3.2.39 \(\int x (d+e x)^{3/2} (a+b \log (c x^n)) \, dx\) [139]

Optimal. Leaf size=163 \[ \frac {8 b d^3 n \sqrt {d+e x}}{35 e^2}+\frac {8 b d^2 n (d+e x)^{3/2}}{105 e^2}+\frac {8 b d n (d+e x)^{5/2}}{175 e^2}-\frac {4 b n (d+e x)^{7/2}}{49 e^2}-\frac {8 b d^{7/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{35 e^2}-\frac {2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2} \]

[Out]

8/105*b*d^2*n*(e*x+d)^(3/2)/e^2+8/175*b*d*n*(e*x+d)^(5/2)/e^2-4/49*b*n*(e*x+d)^(7/2)/e^2-8/35*b*d^(7/2)*n*arct
anh((e*x+d)^(1/2)/d^(1/2))/e^2-2/5*d*(e*x+d)^(5/2)*(a+b*ln(c*x^n))/e^2+2/7*(e*x+d)^(7/2)*(a+b*ln(c*x^n))/e^2+8
/35*b*d^3*n*(e*x+d)^(1/2)/e^2

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Rubi [A]
time = 0.08, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {45, 2392, 12, 81, 52, 65, 214} \begin {gather*} -\frac {2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-\frac {8 b d^{7/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{35 e^2}+\frac {8 b d^3 n \sqrt {d+e x}}{35 e^2}+\frac {8 b d^2 n (d+e x)^{3/2}}{105 e^2}+\frac {8 b d n (d+e x)^{5/2}}{175 e^2}-\frac {4 b n (d+e x)^{7/2}}{49 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b*d^3*n*Sqrt[d + e*x])/(35*e^2) + (8*b*d^2*n*(d + e*x)^(3/2))/(105*e^2) + (8*b*d*n*(d + e*x)^(5/2))/(175*e^
2) - (4*b*n*(d + e*x)^(7/2))/(49*e^2) - (8*b*d^(7/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(35*e^2) - (2*d*(d + e*
x)^(5/2)*(a + b*Log[c*x^n]))/(5*e^2) + (2*(d + e*x)^(7/2)*(a + b*Log[c*x^n]))/(7*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 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 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx &=-\frac {2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-(b n) \int \frac {2 (d+e x)^{5/2} (-2 d+5 e x)}{35 e^2 x} \, dx\\ &=-\frac {2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-\frac {(2 b n) \int \frac {(d+e x)^{5/2} (-2 d+5 e x)}{x} \, dx}{35 e^2}\\ &=-\frac {4 b n (d+e x)^{7/2}}{49 e^2}-\frac {2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}+\frac {(4 b d n) \int \frac {(d+e x)^{5/2}}{x} \, dx}{35 e^2}\\ &=\frac {8 b d n (d+e x)^{5/2}}{175 e^2}-\frac {4 b n (d+e x)^{7/2}}{49 e^2}-\frac {2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}+\frac {\left (4 b d^2 n\right ) \int \frac {(d+e x)^{3/2}}{x} \, dx}{35 e^2}\\ &=\frac {8 b d^2 n (d+e x)^{3/2}}{105 e^2}+\frac {8 b d n (d+e x)^{5/2}}{175 e^2}-\frac {4 b n (d+e x)^{7/2}}{49 e^2}-\frac {2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}+\frac {\left (4 b d^3 n\right ) \int \frac {\sqrt {d+e x}}{x} \, dx}{35 e^2}\\ &=\frac {8 b d^3 n \sqrt {d+e x}}{35 e^2}+\frac {8 b d^2 n (d+e x)^{3/2}}{105 e^2}+\frac {8 b d