4 b^{3/2} \log \left (\frac{2}{1-\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}}\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{(b d-a e)^{5/2}}-\frac{4 b}{3 \sqrt{d+e x} (b d-a e)^2}+\frac{2 b \log (a+b x)}{\sqrt{d+e x} (b d-a e)^2}+\frac{2 \log (a+b x)}{3 (d+e x)^{3/2} (b d-a e)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.25739, antiderivative size = 372, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 14, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.609, Rules used = {2411, 2347, 63, 208, 2348, 12, 1587, 6741, 5984, 5918, 2402, 2315, 2319, 51} \[ -\frac{2 b^{3/2} \text{PolyLog}\left (2,1-\frac{2}{1-\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}}\right )}{(b d-a e)^{5/2}}+\frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )^2}{(b d-a e)^{5/2}}+\frac{16 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{3 (b d-a e)^{5/2}}-\frac{2 b^{3/2} \log (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{(b d-a e)^{5/2}}-\frac{4 b^{3/2} \log \left (\frac{2}{1-\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}}\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{(b d-a e)^{5/2}}-\frac{4 b}{3 \sqrt{d+e x} (b d-a e)^2}+\frac{2 b \log (a+b x)}{\sqrt{d+e x} (b d-a e)^2}+\frac{2 \log (a+b x)}{3 (d+e x)^{3/2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*b)/(3*(b*d - a*e)^2*Sqrt[d + e*x]) + (16*b^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(3*(b*d
 - a*e)^(5/2)) + (2*b^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]]^2)/(b*d - a*e)^(5/2) + (2*Log[a +
 b*x])/(3*(b*d - a*e)*(d + e*x)^(3/2)) + (2*b*Log[a + b*x])/((b*d - a*e)^2*Sqrt[d + e*x]) - (2*b^(3/2)*ArcTanh
[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]]*Log[a + b*x])/(b*d - a*e)^(5/2) - (4*b^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[d
 + e*x])/Sqrt[b*d - a*e]]*Log[2/(1 - (Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e])])/(b*d - a*e)^(5/2) - (2*b^(3/2)
*PolyLog[2, 1 - 2/(1 - (Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e])])/(b*d - a*e)^(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{\log (a+b x)}{(a+b x) (d+e x)^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\log (x)}{x \left (\frac{b d-a e}{b}+\frac{e x}{b}\right )^{5/2}} \, dx,x,a+b x\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\log (x)}{x \left (\frac{b d-a e}{b}+\frac{e x}{b}\right )^{3/2}} \, dx,x,a+b x\right )}{b d-a e}-\frac{e \operatorname{Subst}\left (\int \frac{\log (x)}{\left (\frac{b d-a e}{b}+\frac{e x}{b}\right )^{5/2}} \, dx,x,a+b x\right )}{b (b d-a e)}\\ &=\frac{2 \log (a+b x)}{3 (b d-a e) (d+e x)^{3/2}}+\frac{b \operatorname{Subst}\left (\int \frac{\log (x)}{x \sqrt{\frac{b d-a e}{b}+\frac{e x}{b}}} \, dx,x,a+b x\right )}{(b d-a e)^2}-\frac{e \operatorname{Subst}\left (\int \frac{\log (x)}{\left (\frac{b d-a e}{b}+\frac{e x}{b}\right )^{3/2}} \, dx,x,a+b x\right )}{(b d-a e)^2}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{x \left (\frac{b d-a e}{b}+\frac{e x}{b}\right )^{3/2}} \, dx,x,a+b x\right )}{3 (b d-a e)}\\ &=-\frac{4 b}{3 (b d-a e)^2 \sqrt{d+e x}}+\frac{2 \log (a+b x)}{3 (b d-a e) (d+e x)^{3/2}}+\frac{2 b \log (a+b x)}{(b d-a e)^2 \sqrt{d+e x}}-\frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right ) \log (a+b x)}{(b d-a e)^{5/2}}-\frac{(2 b) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{\frac{b d-a e}{b}+\frac{e x}{b}}} \, dx,x,a+b x\right )}{3 (b d-a e)^2}-\frac{b \operatorname{Subst}\left (\int -\frac{2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d-\frac{a e}{b}+\frac{e x}{b}}}{\sqrt{b d-a e}}\right )}{\sqrt{b d-a e} x} \, dx,x,a+b x\right )}{(b d-a e)^2}-\frac{(2 b) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{\frac{b d-a e}{b}+\frac{e x}{b}}} \, dx,x,a+b