\(\int \frac {x^2 (a+b \log (c x^n))}{\sqrt {d+e x}} \, dx\) [145]

Optimal result

Integrand size = 23, antiderivative size = 169 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=-\frac {32 b d^2 n \sqrt {d+e x}}{15 e^3}+\frac {28 b d n (d+e x)^{3/2}}{45 e^3}-\frac {4 b n (d+e x)^{5/2}}{25 e^3}+\frac {32 b d^{5/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{15 e^3}+\frac {2 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3} \]

[Out]

28/45*b*d*n*(e*x+d)^(3/2)/e^3-4/25*b*n*(e*x+d)^(5/2)/e^3+32/15*b*d^(5/2)*n*arctanh((e*x+d)^(1/2)/d^(1/2))/e^3-
4/3*d*(e*x+d)^(3/2)*(a+b*ln(c*x^n))/e^3+2/5*(e*x+d)^(5/2)*(a+b*ln(c*x^n))/e^3-32/15*b*d^2*n*(e*x+d)^(1/2)/e^3+
2*d^2*(a+b*ln(c*x^n))*(e*x+d)^(1/2)/e^3

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {45, 2392, 12, 911, 1275, 214} \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=\frac {2 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}+\frac {32 b d^{5/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{15 e^3}-\frac {32 b d^2 n \sqrt {d+e x}}{15 e^3}+\frac {28 b d n (d+e x)^{3/2}}{45 e^3}-\frac {4 b n (d+e x)^{5/2}}{25 e^3} \]

[In]

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

[Out]

(-32*b*d^2*n*Sqrt[d + e*x])/(15*e^3) + (28*b*d*n*(d + e*x)^(3/2))/(45*e^3) - (4*b*n*(d + e*x)^(5/2))/(25*e^3)
+ (32*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(15*e^3) + (2*d^2*Sqrt[d + e*x]*(a + b*Log[c*x^n]))/e^3 - (4
*d*(d + e*x)^(3/2)*(a + b*Log[c*x^n]))/(3*e^3) + (2*(d + e*x)^(5/2)*(a + b*Log[c*x^n]))/(5*e^3)

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 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 911

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 1275

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 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*} \text {integral}& = \frac {2 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-(b n) \int \frac {2 \sqrt {d+e x} \left (8 d^2-4 d e x+3 e^2 x^2\right )}{15 e^3 x} \, dx \\ & = \frac {2 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {(2 b n) \int \frac {\sqrt {d+e x} \left (8 d^2-4 d e x+3 e^2 x^2\right )}{x} \, dx}{15 e^3} \\ & = \frac {2 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {(4 b n) \text {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}\right )}{15 e^4} \\ & = \frac {2 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {(4 b n) \text {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}\right )}{15 e^4} \\ & = -\frac {32 b d^2 n \sqrt {d+e x}}{15 e^3}+\frac {28 b d n (d+e x)^{3/2}}{45 e^3}-\frac {4 b n (d+e x)^{5/2}}{25 e^3}+\frac {2 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {\left (32 b d^3 n\right ) \text {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{15 e^4} \\ & = -\frac {32 b d^2 n \sqrt {d+e x}}{15 e^3}+\frac {28 b d n (d+e x)^{3/2}}{45 e^3}-\frac {4 b n (d+e x)^{5/2}}{25 e^3}+\frac {32 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{15 e^3}+\frac {2 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.70 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=\frac {480 b d^{5/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+2 \sqrt {d+e x} \left (15 a \left (8 d^2-4 d e x+3 e^2 x^2\right )-2 b n \left (94 d^2-17 d e x+9 e^2 x^2\right )+15 b \left (8 d^2-4 d e x+3 e^2 x^2\right ) \log \left (c x^n\right )\right )}{225 e^3} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {x^{2} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\sqrt {e x +d}}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.75 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=\left [\frac {2 \, {\left (120 \, b d^{\frac {5}{2}} n \log \left (\frac {e x + 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) - {\left (188 \, b d^{2} n - 120 \, a d^{2} + 9 \, {\left (2 \, b e^{2} n - 5 \, a e^{2}\right )} x^{2} - 2 \, {\left (17 \, b d e n - 30 \, a d e\right )} x - 15 \, {\left (3 \, b e^{2} x^{2} - 4 \, b d e x + 8 \, b d^{2}\right )} \log \left (c\right ) - 15 \, {\left (3 \, b e^{2} n x^{2} - 4 \, b d e n x + 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{225 \, e^{3}}, -\frac {2 \, {\left (240 \, b \sqrt {-d} d^{2} n \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + {\left (188 \, b d^{2} n - 120 \, a d^{2} + 9 \, {\left (2 \, b e^{2} n - 5 \, a e^{2}\right )} x^{2} - 2 \, {\left (17 \, b d e n - 30 \, a d e\right )} x - 15 \, {\left (3 \, b e^{2} x^{2} - 4 \, b d e x + 8 \, b d^{2}\right )} \log \left (c\right ) - 15 \, {\left (3 \, b e^{2} n x^{2} - 4 \, b d e n x + 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{225 \, e^{3}}\right ] \]