n (d+e x)^{5/2}}{175 e^2}-\frac {4 b n (d+e x)^{7/2}}{49 e^2}-\frac {2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}+\frac {\left (4 b d^4 n\right ) \int \frac {1}{x \sqrt {d+e x}} \, dx}{35 e^2}\\ &=\frac {8 b d^3 n \sqrt {d+e x}}{35 e^2}+\frac {8 b d^2 n (d+e x)^{3/2}}{105 e^2}+\frac {8 b d n (d+e x)^{5/2}}{175 e^2}-\frac {4 b n (d+e x)^{7/2}}{49 e^2}-\frac {2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}+\frac {\left (8 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^3}\\ &=\frac {8 b d^3 n \sqrt {d+e x}}{35 e^2}+\frac {8 b d^2 n (d+e x)^{3/2}}{105 e^2}+\frac {8 b d n (d+e x)^{5/2}}{175 e^2}-\frac {4 b n (d+e x)^{7/2}}{49 e^2}-\frac {8 b d^{7/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{35 e^2}-\frac {2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 120, normalized size = 0.74 \begin {gather*} -\frac {2 \left (420 b d^{7/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+\sqrt {d+e x} \left (105 a (2 d-5 e x) (d+e x)^2+2 b n \left (-247 d^3+71 d^2 e x+183 d e^2 x^2+75 e^3 x^3\right )+105 b (2 d-5 e x) (d+e x)^2 \log \left (c x^n\right )\right )\right )}{3675 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(420*b*d^(7/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + Sqrt[d + e*x]*(105*a*(2*d - 5*e*x)*(d + e*x)^2 + 2*b*n*(
-247*d^3 + 71*d^2*e*x + 183*d*e^2*x^2 + 75*e^3*x^3) + 105*b*(2*d - 5*e*x)*(d + e*x)^2*Log[c*x^n])))/(3675*e^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 159, normalized size = 0.98 \begin {gather*} \frac {4}{3675} \, {\left (105 \, d^{\frac {7}{2}} e^{\left (-2\right )} \log \left (\frac {\sqrt {x e + d} - \sqrt {d}}{\sqrt {x e + d} + \sqrt {d}}\right ) - {\left (75 \, {\left (x e + d\right )}^{\frac {7}{2}} - 42 \, {\left (x e + d\right )}^{\frac {5}{2}} d - 70 \, {\left (x e + d\right )}^{\frac {3}{2}} d^{2} - 210 \, \sqrt {x e + d} d^{3}\right )} e^{\left (-2\right )}\right )} b n + \frac {2}{35} \, {\left (5 \, {\left (x e + d\right )}^{\frac {7}{2}} e^{\left (-2\right )} - 7 \, {\left (x e + d\right )}^{\frac {5}{2}} d e^{\left (-2\right )}\right )} b \log \left (c x^{n}\right ) + \frac {2}{35} \, {\left (5 \, {\left (x e + d\right )}^{\frac {7}{2}} e^{\left (-2\right )} - 7 \, {\left (x e + d\right )}^{\frac {5}{2}} d e^{\left (-2\right )}\right )} a \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/3675*(105*d^(7/2)*e^(-2)*log((sqrt(x*e + d) - sqrt(d))/(sqrt(x*e + d) + sqrt(d))) - (75*(x*e + d)^(7/2) - 42
*(x*e + d)^(5/2)*d - 70*(x*e + d)^(3/2)*d^2 - 210*sqrt(x*e + d)*d^3)*e^(-2))*b*n + 2/35*(5*(x*e + d)^(7/2)*e^(
-2) - 7*(x*e + d)^(5/2)*d*e^(-2))*b*log(c*x^n) + 2/35*(5*(x*e + d)^(7/2)*e^(-2) - 7*(x*e + d)^(5/2)*d*e^(-2))*
a

