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

Optimal. Leaf size=182 \[ \frac{d^2 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{8 b d^2 n \sqrt{d+e x^2}}{15 e^3}+\frac{8 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{15 e^3}+\frac{7 b d n \left (d+e x^2\right )^{3/2}}{45 e^3}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 e^3} \]

[Out]

(-8*b*d^2*n*Sqrt[d + e*x^2])/(15*e^3) + (7*b*d*n*(d + e*x^2)^(3/2))/(45*e^3) - (b*n*(d + e*x^2)^(5/2))/(25*e^3
) + (8*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(15*e^3) + (d^2*Sqrt[d + e*x^2]*(a + b*Log[c*x^n]))/e^3 -
(2*d*(d + e*x^2)^(3/2)*(a + b*Log[c*x^n]))/(3*e^3) + ((d + e*x^2)^(5/2)*(a + b*Log[c*x^n]))/(5*e^3)

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Rubi [A]  time = 0.228303, antiderivative size = 182, 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, 208} \[ \frac{d^2 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{8 b d^2 n \sqrt{d+e x^2}}{15 e^3}+\frac{8 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{15 e^3}+\frac{7 b d n \left (d+e x^2\right )^{3/2}}{45 e^3}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*b*d^2*n*Sqrt[d + e*x^2])/(15*e^3) + (7*b*d*n*(d + e*x^2)^(3/2))/(45*e^3) - (b*n*(d + e*x^2)^(5/2))/(25*e^3
) + (8*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(15*e^3) + (d^2*Sqrt[d + e*x^2]*(a + b*Log[c*x^n]))/e^3 -
(2*d*(d + e*x^2)^(3/2)*(a + b*Log[c*x^n]))/(3*e^3) + ((d + e*x^2)^(5/2)*(a + b*Log[c*x^n]))/(5*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 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^5 \left (a+b \log \left (c x^n\right )\right )}{\sqrt{d+e x^2}} \, dx &=\frac{d^2 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-(b n) \int \frac{\sqrt{d+e x^2} \left (8 d^2-4 d e x^2+3 e^2 x^4\right )}{15 e^3 x} \, dx\\ &=\frac{d^2 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{(b n) \int \frac{\sqrt{d+e x^2} \left (8 d^2-4 d e x^2+3 e^2 x^4\right )}{x} \, dx}{15 e^3}\\ &=\frac{d^2 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{(b n) \operatorname{Subst}\left (\int \frac{\sqrt{d+e x} \left (8 d^2-4 d e x+3 e^2 x^2\right )}{x} \, dx,x,x^2\right )}{30 e^3}\\ &=\frac{d^2 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{(b n) \operatorname{Subst}\left (\int \frac{x^2 \left (15 d^2-10 d x^2+3 x^4\right )}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x^2}\right )}{15 e^4}\\ &=\frac{d^2 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{(b n) \operatorname{Subst}\left (\int \left (8 d^2 e-7 d e x^2+3 e x^4+\frac{8 d^3}{-\frac{d}{e}+\frac{x^2}{e}}\right ) \, dx,x,\sqrt{d+e x^2}\right )}{15 e^4}\\ &=-\frac{8 b d^2 n \sqrt{d+e x^2}}{15 e^3}+\frac{7 b d n \left (d+e x^2\right )^{3/2}}{45 e^3}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 e^3}+\frac{d^2 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{\left (8 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 )}{15 e^4}\\ &=-\frac{8 b d^2 n \sqrt{d+e x^2}}{15 e^3}+\frac{7 b d n \left (d+e x^2\right )^{3/2}}{45 e^3}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 e^3}+\frac{8 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{15 e^3}+\frac{d^2 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}\\ \end{align*}

