\(\int \frac {x^3 (a+b \log (c x^n))}{(d+e x)^{3/2}} \, dx\) [151]

Optimal result

Integrand size = 23, antiderivative size = 194 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^{3/2}} \, dx=-\frac {44 b d^2 n \sqrt {d+e x}}{5 e^4}+\frac {16 b d n (d+e x)^{3/2}}{15 e^4}-\frac {4 b n (d+e x)^{5/2}}{25 e^4}+\frac {64 b d^{5/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{5 e^4}+\frac {2 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x}}+\frac {6 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4} \]

[Out]

16/15*b*d*n*(e*x+d)^(3/2)/e^4-4/25*b*n*(e*x+d)^(5/2)/e^4+64/5*b*d^(5/2)*n*arctanh((e*x+d)^(1/2)/d^(1/2))/e^4-2
*d*(e*x+d)^(3/2)*(a+b*ln(c*x^n))/e^4+2/5*(e*x+d)^(5/2)*(a+b*ln(c*x^n))/e^4+2*d^3*(a+b*ln(c*x^n))/e^4/(e*x+d)^(
1/2)-44/5*b*d^2*n*(e*x+d)^(1/2)/e^4+6*d^2*(a+b*ln(c*x^n))*(e*x+d)^(1/2)/e^4

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 194, 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, 1634, 65, 214} \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^{3/2}} \, dx=\frac {2 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x}}+\frac {6 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {64 b d^{5/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{5 e^4}-\frac {44 b d^2 n \sqrt {d+e x}}{5 e^4}+\frac {16 b d n (d+e x)^{3/2}}{15 e^4}-\frac {4 b n (d+e x)^{5/2}}{25 e^4} \]

[In]

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

[Out]

(-44*b*d^2*n*Sqrt[d + e*x])/(5*e^4) + (16*b*d*n*(d + e*x)^(3/2))/(15*e^4) - (4*b*n*(d + e*x)^(5/2))/(25*e^4) +
(64*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(5*e^4) + (2*d^3*(a + b*Log[c*x^n]))/(e^4*Sqrt[d + e*x]) + (6
*d^2*Sqrt[d + e*x]*(a + b*Log[c*x^n]))/e^4 - (2*d*(d + e*x)^(3/2)*(a + b*Log[c*x^n]))/e^4 + (2*(d + e*x)^(5/2)
*(a + b*Log[c*x^n]))/(5*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 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 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x}}+\frac {6 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}-(b n) \int \frac {2 \left (16 d^3+8 d^2 e x-2 d e^2 x^2+e^3 x^3\right )}{5 e^4 x \sqrt {d+e x}} \, dx \\ & = \frac {2 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x}}+\frac {6 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}-\frac {(2 b n) \int \frac {16 d^3+8 d^2 e x-2 d e^2 x^2+e^3 x^3}{x \sqrt {d+e x}} \, dx}{5 e^4} \\ & = \frac {2 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x}}+\frac {6 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}-\frac {(2 b n) \int \left (\frac {11 d^2 e}{\sqrt {d+e x}}+\frac {16 d^3}{x \sqrt {d+e x}}-4 d e \sqrt {d+e x}+e (d+e x)^{3/2}\right ) \, dx}{5 e^4} \\ & = -\frac {44 b d^2 n \sqrt {d+e x}}{5 e^4}+\frac {16 b d n (d+e x)^{3/2}}{15 e^4}-\frac {4 b n (d+e x)^{5/2}}{25 e^4}+\frac {2 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x}}+\frac {6 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}-\frac {\left (32 b d^3 n\right ) \int \frac {1}{x \sqrt {d+e x}} \, dx}{5 e^4} \\ & = -\frac {44 b d^2 n \sqrt {d+e x}}{5 e^4}+\frac {16 b d n (d+e x)^{3/2}}{15 e^4}-\frac {4 b n (d+e x)^{5/2}}{25 e^4}+\frac {2 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x}}+\frac {6 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}-\frac {\left (64 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 )}{5 e^5} \\ & = -\frac {44 b d^2 n \sqrt {d+e x}}{5 e^4}+\frac {16 b d n (d+e x)^{3/2}}{15 e^4}-\frac {4 b n (d+e x)^{5/2}}{25 e^4}+\frac {64 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{5 e^4}+\frac {2 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x}}+\frac {6 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.82 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^{3/2}} \, dx=\frac {480 a d^3-592 b d^3 n+240 a d^2 e x-536 b d^2 e n x-60 a d e^2 x^2+44 b d e^2 n x^2+30 a e^3 x^3-12 b e^3 n x^3+960 b d^{5/2} n \sqrt {d+e x} \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+30 b \left (16 d^3+8 d^2 e x-2 d e^2 x^2+e^3 x^3\right ) \log \left (c x^n\right )}{75 e^4 \sqrt {d+e x}} \]

[In]

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

[Out]

