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

Optimal. Leaf size=102 \[ \frac{\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e}+\frac{b d^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{3 e}-\frac{b d n \sqrt{d+e x^2}}{3 e}-\frac{b n \left (d+e x^2\right )^{3/2}}{9 e} \]

[Out]

-(b*d*n*Sqrt[d + e*x^2])/(3*e) - (b*n*(d + e*x^2)^(3/2))/(9*e) + (b*d^(3/2)*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]
)/(3*e) + ((d + e*x^2)^(3/2)*(a + b*Log[c*x^n]))/(3*e)

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Rubi [A]  time = 0.0878957, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e}+\frac{b d^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{3 e}-\frac{b d n \sqrt{d+e x^2}}{3 e}-\frac{b n \left (d+e x^2\right )^{3/2}}{9 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*d*n*Sqrt[d + e*x^2])/(3*e) - (b*n*(d + e*x^2)^(3/2))/(9*e) + (b*d^(3/2)*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]
)/(3*e) + ((d + e*x^2)^(3/2)*(a + b*Log[c*x^n]))/(3*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 x \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac{\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e}-\frac{(b n) \int \frac{\left (d+e x^2\right )^{3/2}}{x} \, dx}{3 e}\\ &=\frac{\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e}-\frac{(b n) \operatorname{Subst}\left (\int \frac{(d+e x)^{3/2}}{x} \, dx,x,x^2\right )}{6 e}\\ &=-\frac{b n \left (d+e x^2\right )^{3/2}}{9 e}+\frac{\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e}-\frac{(b d n) \operatorname{Subst}\left (\int \frac{\sqrt{d+e x}}{x} \, dx,x,x^2\right )}{6 e}\\ &=-\frac{b d n \sqrt{d+e x^2}}{3 e}-\frac{b n \left (d+e x^2\right )^{3/2}}{9 e}+\frac{\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e}-\frac{\left (b d^2 n\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d+e x}} \, dx,x,x^2\right )}{6 e}\\ &=-\frac{b d n \sqrt{d+e x^2}}{3 e}-\frac{b n \left (d+e x^2\right )^{3/2}}{9 e}+\frac{\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e}-\frac{\left (b d^2 n\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x^2}\right )}{3 e^2}\\ &=-\frac{b d n \sqrt{d+e x^2}}{3 e}-\frac{b n \left (d+e x^2\right )^{3/2}}{9 e}+\frac{b d^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{3 e}+\frac{\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e}\\ \end{align*}

Mathematica [A]  time = 0.104242, size = 136, normalized size = 1.33 \[ \frac{3 a e x^2 \sqrt{d+e x^2}+3 a d \sqrt{d+e x^2}+3 b \left (d+e x^2\right )^{3/2} \log \left (c x^n\right )+3 b d^{3/2} n \log \left (\sqrt{d} \sqrt{d+e x^2}+d\right )-3 b d^{3/2} n \log (x)-b e n x^2 \sqrt{d+e x^2}-4 b d n \sqrt{d+e x^2}}{9 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*d*Sqrt[d + e*x^2] - 4*b*d*n*Sqrt[d + e*x^2] + 3*a*e*x^2*Sqrt[d + e*x^2] - b*e*n*x^2*Sqrt[d + e*x^2] - 3*b
*d^(3/2)*n*Log[x] + 3*b*(d + e*x^2)^(3/2)*Log[c*x^n] + 3*b*d^(3/2)*n*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/(9*e)

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Maple [F]  time = 0.468, size = 0, normalized size = 0. \begin{align*} \int x \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \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 = 1.55026, size = 505, normalized size = 4.95 \begin{align*} \left [\frac{3 \, b d^{\frac{3}{2}} n \log \left (-\frac{e x^{2} + 2 \, \sqrt{e x^{2} + d} \sqrt{d} + 2 \, d}{x^{2}}\right ) - 2 \,{\left (4 \, b d n +{\left (b e n - 3 \, a e\right )} x^{2} - 3 \, a d - 3 \,{\left (b e x^{2} + b d\right )} \log \left (c\right ) - 3 \,{\left (b e n x^{2} + b d n\right )} \log \left (x\right )\right )} \sqrt{e x^{2} + d}}{18 \, e}, -\frac{3 \, b \sqrt{-d} d n \arctan \left (\frac{\sqrt{-d}}{\sqrt{e x^{2} + d}}\right ) +{\left (4 \, b d n +{\left (b e n - 3 \, a e\right )} x^{2} - 3 \, a d - 3 \,{\left (b e x^{2} + b d\right )} \log \left (c\right ) - 3 \,{\left (b e n x^{2} + b d n\right )} \log \left (x\right )\right )} \sqrt{e x^{2} + d}}{9 \, 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]

