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

Optimal. Leaf size=148 \[ -\frac {\left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d-e x} \sqrt {d+e x}}+\frac {b n \left (d^2-e^2 x^2\right )}{e^2 \sqrt {d-e x} \sqrt {d+e x}}-\frac {b d^2 n \sqrt {1-\frac {e^2 x^2}{d^2}} \tanh ^{-1}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right )}{e^2 \sqrt {d-e x} \sqrt {d+e x}} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {2342, 2338, 266, 50, 63, 208} \[ -\frac {\left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d-e x} \sqrt {d+e x}}+\frac {b n \left (d^2-e^2 x^2\right )}{e^2 \sqrt {d-e x} \sqrt {d+e x}}-\frac {b d^2 n \sqrt {1-\frac {e^2 x^2}{d^2}} \tanh ^{-1}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right )}{e^2 \sqrt {d-e x} \sqrt {d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*n*(d^2 - e^2*x^2))/(e^2*Sqrt[d - e*x]*Sqrt[d + e*x]) - (b*d^2*n*Sqrt[1 - (e^2*x^2)/d^2]*ArcTanh[Sqrt[1 - (e
^2*x^2)/d^2]])/(e^2*Sqrt[d - e*x]*Sqrt[d + e*x]) - ((d^2 - e^2*x^2)*(a + b*Log[c*x^n]))/(e^2*Sqrt[d - e*x]*Sqr
t[d + e*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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_))^(q_.), x_Symbol] :
> Simp[(f^m*(d + e*x^r)^(q + 1)*(a + b*Log[c*x^n])^p)/(e*r*(q + 1)), x] - Dist[(b*f^m*n*p)/(e*r*(q + 1)), Int[
((d + e*x^r)^(q + 1)*(a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && EqQ[
m, r - 1] && IGtQ[p, 0] && (IntegerQ[m] || GtQ[f, 0]) && NeQ[r, n] && NeQ[q, -1]

Rule 2342

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

Rubi steps

\begin {align*} \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d-e x} \sqrt {d+e x}} \, dx &=\frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\sqrt {1-\frac {e^2 x^2}{d^2}}} \, dx}{\sqrt {d-e x} \sqrt {d+e x}}\\ &=-\frac {\left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d-e x} \sqrt {d+e x}}+\frac {\left (b d^2 n \sqrt {1-\frac {e^2 x^2}{d^2}}\right ) \int \frac {\sqrt {1-\frac {e^2 x^2}{d^2}}}{x} \, dx}{e^2 \sqrt {d-e x} \sqrt {d+e x}}\\ &=-\frac {\left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d-e x} \sqrt {d+e x}}+\frac {\left (b d^2 n \sqrt {1-\frac {e^2 x^2}{d^2}}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1-\frac {e^2 x}{d^2}}}{x} \, dx,x,x^2\right )}{2 e^2 \sqrt {d-e x} \sqrt {d+e x}}\\ &=\frac {b n \left (d^2-e^2 x^2\right )}{e^2 \sqrt {d-e x} \sqrt {d+e x}}-\frac {\left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d-e x} \sqrt {d+e x}}+\frac {\left (b d^2 n \sqrt {1-\frac {e^2 x^2}{d^2}}\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {e^2 x}{d^2}}} \, dx,x,x^2\right )}{2 e^2 \sqrt {d-e x} \sqrt {d+e x}}\\ &=\frac {b n \left (d^2-e^2 x^2\right )}{e^2 \sqrt {d-e x} \sqrt {d+e x}}-\frac {\left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d-e x} \sqrt {d+e x}}-\frac {\left (b d^4 n \sqrt {1-\frac {e^2 x^2}{d^2}}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {d^2}{e^2}-\frac {d^2 x^2}{e^2}} \, dx,x,\sqrt {1-\frac {e^2 x^2}{d^2}}\right )}{e^4 \sqrt {d-e x} \sqrt {d+e x}}\\ &=\frac {b n \left (d^2-e^2 x^2\right )}{e^2 \sqrt {d-e x} \sqrt {d+e x}}-\frac {b d^2 n \sqrt {1-\frac {e^2 x^2}{d^2}} \tanh ^{-1}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right )}{e^2 \sqrt {d-e x} \sqrt {d+e x}}-\frac {\left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d-e x} \sqrt {d+e x}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 113, normalized size = 0.76 \[ -\frac {\sqrt {d-e x} \sqrt {d+e x} \left (a+b \left (\log \left (c x^n\right )-n \log (x)\right )-b n\right )}{e^2}+\frac {b d n \log (x)}{e^2}-\frac {b n \log (x) \sqrt {d-e x} \sqrt {d+e x}}{e^2}-\frac {b d n \log \left (\sqrt {d-e x} \sqrt {d+e x}+d\right )}{e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d*n*Log[x])/e^2 - (b*n*Sqrt[d - e*x]*Sqrt[d + e*x]*Log[x])/e^2 - (Sqrt[d - e*x]*Sqrt[d + e*x]*(a - b*n + b*
(-(n*Log[x]) + Log[c*x^n])))/e^2 - (b*d*n*Log[d + Sqrt[d - e*x]*Sqrt[d + e*x]])/e^2

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fricas [A]  time = 0.44, size = 66, normalized size = 0.45 \[ \frac {b d n \log \left (\frac {\sqrt {e x + d} \sqrt {-e x + d} - d}{x}\right ) - {\left (b n \log \relax (x) - b n + b \log \relax (c) + a\right )} \sqrt {e x + d} \sqrt {-e x + d}}{e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(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)*x/(sqrt(e*x + d)*sqrt(-e*x + d)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.46, size = 105, normalized size = 0.71 \[ -\frac {{\left (d \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right ) - \sqrt {-e^{2} x^{2} + d^{2}}\right )} b n}{e^{2}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} b \log \left (c x^{n}\right )}{e^{2}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} a}{e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{\sqrt {d+e\,x}\,\sqrt {d-e\,x}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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