\(\int \frac {x^3 (a+b \log (c x^n))}{\sqrt {d+e x}} \, dx\) [144]

Optimal result

Integrand size = 23, antiderivative size = 217 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=\frac {64 b d^3 n \sqrt {d+e x}}{35 e^4}-\frac {76 b d^2 n (d+e x)^{3/2}}{105 e^4}+\frac {64 b d n (d+e x)^{5/2}}{175 e^4}-\frac {4 b n (d+e x)^{7/2}}{49 e^4}-\frac {64 b d^{7/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{35 e^4}-\frac {2 d^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {6 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4} \]

[Out]

-76/105*b*d^2*n*(e*x+d)^(3/2)/e^4+64/175*b*d*n*(e*x+d)^(5/2)/e^4-4/49*b*n*(e*x+d)^(7/2)/e^4-64/35*b*d^(7/2)*n*
arctanh((e*x+d)^(1/2)/d^(1/2))/e^4+2*d^2*(e*x+d)^(3/2)*(a+b*ln(c*x^n))/e^4-6/5*d*(e*x+d)^(5/2)*(a+b*ln(c*x^n))
/e^4+2/7*(e*x+d)^(7/2)*(a+b*ln(c*x^n))/e^4+64/35*b*d^3*n*(e*x+d)^(1/2)/e^4-2*d^3*(a+b*ln(c*x^n))*(e*x+d)^(1/2)
/e^4

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {45, 2392, 12, 1634, 52, 65, 214} \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=-\frac {2 d^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {6 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4}-\frac {64 b d^{7/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{35 e^4}+\frac {64 b d^3 n \sqrt {d+e x}}{35 e^4}-\frac {76 b d^2 n (d+e x)^{3/2}}{105 e^4}+\frac {64 b d n (d+e x)^{5/2}}{175 e^4}-\frac {4 b n (d+e x)^{7/2}}{49 e^4} \]

[In]

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

[Out]

(64*b*d^3*n*Sqrt[d + e*x])/(35*e^4) - (76*b*d^2*n*(d + e*x)^(3/2))/(105*e^4) + (64*b*d*n*(d + e*x)^(5/2))/(175
*e^4) - (4*b*n*(d + e*x)^(7/2))/(49*e^4) - (64*b*d^(7/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(35*e^4) - (2*d^3*S
qrt[d + e*x]*(a + b*Log[c*x^n]))/e^4 + (2*d^2*(d + e*x)^(3/2)*(a + b*Log[c*x^n]))/e^4 - (6*d*(d + e*x)^(5/2)*(
a + b*Log[c*x^n]))/(5*e^4) + (2*(d + e*x)^(7/2)*(a + b*Log[c*x^n]))/(7*e^4)

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 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 1634

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 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^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {6 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4}-(b n) \int \frac {2 \sqrt {d+e x} \left (-16 d^3+8 d^2 e x-6 d e^2 x^2+5 e^3 x^3\right )}{35 e^4 x} \, dx \\ & = -\frac {2 d^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {6 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4}-\frac {(2 b n) \int \frac {\sqrt {d+e x} \left (-16 d^3+8 d^2 e x-6 d e^2 x^2+5 e^3 x^3\right )}{x} \, dx}{35 e^4} \\ & = -\frac {2 d^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {6 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4}-\frac {(2 b n) \int \left (19 d^2 e \sqrt {d+e x}-\frac {16 d^3 \sqrt {d+e x}}{x}-16 d e (d+e x)^{3/2}+5 e (d+e x)^{5/2}\right ) \, dx}{35 e^4} \\ & = -\frac {76 b d^2 n (d+e x)^{3/2}}{105 e^4}+\frac {64 b d n (d+e x)^{5/2}}{175 e^4}-\frac {4 b n (d+e x)^{7/2}}{49 e^4}-\frac {2 d^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {6 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4}+\frac {\left (32 b d^3 n\right ) \int \frac {\sqrt {d+e x}}{x} \, dx}{35 e^4} \\ & = \frac {64 b d^3 n \sqrt {d+e x}}{35 e^4}-\frac {76 b d^2 n (d+e x)^{3/2}}{105 e^4}+\frac {64 b d n (d+e x)^{5/2}}{175 e^4}-\frac {4 b n (d+e x)^{7/2}}{49 e^4}-\frac {2 d^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {6 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4}+\frac {\left (32 b d^4 n\right ) \int \frac {1}{x \sqrt {d+e x}} \, dx}{35 e^4} \\ & = \frac {64 b d^3 n \sqrt {d+e x}}{35 e^4}-\frac {76 b d^2 n (d+e x)^{3/2}}{105 e^4}+\frac {64 b d n (d+e x)^{5/2}}{175 e^4}-\frac {4 b n (d+e x)^{7/2}}{49 e^4}-\frac {2 d^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {6 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4}+\frac {\left (64 b d^4 n\right ) \text {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{35 e^5} \\ & = \frac {64 b d^3 n \sqrt {d+e x}}{35 e^4}-\frac {76 b d^2 n (d+e x)^{3/2}}{105 e^4}+\frac {64 b d n (d+e x)^{5/2}}{175 e^4}-\frac {4 b n (d+e x)^{7/2}}{49 e^4}-\frac {64 b d^{7/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{35 e^4}-\frac {2 d^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {6 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.69 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=-\frac {2 \left (3360 b d^{7/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+\sqrt {d+e x} \left (105 a \left (16 d^3-8 d^2 e x+6 d e^2 x^2-5 e^3 x^3\right )+2 b n \left (-1276 d^3+218 d^2 e x-111 d e^2 x^2+75 e^3 x^3\right )+105 b \left (16 d^3-8 d^2 e x+6 d e^2 x^2-5 e^3 x^3\right ) \log \left (c x^n\right )\right )\right )}{3675 e^4} \]