[1/18*(3*b*d^(3/2)*n*log(-(e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) - 2*(4*b*d*n + (b*e*n - 3*a*e)*x^2 -
3*a*d - 3*(b*e*x^2 + b*d)*log(c) - 3*(b*e*n*x^2 + b*d*n)*log(x))*sqrt(e*x^2 + d))/e, -1/9*(3*b*sqrt(-d)*d*n*ar
ctan(sqrt(-d)/sqrt(e*x^2 + d)) + (4*b*d*n + (b*e*n - 3*a*e)*x^2 - 3*a*d - 3*(b*e*x^2 + b*d)*log(c) - 3*(b*e*n*
x^2 + b*d*n)*log(x))*sqrt(e*x^2 + d))/e]

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Sympy [A]  time = 20.0459, size = 155, normalized size = 1.52 \begin{align*} a \left (\begin{cases} \frac{\sqrt{d} x^{2}}{2} & \text{for}\: e = 0 \\\frac{\left (d + e x^{2}\right )^{\frac{3}{2}}}{3 e} & \text{otherwise} \end{cases}\right ) - b n \left (\begin{cases} \frac{\sqrt{d} x^{2}}{4} & \text{for}\: e = 0 \\\frac{4 d^{\frac{3}{2}} \sqrt{1 + \frac{e x^{2}}{d}}}{9 e} + \frac{d^{\frac{3}{2}} \log{\left (\frac{e x^{2}}{d} \right )}}{6 e} - \frac{d^{\frac{3}{2}} \log{\left (\sqrt{1 + \frac{e x^{2}}{d}} + 1 \right )}}{3 e} + \frac{\sqrt{d} x^{2} \sqrt{1 + \frac{e x^{2}}{d}}}{9} & \text{otherwise} \end{cases}\right ) + b \left (\begin{cases} \frac{\sqrt{d} x^{2}}{2} & \text{for}\: e = 0 \\\frac{\left (d + e x^{2}\right )^{\frac{3}{2}}}{3 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((sqrt(d)*x**2/2, Eq(e, 0)), ((d + e*x**2)**(3/2)/(3*e), True)) - b*n*Piecewise((sqrt(d)*x**2/4, Eq
(e, 0)), (4*d**(3/2)*sqrt(1 + e*x**2/d)/(9*e) + d**(3/2)*log(e*x**2/d)/(6*e) - d**(3/2)*log(sqrt(1 + e*x**2/d)
+ 1)/(3*e) + sqrt(d)*x**2*sqrt(1 + e*x**2/d)/9, True)) + b*Piecewise((sqrt(d)*x**2/2, Eq(e, 0)), ((d + e*x**2
)**(3/2)/(3*e), True))*log(c*x**n)

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Giac [A]  time = 1.40223, size = 196, normalized size = 1.92 \begin{align*} \frac{1}{3} \, \sqrt{x^{2} e + d} b x^{2} \log \left (c\right ) + \frac{1}{3} \, \sqrt{x^{2} e + d} b d e^{\left (-1\right )} \log \left (c\right ) + \frac{1}{3} \, \sqrt{x^{2} e + d} a x^{2} + \frac{1}{3} \, \sqrt{x^{2} e + d} a d e^{\left (-1\right )} + \frac{1}{9} \,{\left (3 \,{\left (x^{2} e + d\right )}^{\frac{3}{2}} e^{\left (-1\right )} \log \left (x\right ) -{\left (\frac{3 \, d^{2} \arctan \left (\frac{\sqrt{x^{2} e + d}}{\sqrt{-d}}\right )}{\sqrt{-d}} +{\left (x^{2} e + d\right )}^{\frac{3}{2}} + 3 \, \sqrt{x^{2} e + d} d\right )} e^{\left (-1\right )}\right )} b n \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]

1/3*sqrt(x^2*e + d)*b*x^2*log(c) + 1/3*sqrt(x^2*e + d)*b*d*e^(-1)*log(c) + 1/3*sqrt(x^2*e + d)*a*x^2 + 1/3*sqr
t(x^2*e + d)*a*d*e^(-1) + 1/9*(3*(x^2*e + d)^(3/2)*e^(-1)*log(x) - (3*d^2*arctan(sqrt(x^2*e + d)/sqrt(-d))/sqr
t(-d) + (x^2*e + d)^(3/2) + 3*sqrt(x^2*e + d)*d)*e^(-1))*b*n