[In]

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

[Out]

[2/225*(120*b*d^(5/2)*n*log((e*x + 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - (188*b*d^2*n - 120*a*d^2 + 9*(2*b*e^2*n
- 5*a*e^2)*x^2 - 2*(17*b*d*e*n - 30*a*d*e)*x - 15*(3*b*e^2*x^2 - 4*b*d*e*x + 8*b*d^2)*log(c) - 15*(3*b*e^2*n*
x^2 - 4*b*d*e*n*x + 8*b*d^2*n)*log(x))*sqrt(e*x + d))/e^3, -2/225*(240*b*sqrt(-d)*d^2*n*arctan(sqrt(e*x + d)*s
qrt(-d)/d) + (188*b*d^2*n - 120*a*d^2 + 9*(2*b*e^2*n - 5*a*e^2)*x^2 - 2*(17*b*d*e*n - 30*a*d*e)*x - 15*(3*b*e^
2*x^2 - 4*b*d*e*x + 8*b*d^2)*log(c) - 15*(3*b*e^2*n*x^2 - 4*b*d*e*n*x + 8*b*d^2*n)*log(x))*sqrt(e*x + d))/e^3]

Sympy [A] (verification not implemented)

Time = 42.28 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.99 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=a \left (\begin {cases} \frac {2 d^{2} \sqrt {d + e x}}{e^{3}} - \frac {4 d \left (d + e x\right )^{\frac {3}{2}}}{3 e^{3}} + \frac {2 \left (d + e x\right )^{\frac {5}{2}}}{5 e^{3}} & \text {for}\: e \neq 0 \\\frac {x^{3}}{3 \sqrt {d}} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} - \frac {524 d^{\frac {5}{2}} \sqrt {1 + \frac {e x}{d}}}{225 e^{3}} - \frac {14 d^{\frac {5}{2}} \log {\left (\frac {e x}{d} \right )}}{15 e^{3}} + \frac {28 d^{\frac {5}{2}} \log {\left (\sqrt {1 + \frac {e x}{d}} + 1 \right )}}{15 e^{3}} - \frac {4 d^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} \sqrt {x}} \right )}}{e^{3}} - \frac {68 d^{\frac {3}{2}} x \sqrt {1 + \frac {e x}{d}}}{225 e^{2}} + \frac {4 \sqrt {d} x^{2} \sqrt {1 + \frac {e x}{d}}}{25 e} + \frac {4 d^{3}}{e^{\frac {7}{2}} \sqrt {x} \sqrt {\frac {d}{e x} + 1}} + \frac {4 d^{2} \sqrt {x}}{e^{\frac {5}{2}} \sqrt {\frac {d}{e x} + 1}} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {x^{3}}{9 \sqrt {d}} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} \frac {2 d^{2} \sqrt {d + e x}}{e^{3}} - \frac {4 d \left (d + e x\right )^{\frac {3}{2}}}{3 e^{3}} + \frac {2 \left (d + e x\right )^{\frac {5}{2}}}{5 e^{3}} & \text {for}\: e \neq 0 \\\frac {x^{3}}{3 \sqrt {d}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]

[In]

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

[Out]

