3.207 \(\int \frac {\log (a+b x)}{(a+b x) (d+e x)^{3/2}} \, dx\)

Optimal. Leaf size=316 \[ -\frac {2 \sqrt {b} \text {Li}_2\left (1-\frac {2}{1-\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}}\right )}{(b d-a e)^{3/2}}+\frac {2 \log (a+b x)}{\sqrt {d+e x} (b d-a e)}+\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )^2}{(b d-a e)^{3/2}}+\frac {4 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}-\frac {2 \sqrt {b} \log (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}-\frac {4 \sqrt {b} \log \left (\frac {2}{1-\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}}\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.94, antiderivative size = 316, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 13, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {2411, 2347, 63, 208, 2348, 12, 1587, 6741, 5984, 5918, 2402, 2315, 2319} \[ -\frac {2 \sqrt {b} \text {PolyLog}\left (2,1-\frac {2}{1-\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}}\right )}{(b d-a e)^{3/2}}+\frac {2 \log (a+b x)}{\sqrt {d+e x} (b d-a e)}+\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )^2}{(b d-a e)^{3/2}}+\frac {4 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}-\frac {2 \sqrt {b} \log (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}-\frac {4 \sqrt {b} \log \left (\frac {2}{1-\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}}\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Log[a + b*x]/((a + b*x)*(d + e*x)^(3/2)),x]

[Out]

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

Rule 12

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

Rule 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 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 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 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 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 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 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 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 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 6741

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)^{3/2}} \, dx &=\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}\\ &=\frac {\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}-\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 (b d-a e)}\\ &=\frac {2 \log (a+b x)}{(b d-a e) \sqrt {d+e x}}-\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right ) \log (a+b x)}{(b d-a e)^{3/2}}-\frac {\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}-\frac {2 \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}\\ &=\frac {2 \log (a+b x)}{(b d-a e) \sqrt {d+e x}}-\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right ) \log (a+b x)}{(b d-a e)^{3/2}}+\frac {\left (2 \sqrt {b}\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)^{3/2}}-\frac {(4 b) \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)}\\ &=\frac {4 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}+\frac {2 \log (a+b x)}{(b d-a e) \sqrt {d+e x}}-\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right ) \log (a+b x)}{(b d-a e)^{3/2}}+\frac {\left (4 b^{3/2}\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)^{3/2}}\\ &=\frac {4 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}+\frac {2 \log (a+b x)}{(b d-a e) \sqrt {d+e x}}-\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right ) \log (a+b x)}{(b d-a e)^{3/2}}+\frac {\left (4 b^{3/2}\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)^{3/2}}\\ &=\frac {4 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}+\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )^2}{(b d-a e)^{3/2}}+\frac {2 \log (a+b x)}{(b d-a e) \sqrt {d+e x}}-\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right ) \log (a+b x)}{(b d-a e)^{3/2}}-\frac {(4 b) \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)^2}\\ &=\frac {4 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}+\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )^2}{(b d-a e)^{3/2}}+\frac {2 \log (a+b x)}{(b d-a e) \sqrt {d+e x}}-\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right ) \log (a+b x)}{(b d-a e)^{3/2}}-\frac {4 \sqrt {b} \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)^{3/2}}+\frac {(4 b) \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)^2}\\ &=\frac {4 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}+\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )^2}{(b d-a e)^{3/2}}+\frac {2 \log (a+b x)}{(b d-a e) \sqrt {d+e x}}-\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right ) \log (a+b x)}{(b d-a e)^{3/2}}-\frac {4 \sqrt {b} \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)^{3/2}}-\frac {\left (4 \sqrt {b}\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)^{3/2}}\\ &=\frac {4 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}+\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )^2}{(b d-a e)^{3/2}}+\frac {2 \log (a+b x)}{(b d-a e) \sqrt {d+e x}}-\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right ) \log (a+b x)}{(b d-a e)^{3/2}}-\frac {4 \sqrt {b} \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)^{3/2}}-\frac {2 \sqrt {b} \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)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.47, size = 550, normalized size = 1.74 \[ \frac {-2 \sqrt {b} \sqrt {d+e x} \text {Li}_2\left (\frac {1}{2}-\frac {\sqrt {b} \sqrt {d+e x}}{2 \sqrt {b d-a e}}\right )+2 \sqrt {b} \sqrt {d+e x} \text {Li}_2\left (\frac {1}{2} \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}+1\right )\right )-\sqrt {b} \sqrt {d+e x} \log ^2\left (\sqrt {b d-a e}-\sqrt {b} \sqrt {d+e x}\right )+\sqrt {b} \sqrt {d+e x} \log ^2\left (\sqrt {b d-a e}+\sqrt {b} \sqrt {d+e x}\right )+2 \sqrt {b} \sqrt {d+e x} \log (a+b x) \log \left (\sqrt {b d-a e}-\sqrt {b} \sqrt {d+e x}\right )-2 \sqrt {b} \sqrt {d+e x} \log \left (\frac {1}{2} \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}+1\right )\right ) \log \left (\sqrt {b d-a e}-\sqrt {b} \sqrt {d+e x}\right )+4 \sqrt {b d-a e} \log (a+b x)-2 \sqrt {b} \sqrt {d+e x} \log (a+b x) \log \left (\sqrt {b d-a e}+\sqrt {b} \sqrt {d+e x}\right )+2 \sqrt {b} \sqrt {d+e x} \log \left (\sqrt {b d-a e}+\sqrt {b} \sqrt {d+e x}\right ) \log \left (\frac {1}{2}-\frac {\sqrt {b} \sqrt {d+e x}}{2 \sqrt {b d-a e}}\right )+8 \sqrt {b} \sqrt {d+e x} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{2 \sqrt {d+e x} (b d-a e)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[a + b*x]/((a + b*x)*(d + e*x)^(3/2)),x]

