3.265 \(\int x (d+e x^2)^{3/2} (a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=125 \[ \frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac{b d^2 n \sqrt{d+e x^2}}{5 e}+\frac{b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{5 e}-\frac{b d n \left (d+e x^2\right )^{3/2}}{15 e}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 e} \]

[Out]

-(b*d^2*n*Sqrt[d + e*x^2])/(5*e) - (b*d*n*(d + e*x^2)^(3/2))/(15*e) - (b*n*(d + e*x^2)^(5/2))/(25*e) + (b*d^(5
/2)*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(5*e) + ((d + e*x^2)^(5/2)*(a + b*Log[c*x^n]))/(5*e)

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Rubi [A]  time = 0.106275, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac{b d^2 n \sqrt{d+e x^2}}{5 e}+\frac{b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{5 e}-\frac{b d n \left (d+e x^2\right )^{3/2}}{15 e}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 e} \]

Antiderivative was successfully verified.

[In]

Int[x*(d + e*x^2)^(3/2)*(a + b*Log[c*x^n]),x]

[Out]

-(b*d^2*n*Sqrt[d + e*x^2])/(5*e) - (b*d*n*(d + e*x^2)^(3/2))/(15*e) - (b*n*(d + e*x^2)^(5/2))/(25*e) + (b*d^(5
/2)*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(5*e) + ((d + e*x^2)^(5/2)*(a + b*Log[c*x^n]))/(5*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 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac{(b n) \int \frac{\left (d+e x^2\right )^{5/2}}{x} \, dx}{5 e}\\ &=\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac{(b n) \operatorname{Subst}\left (\int \frac{(d+e x)^{5/2}}{x} \, dx,x,x^2\right )}{10 e}\\ &=-\frac{b n \left (d+e x^2\right )^{5/2}}{25 e}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac{(b d n) \operatorname{Subst}\left (\int \frac{(d+e x)^{3/2}}{x} \, dx,x,x^2\right )}{10 e}\\ &=-\frac{b d n \left (d+e x^2\right )^{3/2}}{15 e}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 e}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac{\left (b d^2 n\right ) \operatorname{Subst}\left (\int \frac{\sqrt{d+e x}}{x} \, dx,x,x^2\right )}{10 e}\\ &=-\frac{b d^2 n \sqrt{d+e x^2}}{5 e}-\frac{b d n \left (d+e x^2\right )^{3/2}}{15 e}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 e}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac{\left (b d^3 n\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d+e x}} \, dx,x,x^2\right )}{10 e}\\ &=-\frac{b d^2 n \sqrt{d+e x^2}}{5 e}-\frac{b d n \left (d+e x^2\right )^{3/2}}{15 e}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 e}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac{\left (b d^3 n\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x^2}\right )}{5 e^2}\\ &=-\frac{b d^2 n \sqrt{d+e x^2}}{5 e}-\frac{b d n \left (d+e x^2\right )^{3/2}}{15 e}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 e}+\frac{b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{5 e}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}\\ \end{align*}

Mathematica [A]  time = 0.138715, size = 181, normalized size = 1.45 \[ \sqrt{d+e x^2} \left (\frac{d^2 \left (15 a+15 b \left (\log \left (c x^n\right )-n \log (x)\right )-23 b n\right )}{75 e}+\frac{1}{75} d x^2 \left (30 a+30 b \left (\log \left (c x^n\right )-n \log (x)\right )-11 b n\right )+\frac{1}{25} e x^4 \left (5 a+5 b \left (\log \left (c x^n\right )-n \log (x)\right )-b n\right )\right )+\frac{b d^{5/2} n \log \left (\sqrt{d} \sqrt{d+e x^2}+d\right )}{5 e}-\frac{b d^{5/2} n \log (x)}{5 e}+\frac{b n \log (x) \left (d+e x^2\right )^{5/2}}{5 e} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(d + e*x^2)^(3/2)*(a + b*Log[c*x^n]),x]

[Out]

