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

Optimal. Leaf size=155 \[ -\frac{d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac{2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt{d+e x^2}}+\frac{\sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac{b d n}{3 e^3 \sqrt{d+e x^2}}-\frac{b n \sqrt{d+e x^2}}{e^3}+\frac{8 b \sqrt{d} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{3 e^3} \]

[Out]

(b*d*n)/(3*e^3*Sqrt[d + e*x^2]) - (b*n*Sqrt[d + e*x^2])/e^3 + (8*b*Sqrt[d]*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])
/(3*e^3) - (d^2*(a + b*Log[c*x^n]))/(3*e^3*(d + e*x^2)^(3/2)) + (2*d*(a + b*Log[c*x^n]))/(e^3*Sqrt[d + e*x^2])
+ (Sqrt[d + e*x^2]*(a + b*Log[c*x^n]))/e^3

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Rubi [A]  time = 0.234636, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {266, 43, 2350, 12, 1251, 897, 1261, 206} \[ -\frac{d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac{2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt{d+e x^2}}+\frac{\sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac{b d n}{3 e^3 \sqrt{d+e x^2}}-\frac{b n \sqrt{d+e x^2}}{e^3}+\frac{8 b \sqrt{d} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{3 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d*n)/(3*e^3*Sqrt[d + e*x^2]) - (b*n*Sqrt[d + e*x^2])/e^3 + (8*b*Sqrt[d]*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])
/(3*e^3) - (d^2*(a + b*Log[c*x^n]))/(3*e^3*(d + e*x^2)^(3/2)) + (2*d*(a + b*Log[c*x^n]))/(e^3*Sqrt[d + e*x^2])
+ (Sqrt[d + e*x^2]*(a + b*Log[c*x^n]))/e^3

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2350

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = IntHide[(f*x)^m*(d + e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[SimplifyIntegrand[u/x,
x], x], x] /; ((EqQ[r, 1] || EqQ[r, 2]) && IntegerQ[m] && IntegerQ[q - 1/2]) || InverseFunctionFreeQ[u, x]] /
; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && IntegerQ[2*q] && ((IntegerQ[m] && IntegerQ[r]) || IGtQ[q, 0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

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^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx &=-\frac{d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac{2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt{d+e x^2}}+\frac{\sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-(b n) \int \frac{8 d^2+12 d e x^2+3 e^2 x^4}{3 e^3 x \left (d+e x^2\right )^{3/2}} \, dx\\ &=-\frac{d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac{2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt{d+e x^2}}+\frac{\sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{(b n) \int \frac{8 d^2+12 d e x^2+3 e^2 x^4}{x \left (d+e x^2\right )^{3/2}} \, dx}{3 e^3}\\ &=-\frac{d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac{2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt{d+e x^2}}+\frac{\sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{(b n) \operatorname{Subst}\left (\int \frac{8 d^2+12 d e x+3 e^2 x^2}{x (d+e x)^{3/2}} \, dx,x,x^2\right )}{6 e^3}\\ &=-\frac{d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac{2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt{d+e x^2}}+\frac{\sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{(b n) \operatorname{Subst}\left (\int \frac{-d^2+6 d x^2+3 x^4}{x^2 \left (-\frac{d}{e}+\frac{x^2}{e}\right )} \, dx,x,\sqrt{d+e x^2}\right )}{3 e^4}\\ &=-\frac{d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac{2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt{d+e x^2}}+\frac{\sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{(b n) \operatorname{Subst}\left (\int \left (3 e+\frac{d e}{x^2}-\frac{8 d e}{d-x^2}\right ) \, dx,x,\sqrt{d+e x^2}\right )}{3 e^4}\\ &=\frac{b d n}{3 e^3 \sqrt{d+e x^2}}-\frac{b n \sqrt{d+e x^2}}{e^3}-\frac{d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac{2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt{d+e x^2}}+\frac{\sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac{(8 b d n) \operatorname{Subst}\left (\int \frac{1}{d-x^2} \, dx,x,\sqrt{d+e x^2}\right )}{3 e^3}\\ &=\frac{b d n}{3 e^3 \sqrt{d+e x^2}}-\frac{b n \sqrt{d+e x^2}}{e^3}+\frac{8 b \sqrt{d} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{3 e^3}-\frac{d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac{2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt{d+e x^2}}+\frac{\sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}\\ \end{align*}

