3.278 \(\int \frac{x (a+b \log (c x^n))}{\sqrt{d+e x^2}} \, dx\)

Optimal. Leaf size=73 \[ \frac{\sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e}-\frac{b n \sqrt{d+e x^2}}{e}+\frac{b \sqrt{d} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{e} \]

[Out]

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

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Rubi [A]  time = 0.0778711, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {2338, 266, 50, 63, 208} \[ \frac{\sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e}-\frac{b n \sqrt{d+e x^2}}{e}+\frac{b \sqrt{d} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.0849266, size = 91, normalized size = 1.25 \[ \frac{a \sqrt{d+e x^2}+b \sqrt{d+e x^2} \log \left (c x^n\right )-b n \sqrt{d+e x^2}+b \sqrt{d} n \log \left (\sqrt{d} \sqrt{d+e x^2}+d\right )-b \sqrt{d} n \log (x)}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.422, size = 0, normalized size = 0. \begin{align*} \int{x \left ( a+b\ln \left ( c{x}^{n} \right ) \right ){\frac{1}{\sqrt{e{x}^{2}+d}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.33309, size = 317, normalized size = 4.34 \begin{align*} \left [\frac{b \sqrt{d} n \log \left (-\frac{e x^{2} + 2 \, \sqrt{e x^{2} + d} \sqrt{d} + 2 \, d}{x^{2}}\right ) + 2 \, \sqrt{e x^{2} + d}{\left (b n \log \left (x\right ) - b n + b \log \left (c\right ) + a\right )}}{2 \, e}, -\frac{b \sqrt{-d} n \arctan \left (\frac{\sqrt{-d}}{\sqrt{e x^{2} + d}}\right ) - \sqrt{e x^{2} + d}{\left (b n \log \left (x\right ) - b n + b \log \left (c\right ) + a\right )}}{e}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(b*sqrt(d)*n*log(-(e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) + 2*sqrt(e*x^2 + d)*(b*n*log(x) - b*n +
b*log(c) + a))/e, -(b*sqrt(-d)*n*arctan(sqrt(-d)/sqrt(e*x^2 + d)) - sqrt(e*x^2 + d)*(b*n*log(x) - b*n + b*log(
c) + a))/e]

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Sympy [A]  time = 4.70787, size = 126, normalized size = 1.73 \begin{align*} a \left (\begin{cases} \frac{x^{2}}{2 \sqrt{d}} & \text{for}\: e = 0 \\\frac{\sqrt{d + e x^{2}}}{e} & \text{otherwise} \end{cases}\right ) - b n \left (\begin{cases} \frac{x^{2}}{4 \sqrt{d}} & \text{for}\: e = 0 \\- \frac{\sqrt{d} \operatorname{asinh}{\left (\frac{\sqrt{d}}{\sqrt{e} x} \right )}}{e} + \frac{d}{e^{\frac{3}{2}} x \sqrt{\frac{d}{e x^{2}} + 1}} + \frac{x}{\sqrt{e} \sqrt{\frac{d}{e x^{2}} + 1}} & \text{otherwise} \end{cases}\right ) + b \left (\begin{cases} \frac{x^{2}}{2 \sqrt{d}} & \text{for}\: e = 0 \\\frac{\sqrt{d + e x^{2}}}{e} & \text{otherwise} \end{cases}\right ) \log{\left (c x^{n} \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*Piecewise((x**2/(2*sqrt(d)), Eq(e, 0)), (sqrt(d + e*x**2)/e, True)) - b*n*Piecewise((x**2/(4*sqrt(d)), Eq(e,
0)), (-sqrt(d)*asinh(sqrt(d)/(sqrt(e)*x))/e + d/(e**(3/2)*x*sqrt(d/(e*x**2) + 1)) + x/(sqrt(e)*sqrt(d/(e*x**2
) + 1)), True)) + b*Piecewise((x**2/(2*sqrt(d)), Eq(e, 0)), (sqrt(d + e*x**2)/e, True))*log(c*x**n)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \log \left (c x^{n}\right ) + a\right )} x}{\sqrt{e x^{2} + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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