3.263 \(\int x^5 (d+e x^2)^{3/2} (a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=231 \[ \frac {d^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {2 d \left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}+\frac {\left (d+e x^2\right )^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}+\frac {8 b d^{9/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{315 e^3}-\frac {8 b d^4 n \sqrt {d+e x^2}}{315 e^3}-\frac {8 b d^3 n \left (d+e x^2\right )^{3/2}}{945 e^3}-\frac {8 b d^2 n \left (d+e x^2\right )^{5/2}}{1575 e^3}+\frac {11 b d n \left (d+e x^2\right )^{7/2}}{441 e^3}-\frac {b n \left (d+e x^2\right )^{9/2}}{81 e^3} \]

[Out]

-8/945*b*d^3*n*(e*x^2+d)^(3/2)/e^3-8/1575*b*d^2*n*(e*x^2+d)^(5/2)/e^3+11/441*b*d*n*(e*x^2+d)^(7/2)/e^3-1/81*b*
n*(e*x^2+d)^(9/2)/e^3+8/315*b*d^(9/2)*n*arctanh((e*x^2+d)^(1/2)/d^(1/2))/e^3+1/5*d^2*(e*x^2+d)^(5/2)*(a+b*ln(c
*x^n))/e^3-2/7*d*(e*x^2+d)^(7/2)*(a+b*ln(c*x^n))/e^3+1/9*(e*x^2+d)^(9/2)*(a+b*ln(c*x^n))/e^3-8/315*b*d^4*n*(e*
x^2+d)^(1/2)/e^3

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Rubi [A]  time = 0.28, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {266, 43, 2350, 12, 1251, 897, 1261, 208} \[ \frac {d^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {2 d \left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}+\frac {\left (d+e x^2\right )^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac {8 b d^4 n \sqrt {d+e x^2}}{315 e^3}-\frac {8 b d^3 n \left (d+e x^2\right )^{3/2}}{945 e^3}-\frac {8 b d^2 n \left (d+e x^2\right )^{5/2}}{1575 e^3}+\frac {8 b d^{9/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{315 e^3}+\frac {11 b d n \left (d+e x^2\right )^{7/2}}{441 e^3}-\frac {b n \left (d+e x^2\right )^{9/2}}{81 e^3} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(d + e*x^2)^(3/2)*(a + b*Log[c*x^n]),x]

[Out]

(-8*b*d^4*n*Sqrt[d + e*x^2])/(315*e^3) - (8*b*d^3*n*(d + e*x^2)^(3/2))/(945*e^3) - (8*b*d^2*n*(d + e*x^2)^(5/2
))/(1575*e^3) + (11*b*d*n*(d + e*x^2)^(7/2))/(441*e^3) - (b*n*(d + e*x^2)^(9/2))/(81*e^3) + (8*b*d^(9/2)*n*Arc
Tanh[Sqrt[d + e*x^2]/Sqrt[d]])/(315*e^3) + (d^2*(d + e*x^2)^(5/2)*(a + b*Log[c*x^n]))/(5*e^3) - (2*d*(d + e*x^
2)^(7/2)*(a + b*Log[c*x^n]))/(7*e^3) + ((d + e*x^2)^(9/2)*(a + b*Log[c*x^n]))/(9*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 43

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 897

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 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 1261

