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

Optimal. Leaf size=84 \[ -\frac{a+b \log \left (c x^n\right )}{3 e \left (d+e x^2\right )^{3/2}}-\frac{b n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{3 d^{3/2} e}+\frac{b n}{3 d e \sqrt{d+e x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0879107, antiderivative size = 84, 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, 51, 63, 208} \[ -\frac{a+b \log \left (c x^n\right )}{3 e \left (d+e x^2\right )^{3/2}}-\frac{b n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{3 d^{3/2} e}+\frac{b n}{3 d e \sqrt{d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 )}{\left (d+e x^2\right )^{5/2}} \, dx &=-\frac{a+b \log \left (c x^n\right )}{3 e \left (d+e x^2\right )^{3/2}}+\frac{(b n) \int \frac{1}{x \left (d+e x^2\right )^{3/2}} \, dx}{3 e}\\ &=-\frac{a+b \log \left (c x^n\right )}{3 e \left (d+e x^2\right )^{3/2}}+\frac{(b n) \operatorname{Subst}\left (\int \frac{1}{x (d+e x)^{3/2}} \, dx,x,x^2\right )}{6 e}\\ &=\frac{b n}{3 d e \sqrt{d+e x^2}}-\frac{a+b \log \left (c x^n\right )}{3 e \left (d+e x^2\right )^{3/2}}+\frac{(b n) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d+e x}} \, dx,x,x^2\right )}{6 d e}\\ &=\frac{b n}{3 d e \sqrt{d+e x^2}}-\frac{a+b \log \left (c x^n\right )}{3 e \left (d+e x^2\right )^{3/2}}+\frac{(b n) \operatorname{Subst}\left (\int \frac{1}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x^2}\right )}{3 d e^2}\\ &=\frac{b n}{3 d e \sqrt{d+e x^2}}-\frac{b n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{3 d^{3/2} e}-\frac{a+b \log \left (c x^n\right )}{3 e \left (d+e x^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.230545, size = 97, normalized size = 1.15 \[ -\frac{\frac{a}{\left (d+e x^2\right )^{3/2}}+\frac{b \log \left (c x^n\right )}{\left (d+e x^2\right )^{3/2}}+\frac{b n \log \left (\sqrt{d} \sqrt{d+e x^2}+d\right )}{d^{3/2}}-\frac{b n \log (x)}{d^{3/2}}-\frac{b n}{d \sqrt{d+e x^2}}}{3 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.436, size = 0, normalized size = 0. \begin{align*} \int{x \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*(a+b*ln(c*x^n))/(e*x^2+d)^(5/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.5408, size = 597, normalized size = 7.11 \begin{align*} \left [\frac{{\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \sqrt{d} \log \left (-\frac{e x^{2} - 2 \, \sqrt{e x^{2} + d} \sqrt{d} + 2 \, d}{x^{2}}\right ) + 2 \,{\left (b d e n x^{2} - b d^{2} n \log \left (x\right ) + b d^{2} n - b d^{2} \log \left (c\right ) - a d^{2}\right )} \sqrt{e x^{2} + d}}{6 \,{\left (d^{2} e^{3} x^{4} + 2 \, d^{3} e^{2} x^{2} + d^{4} e\right )}}, \frac{{\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{-d}}{\sqrt{e x^{2} + d}}\right ) +{\left (b d e n x^{2} - b d^{2} n \log \left (x\right ) + b d^{2} n - b d^{2} \log \left (c\right ) - a d^{2}\right )} \sqrt{e x^{2} + d}}{3 \,{\left (d^{2} e^{3} x^{4} + 2 \, d^{3} e^{2} x^{2} + d^{4} e\right )}}\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)^(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(d)*log(-(e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) + 2*
(b*d*e*n*x^2 - b*d^2*n*log(x) + b*d^2*n - b*d^2*log(c) - a*d^2)*sqrt(e*x^2 + d))/(d^2*e^3*x^4 + 2*d^3*e^2*x^2
+ d^4*e), 1/3*((b*e^2*n*x^4 + 2*b*d*e*n*x^2 + b*d^2*n)*sqrt(-d)*arctan(sqrt(-d)/sqrt(e*x^2 + d)) + (b*d*e*n*x^
2 - b*d^2*n*log(x) + b*d^2*n - b*d^2*log(c) - a*d^2)*sqrt(e*x^2 + d))/(d^2*e^3*x^4 + 2*d^3*e^2*x^2 + d^4*e)]

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Sympy [B]  time = 54.2549, size = 245, normalized size = 2.92 \begin{align*} - \frac{a}{3 e \left (d + e x^{2}\right )^{\frac{3}{2}}} + \frac{2 b d^{3} n \sqrt{1 + \frac{e x^{2}}{d}}}{6 d^{\frac{9}{2}} e + 6 d^{\frac{7}{2}} e^{2} x^{2}} + \frac{b d^{3} n \log{\left (\frac{e x^{2}}{d} \right )}}{6 d^{\frac{9}{2}} e + 6 d^{\frac{7}{2}} e^{2} x^{2}} - \frac{2 b d^{3} n \log{\left (\sqrt{1 + \frac{e x^{2}}{d}} + 1 \right )}}{6 d^{\frac{9}{2}} e + 6 d^{\frac{7}{2}} e^{2} x^{2}} + \frac{b d^{2} n x^{2} \log{\left (\frac{e x^{2}}{d} \right )}}{3 \left (2 d^{\frac{9}{2}} + 2 d^{\frac{7}{2}} e x^{2}\right )} - \frac{2 b d^{2} n x^{2} \log{\left (\sqrt{1 + \frac{e x^{2}}{d}} + 1 \right )}}{3 \left (2 d^{\frac{9}{2}} + 2 d^{\frac{7}{2}} e x^{2}\right )} - \frac{b \log{\left (c x^{n} \right )}}{3 e \left (d + e x^{2}\right )^{\frac{3}{2}}} \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)**(5/2),x)

[Out]

-a/(3*e*(d + e*x**2)**(3/2)) + 2*b*d**3*n*sqrt(1 + e*x**2/d)/(6*d**(9/2)*e + 6*d**(7/2)*e**2*x**2) + b*d**3*n*
log(e*x**2/d)/(6*d**(9/2)*e + 6*d**(7/2)*e**2*x**2) - 2*b*d**3*n*log(sqrt(1 + e*x**2/d) + 1)/(6*d**(9/2)*e + 6
*d**(7/2)*e**2*x**2) + b*d**2*n*x**2*log(e*x**2/d)/(3*(2*d**(9/2) + 2*d**(7/2)*e*x**2)) - 2*b*d**2*n*x**2*log(
sqrt(1 + e*x**2/d) + 1)/(3*(2*d**(9/2) + 2*d**(7/2)*e*x**2)) - b*log(c*x**n)/(3*e*(d + e*x**2)**(3/2))

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

[Out]

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