(480*a*d^3 - 592*b*d^3*n + 240*a*d^2*e*x - 536*b*d^2*e*n*x - 60*a*d*e^2*x^2 + 44*b*d*e^2*n*x^2 + 30*a*e^3*x^3
- 12*b*e^3*n*x^3 + 960*b*d^(5/2)*n*Sqrt[d + e*x]*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + 30*b*(16*d^3 + 8*d^2*e*x - 2
*d*e^2*x^2 + e^3*x^3)*Log[c*x^n])/(75*e^4*Sqrt[d + e*x])

Maple [F]

\[\int \frac {x^{3} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\left (e x +d \right )^{\frac {3}{2}}}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 435, normalized size of antiderivative = 2.24 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^{3/2}} \, dx=\left [\frac {2 \, {\left (240 \, {\left (b d^{2} e n x + b d^{3} n\right )} \sqrt {d} \log \left (\frac {e x + 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) - {\left (296 \, b d^{3} n - 240 \, a d^{3} + 3 \, {\left (2 \, b e^{3} n - 5 \, a e^{3}\right )} x^{3} - 2 \, {\left (11 \, b d e^{2} n - 15 \, a d e^{2}\right )} x^{2} + 4 \, {\left (67 \, b d^{2} e n - 30 \, a d^{2} e\right )} x - 15 \, {\left (b e^{3} x^{3} - 2 \, b d e^{2} x^{2} + 8 \, b d^{2} e x + 16 \, b d^{3}\right )} \log \left (c\right ) - 15 \, {\left (b e^{3} n x^{3} - 2 \, b d e^{2} n x^{2} + 8 \, b d^{2} e n x + 16 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{75 \, {\left (e^{5} x + d e^{4}\right )}}, -\frac {2 \, {\left (480 \, {\left (b d^{2} e n x + b d^{3} n\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + {\left (296 \, b d^{3} n - 240 \, a d^{3} + 3 \, {\left (2 \, b e^{3} n - 5 \, a e^{3}\right )} x^{3} - 2 \, {\left (11 \, b d e^{2} n - 15 \, a d e^{2}\right )} x^{2} + 4 \, {\left (67 \, b d^{2} e n - 30 \, a d^{2} e\right )} x - 15 \, {\left (b e^{3} x^{3} - 2 \, b d e^{2} x^{2} + 8 \, b d^{2} e x + 16 \, b d^{3}\right )} \log \left (c\right ) - 15 \, {\left (b e^{3} n x^{3} - 2 \, b d e^{2} n x^{2} + 8 \, b d^{2} e n x + 16 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{75 \, {\left (e^{5} x + d e^{4}\right )}}\right ] \]

[In]

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

[Out]

[2/75*(240*(b*d^2*e*n*x + b*d^3*n)*sqrt(d)*log((e*x + 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - (296*b*d^3*n - 240*a
*d^3 + 3*(2*b*e^3*n - 5*a*e^3)*x^3 - 2*(11*b*d*e^2*n - 15*a*d*e^2)*x^2 + 4*(67*b*d^2*e*n - 30*a*d^2*e)*x - 15*
(b*e^3*x^3 - 2*b*d*e^2*x^2 + 8*b*d^2*e*x + 16*b*d^3)*log(c) - 15*(b*e^3*n*x^3 - 2*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^5*x + d*e^4), -2/75*(480*(b*d^2*e*n*x + b*d^3*n)*sqrt(-d)*arctan(sqr
t(e*x + d)*sqrt(-d)/d) + (296*b*d^3*n - 240*a*d^3 + 3*(2*b*e^3*n - 5*a*e^3)*x^3 - 2*(11*b*d*e^2*n - 15*a*d*e^2
)*x^2 + 4*(67*b*d^2*e*n - 30*a*d^2*e)*x - 15*(b*e^3*x^3 - 2*b*d*e^2*x^2 + 8*b*d^2*e*x + 16*b*d^3)*log(c) - 15*
(b*e^3*n*x^3 - 2*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^5*x + d*e^4)]

Sympy [A] (verification not implemented)