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 2350

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^5 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac {d^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {2 d \left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}+\frac {\left (d+e x^2\right )^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-(b n) \int \frac {\left (d+e x^2\right )^{5/2} \left (8 d^2-20 d e x^2+35 e^2 x^4\right )}{315 e^3 x} \, dx\\ &=\frac {d^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {2 d \left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}+\frac {\left (d+e x^2\right )^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac {(b n) \int \frac {\left (d+e x^2\right )^{5/2} \left (8 d^2-20 d e x^2+35 e^2 x^4\right )}{x} \, dx}{315 e^3}\\ &=\frac {d^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {2 d \left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}+\frac {\left (d+e x^2\right )^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac {(b n) \operatorname {Subst}\left (\int \frac {(d+e x)^{5/2} \left (8 d^2-20 d e x+35 e^2 x^2\right )}{x} \, dx,x,x^2\right )}{630 e^3}\\ &=\frac {d^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {2 d \left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}+\frac {\left (d+e x^2\right )^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac {(b n) \operatorname {Subst}\left (\int \frac {x^6 \left (63 d^2-90 d x^2+35 x^4\right )}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{315 e^4}\\ &=\frac {d^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {2 d \left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}+\frac {\left (d+e x^2\right )^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac {(b n) \operatorname {Subst}\left (\int \left (8 d^4 e+8 d^3 e x^2+8 d^2 e x^4-55 d e x^6+35 e x^8+\frac {8 d^5}{-\frac {d}{e}+\frac {x^2}{e}}\right ) \, dx,x,\sqrt {d+e x^2}\right )}{315 e^4}\\ &=-\frac {8 b d^4 n \sqrt {d+e x^2}}{315 e^3}-\frac {8 b d^3 n \left (d+e x^2\right )^{3/2}}{945 e^3}-\frac {8 b d^2 n \left (d+e x^2\right )^{5/2}}{1575 e^3}+\frac {11 b d n \left (d+e x^2\right )^{7/2}}{441 e^3}-\frac {b n \left (d+e x^2\right )^{9/2}}{81 e^3}+\frac {d^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {2 d \left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}+\frac {\left (d+e x^2\right )^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac {\left (8 b d^5 n\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{315 e^4}\\ &=-\frac {8 b d^4 n \sqrt {d+e x^2}}{315 e^3}-\frac {8 b d^3 n \left (d+e x^2\right )^{3/2}}{945 e^3}-\frac {8 b d^2 n \left (d+e x^2\right )^{5/2}}{1575 e^3}+\frac {11 b d n \left (d+e x^2\right )^{7/2}}{441 e^3}-\frac {b n \left (d+e x^2\right )^{9/2}}{81 e^3}+\frac {8 b d^{9/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{315 e^3}+\frac {d^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {2 d \left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}+\frac {\left (d+e x^2\right )^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}\\ \end {align*}

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Mathematica [A]  time = 0.36, size = 256, normalized size = 1.11 \[ \frac {\sqrt {d+e x^2} \left (2 d^4 \left (1260 a+1260 b \left (\log \left (c x^n\right )-n \log (x)\right )-1307 b n\right )-d^3 e x^2 \left (1260 a+1260 b \left (\log \left (c x^n\right )-n \log (x)\right )-677 b n\right )+3 d^2 e^2 x^4 \left (315 a+315 b \left (\log \left (c x^n\right )-n \log (x)\right )-143 b n\right )+25 d e^3 x^6 \left (630 a+630 b \left (\log \left (c x^n\right )-n \log (x)\right )-97 b n\right )+1225 e^4 x^8 \left (9 a+9 b \log \left (c x^n\right )-9 b n \log (x)-b n\right )\right )+2520 b d^{9/2} n \log \left (\sqrt {d} \sqrt {d+e x^2}+d\right )-2520 b d^{9/2} n \log (x)+315 b n \log (x) \left (d+e x^2\right )^{5/2} \left (8 d^2-20 d e x^2+35 e^2 x^4\right )}{99225 e^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(d + e*x^2)^(3/2)*(a + b*Log[c*x^n]),x]

[Out]

(-2520*b*d^(9/2)*n*Log[x] + 315*b*n*(d + e*x^2)^(5/2)*(8*d^2 - 20*d*e*x^2 + 35*e^2*x^4)*Log[x] + Sqrt[d + e*x^
2]*(1225*e^4*x^8*(9*a - b*n - 9*b*n*Log[x] + 9*b*Log[c*x^n]) + 3*d^2*e^2*x^4*(315*a - 143*b*n + 315*b*(-(n*Log
[x]) + Log[c*x^n])) + 25*d*e^3*x^6*(630*a - 97*b*n + 630*b*(-(n*Log[x]) + Log[c*x^n])) + 2*d^4*(1260*a - 1307*
b*n + 1260*b*(-(n*Log[x]) + Log[c*x^n])) - d^3*e*x^2*(1260*a - 677*b*n + 1260*b*(-(n*Log[x]) + Log[c*x^n]))) +
2520*b*d^(9/2)*n*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/(99225*e^3)