a*Piecewise((2*d**2*sqrt(d + e*x)/e**3 - 4*d*(d + e*x)**(3/2)/(3*e**3) + 2*(d + e*x)**(5/2)/(5*e**3), Ne(e, 0)
), (x**3/(3*sqrt(d)), True)) - b*n*Piecewise((-524*d**(5/2)*sqrt(1 + e*x/d)/(225*e**3) - 14*d**(5/2)*log(e*x/d
)/(15*e**3) + 28*d**(5/2)*log(sqrt(1 + e*x/d) + 1)/(15*e**3) - 4*d**(5/2)*asinh(sqrt(d)/(sqrt(e)*sqrt(x)))/e**
3 - 68*d**(3/2)*x*sqrt(1 + e*x/d)/(225*e**2) + 4*sqrt(d)*x**2*sqrt(1 + e*x/d)/(25*e) + 4*d**3/(e**(7/2)*sqrt(x
)*sqrt(d/(e*x) + 1)) + 4*d**2*sqrt(x)/(e**(5/2)*sqrt(d/(e*x) + 1)), (e > -oo) & (e < oo) & Ne(e, 0)), (x**3/(9
*sqrt(d)), True)) + b*Piecewise((2*d**2*sqrt(d + e*x)/e**3 - 4*d*(d + e*x)**(3/2)/(3*e**3) + 2*(d + e*x)**(5/2
)/(5*e**3), Ne(e, 0)), (x**3/(3*sqrt(d)), True))*log(c*x**n)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.02 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=-\frac {4}{225} \, b n {\left (\frac {60 \, d^{\frac {5}{2}} \log \left (\frac {\sqrt {e x + d} - \sqrt {d}}{\sqrt {e x + d} + \sqrt {d}}\right )}{e^{3}} + \frac {9 \, {\left (e x + d\right )}^{\frac {5}{2}} - 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 120 \, \sqrt {e x + d} d^{2}}{e^{3}}\right )} + \frac {2}{15} \, b {\left (\frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}}}{e^{3}} - \frac {10 \, {\left (e x + d\right )}^{\frac {3}{2}} d}{e^{3}} + \frac {15 \, \sqrt {e x + d} d^{2}}{e^{3}}\right )} \log \left (c x^{n}\right ) + \frac {2}{15} \, a {\left (\frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}}}{e^{3}} - \frac {10 \, {\left (e x + d\right )}^{\frac {3}{2}} d}{e^{3}} + \frac {15 \, \sqrt {e x + d} d^{2}}{e^{3}}\right )} \]

[In]

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

[Out]

-4/225*b*n*(60*d^(5/2)*log((sqrt(e*x + d) - sqrt(d))/(sqrt(e*x + d) + sqrt(d)))/e^3 + (9*(e*x + d)^(5/2) - 35*
(e*x + d)^(3/2)*d + 120*sqrt(e*x + d)*d^2)/e^3) + 2/15*b*(3*(e*x + d)^(5/2)/e^3 - 10*(e*x + d)^(3/2)*d/e^3 + 1
5*sqrt(e*x + d)*d^2/e^3)*log(c*x^n) + 2/15*a*(3*(e*x + d)^(5/2)/e^3 - 10*(e*x + d)^(3/2)*d/e^3 + 15*sqrt(e*x +
d)*d^2/e^3)

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.14 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=-\frac {32 \, b d^{3} n \arctan \left (\frac {\sqrt {e x + d}}{\sqrt {-d}}\right )}{15 \, \sqrt {-d} e^{3}} + \frac {2}{15} \, {\left (\frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}} b n}{e^{3}} - \frac {10 \, {\left (e x + d\right )}^{\frac {3}{2}} b d n}{e^{3}} + \frac {15 \, \sqrt {e x + d} b d^{2} n}{e^{3}}\right )} \log \left (e x\right ) - \frac {2 \, {\left (5 \, b n \log \left (e\right ) + 2 \, b n - 5 \, b \log \left (c\right ) - 5 \, a\right )} {\left (e x + d\right )}^{\frac {5}{2}}}{25 \, e^{3}} + \frac {4 \, {\left (15 \, b d n \log \left (e\right ) + 7 \, b d n - 15 \, b d \log \left (c\right ) - 15 \, a d\right )} {\left (e x + d\right )}^{\frac {3}{2}}}{45 \, e^{3}} - \frac {2 \, {\left (15 \, b d^{2} n \log \left (e\right ) + 16 \, b d^{2} n - 15 \, b d^{2} \log \left (c\right ) - 15 \, a d^{2}\right )} \sqrt {e x + d}}{15 \, e^{3}} \]

[In]

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

[Out]

-32/15*b*d^3*n*arctan(sqrt(e*x + d)/sqrt(-d))/(sqrt(-d)*e^3) + 2/15*(3*(e*x + d)^(5/2)*b*n/e^3 - 10*(e*x + d)^
(3/2)*b*d*n/e^3 + 15*sqrt(e*x + d)*b*d^2*n/e^3)*log(e*x) - 2/25*(5*b*n*log(e) + 2*b*n - 5*b*log(c) - 5*a)*(e*x
+ d)^(5/2)/e^3 + 4/45*(15*b*d*n*log(e) + 7*b*d*n - 15*b*d*log(c) - 15*a*d)*(e*x + d)^(3/2)/e^3 - 2/15*(15*b*d
^2*n*log(e) + 16*b*d^2*n - 15*b*d^2*log(c) - 15*a*d^2)*sqrt(e*x + d)/e^3

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=\int \frac {x^2\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{\sqrt {d+e\,x}} \,d x \]

[In]

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

[Out]

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