[In]

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

[Out]

(-2*(3360*b*d^(7/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + Sqrt[d + e*x]*(105*a*(16*d^3 - 8*d^2*e*x + 6*d*e^2*x^2
- 5*e^3*x^3) + 2*b*n*(-1276*d^3 + 218*d^2*e*x - 111*d*e^2*x^2 + 75*e^3*x^3) + 105*b*(16*d^3 - 8*d^2*e*x + 6*d*
e^2*x^2 - 5*e^3*x^3)*Log[c*x^n])))/(3675*e^4)

Maple [F]

\[\int \frac {x^{3} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\sqrt {e x +d}}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.82 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=\left [\frac {2 \, {\left (1680 \, b d^{\frac {7}{2}} n \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + {\left (2552 \, b d^{3} n - 1680 \, a d^{3} - 75 \, {\left (2 \, b e^{3} n - 7 \, a e^{3}\right )} x^{3} + 6 \, {\left (37 \, b d e^{2} n - 105 \, a d e^{2}\right )} x^{2} - 4 \, {\left (109 \, b d^{2} e n - 210 \, a d^{2} e\right )} x + 105 \, {\left (5 \, b e^{3} x^{3} - 6 \, b d e^{2} x^{2} + 8 \, b d^{2} e x - 16 \, b d^{3}\right )} \log \left (c\right ) + 105 \, {\left (5 \, b e^{3} n x^{3} - 6 \, b d e^{2} n x^{2} + 8 \, b d^{2} e n x - 16 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{3675 \, e^{4}}, \frac {2 \, {\left (3360 \, b \sqrt {-d} d^{3} n \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + {\left (2552 \, b d^{3} n - 1680 \, a d^{3} - 75 \, {\left (2 \, b e^{3} n - 7 \, a e^{3}\right )} x^{3} + 6 \, {\left (37 \, b d e^{2} n - 105 \, a d e^{2}\right )} x^{2} - 4 \, {\left (109 \, b d^{2} e n - 210 \, a d^{2} e\right )} x + 105 \, {\left (5 \, b e^{3} x^{3} - 6 \, b d e^{2} x^{2} + 8 \, b d^{2} e x - 16 \, b d^{3}\right )} \log \left (c\right ) + 105 \, {\left (5 \, b e^{3} n x^{3} - 6 \, b d e^{2} n x^{2} + 8 \, b d^{2} e n x - 16 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{3675 \, e^{4}}\right ] \]

[In]

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

[Out]

[2/3675*(1680*b*d^(7/2)*n*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + (2552*b*d^3*n - 1680*a*d^3 - 75*(2*b*
e^3*n - 7*a*e^3)*x^3 + 6*(37*b*d*e^2*n - 105*a*d*e^2)*x^2 - 4*(109*b*d^2*e*n - 210*a*d^2*e)*x + 105*(5*b*e^3*x
^3 - 6*b*d*e^2*x^2 + 8*b*d^2*e*x - 16*b*d^3)*log(c) + 105*(5*b*e^3*n*x^3 - 6*b*d*e^2*n*x^2 + 8*b*d^2*e*n*x - 1
6*b*d^3*n)*log(x))*sqrt(e*x + d))/e^4, 2/3675*(3360*b*sqrt(-d)*d^3*n*arctan(sqrt(e*x + d)*sqrt(-d)/d) + (2552*
b*d^3*n - 1680*a*d^3 - 75*(2*b*e^3*n - 7*a*e^3)*x^3 + 6*(37*b*d*e^2*n - 105*a*d*e^2)*x^2 - 4*(109*b*d^2*e*n -
210*a*d^2*e)*x + 105*(5*b*e^3*x^3 - 6*b*d*e^2*x^2 + 8*b*d^2*e*x - 16*b*d^3)*log(c) + 105*(5*b*e^3*n*x^3 - 6*b*
d*e^2*n*x^2 + 8*b*d^2*e*n*x - 16*b*d^3*n)*log(x))*sqrt(e*x + d))/e^4]

