3.3.52 \(\int x^3 \sqrt {d+e x^2} (a+b \log (c x^n)) \, dx\) [252]

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

[Out]

2/45*b*d*n*(e*x^2+d)^(3/2)/e^2-1/25*b*n*(e*x^2+d)^(5/2)/e^2-2/15*b*d^(5/2)*n*arctanh((e*x^2+d)^(1/2)/d^(1/2))/
e^2-1/3*d*(e*x^2+d)^(3/2)*(a+b*ln(c*x^n))/e^2+1/5*(e*x^2+d)^(5/2)*(a+b*ln(c*x^n))/e^2+2/15*b*d^2*n*(e*x^2+d)^(
1/2)/e^2

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Rubi [A]
time = 0.12, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {272, 45, 2392, 12, 457, 81, 52, 65, 214} \begin {gather*} -\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^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\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} \end {gather*}

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 12

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

Rule 45

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 52

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 65

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 81

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 214

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

Rule 272

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 457

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 2392

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])

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) \text {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) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.09, size = 204, normalized size = 1.32 \begin {gather*} \frac {2 b d^{5/2} n \log (x)}{15 e^2}-\frac {b n \sqrt {d+e x^2} \left (2 d^2-d e x^2-3 e^2 x^4\right ) \log (x)}{15 e^2}+\sqrt {d+e x^2} \left (\frac {1}{25} x^4 \left (5 a-b n+5 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )+\frac {d x^2 \left (15 a-8 b n+15 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{225 e}-\frac {d^2 \left (30 a-31 b n+30 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{225 e^2}\right )-\frac {2 b d^{5/2} n \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{15 e^2} \end {gather*}

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 = 0.02, size = 0, normalized size = 0.00 \[\int x^{3} \left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {e \,x^{2}+d}\, dx\]

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 [A]
time = 0.51, size = 170, normalized size = 1.10 \begin {gather*} \frac {1}{225} \, {\left (15 \, d^{\frac {5}{2}} e^{\left (-2\right )} \log \left (\frac {\sqrt {x^{2} e + d} - \sqrt {d}}{\sqrt {x^{2} e + d} + \sqrt {d}}\right ) - {\left (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 + \frac {1}{15} \, {\left (3 \, {\left (x^{2} e + d\right )}^{\frac {3}{2}} x^{2} e^{\left (-1\right )} - 2 \, {\left (x^{2} e + d\right )}^{\frac {3}{2}} d e^{\left (-2\right )}\right )} b \log \left (c x^{n}\right ) + \frac {1}{15} \, {\left (3 \, {\left (x^{2} e + d\right )}^{\frac {3}{2}} x^{2} e^{\left (-1\right )} - 2 \, {\left (x^{2} e + d\right )}^{\frac {3}{2}} d e^{\left (-2\right )}\right )} a \end {gather*}

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]

1/225*(15*d^(5/2)*e^(-2)*log((sqrt(x^2*e + d) - sqrt(d))/(sqrt(x^2*e + 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 + 1/15*(3*(x^2*e + d)^(3/2)*x^2*e^(-1) - 2*(x^2*
e + d)^(3/2)*d*e^(-2))*b*log(c*x^n) + 1/15*(3*(x^2*e + d)^(3/2)*x^2*e^(-1) - 2*(x^2*e + d)^(3/2)*d*e^(-2))*a

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Fricas [A]
time = 0.40, size = 304, normalized size = 1.97 \begin {gather*} \left [\frac {1}{225} \, {\left (15 \, b d^{\frac {5}{2}} n \log \left (-\frac {x^{2} e - 2 \, \sqrt {x^{2} e + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - {\left (9 \, {\left (b n - 5 \, a\right )} x^{4} e^{2} - 31 \, b d^{2} n + {\left (8 \, b d n - 15 \, a d\right )} x^{2} e + 30 \, a d^{2} - 15 \, {\left (3 \, b x^{4} e^{2} + b d x^{2} e - 2 \, b d^{2}\right )} \log \left (c\right ) - 15 \, {\left (3 \, b n x^{4} e^{2} + b d n x^{2} e - 2 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {x^{2} e + d}\right )} e^{\left (-2\right )}, \frac {1}{225} \, {\left (30 \, b \sqrt {-d} d^{2} n \arctan \left (\frac {\sqrt {-d}}{\sqrt {x^{2} e + d}}\right ) - {\left (9 \, {\left (b n - 5 \, a\right )} x^{4} e^{2} - 31 \, b d^{2} n + {\left (8 \, b d n - 15 \, a d\right )} x^{2} e + 30 \, a d^{2} - 15 \, {\left (3 \, b x^{4} e^{2} + b d x^{2} e - 2 \, b d^{2}\right )} \log \left (c\right ) - 15 \, {\left (3 \, b n x^{4} e^{2} + b d n x^{2} e - 2 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {x^{2} e + d}\right )} e^{\left (-2\right )}\right ] \end {gather*}

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(-(x^2*e - 2*sqrt(x^2*e + d)*sqrt(d) + 2*d)/x^2) - (9*(b*n - 5*a)*x^4*e^2 - 31*b*d^2
*n + (8*b*d*n - 15*a*d)*x^2*e + 30*a*d^2 - 15*(3*b*x^4*e^2 + b*d*x^2*e - 2*b*d^2)*log(c) - 15*(3*b*n*x^4*e^2 +
b*d*n*x^2*e - 2*b*d^2*n)*log(x))*sqrt(x^2*e + d))*e^(-2), 1/225*(30*b*sqrt(-d)*d^2*n*arctan(sqrt(-d)/sqrt(x^2
*e + d)) - (9*(b*n - 5*a)*x^4*e^2 - 31*b*d^2*n + (8*b*d*n - 15*a*d)*x^2*e + 30*a*d^2 - 15*(3*b*x^4*e^2 + b*d*x
^2*e - 2*b*d^2)*log(c) - 15*(3*b*n*x^4*e^2 + b*d*n*x^2*e - 2*b*d^2*n)*log(x))*sqrt(x^2*e + d))*e^(-2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \left (a + b \log {\left (c x^{n} \right )}\right ) \sqrt {d + e x^{2}}\, dx \end {gather*}

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 = 2.64, size = 221, normalized size = 1.44 \begin {gather*} \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 {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,\sqrt {e\,x^2+d}\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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