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Fricas [A]
time = 0.40, size = 374, normalized size = 2.29 \begin {gather*} \left [\frac {2}{3675} \, {\left (210 \, b d^{\frac {7}{2}} n \log \left (\frac {x e - 2 \, \sqrt {x e + d} \sqrt {d} + 2 \, d}{x}\right ) + {\left (494 \, b d^{3} n - 75 \, {\left (2 \, b n - 7 \, a\right )} x^{3} e^{3} - 210 \, a d^{3} - 6 \, {\left (61 \, b d n - 140 \, a d\right )} x^{2} e^{2} - {\left (142 \, b d^{2} n - 105 \, a d^{2}\right )} x e + 105 \, {\left (5 \, b x^{3} e^{3} + 8 \, b d x^{2} e^{2} + b d^{2} x e - 2 \, b d^{3}\right )} \log \left (c\right ) + 105 \, {\left (5 \, b n x^{3} e^{3} + 8 \, b d n x^{2} e^{2} + b d^{2} n x e - 2 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {x e + d}\right )} e^{\left (-2\right )}, \frac {2}{3675} \, {\left (420 \, b \sqrt {-d} d^{3} n \arctan \left (\frac {\sqrt {x e + d} \sqrt {-d}}{d}\right ) + {\left (494 \, b d^{3} n - 75 \, {\left (2 \, b n - 7 \, a\right )} x^{3} e^{3} - 210 \, a d^{3} - 6 \, {\left (61 \, b d n - 140 \, a d\right )} x^{2} e^{2} - {\left (142 \, b d^{2} n - 105 \, a d^{2}\right )} x e + 105 \, {\left (5 \, b x^{3} e^{3} + 8 \, b d x^{2} e^{2} + b d^{2} x e - 2 \, b d^{3}\right )} \log \left (c\right ) + 105 \, {\left (5 \, b n x^{3} e^{3} + 8 \, b d n x^{2} e^{2} + b d^{2} n x e - 2 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {x e + d}\right )} e^{\left (-2\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/3675*(210*b*d^(7/2)*n*log((x*e - 2*sqrt(x*e + d)*sqrt(d) + 2*d)/x) + (494*b*d^3*n - 75*(2*b*n - 7*a)*x^3*e^
3 - 210*a*d^3 - 6*(61*b*d*n - 140*a*d)*x^2*e^2 - (142*b*d^2*n - 105*a*d^2)*x*e + 105*(5*b*x^3*e^3 + 8*b*d*x^2*
e^2 + b*d^2*x*e - 2*b*d^3)*log(c) + 105*(5*b*n*x^3*e^3 + 8*b*d*n*x^2*e^2 + b*d^2*n*x*e - 2*b*d^3*n)*log(x))*sq
rt(x*e + d))*e^(-2), 2/3675*(420*b*sqrt(-d)*d^3*n*arctan(sqrt(x*e + d)*sqrt(-d)/d) + (494*b*d^3*n - 75*(2*b*n
- 7*a)*x^3*e^3 - 210*a*d^3 - 6*(61*b*d*n - 140*a*d)*x^2*e^2 - (142*b*d^2*n - 105*a*d^2)*x*e + 105*(5*b*x^3*e^3
+ 8*b*d*x^2*e^2 + b*d^2*x*e - 2*b*d^3)*log(c) + 105*(5*b*n*x^3*e^3 + 8*b*d*n*x^2*e^2 + b*d^2*n*x*e - 2*b*d^3*
n)*log(x))*sqrt(x*e + d))*e^(-2)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 583 vs. \(2 (165) = 330\).
time = 33.33, size = 583, normalized size = 3.58 \begin {gather*} \frac {2 a d \left (- \frac {d \left (d + e x\right )^{\frac {3}{2}}}{3} + \frac {\left (d + e x\right )^{\frac {5}{2}}}{5}\right )}{e^{2}} + \frac {2 a \left (\frac {d^{2} \left (d + e x\right )^{\frac {3}{2}}}{3} - \frac {2 d \left (d + e x\right )^{\frac {5}{2}}}{5} + \frac {\left (d + e x\right )^{\frac {7}{2}}}{7}\right )}{e^{2}} + \frac {2 b d \left (- 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 ) + \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^{2}} + \frac {2 b \left (d^{2} \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 d \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 ) + \frac {\left (d + e x\right )^{\frac {7}{2}} \log {\left (c \left (- \frac {d}{e} + \frac {d + e x}{e}\right )^{n} \right )}}{7} - \frac {2 n \left (\frac {d^{4} e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{\sqrt {- d}} + d^{3} e \sqrt {d + e x} + \frac {d^{2} e \left (d + e x\right )^{\frac {3}{2}}}{3} + \frac {d e \left (d + e x\right )^{\frac {5}{2}}}{5} + \frac {e \left (d + e x\right )^{\frac {7}{2}}}{7}\right )}{7 e}\right )}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*d*(-d*(d + e*x)**(3/2)/3 + (d + e*x)**(5/2)/5)/e**2 + 2*a*(d**2*(d + e*x)**(3/2)/3 - 2*d*(d + e*x)**(5/2)/
5 + (d + e*x)**(7/2)/7)/e**2 + 2*b*d*(-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)) + (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**2 + 2*b*(d**2*((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*d*((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)) + (d + e*x)**(7/2)*log(c*(-d/e +
(d + e*x)/e)**n)/7 - 2*n*(d**4*e*atan(sqrt(d + e*x)/sqrt(-d))/sqrt(-d) + d**3*e*sqrt(d + e*x) + d**2*e*(d + e*
x)**(3/2)/3 + d*e*(d + e*x)**(5/2)/5 + e*(d + e*x)**(7/2)/7)/(7*e))/e**2

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

[Out]

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

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

[Out]

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

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