Time = 116.99 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.90 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^{3/2}} \, dx=a \left (\begin {cases} \frac {2 d^{3}}{e^{4} \sqrt {d + e x}} + \frac {6 d^{2} \sqrt {d + e x}}{e^{4}} - \frac {2 d \left (d + e x\right )^{\frac {3}{2}}}{e^{4}} + \frac {2 \left (d + e x\right )^{\frac {5}{2}}}{5 e^{4}} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 d^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} - \frac {308 d^{\frac {5}{2}} \sqrt {1 + \frac {e x}{d}}}{75 e^{4}} - \frac {8 d^{\frac {5}{2}} \log {\left (\frac {e x}{d} \right )}}{5 e^{4}} + \frac {16 d^{\frac {5}{2}} \log {\left (\sqrt {1 + \frac {e x}{d}} + 1 \right )}}{5 e^{4}} - \frac {16 d^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} \sqrt {x}} \right )}}{e^{4}} - \frac {56 d^{\frac {3}{2}} x \sqrt {1 + \frac {e x}{d}}}{75 e^{3}} + \frac {4 \sqrt {d} x^{2} \sqrt {1 + \frac {e x}{d}}}{25 e^{2}} + \frac {12 d^{3}}{e^{\frac {9}{2}} \sqrt {x} \sqrt {\frac {d}{e x} + 1}} + \frac {12 d^{2} \sqrt {x}}{e^{\frac {7}{2}} \sqrt {\frac {d}{e x} + 1}} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {x^{4}}{16 d^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} \frac {2 d^{3}}{e^{4} \sqrt {d + e x}} + \frac {6 d^{2} \sqrt {d + e x}}{e^{4}} - \frac {2 d \left (d + e x\right )^{\frac {3}{2}}}{e^{4}} + \frac {2 \left (d + e x\right )^{\frac {5}{2}}}{5 e^{4}} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 d^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]

[In]

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

[Out]

a*Piecewise((2*d**3/(e**4*sqrt(d + e*x)) + 6*d**2*sqrt(d + e*x)/e**4 - 2*d*(d + e*x)**(3/2)/e**4 + 2*(d + e*x)
**(5/2)/(5*e**4), Ne(e, 0)), (x**4/(4*d**(3/2)), True)) - b*n*Piecewise((-308*d**(5/2)*sqrt(1 + e*x/d)/(75*e**
4) - 8*d**(5/2)*log(e*x/d)/(5*e**4) + 16*d**(5/2)*log(sqrt(1 + e*x/d) + 1)/(5*e**4) - 16*d**(5/2)*asinh(sqrt(d
)/(sqrt(e)*sqrt(x)))/e**4 - 56*d**(3/2)*x*sqrt(1 + e*x/d)/(75*e**3) + 4*sqrt(d)*x**2*sqrt(1 + e*x/d)/(25*e**2)
+ 12*d**3/(e**(9/2)*sqrt(x)*sqrt(d/(e*x) + 1)) + 12*d**2*sqrt(x)/(e**(7/2)*sqrt(d/(e*x) + 1)), (e > -oo) & (e
< oo) & Ne(e, 0)), (x**4/(16*d**(3/2)), True)) + b*Piecewise((2*d**3/(e**4*sqrt(d + e*x)) + 6*d**2*sqrt(d + e
*x)/e**4 - 2*d*(d + e*x)**(3/2)/e**4 + 2*(d + e*x)**(5/2)/(5*e**4), Ne(e, 0)), (x**4/(4*d**(3/2)), True))*log(
c*x**n)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.03 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^{3/2}} \, dx=-\frac {4}{75} \, b n {\left (\frac {120 \, d^{\frac {5}{2}} \log \left (\frac {\sqrt {e x + d} - \sqrt {d}}{\sqrt {e x + d} + \sqrt {d}}\right )}{e^{4}} + \frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 20 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 165 \, \sqrt {e x + d} d^{2}}{e^{4}}\right )} + \frac {2}{5} \, b {\left (\frac {{\left (e x + d\right )}^{\frac {5}{2}}}{e^{4}} - \frac {5 \, {\left (e x + d\right )}^{\frac {3}{2}} d}{e^{4}} + \frac {15 \, \sqrt {e x + d} d^{2}}{e^{4}} + \frac {5 \, d^{3}}{\sqrt {e x + d} e^{4}}\right )} \log \left (c x^{n}\right ) + \frac {2}{5} \, a {\left (\frac {{\left (e x + d\right )}^{\frac {5}{2}}}{e^{4}} - \frac {5 \, {\left (e x + d\right )}^{\frac {3}{2}} d}{e^{4}} + \frac {15 \, \sqrt {e x + d} d^{2}}{e^{4}} + \frac {5 \, d^{3}}{\sqrt {e x + d} e^{4}}\right )} \]

[In]

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

[Out]

-4/75*b*n*(120*d^(5/2)*log((sqrt(e*x + d) - sqrt(d))/(sqrt(e*x + d) + sqrt(d)))/e^4 + (3*(e*x + d)^(5/2) - 20*
(e*x + d)^(3/2)*d + 165*sqrt(e*x + d)*d^2)/e^4) + 2/5*b*((e*x + d)^(5/2)/e^4 - 5*(e*x + d)^(3/2)*d/e^4 + 15*sq
rt(e*x + d)*d^2/e^4 + 5*d^3/(sqrt(e*x + d)*e^4))*log(c*x^n) + 2/5*a*((e*x + d)^(5/2)/e^4 - 5*(e*x + d)^(3/2)*d
/e^4 + 15*sqrt(e*x + d)*d^2/e^4 + 5*d^3/(sqrt(e*x + d)*e^4))

Giac [F]

\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^{3/2}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{{\left (e x + d\right )}^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^{3/2}} \, dx=\int \frac {x^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (d+e\,x\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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