3.314 \(\int \frac {a+b \log (c x^n)}{\sqrt {d-e x} \sqrt {d+e x}} \, dx\)

Optimal. Leaf size=248 \[ \frac {d \sqrt {1-\frac {e^2 x^2}{d^2}} \sin ^{-1}\left (\frac {e x}{d}\right ) \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {d-e x} \sqrt {d+e x}}+\frac {i b d n \sqrt {1-\frac {e^2 x^2}{d^2}} \text {Li}_2\left (e^{2 i \sin ^{-1}\left (\frac {e x}{d}\right )}\right )}{2 e \sqrt {d-e x} \sqrt {d+e x}}+\frac {i b d n \sqrt {1-\frac {e^2 x^2}{d^2}} \sin ^{-1}\left (\frac {e x}{d}\right )^2}{2 e \sqrt {d-e x} \sqrt {d+e x}}-\frac {b d n \sqrt {1-\frac {e^2 x^2}{d^2}} \sin ^{-1}\left (\frac {e x}{d}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac {e x}{d}\right )}\right )}{e \sqrt {d-e x} \sqrt {d+e x}} \]

[Out]

1/2*I*b*d*n*arcsin(e*x/d)^2*(1-e^2*x^2/d^2)^(1/2)/e/(-e*x+d)^(1/2)/(e*x+d)^(1/2)-b*d*n*arcsin(e*x/d)*ln(1-(I*e
*x/d+(1-e^2*x^2/d^2)^(1/2))^2)*(1-e^2*x^2/d^2)^(1/2)/e/(-e*x+d)^(1/2)/(e*x+d)^(1/2)+d*arcsin(e*x/d)*(a+b*ln(c*
x^n))*(1-e^2*x^2/d^2)^(1/2)/e/(-e*x+d)^(1/2)/(e*x+d)^(1/2)+1/2*I*b*d*n*polylog(2,(I*e*x/d+(1-e^2*x^2/d^2)^(1/2
))^2)*(1-e^2*x^2/d^2)^(1/2)/e/(-e*x+d)^(1/2)/(e*x+d)^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {2328, 2326, 4625, 3717, 2190, 2279, 2391} \[ \frac {i b d n \sqrt {1-\frac {e^2 x^2}{d^2}} \text {PolyLog}\left (2,e^{2 i \sin ^{-1}\left (\frac {e x}{d}\right )}\right )}{2 e \sqrt {d-e x} \sqrt {d+e x}}+\frac {d \sqrt {1-\frac {e^2 x^2}{d^2}} \sin ^{-1}\left (\frac {e x}{d}\right ) \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {d-e x} \sqrt {d+e x}}+\frac {i b d n \sqrt {1-\frac {e^2 x^2}{d^2}} \sin ^{-1}\left (\frac {e x}{d}\right )^2}{2 e \sqrt {d-e x} \sqrt {d+e x}}-\frac {b d n \sqrt {1-\frac {e^2 x^2}{d^2}} \sin ^{-1}\left (\frac {e x}{d}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac {e x}{d}\right )}\right )}{e \sqrt {d-e x} \sqrt {d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*b*d*n*Sqrt[1 - (e^2*x^2)/d^2]*ArcSin[(e*x)/d]^2)/(e*Sqrt[d - e*x]*Sqrt[d + e*x]) - (b*d*n*Sqrt[1 - (e^2
*x^2)/d^2]*ArcSin[(e*x)/d]*Log[1 - E^((2*I)*ArcSin[(e*x)/d])])/(e*Sqrt[d - e*x]*Sqrt[d + e*x]) + (d*Sqrt[1 - (
e^2*x^2)/d^2]*ArcSin[(e*x)/d]*(a + b*Log[c*x^n]))/(e*Sqrt[d - e*x]*Sqrt[d + e*x]) + ((I/2)*b*d*n*Sqrt[1 - (e^2
*x^2)/d^2]*PolyLog[2, E^((2*I)*ArcSin[(e*x)/d])])/(e*Sqrt[d - e*x]*Sqrt[d + e*x])