Mathematica [A]  time = 0.207138, size = 205, normalized size = 1.32 \[ \sqrt{d+e x^2} \left (-\frac{d^2 \left (a+b \left (\log \left (c x^n\right )-n \log (x)\right )\right )}{3 e^3 \left (d+e x^2\right )^2}+\frac{d \left (6 a+6 b \left (\log \left (c x^n\right )-n \log (x)\right )+b n\right )}{3 e^3 \left (d+e x^2\right )}+\frac{a+b \left (\log \left (c x^n\right )-n \log (x)\right )-b n}{e^3}\right )+\frac{b n \log (x) \left (8 d^2+12 d e x^2+3 e^2 x^4\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac{8 b \sqrt{d} n \log \left (\sqrt{d} \sqrt{d+e x^2}+d\right )}{3 e^3}-\frac{8 b \sqrt{d} n \log (x)}{3 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*b*Sqrt[d]*n*Log[x])/(3*e^3) + (b*n*(8*d^2 + 12*d*e*x^2 + 3*e^2*x^4)*Log[x])/(3*e^3*(d + e*x^2)^(3/2)) + Sq
rt[d + e*x^2]*(-(d^2*(a + b*(-(n*Log[x]) + Log[c*x^n])))/(3*e^3*(d + e*x^2)^2) + (a - b*n + b*(-(n*Log[x]) + L
og[c*x^n]))/e^3 + (d*(6*a + b*n + 6*b*(-(n*Log[x]) + Log[c*x^n])))/(3*e^3*(d + e*x^2))) + (8*b*Sqrt[d]*n*Log[d
+ Sqrt[d]*Sqrt[d + e*x^2]])/(3*e^3)

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Maple [F]  time = 0.41, size = 0, normalized size = 0. \begin{align*} \int{{x}^{5} \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^5*(a+b*ln(c*x^n))/(e*x^2+d)^(5/2),x)

[Out]

int(x^5*(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^5*(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 = 1.74264, size = 900, normalized size = 5.81 \begin{align*} \left [\frac{4 \,{\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 ) -{\left (3 \,{\left (b e^{2} n - a e^{2}\right )} x^{4} + 2 \, b d^{2} n - 8 \, a d^{2} +{\left (5 \, b d e n - 12 \, a d e\right )} x^{2} -{\left (3 \, b e^{2} x^{4} + 12 \, b d e x^{2} + 8 \, b d^{2}\right )} \log \left (c\right ) -{\left (3 \, b e^{2} n x^{4} + 12 \, b d e n x^{2} + 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt{e x^{2} + d}}{3 \,{\left (e^{5} x^{4} + 2 \, d e^{4} x^{2} + d^{2} e^{3}\right )}}, -\frac{8 \,{\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 (3 \,{\left (b e^{2} n - a e^{2}\right )} x^{4} + 2 \, b d^{2} n - 8 \, a d^{2} +{\left (5 \, b d e n - 12 \, a d e\right )} x^{2} -{\left (3 \, b e^{2} x^{4} + 12 \, b d e x^{2} + 8 \, b d^{2}\right )} \log \left (c\right ) -{\left (3 \, b e^{2} n x^{4} + 12 \, b d e n x^{2} + 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt{e x^{2} + d}}{3 \,{\left (e^{5} x^{4} + 2 \, d e^{4} x^{2} + d^{2} e^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3*(4*(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) -
(3*(b*e^2*n - a*e^2)*x^4 + 2*b*d^2*n - 8*a*d^2 + (5*b*d*e*n - 12*a*d*e)*x^2 - (3*b*e^2*x^4 + 12*b*d*e*x^2 + 8*
b*d^2)*log(c) - (3*b*e^2*n*x^4 + 12*b*d*e*n*x^2 + 8*b*d^2*n)*log(x))*sqrt(e*x^2 + d))/(e^5*x^4 + 2*d*e^4*x^2 +
d^2*e^3), -1/3*(8*(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)) + (3*(b*e
^2*n - a*e^2)*x^4 + 2*b*d^2*n - 8*a*d^2 + (5*b*d*e*n - 12*a*d*e)*x^2 - (3*b*e^2*x^4 + 12*b*d*e*x^2 + 8*b*d^2)*
log(c) - (3*b*e^2*n*x^4 + 12*b*d*e*n*x^2 + 8*b*d^2*n)*log(x))*sqrt(e*x^2 + d))/(e^5*x^4 + 2*d*e^4*x^2 + d^2*e^
3)]

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

[Out]

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