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

Optimal. Leaf size=154 \[ -\frac{d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac{2 b d^2 n \sqrt{d+e x^2}}{15 e^2}-\frac{2 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{15 e^2}+\frac{2 b d n \left (d+e x^2\right )^{3/2}}{45 e^2}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 e^2} \]

[Out]

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

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Rubi [A]  time = 0.175445, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {266, 43, 2350, 12, 446, 80, 50, 63, 208} \[ -\frac{d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac{2 b d^2 n \sqrt{d+e x^2}}{15 e^2}-\frac{2 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{15 e^2}+\frac{2 b d n \left (d+e x^2\right )^{3/2}}{45 e^2}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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

Mathematica [A]  time = 0.147632, size = 204, normalized size = 1.32 \[ \sqrt{d+e x^2} \left (-\frac{d^2 \left (30 a+30 b \left (\log \left (c x^n\right )-n \log (x)\right )-31 b n\right )}{225 e^2}+\frac{d x^2 \left (15 a+15 b \left (\log \left (c x^n\right )-n \log (x)\right )-8 b n\right )}{225 e}+\frac{1}{25} x^4 \left (5 a+5 b \left (\log \left (c x^n\right )-n \log (x)\right )-b n\right )\right )-\frac{2 b d^{5/2} n \log \left (\sqrt{d} \sqrt{d+e x^2}+d\right )}{15 e^2}-\frac{b n \log (x) \sqrt{d+e x^2} \left (2 d^2-d e x^2-3 e^2 x^4\right )}{15 e^2}+\frac{2 b d^{5/2} n \log (x)}{15 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*d^(5/2)*n*Log[x])/(15*e^2) - (b*n*Sqrt[d + e*x^2]*(2*d^2 - d*e*x^2 - 3*e^2*x^4)*Log[x])/(15*e^2) + Sqrt[d
+ e*x^2]*((x^4*(5*a - b*n + 5*b*(-(n*Log[x]) + Log[c*x^n])))/25 + (d*x^2*(15*a - 8*b*n + 15*b*(-(n*Log[x]) +
Log[c*x^n])))/(225*e) - (d^2*(30*a - 31*b*n + 30*b*(-(n*Log[x]) + Log[c*x^n])))/(225*e^2)) - (2*b*d^(5/2)*n*Lo
g[d + Sqrt[d]*Sqrt[d + e*x^2]])/(15*e^2)

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Maple [F]  time = 0.464, size = 0, normalized size = 0. \begin{align*} \int{x}^{3} \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \sqrt{e{x}^{2}+d}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*ln(c*x^n))*(e*x^2+d)^(1/2),x)

[Out]

int(x^3*(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^3*(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 = 1.58401, size = 745, normalized size = 4.84 \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 ) -{\left (9 \,{\left (b e^{2} n - 5 \, a e^{2}\right )} x^{4} - 31 \, b d^{2} n + 30 \, a d^{2} +{\left (8 \, b d e n - 15 \, a d e\right )} x^{2} - 15 \,{\left (3 \, b e^{2} x^{4} + b d e x^{2} - 2 \, b d^{2}\right )} \log \left (c\right ) - 15 \,{\left (3 \, b e^{2} n x^{4} + b d e n x^{2} - 2 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt{e x^{2} + d}}{225 \, e^{2}}, \frac{30 \, 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} - 31 \, b d^{2} n + 30 \, a d^{2} +{\left (8 \, b d e n - 15 \, a d e\right )} x^{2} - 15 \,{\left (3 \, b e^{2} x^{4} + b d e x^{2} - 2 \, b d^{2}\right )} \log \left (c\right ) - 15 \,{\left (3 \, b e^{2} n x^{4} + b d e n x^{2} - 2 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt{e x^{2} + d}}{225 \, e^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*log(c*x^n))*(e*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

[1/225*(15*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 - 31*b
*d^2*n + 30*a*d^2 + (8*b*d*e*n - 15*a*d*e)*x^2 - 15*(3*b*e^2*x^4 + b*d*e*x^2 - 2*b*d^2)*log(c) - 15*(3*b*e^2*n
*x^4 + b*d*e*n*x^2 - 2*b*d^2*n)*log(x))*sqrt(e*x^2 + d))/e^2, 1/225*(30*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 - 31*b*d^2*n + 30*a*d^2 + (8*b*d*e*n - 15*a*d*e)*x^2 - 15*(3*b*e^2*x^
4 + b*d*e*x^2 - 2*b*d^2)*log(c) - 15*(3*b*e^2*n*x^4 + b*d*e*n*x^2 - 2*b*d^2*n)*log(x))*sqrt(e*x^2 + d))/e^2]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \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**3*(a+b*ln(c*x**n))*(e*x**2+d)**(1/2),x)

[Out]

Integral(x**3*(a + b*log(c*x**n))*sqrt(d + e*x**2), x)

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Giac [A]  time = 1.78015, size = 298, normalized size = 1.94 \begin{align*} \frac{1}{5} \, \sqrt{x^{2} e + d} b x^{4} \log \left (c\right ) + \frac{1}{15} \, \sqrt{x^{2} e + d} b d x^{2} e^{\left (-1\right )} \log \left (c\right ) + \frac{1}{5} \, \sqrt{x^{2} e + d} a x^{4} + \frac{1}{15} \, \sqrt{x^{2} e + d} a d x^{2} e^{\left (-1\right )} - \frac{2}{15} \, \sqrt{x^{2} e + d} b d^{2} e^{\left (-2\right )} \log \left (c\right ) - \frac{2}{15} \, \sqrt{x^{2} e + d} a d^{2} e^{\left (-2\right )} + \frac{1}{225} \,{\left (15 \,{\left (3 \,{\left (x^{2} e + d\right )}^{\frac{5}{2}} - 5 \,{\left (x^{2} e + d\right )}^{\frac{3}{2}} d\right )} e^{\left (-2\right )} \log \left (x\right ) +{\left (\frac{30 \, d^{3} \arctan \left (\frac{\sqrt{x^{2} e + d}}{\sqrt{-d}}\right )}{\sqrt{-d}} - 9 \,{\left (x^{2} e + d\right )}^{\frac{5}{2}} + 10 \,{\left (x^{2} e + d\right )}^{\frac{3}{2}} d + 30 \, \sqrt{x^{2} e + d} d^{2}\right )} e^{\left (-2\right )}\right )} b n \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*log(c*x^n))*(e*x^2+d)^(1/2),x, algorithm="giac")

[Out]

1/5*sqrt(x^2*e + d)*b*x^4*log(c) + 1/15*sqrt(x^2*e + d)*b*d*x^2*e^(-1)*log(c) + 1/5*sqrt(x^2*e + d)*a*x^4 + 1/
15*sqrt(x^2*e + d)*a*d*x^2*e^(-1) - 2/15*sqrt(x^2*e + d)*b*d^2*e^(-2)*log(c) - 2/15*sqrt(x^2*e + d)*a*d^2*e^(-
2) + 1/225*(15*(3*(x^2*e + d)^(5/2) - 5*(x^2*e + d)^(3/2)*d)*e^(-2)*log(x) + (30*d^3*arctan(sqrt(x^2*e + d)/sq
rt(-d))/sqrt(-d) - 9*(x^2*e + d)^(5/2) + 10*(x^2*e + d)^(3/2)*d + 30*sqrt(x^2*e + d)*d^2)*e^(-2))*b*n