[Out]

(8*Sqrt[b]*Sqrt[d + e*x]*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]] + 4*Sqrt[b*d - a*e]*Log[a + b*x] + 2
*Sqrt[b]*Sqrt[d + e*x]*Log[a + b*x]*Log[Sqrt[b*d - a*e] - Sqrt[b]*Sqrt[d + e*x]] - Sqrt[b]*Sqrt[d + e*x]*Log[S
qrt[b*d - a*e] - Sqrt[b]*Sqrt[d + e*x]]^2 - 2*Sqrt[b]*Sqrt[d + e*x]*Log[a + b*x]*Log[Sqrt[b*d - a*e] + Sqrt[b]
*Sqrt[d + e*x]] + Sqrt[b]*Sqrt[d + e*x]*Log[Sqrt[b*d - a*e] + Sqrt[b]*Sqrt[d + e*x]]^2 + 2*Sqrt[b]*Sqrt[d + e*
x]*Log[Sqrt[b*d - a*e] + Sqrt[b]*Sqrt[d + e*x]]*Log[1/2 - (Sqrt[b]*Sqrt[d + e*x])/(2*Sqrt[b*d - a*e])] - 2*Sqr
t[b]*Sqrt[d + e*x]*Log[Sqrt[b*d - a*e] - Sqrt[b]*Sqrt[d + e*x]]*Log[(1 + (Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*
e])/2] - 2*Sqrt[b]*Sqrt[d + e*x]*PolyLog[2, 1/2 - (Sqrt[b]*Sqrt[d + e*x])/(2*Sqrt[b*d - a*e])] + 2*Sqrt[b]*Sqr
t[d + e*x]*PolyLog[2, (1 + (Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e])/2])/(2*(b*d - a*e)^(3/2)*Sqrt[d + e*x])

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e x + d} \log \left (b x + a\right )}{b e^{2} x^{3} + a d^{2} + {\left (2 \, b d e + a e^{2}\right )} x^{2} + {\left (b d^{2} + 2 \, a d e\right )} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(b*x+a)/(b*x+a)/(e*x+d)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left (b x + a\right )}{{\left (b x + a\right )} {\left (e x + d\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(b*x+a)/(b*x+a)/(e*x+d)^(3/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.50, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (b x +a \right )}{\left (b x +a \right ) \left (e x +d \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(b*x+a)/(b*x+a)/(e*x+d)^(3/2),x)

[Out]

int(ln(b*x+a)/(b*x+a)/(e*x+d)^(3/2),x)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(b*x+a)/(b*x+a)/(e*x+d)^(3/2),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(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\ln \left (a+b\,x\right )}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(b*x+a)/(b*x+a)/(e*x+d)**(3/2),x)

[Out]

Timed out

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