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fricas [A]  time = 0.53, size = 514, normalized size = 2.23 \[ \left [\frac {1260 \, b d^{\frac {9}{2}} n \log \left (-\frac {e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - {\left (1225 \, {\left (b e^{4} n - 9 \, a e^{4}\right )} x^{8} + 25 \, {\left (97 \, b d e^{3} n - 630 \, a d e^{3}\right )} x^{6} + 2614 \, b d^{4} n - 2520 \, a d^{4} + 3 \, {\left (143 \, b d^{2} e^{2} n - 315 \, a d^{2} e^{2}\right )} x^{4} - {\left (677 \, b d^{3} e n - 1260 \, a d^{3} e\right )} x^{2} - 315 \, {\left (35 \, b e^{4} x^{8} + 50 \, b d e^{3} x^{6} + 3 \, b d^{2} e^{2} x^{4} - 4 \, b d^{3} e x^{2} + 8 \, b d^{4}\right )} \log \relax (c) - 315 \, {\left (35 \, b e^{4} n x^{8} + 50 \, b d e^{3} n x^{6} + 3 \, b d^{2} e^{2} n x^{4} - 4 \, b d^{3} e n x^{2} + 8 \, b d^{4} n\right )} \log \relax (x)\right )} \sqrt {e x^{2} + d}}{99225 \, e^{3}}, -\frac {2520 \, b \sqrt {-d} d^{4} n \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x^{2} + d}}\right ) + {\left (1225 \, {\left (b e^{4} n - 9 \, a e^{4}\right )} x^{8} + 25 \, {\left (97 \, b d e^{3} n - 630 \, a d e^{3}\right )} x^{6} + 2614 \, b d^{4} n - 2520 \, a d^{4} + 3 \, {\left (143 \, b d^{2} e^{2} n - 315 \, a d^{2} e^{2}\right )} x^{4} - {\left (677 \, b d^{3} e n - 1260 \, a d^{3} e\right )} x^{2} - 315 \, {\left (35 \, b e^{4} x^{8} + 50 \, b d e^{3} x^{6} + 3 \, b d^{2} e^{2} x^{4} - 4 \, b d^{3} e x^{2} + 8 \, b d^{4}\right )} \log \relax (c) - 315 \, {\left (35 \, b e^{4} n x^{8} + 50 \, b d e^{3} n x^{6} + 3 \, b d^{2} e^{2} n x^{4} - 4 \, b d^{3} e n x^{2} + 8 \, b d^{4} n\right )} \log \relax (x)\right )} \sqrt {e x^{2} + d}}{99225 \, e^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*x^2+d)^(3/2)*(a+b*log(c*x^n)),x, algorithm="fricas")

[Out]

[1/99225*(1260*b*d^(9/2)*n*log(-(e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) - (1225*(b*e^4*n - 9*a*e^4)*x^8
+ 25*(97*b*d*e^3*n - 630*a*d*e^3)*x^6 + 2614*b*d^4*n - 2520*a*d^4 + 3*(143*b*d^2*e^2*n - 315*a*d^2*e^2)*x^4 -
(677*b*d^3*e*n - 1260*a*d^3*e)*x^2 - 315*(35*b*e^4*x^8 + 50*b*d*e^3*x^6 + 3*b*d^2*e^2*x^4 - 4*b*d^3*e*x^2 + 8
*b*d^4)*log(c) - 315*(35*b*e^4*n*x^8 + 50*b*d*e^3*n*x^6 + 3*b*d^2*e^2*n*x^4 - 4*b*d^3*e*n*x^2 + 8*b*d^4*n)*log
(x))*sqrt(e*x^2 + d))/e^3, -1/99225*(2520*b*sqrt(-d)*d^4*n*arctan(sqrt(-d)/sqrt(e*x^2 + d)) + (1225*(b*e^4*n -
9*a*e^4)*x^8 + 25*(97*b*d*e^3*n - 630*a*d*e^3)*x^6 + 2614*b*d^4*n - 2520*a*d^4 + 3*(143*b*d^2*e^2*n - 315*a*d
^2*e^2)*x^4 - (677*b*d^3*e*n - 1260*a*d^3*e)*x^2 - 315*(35*b*e^4*x^8 + 50*b*d*e^3*x^6 + 3*b*d^2*e^2*x^4 - 4*b*
d^3*e*x^2 + 8*b*d^4)*log(c) - 315*(35*b*e^4*n*x^8 + 50*b*d*e^3*n*x^6 + 3*b*d^2*e^2*n*x^4 - 4*b*d^3*e*n*x^2 + 8
*b*d^4*n)*log(x))*sqrt(e*x^2 + d))/e^3]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x^{2} + d\right )}^{\frac {3}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )} x^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*x^2+d)^(3/2)*(a+b*log(c*x^n)),x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^(3/2)*(b*log(c*x^n) + a)*x^5, x)