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2326

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(ArcSin[(Rt[-e, 2]*x)/S
qrt[d]]*(a + b*Log[c*x^n]))/Rt[-e, 2], x] - Dist[(b*n)/Rt[-e, 2], Int[ArcSin[(Rt[-e, 2]*x)/Sqrt[d]]/x, x], x]
/; FreeQ[{a, b, c, d, e, n}, x] && GtQ[d, 0] && NegQ[e]

Rule 2328

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_.)*(x_)]), x_Symbol] :>
Dist[Sqrt[1 + (e1*e2*x^2)/(d1*d2)]/(Sqrt[d1 + e1*x]*Sqrt[d2 + e2*x]), Int[(a + b*Log[c*x^n])/Sqrt[1 + (e1*e2*x
^2)/(d1*d2)], x], x] /; FreeQ[{a, b, c, d1, e1, d2, e2, n}, x] && EqQ[d2*e1 + d1*e2, 0]

Rule 2391

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

Rule 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*
(m + 1)), x] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 4625

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n/Tan[x], x], x, ArcSin[c*
x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {a+b \log \left (c x^n\right )}{\sqrt {d-e x} \sqrt {d+e x}} \, dx &=\frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \int \frac {a+b \log \left (c x^n\right )}{\sqrt {1-\frac {e^2 x^2}{d^2}}} \, dx}{\sqrt {d-e x} \sqrt {d+e x}}\\ &=\frac {d \sqrt {1-\frac {e^2 x^2}{d^2}} \sin ^{-1}\left (\frac {e x}{d}\right ) \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {d-e x} \sqrt {d+e x}}-\frac {\left (b d n \sqrt {1-\frac {e^2 x^2}{d^2}}\right ) \int \frac {\sin ^{-1}\left (\frac {e x}{d}\right )}{x} \, dx}{e \sqrt {d-e x} \sqrt {d+e x}}\\ &=\frac {d \sqrt {1-\frac {e^2 x^2}{d^2}} \sin ^{-1}\left (\frac {e x}{d}\right ) \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {d-e x} \sqrt {d+e x}}-\frac {\left (b d n \sqrt {1-\frac {e^2 x^2}{d^2}}\right ) \operatorname {Subst}\left (\int x \cot (x) \, dx,x,\sin ^{-1}\left (\frac {e x}{d}\right )\right )}{e \sqrt {d-e x} \sqrt {d+e x}}\\ &=\frac {i b d n \sqrt {1-\frac {e^2 x^2}{d^2}} \sin ^{-1}\left (\frac {e x}{d}\right )^2}{2 e \sqrt {d-e x} \sqrt {d+e x}}+\frac {d \sqrt {1-\frac {e^2 x^2}{d^2}} \sin ^{-1}\left (\frac {e x}{d}\right ) \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {d-e x} \sqrt {d+e x}}+\frac {\left (2 i b d n \sqrt {1-\frac {e^2 x^2}{d^2}}\right ) \operatorname {Subst}\left (\int \frac {e^{2 i x} x}{1-e^{2 i x}} \, dx,x,\sin ^{-1}\left (\frac {e x}{d}\right )\right )}{e \sqrt {d-e x} \sqrt {d+e x}}\\ &=\frac {i b d n \sqrt {1-\frac {e^2 x^2}{d^2}} \sin ^{-1}\left (\frac {e x}{d}\right )^2}{2 e \sqrt {d-e x} \sqrt {d+e x}}-\frac {b d n \sqrt {1-\frac {e^2 x^2}{d^2}} \sin ^{-1}\left (\frac {e x}{d}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac {e x}{d}\right )}\right )}{e \sqrt {d-e x} \sqrt {d+e x}}+\frac {d \sqrt {1-\frac {e^2 x^2}{d^2}} \sin ^{-1}\left (\frac {e x}{d}\right ) \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {d-e x} \sqrt {d+e x}}+\frac {\left (b d n \sqrt {1-\frac {e^2 x^2}{d^2}}\right ) \operatorname {Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}\left (\frac {e x}{d}\right )\right )}{e \sqrt {d-e x} \sqrt {d+e x}}\\ &=\frac {i b d n \sqrt {1-\frac {e^2 x^2}{d^2}} \sin ^{-1}\left (\frac {e x}{d}\right )^2}{2 e \sqrt {d-e x} \sqrt {d+e x}}-\frac {b d n \sqrt {1-\frac {e^2 x^2}{d^2}} \sin ^{-1}\left (\frac {e x}{d}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac {e x}{d}\right )}\right )}{e \sqrt {d-e x} \sqrt {d+e x}}+\frac {d \sqrt {1-\frac {e^2 x^2}{d^2}} \sin ^{-1}\left (\frac {e x}{d}\right ) \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {d-e x} \sqrt {d+e x}}-\frac {\left (i b d n \sqrt {1-\frac {e^2 x^2}{d^2}}\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i \sin ^{-1}\left (\frac {e x}{d}\right )}\right )}{2 e \sqrt {d-e x} \sqrt {d+e x}}\\ &=\frac {i b d n \sqrt {1-\frac {e^2 x^2}{d^2}} \sin ^{-1}\left (\frac {e x}{d}\right )^2}{2 e \sqrt {d-e x} \sqrt {d+e x}}-\frac {b d n \sqrt {1-\frac {e^2 x^2}{d^2}} \sin ^{-1}\left (\frac {e x}{d}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac {e x}{d}\right )}\right )}{e \sqrt {d-e x} \sqrt {d+e x}}+\frac {d \sqrt {1-\frac {e^2 x^2}{d^2}} \sin ^{-1}\left (\frac {e x}{d}\right ) \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {d-e x} \sqrt {d+e x}}+\frac {i b d n \sqrt {1-\frac {e^2 x^2}{d^2}} \text {Li}_2\left (e^{2 i \sin ^{-1}\left (\frac {e x}{d}\right )}\right )}{2 e \sqrt {d-e x} \sqrt {d+e x}}\\ \end {align*}

