3.3.8 \(\int \frac {\log (a+b x)}{(a+b x) (d+e x)^{5/2}} \, dx\) [208]

Optimal. Leaf size=372 \[ -\frac {4 b}{3 (b d-a e)^2 \sqrt {d+e x}}+\frac {16 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{3 (b d-a e)^{5/2}}+\frac {2 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )^2}{(b d-a e)^{5/2}}+\frac {2 \log (a+b x)}{3 (b d-a e) (d+e x)^{3/2}}+\frac {2 b \log (a+b x)}{(b d-a e)^2 \sqrt {d+e x}}-\frac {2 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right ) \log (a+b x)}{(b d-a e)^{5/2}}-\frac {4 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right ) \log \left (\frac {2}{1-\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}}\right )}{(b d-a e)^{5/2}}-\frac {2 b^{3/2} \text {Li}_2\left (1-\frac {2}{1-\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}}\right )}{(b d-a e)^{5/2}} \]

[Out]

16/3*b^(3/2)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/(-a*e+b*d)^(5/2)+2*b^(3/2)*arctanh(b^(1/2)*(e*x+d
)^(1/2)/(-a*e+b*d)^(1/2))^2/(-a*e+b*d)^(5/2)+2/3*ln(b*x+a)/(-a*e+b*d)/(e*x+d)^(3/2)-2*b^(3/2)*arctanh(b^(1/2)*
(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))*ln(b*x+a)/(-a*e+b*d)^(5/2)-4*b^(3/2)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^
(1/2))*ln(2/(1-b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2)))/(-a*e+b*d)^(5/2)-2*b^(3/2)*polylog(2,1-2/(1-b^(1/2)*(e
*x+d)^(1/2)/(-a*e+b*d)^(1/2)))/(-a*e+b*d)^(5/2)-4/3*b/(-a*e+b*d)^2/(e*x+d)^(1/2)+2*b*ln(b*x+a)/(-a*e+b*d)^2/(e
*x+d)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.88, antiderivative size = 372, normalized size of antiderivative = 1.00, 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 = {2458, 2389, 65, 214, 2390, 12, 1601, 6873, 6131, 6055, 2449, 2352, 2356, 53} \begin {gather*} -\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)} \end {gather*}

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 12

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

Rule 53

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]

Rule 65

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 214

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

Rule 1601

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]/(q*Coeff[Qq, x, q]))
*D[Qq, x]]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

Rule 2352

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

Rule 2356

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 2389

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 2390

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 2449

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 2458

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 6055

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 6131

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 6873

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

Rubi steps

\begin {align*} \int \frac {\log (a+b x)}{(a+b x) (d+e x)^{5/2}} \, dx &=\frac {\text {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 {\text {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 \text {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 \text {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 \text {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 \text {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) \text {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 \text {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) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 2.95, size = 197, normalized size = 0.53 \begin {gather*} \frac {2 \left (-\frac {6 \left (\frac {b (d+e x)}{e (a+b x)}\right )^{3/2} \, _3F_2\left (\frac {5}{2},\frac {5}{2},\frac {5}{2};\frac {7}{2},\frac {7}{2};\frac {-b d+a e}{e (a+b x)}\right )}{e (a+b x)}+\frac {25 \left (\sqrt {b d-a e} (4 b d-a e+3 b e x)-3 e^{3/2} (a+b x)^{3/2} \left (\frac {b (d+e x)}{e (a+b x)}\right )^{3/2} \sinh ^{-1}\left (\frac {\sqrt {b d-a e}}{\sqrt {e} \sqrt {a+b x}}\right )\right ) \log (a+b x)}{(b d-a e)^{5/2}}\right )}{75 (d+e x)^{3/2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*((-6*((b*(d + e*x))/(e*(a + b*x)))^(3/2)*HypergeometricPFQ[{5/2, 5/2, 5/2}, {7/2, 7/2}, (-(b*d) + a*e)/(e*(
a + b*x))])/(e*(a + b*x)) + (25*(Sqrt[b*d - a*e]*(4*b*d - a*e + 3*b*e*x) - 3*e^(3/2)*(a + b*x)^(3/2)*((b*(d +
e*x))/(e*(a + b*x)))^(3/2)*ArcSinh[Sqrt[b*d - a*e]/(Sqrt[e]*Sqrt[a + b*x])])*Log[a + b*x])/(b*d - a*e)^(5/2)))
/(75*(d + e*x)^(3/2))

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Maple [F]
time = 0.37, size = 0, normalized size = 0.00 \[\int \frac {\ln \left (b x +a \right )}{\left (b x +a \right ) \left (e x +d \right )^{\frac {5}{2}}}\, dx\]

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)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b*d-%e*a>0)', see `assume?` fo
r more detai

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)*(x*e + d)^(5/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\ln \left (a+b\,x\right )}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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