3.305 \(\int \frac{x^2 (a+b \log (c x^n))}{(d+e x^2)^{5/2}} \, dx\)

Optimal. Leaf size=89 \[ \frac{x^3 \left (a+b \log \left (c x^n\right )\right )}{3 d \left (d+e x^2\right )^{3/2}}-\frac{b n \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{3 d e^{3/2}}+\frac{b n x}{3 d e \sqrt{d+e x^2}} \]

[Out]

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

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Rubi [A]  time = 0.108828, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2335, 288, 217, 206} \[ \frac{x^3 \left (a+b \log \left (c x^n\right )\right )}{3 d \left (d+e x^2\right )^{3/2}}-\frac{b n \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{3 d e^{3/2}}+\frac{b n x}{3 d e \sqrt{d+e x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*Log[c*x^n]))/(d + e*x^2)^(5/2),x]

[Out]

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

Rule 2335

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

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.14489, size = 101, normalized size = 1.13 \[ \frac{\sqrt{e} x \left (a e x^2+b n \left (d+e x^2\right )\right )+b e^{3/2} x^3 \log \left (c x^n\right )-b n \left (d+e x^2\right )^{3/2} \log \left (\sqrt{e} \sqrt{d+e x^2}+e x\right )}{3 d e^{3/2} \left (d+e x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(a + b*Log[c*x^n]))/(d + e*x^2)^(5/2),x]

[Out]

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

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Maple [F]  time = 0.411, size = 0, normalized size = 0. \begin{align*} \int{{x}^{2} \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \left ( e{x}^{2}+d \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*ln(c*x^n))/(e*x^2+d)^(5/2),x)

[Out]

int(x^2*(a+b*ln(c*x^n))/(e*x^2+d)^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{3} \, a{\left (\frac{x}{{\left (e x^{2} + d\right )}^{\frac{3}{2}} e} - \frac{x}{\sqrt{e x^{2} + d} d e}\right )} + b \int \frac{x^{2} \log \left (c\right ) + x^{2} \log \left (x^{n}\right )}{{\left (e^{2} x^{4} + 2 \, d e x^{2} + d^{2}\right )} \sqrt{e x^{2} + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*log(c*x^n))/(e*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

-1/3*a*(x/((e*x^2 + d)^(3/2)*e) - x/(sqrt(e*x^2 + d)*d*e)) + b*integrate((x^2*log(c) + x^2*log(x^n))/((e^2*x^4
+ 2*d*e*x^2 + d^2)*sqrt(e*x^2 + d)), x)

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Fricas [A]  time = 1.47416, size = 626, normalized size = 7.03 \begin{align*} \left [\frac{{\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \sqrt{e} \log \left (-2 \, e x^{2} + 2 \, \sqrt{e x^{2} + d} \sqrt{e} x - d\right ) + 2 \,{\left (b e^{2} n x^{3} \log \left (x\right ) + b e^{2} x^{3} \log \left (c\right ) + b d e n x +{\left (b e^{2} n + a e^{2}\right )} x^{3}\right )} \sqrt{e x^{2} + d}}{6 \,{\left (d e^{4} x^{4} + 2 \, d^{2} e^{3} x^{2} + d^{3} e^{2}\right )}}, \frac{{\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \sqrt{-e} \arctan \left (\frac{\sqrt{-e} x}{\sqrt{e x^{2} + d}}\right ) +{\left (b e^{2} n x^{3} \log \left (x\right ) + b e^{2} x^{3} \log \left (c\right ) + b d e n x +{\left (b e^{2} n + a e^{2}\right )} x^{3}\right )} \sqrt{e x^{2} + d}}{3 \,{\left (d e^{4} x^{4} + 2 \, d^{2} e^{3} x^{2} + d^{3} e^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*log(c*x^n))/(e*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

[1/6*((b*e^2*n*x^4 + 2*b*d*e*n*x^2 + b*d^2*n)*sqrt(e)*log(-2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) + 2*(b*e
^2*n*x^3*log(x) + b*e^2*x^3*log(c) + b*d*e*n*x + (b*e^2*n + a*e^2)*x^3)*sqrt(e*x^2 + d))/(d*e^4*x^4 + 2*d^2*e^
3*x^2 + d^3*e^2), 1/3*((b*e^2*n*x^4 + 2*b*d*e*n*x^2 + b*d^2*n)*sqrt(-e)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) + (
b*e^2*n*x^3*log(x) + b*e^2*x^3*log(c) + b*d*e*n*x + (b*e^2*n + a*e^2)*x^3)*sqrt(e*x^2 + d))/(d*e^4*x^4 + 2*d^2
*e^3*x^2 + d^3*e^2)]

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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**2*(a+b*ln(c*x**n))/(e*x**2+d)**(5/2),x)

[Out]

Timed out

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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^{2}}{{\left (e x^{2} + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*log(c*x^n))/(e*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)*x^2/(e*x^2 + d)^(5/2), x)