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maple [F]  time = 0.39, size = 0, normalized size = 0.00 \[ \int \left (e \,x^{2}+d \right )^{\frac {3}{2}} \left (b \ln \left (c \,x^{n}\right )+a \right ) x^{5}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(e*x^2+d)^(3/2)*(b*ln(c*x^n)+a),x)

[Out]

int(x^5*(e*x^2+d)^(3/2)*(b*ln(c*x^n)+a),x)

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maxima [A]  time = 1.45, size = 234, normalized size = 1.01 \[ -\frac {1}{99225} \, {\left (\frac {1260 \, d^{\frac {9}{2}} \log \left (\frac {\sqrt {e x^{2} + d} - \sqrt {d}}{\sqrt {e x^{2} + d} + \sqrt {d}}\right )}{e^{3}} + \frac {1225 \, {\left (e x^{2} + d\right )}^{\frac {9}{2}} - 2475 \, {\left (e x^{2} + d\right )}^{\frac {7}{2}} d + 504 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d^{2} + 840 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d^{3} + 2520 \, \sqrt {e x^{2} + d} d^{4}}{e^{3}}\right )} b n + \frac {1}{315} \, {\left (\frac {35 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} x^{4}}{e} - \frac {20 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d x^{2}}{e^{2}} + \frac {8 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d^{2}}{e^{3}}\right )} b \log \left (c x^{n}\right ) + \frac {1}{315} \, {\left (\frac {35 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} x^{4}}{e} - \frac {20 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d x^{2}}{e^{2}} + \frac {8 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d^{2}}{e^{3}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*x^2+d)^(3/2)*(a+b*log(c*x^n)),x, algorithm="maxima")

[Out]

-1/99225*(1260*d^(9/2)*log((sqrt(e*x^2 + d) - sqrt(d))/(sqrt(e*x^2 + d) + sqrt(d)))/e^3 + (1225*(e*x^2 + d)^(9
/2) - 2475*(e*x^2 + d)^(7/2)*d + 504*(e*x^2 + d)^(5/2)*d^2 + 840*(e*x^2 + d)^(3/2)*d^3 + 2520*sqrt(e*x^2 + d)*
d^4)/e^3)*b*n + 1/315*(35*(e*x^2 + d)^(5/2)*x^4/e - 20*(e*x^2 + d)^(5/2)*d*x^2/e^2 + 8*(e*x^2 + d)^(5/2)*d^2/e
^3)*b*log(c*x^n) + 1/315*(35*(e*x^2 + d)^(5/2)*x^4/e - 20*(e*x^2 + d)^(5/2)*d*x^2/e^2 + 8*(e*x^2 + d)^(5/2)*d^
2/e^3)*a

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^5\,{\left (e\,x^2+d\right )}^{3/2}\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(d + e*x^2)^(3/2)*(a + b*log(c*x^n)),x)

[Out]

int(x^5*(d + e*x^2)^(3/2)*(a + b*log(c*x^n)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(e*x**2+d)**(3/2)*(a+b*ln(c*x**n)),x)

[Out]

Timed out

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