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

Optimal. Leaf size=194 \[ -\frac {44 b d^2 n \sqrt {d+e x}}{5 e^4}+\frac {16 b d n (d+e x)^{3/2}}{15 e^4}-\frac {4 b n (d+e x)^{5/2}}{25 e^4}+\frac {64 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{5 e^4}+\frac {2 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x}}+\frac {6 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4} \]

[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

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Rubi [A]
time = 0.14, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {45, 2392, 12, 1634, 65, 214} \begin {gather*} \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 \tanh ^{-1}\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} \end {gather*}

Antiderivative was successfully verified.

[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*} \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}-(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*}

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Mathematica [A]
time = 0.08, size = 159, normalized size = 0.82 \begin {gather*} \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} \tanh ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[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])

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [A]
time = 0.50, size = 203, normalized size = 1.05 \begin {gather*} -\frac {4}{75} \, {\left (120 \, d^{\frac {5}{2}} e^{\left (-4\right )} \log \left (\frac {\sqrt {x e + d} - \sqrt {d}}{\sqrt {x e + d} + \sqrt {d}}\right ) + {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} - 20 \, {\left (x e + d\right )}^{\frac {3}{2}} d + 165 \, \sqrt {x e + d} d^{2}\right )} e^{\left (-4\right )}\right )} b n + \frac {2}{5} \, {\left ({\left (x e + d\right )}^{\frac {5}{2}} e^{\left (-4\right )} - 5 \, {\left (x e + d\right )}^{\frac {3}{2}} d e^{\left (-4\right )} + 15 \, \sqrt {x e + d} d^{2} e^{\left (-4\right )} + \frac {5 \, d^{3} e^{\left (-4\right )}}{\sqrt {x e + d}}\right )} b \log \left (c x^{n}\right ) + \frac {2}{5} \, {\left ({\left (x e + d\right )}^{\frac {5}{2}} e^{\left (-4\right )} - 5 \, {\left (x e + d\right )}^{\frac {3}{2}} d e^{\left (-4\right )} + 15 \, \sqrt {x e + d} d^{2} e^{\left (-4\right )} + \frac {5 \, d^{3} e^{\left (-4\right )}}{\sqrt {x e + d}}\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+d)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.43, size = 418, normalized size = 2.15 \begin {gather*} \left [\frac {2 \, {\left (240 \, {\left (b d^{2} n x e + b d^{3} n\right )} \sqrt {d} \log \left (\frac {x e + 2 \, \sqrt {x e + d} \sqrt {d} + 2 \, d}{x}\right ) - {\left (296 \, b d^{3} n + 3 \, {\left (2 \, b n - 5 \, a\right )} x^{3} e^{3} - 240 \, a d^{3} - 2 \, {\left (11 \, b d n - 15 \, a d\right )} x^{2} e^{2} + 4 \, {\left (67 \, b d^{2} n - 30 \, a d^{2}\right )} x e - 15 \, {\left (b x^{3} e^{3} - 2 \, b d x^{2} e^{2} + 8 \, b d^{2} x e + 16 \, b d^{3}\right )} \log \left (c\right ) - 15 \, {\left (b n x^{3} e^{3} - 2 \, b d n x^{2} e^{2} + 8 \, b d^{2} n x e + 16 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {x e + d}\right )}}{75 \, {\left (x e^{5} + d e^{4}\right )}}, -\frac {2 \, {\left (480 \, {\left (b d^{2} n x e + b d^{3} n\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {x e + d} \sqrt {-d}}{d}\right ) + {\left (296 \, b d^{3} n + 3 \, {\left (2 \, b n - 5 \, a\right )} x^{3} e^{3} - 240 \, a d^{3} - 2 \, {\left (11 \, b d n - 15 \, a d\right )} x^{2} e^{2} + 4 \, {\left (67 \, b d^{2} n - 30 \, a d^{2}\right )} x e - 15 \, {\left (b x^{3} e^{3} - 2 \, b d x^{2} e^{2} + 8 \, b d^{2} x e + 16 \, b d^{3}\right )} \log \left (c\right ) - 15 \, {\left (b n x^{3} e^{3} - 2 \, b d n x^{2} e^{2} + 8 \, b d^{2} n x e + 16 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {x e + d}\right )}}{75 \, {\left (x e^{5} + d e^{4}\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+d)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 26.84, size = 384, normalized size = 1.98 \begin {gather*} \frac {\frac {2 a d^{3}}{\sqrt {d + e x}} + 6 a d^{2} \sqrt {d + e x} - 2 a d \left (d + e x\right )^{\frac {3}{2}} + \frac {2 a \left (d + e x\right )^{\frac {5}{2}}}{5} - 2 b d^{3} \cdot \left (\frac {2 n \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{\sqrt {- d}} - \frac {\log {\left (c \left (- \frac {d}{e} + \frac {d + e x}{e}\right )^{n} \right )}}{\sqrt {d + e x}}\right ) + 6 b d^{2} \left (\sqrt {d + e x} \log {\left (c \left (- \frac {d}{e} + \frac {d + e x}{e}\right )^{n} \right )} - \frac {2 n \left (\frac {d e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{\sqrt {- d}} + e \sqrt {d + e x}\right )}{e}\right ) - 6 b d \left (\frac {\left (d + e x\right )^{\frac {3}{2}} \log {\left (c \left (- \frac {d}{e} + \frac {d + e x}{e}\right )^{n} \right )}}{3} - \frac {2 n \left (\frac {d^{2} e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{\sqrt {- d}} + d e \sqrt {d + e x} + \frac {e \left (d + e x\right )^{\frac {3}{2}}}{3}\right )}{3 e}\right ) + 2 b \left (\frac {\left (d + e x\right )^{\frac {5}{2}} \log {\left (c \left (- \frac {d}{e} + \frac {d + e x}{e}\right )^{n} \right )}}{5} - \frac {2 n \left (\frac {d^{3} e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{\sqrt {- d}} + d^{2} e \sqrt {d + e x} + \frac {d e \left (d + e x\right )^{\frac {3}{2}}}{3} + \frac {e \left (d + e x\right )^{\frac {5}{2}}}{5}\right )}{5 e}\right )}{e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*d**3/sqrt(d + e*x) + 6*a*d**2*sqrt(d + e*x) - 2*a*d*(d + e*x)**(3/2) + 2*a*(d + e*x)**(5/2)/5 - 2*b*d**3*
(2*n*atan(sqrt(d + e*x)/sqrt(-d))/sqrt(-d) - log(c*(-d/e + (d + e*x)/e)**n)/sqrt(d + e*x)) + 6*b*d**2*(sqrt(d
+ e*x)*log(c*(-d/e + (d + e*x)/e)**n) - 2*n*(d*e*atan(sqrt(d + e*x)/sqrt(-d))/sqrt(-d) + e*sqrt(d + e*x))/e) -
6*b*d*((d + e*x)**(3/2)*log(c*(-d/e + (d + e*x)/e)**n)/3 - 2*n*(d**2*e*atan(sqrt(d + e*x)/sqrt(-d))/sqrt(-d)
+ d*e*sqrt(d + e*x) + e*(d + e*x)**(3/2)/3)/(3*e)) + 2*b*((d + e*x)**(5/2)*log(c*(-d/e + (d + e*x)/e)**n)/5 -
2*n*(d**3*e*atan(sqrt(d + e*x)/sqrt(-d))/sqrt(-d) + d**2*e*sqrt(d + e*x) + d*e*(d + e*x)**(3/2)/3 + e*(d + e*x
)**(5/2)/5)/(5*e)))/e**4

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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/(x*e + d)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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