3.199 \(\int \frac {(f+g x)^{3/2} (a+b \log (c (d+e x)^n))}{d+e x} \, dx\)

Optimal. Leaf size=417 \[ -\frac {2 (e f-d g)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{e^{5/2}}+\frac {2 \sqrt {f+g x} (e f-d g) \left (a+b \log \left (c (d+e x)^n\right )\right )}{e^2}+\frac {2 (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 e}-\frac {2 b n (e f-d g)^{3/2} \text {Li}_2\left (1-\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right )}{e^{5/2}}+\frac {2 b n (e f-d g)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )^2}{e^{5/2}}+\frac {16 b n (e f-d g)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{3 e^{5/2}}-\frac {4 b n (e f-d g)^{3/2} \log \left (\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{5/2}}-\frac {16 b n \sqrt {f+g x} (e f-d g)}{3 e^2}-\frac {4 b n (f+g x)^{3/2}}{9 e} \]

[Out]

-4/9*b*n*(g*x+f)^(3/2)/e+16/3*b*(-d*g+e*f)^(3/2)*n*arctanh(e^(1/2)*(g*x+f)^(1/2)/(-d*g+e*f)^(1/2))/e^(5/2)+2*b
*(-d*g+e*f)^(3/2)*n*arctanh(e^(1/2)*(g*x+f)^(1/2)/(-d*g+e*f)^(1/2))^2/e^(5/2)+2/3*(g*x+f)^(3/2)*(a+b*ln(c*(e*x
+d)^n))/e-2*(-d*g+e*f)^(3/2)*arctanh(e^(1/2)*(g*x+f)^(1/2)/(-d*g+e*f)^(1/2))*(a+b*ln(c*(e*x+d)^n))/e^(5/2)-4*b
*(-d*g+e*f)^(3/2)*n*arctanh(e^(1/2)*(g*x+f)^(1/2)/(-d*g+e*f)^(1/2))*ln(2/(1-e^(1/2)*(g*x+f)^(1/2)/(-d*g+e*f)^(
1/2)))/e^(5/2)-2*b*(-d*g+e*f)^(3/2)*n*polylog(2,1-2/(1-e^(1/2)*(g*x+f)^(1/2)/(-d*g+e*f)^(1/2)))/e^(5/2)-16/3*b
*(-d*g+e*f)*n*(g*x+f)^(1/2)/e^2+2*(-d*g+e*f)*(a+b*ln(c*(e*x+d)^n))*(g*x+f)^(1/2)/e^2

________________________________________________________________________________________

Rubi [A]  time = 1.38, antiderivative size = 417, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 14, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {2411, 2346, 63, 208, 2348, 12, 1587, 6741, 5984, 5918, 2402, 2315, 2319, 50} \[ -\frac {2 b n (e f-d g)^{3/2} \text {PolyLog}\left (2,1-\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right )}{e^{5/2}}+\frac {2 \sqrt {f+g x} (e f-d g) \left (a+b \log \left (c (d+e x)^n\right )\right )}{e^2}-\frac {2 (e f-d g)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{e^{5/2}}+\frac {2 (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 e}-\frac {16 b n \sqrt {f+g x} (e f-d g)}{3 e^2}+\frac {2 b n (e f-d g)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )^2}{e^{5/2}}+\frac {16 b n (e f-d g)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{3 e^{5/2}}-\frac {4 b n (e f-d g)^{3/2} \log \left (\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{5/2}}-\frac {4 b n (f+g x)^{3/2}}{9 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*b*(e*f - d*g)*n*Sqrt[f + g*x])/(3*e^2) - (4*b*n*(f + g*x)^(3/2))/(9*e) + (16*b*(e*f - d*g)^(3/2)*n*ArcTan
h[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/(3*e^(5/2)) + (2*b*(e*f - d*g)^(3/2)*n*ArcTanh[(Sqrt[e]*Sqrt[f + g
*x])/Sqrt[e*f - d*g]]^2)/e^(5/2) + (2*(e*f - d*g)*Sqrt[f + g*x]*(a + b*Log[c*(d + e*x)^n]))/e^2 + (2*(f + g*x)
^(3/2)*(a + b*Log[c*(d + e*x)^n]))/(3*e) - (2*(e*f - d*g)^(3/2)*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g
]]*(a + b*Log[c*(d + e*x)^n]))/e^(5/2) - (4*b*(e*f - d*g)^(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^(5/2) - (2*b*(e*f - d*g)^(3/2)*n*PolyLog[2, 1 - 2
/(1 - (Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g])])/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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, 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 2346