-(b*d^(5/2)*n*Log[x])/(5*e) + (b*n*(d + e*x^2)^(5/2)*Log[x])/(5*e) + Sqrt[d + e*x^2]*((e*x^4*(5*a - b*n + 5*b*
(-(n*Log[x]) + Log[c*x^n])))/25 + (d^2*(15*a - 23*b*n + 15*b*(-(n*Log[x]) + Log[c*x^n])))/(75*e) + (d*x^2*(30*
a - 11*b*n + 30*b*(-(n*Log[x]) + Log[c*x^n])))/75) + (b*d^(5/2)*n*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/(5*e)

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Maple [F]  time = 0.459, size = 0, normalized size = 0. \begin{align*} \int x \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}} \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(e*x^2+d)^(3/2)*(a+b*ln(c*x^n)),x)

[Out]

int(x*(e*x^2+d)^(3/2)*(a+b*ln(c*x^n)),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*(e*x^2+d)^(3/2)*(a+b*log(c*x^n)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.58339, size = 734, normalized size = 5.87 \begin{align*} \left [\frac{15 \, b d^{\frac{5}{2}} n \log \left (-\frac{e x^{2} + 2 \, \sqrt{e x^{2} + d} \sqrt{d} + 2 \, d}{x^{2}}\right ) - 2 \,{\left (3 \,{\left (b e^{2} n - 5 \, a e^{2}\right )} x^{4} + 23 \, b d^{2} n - 15 \, a d^{2} +{\left (11 \, b d e n - 30 \, a d e\right )} x^{2} - 15 \,{\left (b e^{2} x^{4} + 2 \, b d e x^{2} + b d^{2}\right )} \log \left (c\right ) - 15 \,{\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \log \left (x\right )\right )} \sqrt{e x^{2} + d}}{150 \, e}, -\frac{15 \, b \sqrt{-d} d^{2} n \arctan \left (\frac{\sqrt{-d}}{\sqrt{e x^{2} + d}}\right ) +{\left (3 \,{\left (b e^{2} n - 5 \, a e^{2}\right )} x^{4} + 23 \, b d^{2} n - 15 \, a d^{2} +{\left (11 \, b d e n - 30 \, a d e\right )} x^{2} - 15 \,{\left (b e^{2} x^{4} + 2 \, b d e x^{2} + b d^{2}\right )} \log \left (c\right ) - 15 \,{\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \log \left (x\right )\right )} \sqrt{e x^{2} + d}}{75 \, e}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x^2+d)^(3/2)*(a+b*log(c*x^n)),x, algorithm="fricas")

[Out]

[1/150*(15*b*d^(5/2)*n*log(-(e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) - 2*(3*(b*e^2*n - 5*a*e^2)*x^4 + 23
*b*d^2*n - 15*a*d^2 + (11*b*d*e*n - 30*a*d*e)*x^2 - 15*(b*e^2*x^4 + 2*b*d*e*x^2 + b*d^2)*log(c) - 15*(b*e^2*n*
x^4 + 2*b*d*e*n*x^2 + b*d^2*n)*log(x))*sqrt(e*x^2 + d))/e, -1/75*(15*b*sqrt(-d)*d^2*n*arctan(sqrt(-d)/sqrt(e*x
^2 + d)) + (3*(b*e^2*n - 5*a*e^2)*x^4 + 23*b*d^2*n - 15*a*d^2 + (11*b*d*e*n - 30*a*d*e)*x^2 - 15*(b*e^2*x^4 +
2*b*d*e*x^2 + b*d^2)*log(c) - 15*(b*e^2*n*x^4 + 2*b*d*e*n*x^2 + b*d^2*n)*log(x))*sqrt(e*x^2 + d))/e]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x**2+d)**(3/2)*(a+b*ln(c*x**n)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x^{2} + d\right )}^{\frac{3}{2}}{\left (b \log \left (c x^{n}\right ) + a\right )} x\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x^2+d)^(3/2)*(a+b*log(c*x^n)),x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^(3/2)*(b*log(c*x^n) + a)*x, x)