Sympy [A] (verification not implemented)

Time = 57.10 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.82 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=a \left (\begin {cases} - \frac {2 d^{3} \sqrt {d + e x}}{e^{4}} + \frac {2 d^{2} \left (d + e x\right )^{\frac {3}{2}}}{e^{4}} - \frac {6 d \left (d + e x\right )^{\frac {5}{2}}}{5 e^{4}} + \frac {2 \left (d + e x\right )^{\frac {7}{2}}}{7 e^{4}} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 \sqrt {d}} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} \frac {9596 d^{\frac {7}{2}} \sqrt {1 + \frac {e x}{d}}}{3675 e^{4}} + \frac {38 d^{\frac {7}{2}} \log {\left (\frac {e x}{d} \right )}}{35 e^{4}} - \frac {76 d^{\frac {7}{2}} \log {\left (\sqrt {1 + \frac {e x}{d}} + 1 \right )}}{35 e^{4}} + \frac {4 d^{\frac {7}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} \sqrt {x}} \right )}}{e^{4}} + \frac {872 d^{\frac {5}{2}} x \sqrt {1 + \frac {e x}{d}}}{3675 e^{3}} - \frac {148 d^{\frac {3}{2}} x^{2} \sqrt {1 + \frac {e x}{d}}}{1225 e^{2}} + \frac {4 \sqrt {d} x^{3} \sqrt {1 + \frac {e x}{d}}}{49 e} - \frac {4 d^{4}}{e^{\frac {9}{2}} \sqrt {x} \sqrt {\frac {d}{e x} + 1}} - \frac {4 d^{3} \sqrt {x}}{e^{\frac {7}{2}} \sqrt {\frac {d}{e x} + 1}} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {x^{4}}{16 \sqrt {d}} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} - \frac {2 d^{3} \sqrt {d + e x}}{e^{4}} + \frac {2 d^{2} \left (d + e x\right )^{\frac {3}{2}}}{e^{4}} - \frac {6 d \left (d + e x\right )^{\frac {5}{2}}}{5 e^{4}} + \frac {2 \left (d + e x\right )^{\frac {7}{2}}}{7 e^{4}} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 \sqrt {d}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]

[In]

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

[Out]

a*Piecewise((-2*d**3*sqrt(d + e*x)/e**4 + 2*d**2*(d + e*x)**(3/2)/e**4 - 6*d*(d + e*x)**(5/2)/(5*e**4) + 2*(d
+ e*x)**(7/2)/(7*e**4), Ne(e, 0)), (x**4/(4*sqrt(d)), True)) - b*n*Piecewise((9596*d**(7/2)*sqrt(1 + e*x/d)/(3
675*e**4) + 38*d**(7/2)*log(e*x/d)/(35*e**4) - 76*d**(7/2)*log(sqrt(1 + e*x/d) + 1)/(35*e**4) + 4*d**(7/2)*asi
nh(sqrt(d)/(sqrt(e)*sqrt(x)))/e**4 + 872*d**(5/2)*x*sqrt(1 + e*x/d)/(3675*e**3) - 148*d**(3/2)*x**2*sqrt(1 + e
*x/d)/(1225*e**2) + 4*sqrt(d)*x**3*sqrt(1 + e*x/d)/(49*e) - 4*d**4/(e**(9/2)*sqrt(x)*sqrt(d/(e*x) + 1)) - 4*d*
*3*sqrt(x)/(e**(7/2)*sqrt(d/(e*x) + 1)), (e > -oo) & (e < oo) & Ne(e, 0)), (x**4/(16*sqrt(d)), True)) + b*Piec
ewise((-2*d**3*sqrt(d + e*x)/e**4 + 2*d**2*(d + e*x)**(3/2)/e**4 - 6*d*(d + e*x)**(5/2)/(5*e**4) + 2*(d + e*x)
**(7/2)/(7*e**4), Ne(e, 0)), (x**4/(4*sqrt(d)), True))*log(c*x**n)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.99 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=\frac {4}{3675} \, b n {\left (\frac {840 \, d^{\frac {7}{2}} \log \left (\frac {\sqrt {e x + d} - \sqrt {d}}{\sqrt {e x + d} + \sqrt {d}}\right )}{e^{4}} - \frac {75 \, {\left (e x + d\right )}^{\frac {7}{2}} - 336 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 665 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 1680 \, \sqrt {e x + d} d^{3}}{e^{4}}\right )} + \frac {2}{35} \, b {\left (\frac {5 \, {\left (e x + d\right )}^{\frac {7}{2}}}{e^{4}} - \frac {21 \, {\left (e x + d\right )}^{\frac {5}{2}} d}{e^{4}} + \frac {35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2}}{e^{4}} - \frac {35 \, \sqrt {e x + d} d^{3}}{e^{4}}\right )} \log \left (c x^{n}\right ) + \frac {2}{35} \, a {\left (\frac {5 \, {\left (e x + d\right )}^{\frac {7}{2}}}{e^{4}} - \frac {21 \, {\left (e x + d\right )}^{\frac {5}{2}} d}{e^{4}} + \frac {35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2}}{e^{4}} - \frac {35 \, \sqrt {e x + d} d^{3}}{e^{4}}\right )} \]