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.))/(x_), x_Symbol] :> Dist[d, Int[((d
+ e*x)^(q - 1)*(a + b*Log[c*x^n])^p)/x, x], x] + Dist[e, Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^p, x], x] /
; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && GtQ[q, 0] && 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 {(f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{d+e x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (\frac {e f-d g}{e}+\frac {g x}{e}\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x} \, dx,x,d+e x\right )}{e}\\ &=\frac {g \operatorname {Subst}\left (\int \sqrt {\frac {e f-d g}{e}+\frac {g x}{e}} \left (a+b \log \left (c x^n\right )\right ) \, dx,x,d+e x\right )}{e^2}+\frac {(e f-d g) \operatorname {Subst}\left (\int \frac {\sqrt {\frac {e f-d g}{e}+\frac {g x}{e}} \left (a+b \log \left (c x^n\right )\right )}{x} \, dx,x,d+e x\right )}{e^2}\\ &=\frac {2 (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 e}+\frac {(g (e f-d g)) \operatorname {Subst}\left (\int \frac {a+b \log \left (c x^n\right )}{\sqrt {\frac {e f-d g}{e}+\frac {g x}{e}}} \, dx,x,d+e x\right )}{e^3}+\frac {(e f-d g)^2 \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^3}-\frac {(2 b n) \operatorname {Subst}\left (\int \frac {\left (\frac {e f-d g}{e}+\frac {g x}{e}\right )^{3/2}}{x} \, dx,x,d+e x\right )}{3 e}\\ &=-\frac {4 b n (f+g x)^{3/2}}{9 e}+\frac {2 (e f-d g) \sqrt {f+g x} \left (a+b \log \left (c (d+e x)^n\right )\right )}{e^2}+\frac {2 (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 e}-\frac {2 (e f-d g)^{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^{5/2}}-\frac {(2 b (e f-d g) n) \operatorname {Subst}\left (\int \frac {\sqrt {\frac {e f-d g}{e}+\frac {g x}{e}}}{x} \, dx,x,d+e x\right )}{3 e^2}-\frac {(2 b (e f-d g) n) \operatorname {Subst}\left (\int \frac {\sqrt {\frac {e f-d g}{e}+\frac {g x}{e}}}{x} \, dx,x,d+e x\right )}{e^2}-\frac {\left (b (e f-d g)^2 n\right ) \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^3}\\ &=-\frac {16 b (e f-d g) n \sqrt {f+g x}}{3 e^2}-\frac {4 b n (f+g x)^{3/2}}{9 e}+\frac {2 (e f-d g) \sqrt {f+g x} \left (a+b \log \left (c (d+e x)^n\right )\right )}{e^2}+\frac {2 (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 e}-\frac {2 (e f-d g)^{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^{5/2}}+\frac {\left (2 b (e f-d g)^{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^{5/2}}-\frac {\left (2 b (e f-d g)^2 n\right ) \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^3}-\frac {\left (2 b (e f-d g)^2 n\right ) \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^3}\\ &=-\frac {16 b (e f-d g) n \sqrt {f+g x}}{3 e^2}-\frac {4 b n (f+g x)^{3/2}}{9 e}+\frac {2 (e f-d g) \sqrt {f+g x} \left (a+b \log \left (c (d+e x)^n\right )\right )}{e^2}+\frac {2 (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 e}-\frac {2 (e f-d g)^{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^{5/2}}+\frac {\left (4 b (e f-d g)^{3/2} 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^{3/2}}-\frac {\left (4 b (e f-d g)^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 e^2 g}-\frac {\left (4 b (e f-d g)^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 )}{e^2 g}\\ &=-\frac {16 b (e f-d g) n \sqrt {f+g x}}{3 e^2}-\frac {4 b n (f+g x)^{3/2}}{9 e}+\frac {16 b (e f-d g)^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{3 e^{5/2}}+\frac {2 (e f-d g) \sqrt {f+g x} \left (a+b \log \left (c (d+e x)^n\right )\right )}{e^2}+\frac {2 (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 e}-\frac {2 (e f-d g)^{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^{5/2}}+\frac {\left (4 b (e f-d g)^{3/2} 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^{3/2}}\\ &=-\frac {16 b (e f-d g) n \sqrt {f+g x}}{3 e^2}-\frac {4 b n (f+g x)^{3/2}}{9 e}+\frac {16 b (e f-d g)^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{3 e^{5/2}}+\frac {2 b (e f-d g)^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )^2}{e^{5/2}}+\frac {2 (e f-d g) \sqrt {f+g x} \left (a+b \log \left (c (d+e x)^n\right )\right )}{e^2}+\frac {2 (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 e}-\frac {2 (e f-d g)^{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^{5/2}}-\frac {(4 b (e f-d g) n) \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^2}\\ &=-\frac {16 b (e f-d g) n \sqrt {f+g x}}{3 e^2}-\frac {4 b n (f+g x)^{3/2}}{9 e}+\frac {16 b (e f-d g)^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{3 e^{5/2}}+\frac {2 b (e f-d