\(\int \frac {a+b \log (c (d+e x)^n)}{(d+e x) \sqrt {f+g x}} \, dx\) [201]

Optimal result

Integrand size = 31, antiderivative size = 256 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(d+e x) \sqrt {f+g x}} \, dx=\frac {2 b n \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )^2}{\sqrt {e} \sqrt {e f-d g}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{\sqrt {e} \sqrt {e f-d g}}-\frac {4 b n \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \log \left (\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right )}{\sqrt {e} \sqrt {e f-d g}}-\frac {2 b n \operatorname {PolyLog}\left (2,1-\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right )}{\sqrt {e} \sqrt {e f-d g}} \]

[Out]

2*b*n*arctanh(e^(1/2)*(g*x+f)^(1/2)/(-d*g+e*f)^(1/2))^2/e^(1/2)/(-d*g+e*f)^(1/2)-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^(1/2)/(-d*g+e*f)^(1/2)-4*b*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^(1/2)/(-d*g+e*f)^(1/2)-2*b*n*polylog(2,1-2/(1-e
^(1/2)*(g*x+f)^(1/2)/(-d*g+e*f)^(1/2)))/e^(1/2)/(-d*g+e*f)^(1/2)

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {2458, 65, 214, 2390, 12, 1601, 6873, 6131, 6055, 2449, 2352} \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(d+e x) \sqrt {f+g x}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{\sqrt {e} \sqrt {e f-d g}}+\frac {2 b n \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )^2}{\sqrt {e} \sqrt {e f-d g}}-\frac {4 b n \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \log \left (\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right )}{\sqrt {e} \sqrt {e f-d g}}-\frac {2 b n \operatorname {PolyLog}\left (2,1-\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right )}{\sqrt {e} \sqrt {e f-d g}} \]

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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*} \text {integral}& = \frac {\text {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} \\ & = -\frac {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 )}{\sqrt {e} \sqrt {e f-d g}}-\frac {(b n) \text {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} \\ & = -\frac {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 )}{\sqrt {e} \sqrt {e f-d g}}+\frac {(2 b n) \text {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 )}{\sqrt {e} \sqrt {e f-d g}} \\ & = -\frac {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 )}{\sqrt {e} \sqrt {e f-d g}}+\frac {\left (4 b \sqrt {e} n\right ) \text {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 )}{\sqrt {e f-d g}} \\ & = -\frac {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 )}{\sqrt {e} \sqrt {e f-d g}}+\frac {\left (4 b \sqrt {e} n\right ) \text {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 )}{\sqrt {e f-d g}} \\ & = \frac {2 b n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )^2}{\sqrt {e} \sqrt {e f-d g}}-\frac {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 )}{\sqrt {e} \sqrt {e f-d g}}-\frac {(4 b n) \text {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} \\ & = \frac {2 b n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )^2}{\sqrt {e} \sqrt {e f-d g}}-\frac {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 )}{\sqrt {e} \sqrt {e f-d g}}-\frac {4 b 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 )}{\sqrt {e} \sqrt {e f-d g}}+\frac {(4 b n) \text {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} \\ & = \frac {2 b n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )^2}{\sqrt {e} \sqrt {e f-d g}}-\frac {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 )}{\sqrt {e} \sqrt {e f-d g}}-\frac {4 b 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 )}{\sqrt {e} \sqrt {e f-d g}}-\frac {(4 b n) \text {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 )}{\sqrt {e} \sqrt {e f-d g}} \\ & = \frac {2 b n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )^2}{\sqrt {e} \sqrt {e f-d g}}-\frac {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 )}{\sqrt {e} \sqrt {e f-d g}}-\frac {4 b 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 )}{\sqrt {e} \sqrt {e f-d g}}-\frac {2 b n \text {Li}_2\left (1-\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right )}{\sqrt {e} \sqrt {e f-d g}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.23 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.10 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(d+e x) \sqrt {f+g x}} \, dx=\frac {-2 a \sqrt {-e f+d g} \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )+2 b \sqrt {e f-d g} \arctan \left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {-e f+d g}}\right ) \left (i n \arctan \left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {-e f+d g}}\right )+\log \left (c (d+e x)^n\right )+2 n \log \left (\frac {2 i}{i-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {-e f+d g}}}\right )\right )+2 i b \sqrt {e f-d g} n \operatorname {PolyLog}\left (2,-\frac {\sqrt {-e f+d g}-i \sqrt {e} \sqrt {f+g x}}{\sqrt {-e f+d g}+i \sqrt {e} \sqrt {f+g x}}\right )}{\sqrt {e} \sqrt {-(e f-d g)^2}} \]

[In]

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

[Out]

(-2*a*Sqrt[-(e*f) + d*g]*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]] + 2*b*Sqrt[e*f - d*g]*ArcTan[(Sqrt[e
]*Sqrt[f + g*x])/Sqrt[-(e*f) + d*g]]*(I*n*ArcTan[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[-(e*f) + d*g]] + Log[c*(d + e*x)
^n] + 2*n*Log[(2*I)/(I - (Sqrt[e]*Sqrt[f + g*x])/Sqrt[-(e*f) + d*g])]) + (2*I)*b*Sqrt[e*f - d*g]*n*PolyLog[2,
-((Sqrt[-(e*f) + d*g] - I*Sqrt[e]*Sqrt[f + g*x])/(Sqrt[-(e*f) + d*g] + I*Sqrt[e]*Sqrt[f + g*x]))])/(Sqrt[e]*Sq
rt[-(e*f - d*g)^2])

Maple [F]

\[\int \frac {a +b \ln \left (c \left (e x +d \right )^{n}\right )}{\left (e x +d \right ) \sqrt {g x +f}}d x\]

[In]

int((a+b*ln(c*(e*x+d)^n))/(e*x+d)/(g*x+f)^(1/2),x)

[Out]

int((a+b*ln(c*(e*x+d)^n))/(e*x+d)/(g*x+f)^(1/2),x)

Fricas [F]

\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(d+e x) \sqrt {f+g x}} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (e x + d\right )} \sqrt {g x + f}} \,d x } \]

[In]

integrate((a+b*log(c*(e*x+d)^n))/(e*x+d)/(g*x+f)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(d+e x) \sqrt {f+g x}} \, dx=\int \frac {a + b \log {\left (c \left (d + e x\right )^{n} \right )}}{\left (d + e x\right ) \sqrt {f + g x}}\, dx \]

[In]

integrate((a+b*ln(c*(e*x+d)**n))/(e*x+d)/(g*x+f)**(1/2),x)

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(d+e x) \sqrt {f+g x}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((a+b*log(c*(e*x+d)^n))/(e*x+d)/(g*x+f)^(1/2),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(e*(d*g-e*f)>0)', see `assume?`
for more de

Giac [F]

\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(d+e x) \sqrt {f+g x}} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (e x + d\right )} \sqrt {g x + f}} \,d x } \]

[In]

integrate((a+b*log(c*(e*x+d)^n))/(e*x+d)/(g*x+f)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(d+e x) \sqrt {f+g x}} \, dx=\int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{\sqrt {f+g\,x}\,\left (d+e\,x\right )} \,d x \]

[In]

int((a + b*log(c*(d + e*x)^n))/((f + g*x)^(1/2)*(d + e*x)),x)

[Out]

int((a + b*log(c*(d + e*x)^n))/((f + g*x)^(1/2)*(d + e*x)), x)