x\right )}{(b d-a e)^2}\\ &=-\frac{4 b}{3 (b d-a e)^2 \sqrt{d+e x}}+\frac{2 \log (a+b x)}{3 (b d-a e) (d+e x)^{3/2}}+\frac{2 b \log (a+b x)}{(b d-a e)^2 \sqrt{d+e x}}-\frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right ) \log (a+b x)}{(b d-a e)^{5/2}}+\frac{\left (2 b^{3/2}\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d-\frac{a e}{b}+\frac{e x}{b}}}{\sqrt{b d-a e}}\right )}{x} \, dx,x,a+b x\right )}{(b d-a e)^{5/2}}-\frac{\left (4 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{b d-a e}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{3 e (b d-a e)^2}-\frac{\left (4 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{b d-a e}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{e (b d-a e)^2}\\ &=-\frac{4 b}{3 (b d-a e)^2 \sqrt{d+e x}}+\frac{16 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{3 (b d-a e)^{5/2}}+\frac{2 \log (a+b x)}{3 (b d-a e) (d+e x)^{3/2}}+\frac{2 b \log (a+b x)}{(b d-a e)^2 \sqrt{d+e x}}-\frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right ) \log (a+b x)}{(b d-a e)^{5/2}}+\frac{\left (4 b^{5/2}\right ) \operatorname{Subst}\left (\int \frac{x \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b d-a e}}\right )}{a e+b \left (-d+x^2\right )} \, dx,x,\sqrt{d+e x}\right )}{(b d-a e)^{5/2}}\\ &=-\frac{4 b}{3 (b d-a e)^2 \sqrt{d+e x}}+\frac{16 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{3 (b d-a e)^{5/2}}+\frac{2 \log (a+b x)}{3 (b d-a e) (d+e x)^{3/2}}+\frac{2 b \log (a+b x)}{(b d-a e)^2 \sqrt{d+e x}}-\frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right ) \log (a+b x)}{(b d-a e)^{5/2}}+\frac{\left (4 b^{5/2}\right ) \operatorname{Subst}\left (\int \frac{x \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b d-a e}}\right )}{-b d+a e+b x^2} \, dx,x,\sqrt{d+e x}\right )}{(b d-a e)^{5/2}}\\ &=-\frac{4 b}{3 (b d-a e)^2 \sqrt{d+e x}}+\frac{16 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{3 (b d-a e)^{5/2}}+\frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )^2}{(b d-a e)^{5/2}}+\frac{2 \log (a+b x)}{3 (b d-a e) (d+e x)^{3/2}}+\frac{2 b \log (a+b x)}{(b d-a e)^2 \sqrt{d+e x}}-\frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right ) \log (a+b x)}{(b d-a e)^{5/2}}-\frac{\left (4 b^2\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b d-a e}}\right )}{1-\frac{\sqrt{b} x}{\sqrt{b d-a e}}} \, dx,x,\sqrt{d+e x}\right )}{(b d-a e)^3}\\ &=-\frac{4 b}{3 (b d-a e)^2 \sqrt{d+e x}}+\frac{16 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{3 (b d-a e)^{5/2}}+\frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )^2}{(b d-a e)^{5/2}}+\frac{2 \log (a+b x)}{3 (b d-a e) (d+e x)^{3/2}}+\frac{2 b \log (a+b x)}{(b d-a e)^2 \sqrt{d+e x}}-\frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right ) \log (a+b x)}{(b d-a e)^{5/2}}-\frac{4 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right ) \log \left (\frac{2}{1-\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}}\right )}{(b d-a e)^{5/2}}+\frac{\left (4 b^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{2}{1-\frac{\sqrt{b} x}{\sqrt{b d-a e}}}\right )}{1-\frac{b x^2}{b d-a e}} \, dx,x,\sqrt{d+e x}\right )}{(b d-a e)^3}\\ &=-\frac{4 b}{3 (b d-a e)^2 \sqrt{d+e x}}+\frac{16 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{3 (b d-a e)^{5/2}}+\frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )^2}{(b d-a e)^{5/2}}+\frac{2 \log (a+b x)}{3 (b d-a e) (d+e x)^{3/2}}+\frac{2 b \log (a+b x)}{(b d-a e)^2 \sqrt{d+e x}}-\frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right ) \log (a+b x)}{(b d-a e)^{5/2}}-\frac{4 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right ) \log \left (\frac{2}{1-\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}}\right )}{(b d-a e)^{5/2}}-\frac{\left (4 b^{3/2}\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}}\right )}{(b d-a e)^{5/2}}\\ &=-\frac{4 b}{3 (b d-a e)^2 \sqrt{d+e x}}+\frac{16 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{3 (b d-a e)^{5/2}}+\frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )^2}{(b d-a e)^{5/2}}+\frac{2 \log (a+b x)}{3 (b d-a e) (d+e x)^{3/2}}+\frac{2 b \log (a+b x)}{(b d-a e)^2 \sqrt{d+e x}}-\frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right ) \log (a+b x)}{(b d-a e)^{5/2}}-\frac{4 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right ) \log \left (\frac{2}{1-\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}}\right )}{(b d-a e)^{5/2}}-\frac{2 b^{3/2} \text{Li}_2\left (1-\frac{2}{1-\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}}\right )}{(b d-a e)^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.865715, size = 568, normalized size = 1.53 \[ \frac{-3 b^{3/2} (d+e x)^{3/2} \left (2 \text{PolyLog}\left (2,\frac{1}{2}-\frac{\sqrt{b} \sqrt{d+e x}}{2 \sqrt{b d-a e}}\right )+\log \left (\sqrt{b d-a e}-\sqrt{b} \sqrt{d+e x}\right ) \left (\log \left (\sqrt{b d-a e}-\sqrt{b} \sqrt{d+e x}\right )+2 \log \left (\frac{1}{2} \left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}+1\right )\right )\right )\right )+3 b^{3/2} (d+e x)^{3/2} \left (2 \text{PolyLog}\left (2,\frac{1}{2} \left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}+1\right )\right )+\log \left (\sqrt{b d-a e}+\sqrt{b} \sqrt{d+e x}\right ) \left (\log \left (\sqrt{b d-a e}+\sqrt{b} \sqrt{d+e x}\right )+2 \log \left (\frac{1}{2}-\frac{\sqrt{b} \sqrt{d+e x}}{2 \sqrt{b d-a e}}\right )\right )\right )+6 b^{3/2} (d+e x)^{3/2} \log (a+b x) \log \left (\sqrt{b d-a e}-\sqrt{b} \sqrt{d+e x}\right )-6 b^{3/2} (d+e x)^{3/2} \log (a+b x) \log \left (\sqrt{b d-a e}+\sqrt{b} \sqrt{d+e x}\right )+24 b^{3/2} (d+e x)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )-8 b (d+e x) \sqrt{b d-a e} \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b (d+e x)}{b d-a e}\right )+4 (b d-a e)^{3/2} \log (a+b x)+12 b (d+e x) \sqrt{b d-a e} \log (a+b x)}{6 (d+e x)^{3/2} (b d-a e)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(24*b^(3/2)*(d + e*x)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]] - 8*b*Sqrt[b*d - a*e]*(d + e*x)*H
ypergeometric2F1[-1/2, 1, 1/2, (b*(d + e*x))/(b*d - a*e)] + 4*(b*d - a*e)^(3/2)*Log[a + b*x] + 12*b*Sqrt[b*d -
 a*e]*(d + e*x)*Log[a + b*x] + 6*b^(3/2)*(d + e*x)^(3/2)*Log[a + b*x]*Log[Sqrt[b*d - a*e] - Sqrt[b]*Sqrt[d + e
*x]] - 6*b^(3/2)*(d + e*x)^(3/2)*Log[a + b*x]*Log[Sqrt[b*d - a*e] + Sqrt[b]*Sqrt[d + e*x]] - 3*b^(3/2)*(d + e*
x)^(3/2)*(Log[Sqrt[b*d - a*e] - Sqrt[b]*Sqrt[d + e*x]]*(Log[Sqrt[b*d - a*e] - Sqrt[b]*Sqrt[d + e*x]] + 2*Log[(
1 + (Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e])/2]) + 2*PolyLog[2, 1/2 - (Sqrt[b]*Sqrt[d + e*x])/(2*Sqrt[b*d - a*
e])]) + 3*b^(3/2)*(d + e*x)^(3/2)*(Log[Sqrt[b*d - a*e] + Sqrt[b]*Sqrt[d + e*x]]*(Log[Sqrt[b*d - a*e] + Sqrt[b]
*Sqrt[d + e*x]] + 2*Log[1/2 - (Sqrt[b]*Sqrt[d + e*x])/(2*Sqrt[b*d - a*e])]) + 2*PolyLog[2, (1 + (Sqrt[b]*Sqrt[
d + e*x])/Sqrt[b*d - a*e])/2]))/(6*(b*d - a*e)^(5/2)*(d + e*x)^(3/2))

________________________________________________________________________________________

Maple [F]  time = 0.865, size = 0, normalized size = 0. \begin{align*} \int{\frac{\ln \left ( bx+a \right ) }{bx+a} \left ( ex+d \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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(log(b*x+a)/(b*x+a)/(e*x+d)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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(ln(b*x+a)/(b*x+a)/(e*x+d)**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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