g)^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )^2}{e^{5/2}}+\frac {2 (e f-d g) \sqrt {f+g x} \left (a+b \log \left (c (d+e x)^n\right )\right )}{e^2}+\frac {2 (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 e}-\frac {2 (e f-d g)^{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^{5/2}}-\frac {4 b (e f-d g)^{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^{5/2}}+\frac {(4 b (e f-d g) n) \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^2}\\ &=-\frac {16 b (e f-d g) n \sqrt {f+g x}}{3 e^2}-\frac {4 b n (f+g x)^{3/2}}{9 e}+\frac {16 b (e f-d g)^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{3 e^{5/2}}+\frac {2 b (e f-d g)^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )^2}{e^{5/2}}+\frac {2 (e f-d g) \sqrt {f+g x} \left (a+b \log \left (c (d+e x)^n\right )\right )}{e^2}+\frac {2 (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 e}-\frac {2 (e f-d g)^{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^{5/2}}-\frac {4 b (e f-d g)^{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^{5/2}}-\frac {\left (4 b (e f-d g)^{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^{5/2}}\\ &=-\frac {16 b (e f-d g) n \sqrt {f+g x}}{3 e^2}-\frac {4 b n (f+g x)^{3/2}}{9 e}+\frac {16 b (e f-d g)^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{3 e^{5/2}}+\frac {2 b (e f-d g)^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )^2}{e^{5/2}}+\frac {2 (e f-d g) \sqrt {f+g x} \left (a+b \log \left (c (d+e x)^n\right )\right )}{e^2}+\frac {2 (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 e}-\frac {2 (e f-d g)^{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^{5/2}}-\frac {4 b (e f-d g)^{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^{5/2}}-\frac {2 b (e f-d g)^{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^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 1.14, size = 607, normalized size = 1.46 \[ \frac {12 e^{3/2} (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )+18 (e f-d g)^{3/2} \log \left (\sqrt {e f-d g}-\sqrt {e} \sqrt {f+g x}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-18 (e f-d g)^{3/2} \log \left (\sqrt {e f-d g}+\sqrt {e} \sqrt {f+g x}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+36 a \sqrt {e} \sqrt {f+g x} (e f-d g)+36 b \sqrt {e} \sqrt {f+g x} (e f-d g) \log \left (c (d+e x)^n\right )-9 b n (e f-d g)^{3/2} \left (2 \text {Li}_2\left (\frac {1}{2}-\frac {\sqrt {e} \sqrt {f+g x}}{2 \sqrt {e f-d g}}\right )+\log \left (\sqrt {e f-d g}-\sqrt {e} \sqrt {f+g x}\right ) \left (\log \left (\sqrt {e f-d g}-\sqrt {e} \sqrt {f+g x}\right )+2 \log \left (\frac {1}{2} \left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}+1\right )\right )\right )\right )+9 b n (e f-d g)^{3/2} \left (2 \text {Li}_2\left (\frac {1}{2} \left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}+1\right )\right )+\log \left (\sqrt {e f-d g}+\sqrt {e} \sqrt {f+g x}\right ) \left (\log \left (\sqrt {e f-d g}+\sqrt {e} \sqrt {f+g x}\right )+2 \log \left (\frac {1}{2}-\frac {\sqrt {e} \sqrt {f+g x}}{2 \sqrt {e f-d g}}\right )\right )\right )-96 b n (e f-d g) \left (\sqrt {e} \sqrt {f+g x}-\sqrt {e f-d g} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )\right )-8 b e^{3/2} n (f+g x)^{3/2}}{18 e^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(36*a*Sqrt[e]*(e*f - d*g)*Sqrt[f + g*x] - 8*b*e^(3/2)*n*(f + g*x)^(3/2) - 96*b*(e*f - d*g)*n*(Sqrt[e]*Sqrt[f +
g*x] - Sqrt[e*f - d*g]*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]]) + 36*b*Sqrt[e]*(e*f - d*g)*Sqrt[f +
g*x]*Log[c*(d + e*x)^n] + 12*e^(3/2)*(f + g*x)^(3/2)*(a + b*Log[c*(d + e*x)^n]) + 18*(e*f - d*g)^(3/2)*(a + b*
Log[c*(d + e*x)^n])*Log[Sqrt[e*f - d*g] - Sqrt[e]*Sqrt[f + g*x]] - 18*(e*f - d*g)^(3/2)*(a + b*Log[c*(d + e*x)
^n])*Log[Sqrt[e*f - d*g] + Sqrt[e]*Sqrt[f + g*x]] - 9*b*(e*f - d*g)^(3/2)*n*(Log[Sqrt[e*f - d*g] - Sqrt[e]*Sqr
t[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])]) + 9*b*(e*f - d*g)^(3/2)*n*(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]))/(18*e^(5/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.63, size = 0, normalized size = 0.00 \[ \int \frac {\left (g x +f \right )^{\frac {3}{2}} \left (b \ln \left (c \left (e x +d \right )^{n}\right )+a \right )}{e x +d}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(((f + g*x)^(3/2)*(a + b*log(c*(d + e*x)^n)))/(d + e*x), 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((g*x+f)**(3/2)*(a+b*ln(c*(e*x+d)**n))/(e*x+d),x)

[Out]

Timed out

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