[In]

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

[Out]

4/3675*b*n*(840*d^(7/2)*log((sqrt(e*x + d) - sqrt(d))/(sqrt(e*x + d) + sqrt(d)))/e^4 - (75*(e*x + d)^(7/2) - 3
36*(e*x + d)^(5/2)*d + 665*(e*x + d)^(3/2)*d^2 - 1680*sqrt(e*x + d)*d^3)/e^4) + 2/35*b*(5*(e*x + d)^(7/2)/e^4
- 21*(e*x + d)^(5/2)*d/e^4 + 35*(e*x + d)^(3/2)*d^2/e^4 - 35*sqrt(e*x + d)*d^3/e^4)*log(c*x^n) + 2/35*a*(5*(e*
x + d)^(7/2)/e^4 - 21*(e*x + d)^(5/2)*d/e^4 + 35*(e*x + d)^(3/2)*d^2/e^4 - 35*sqrt(e*x + d)*d^3/e^4)

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.16 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=\frac {64 \, b d^{4} n \arctan \left (\frac {\sqrt {e x + d}}{\sqrt {-d}}\right )}{35 \, \sqrt {-d} e^{4}} + \frac {2}{35} \, {\left (\frac {5 \, {\left (e x + d\right )}^{\frac {7}{2}} b n}{e^{4}} - \frac {21 \, {\left (e x + d\right )}^{\frac {5}{2}} b d n}{e^{4}} + \frac {35 \, {\left (e x + d\right )}^{\frac {3}{2}} b d^{2} n}{e^{4}} - \frac {35 \, \sqrt {e x + d} b d^{3} n}{e^{4}}\right )} \log \left (e x\right ) - \frac {2 \, {\left (7 \, b n \log \left (e\right ) + 2 \, b n - 7 \, b \log \left (c\right ) - 7 \, a\right )} {\left (e x + d\right )}^{\frac {7}{2}}}{49 \, e^{4}} + \frac {2 \, {\left (105 \, b d n \log \left (e\right ) + 32 \, b d n - 105 \, b d \log \left (c\right ) - 105 \, a d\right )} {\left (e x + d\right )}^{\frac {5}{2}}}{175 \, e^{4}} - \frac {2 \, {\left (105 \, b d^{2} n \log \left (e\right ) + 38 \, b d^{2} n - 105 \, b d^{2} \log \left (c\right ) - 105 \, a d^{2}\right )} {\left (e x + d\right )}^{\frac {3}{2}}}{105 \, e^{4}} + \frac {2 \, {\left (35 \, b d^{3} n \log \left (e\right ) + 32 \, b d^{3} n - 35 \, b d^{3} \log \left (c\right ) - 35 \, a d^{3}\right )} \sqrt {e x + d}}{35 \, e^{4}} \]

[In]

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

[Out]

64/35*b*d^4*n*arctan(sqrt(e*x + d)/sqrt(-d))/(sqrt(-d)*e^4) + 2/35*(5*(e*x + d)^(7/2)*b*n/e^4 - 21*(e*x + d)^(
5/2)*b*d*n/e^4 + 35*(e*x + d)^(3/2)*b*d^2*n/e^4 - 35*sqrt(e*x + d)*b*d^3*n/e^4)*log(e*x) - 2/49*(7*b*n*log(e)
+ 2*b*n - 7*b*log(c) - 7*a)*(e*x + d)^(7/2)/e^4 + 2/175*(105*b*d*n*log(e) + 32*b*d*n - 105*b*d*log(c) - 105*a*
d)*(e*x + d)^(5/2)/e^4 - 2/105*(105*b*d^2*n*log(e) + 38*b*d^2*n - 105*b*d^2*log(c) - 105*a*d^2)*(e*x + d)^(3/2
)/e^4 + 2/35*(35*b*d^3*n*log(e) + 32*b*d^3*n - 35*b*d^3*log(c) - 35*a*d^3)*sqrt(e*x + d)/e^4

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=\int \frac {x^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{\sqrt {d+e\,x}} \,d x \]

[In]

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

[Out]

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