Mathematica [A]  time = 0.190928, size = 204, normalized size = 1.12 \[ \frac{120 a d^2 \sqrt{d+e x^2}+45 a e^2 x^4 \sqrt{d+e x^2}-60 a d e x^2 \sqrt{d+e x^2}+15 b \sqrt{d+e x^2} \left (8 d^2-4 d e x^2+3 e^2 x^4\right ) \log \left (c x^n\right )-94 b d^2 n \sqrt{d+e x^2}+120 b d^{5/2} n \log \left (\sqrt{d} \sqrt{d+e x^2}+d\right )-120 b d^{5/2} n \log (x)-9 b e^2 n x^4 \sqrt{d+e x^2}+17 b d e n x^2 \sqrt{d+e x^2}}{225 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(120*a*d^2*Sqrt[d + e*x^2] - 94*b*d^2*n*Sqrt[d + e*x^2] - 60*a*d*e*x^2*Sqrt[d + e*x^2] + 17*b*d*e*n*x^2*Sqrt[d
+ e*x^2] + 45*a*e^2*x^4*Sqrt[d + e*x^2] - 9*b*e^2*n*x^4*Sqrt[d + e*x^2] - 120*b*d^(5/2)*n*Log[x] + 15*b*Sqrt[
d + e*x^2]*(8*d^2 - 4*d*e*x^2 + 3*e^2*x^4)*Log[c*x^n] + 120*b*d^(5/2)*n*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/(225
*e^3)

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Maple [F]  time = 0.434, size = 0, normalized size = 0. \begin{align*} \int{{x}^{5} \left ( a+b\ln \left ( c{x}^{n} \right ) \right ){\frac{1}{\sqrt{e{x}^{2}+d}}}}\, 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)^(1/2),x)

[Out]

int(x^5*(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^5*(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 = 1.57755, size = 764, normalized size = 4.2 \begin{align*} \left [\frac{60 \, 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 ) -{\left (9 \,{\left (b e^{2} n - 5 \, a e^{2}\right )} x^{4} + 94 \, b d^{2} n - 120 \, a d^{2} -{\left (17 \, b d e n - 60 \, a d e\right )} x^{2} - 15 \,{\left (3 \, b e^{2} x^{4} - 4 \, b d e x^{2} + 8 \, b d^{2}\right )} \log \left (c\right ) - 15 \,{\left (3 \, b e^{2} n x^{4} - 4 \, b d e n x^{2} + 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt{e x^{2} + d}}{225 \, e^{3}}, -\frac{120 \, b \sqrt{-d} d^{2} n \arctan \left (\frac{\sqrt{-d}}{\sqrt{e x^{2} + d}}\right ) +{\left (9 \,{\left (b e^{2} n - 5 \, a e^{2}\right )} x^{4} + 94 \, b d^{2} n - 120 \, a d^{2} -{\left (17 \, b d e n - 60 \, a d e\right )} x^{2} - 15 \,{\left (3 \, b e^{2} x^{4} - 4 \, b d e x^{2} + 8 \, b d^{2}\right )} \log \left (c\right ) - 15 \,{\left (3 \, b e^{2} n x^{4} - 4 \, b d e n x^{2} + 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt{e x^{2} + d}}{225 \, e^{3}}\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)^(1/2),x, algorithm="fricas")

[Out]

[1/225*(60*b*d^(5/2)*n*log(-(e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) - (9*(b*e^2*n - 5*a*e^2)*x^4 + 94*b
*d^2*n - 120*a*d^2 - (17*b*d*e*n - 60*a*d*e)*x^2 - 15*(3*b*e^2*x^4 - 4*b*d*e*x^2 + 8*b*d^2)*log(c) - 15*(3*b*e
^2*n*x^4 - 4*b*d*e*n*x^2 + 8*b*d^2*n)*log(x))*sqrt(e*x^2 + d))/e^3, -1/225*(120*b*sqrt(-d)*d^2*n*arctan(sqrt(-
d)/sqrt(e*x^2 + d)) + (9*(b*e^2*n - 5*a*e^2)*x^4 + 94*b*d^2*n - 120*a*d^2 - (17*b*d*e*n - 60*a*d*e)*x^2 - 15*(
3*b*e^2*x^4 - 4*b*d*e*x^2 + 8*b*d^2)*log(c) - 15*(3*b*e^2*n*x^4 - 4*b*d*e*n*x^2 + 8*b*d^2*n)*log(x))*sqrt(e*x^
2 + d))/e^3]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5} \left (a + b \log{\left (c x^{n} \right )}\right )}{\sqrt{d + e x^{2}}}\, dx \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)**(1/2),x)

[Out]

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

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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}}{\sqrt{e x^{2} + d}}\,{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)^(1/2),x, algorithm="giac")

[Out]

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