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Mathematica [A]  time = 0.55, size = 217, normalized size = 0.88 \[ \frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d-e x} \sqrt {d+e x}}\right ) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{e}-\frac {b n \sqrt {1-\frac {e^2 x^2}{d^2}} \left (-\text {Li}_2\left (e^{-2 \sinh ^{-1}\left (\sqrt {-\frac {e^2}{d^2}} x\right )}\right )-2 \log (x) \log \left (\sqrt {1-\frac {e^2 x^2}{d^2}}+x \sqrt {-\frac {e^2}{d^2}}\right )+\sinh ^{-1}\left (x \sqrt {-\frac {e^2}{d^2}}\right )^2+2 \sinh ^{-1}\left (x \sqrt {-\frac {e^2}{d^2}}\right ) \log \left (1-e^{-2 \sinh ^{-1}\left (x \sqrt {-\frac {e^2}{d^2}}\right )}\right )\right )}{2 \sqrt {-\frac {e^2}{d^2}} \sqrt {d-e x} \sqrt {d+e x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(ArcTan[(e*x)/(Sqrt[d - e*x]*Sqrt[d + e*x])]*(a - b*n*Log[x] + b*Log[c*x^n]))/e - (b*n*Sqrt[1 - (e^2*x^2)/d^2]
*(ArcSinh[Sqrt[-(e^2/d^2)]*x]^2 + 2*ArcSinh[Sqrt[-(e^2/d^2)]*x]*Log[1 - E^(-2*ArcSinh[Sqrt[-(e^2/d^2)]*x])] -
2*Log[x]*Log[Sqrt[-(e^2/d^2)]*x + Sqrt[1 - (e^2*x^2)/d^2]] - PolyLog[2, E^(-2*ArcSinh[Sqrt[-(e^2/d^2)]*x])]))/
(2*Sqrt[-(e^2/d^2)]*Sqrt[d - e*x]*Sqrt[d + e*x])

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {e x + d} \sqrt {-e x + d} b \log \left (c x^{n}\right ) + \sqrt {e x + d} \sqrt {-e x + d} a}{e^{2} x^{2} - d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.48, size = 0, normalized size = 0.00 \[ \int \frac {b \ln \left (c \,x^{n}\right )+a}{\sqrt {-e x +d}\, \sqrt {e x +d}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b \int \frac {\log \relax (c) + \log \left (x^{n}\right )}{\sqrt {e x + d} \sqrt {-e x + d}}\,{d x} + \frac {a \arcsin \left (\frac {e x}{d}\right )}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \log {\left (c x^{n} \right )}}{\sqrt {d - e x} \sqrt {d + e x}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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