\(\int \frac {x^3 (a+b \arctan (c x))}{\sqrt {d+e x^2}} \, dx\) [1201]

Optimal result

Integrand size = 23, antiderivative size = 176 \[ \int \frac {x^3 (a+b \arctan (c x))}{\sqrt {d+e x^2}} \, dx=-\frac {b x \sqrt {d+e x^2}}{6 c e}-\frac {d \sqrt {d+e x^2} (a+b \arctan (c x))}{e^2}+\frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{3 e^2}+\frac {b \sqrt {c^2 d-e} \left (2 c^2 d+e\right ) \arctan \left (\frac {\sqrt {c^2 d-e} x}{\sqrt {d+e x^2}}\right )}{3 c^3 e^2}+\frac {b \left (3 c^2 d+2 e\right ) \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{6 c^3 e^{3/2}} \]

[Out]

1/3*(e*x^2+d)^(3/2)*(a+b*arctan(c*x))/e^2+1/6*b*(3*c^2*d+2*e)*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/c^3/e^(3/2)+1
/3*b*(2*c^2*d+e)*arctan(x*(c^2*d-e)^(1/2)/(e*x^2+d)^(1/2))*(c^2*d-e)^(1/2)/c^3/e^2-1/6*b*x*(e*x^2+d)^(1/2)/c/e
-d*(a+b*arctan(c*x))*(e*x^2+d)^(1/2)/e^2

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {272, 45, 5096, 12, 542, 537, 223, 212, 385, 209} \[ \int \frac {x^3 (a+b \arctan (c x))}{\sqrt {d+e x^2}} \, dx=\frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{3 e^2}-\frac {d \sqrt {d+e x^2} (a+b \arctan (c x))}{e^2}+\frac {b \sqrt {c^2 d-e} \left (2 c^2 d+e\right ) \arctan \left (\frac {x \sqrt {c^2 d-e}}{\sqrt {d+e x^2}}\right )}{3 c^3 e^2}+\frac {b \left (3 c^2 d+2 e\right ) \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{6 c^3 e^{3/2}}-\frac {b x \sqrt {d+e x^2}}{6 c e} \]

[In]

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

[Out]

-1/6*(b*x*Sqrt[d + e*x^2])/(c*e) - (d*Sqrt[d + e*x^2]*(a + b*ArcTan[c*x]))/e^2 + ((d + e*x^2)^(3/2)*(a + b*Arc
Tan[c*x]))/(3*e^2) + (b*Sqrt[c^2*d - e]*(2*c^2*d + e)*ArcTan[(Sqrt[c^2*d - e]*x)/Sqrt[d + e*x^2]])/(3*c^3*e^2)
+ (b*(3*c^2*d + 2*e)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(6*c^3*e^(3/2))

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 272

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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 537

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 542

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[
f*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(b*(n*(p + q + 1) + 1))), x] + Dist[1/(b*(n*(p + q + 1) + 1)), Int[(a +
b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f + b*e*n*(p + q + 1)) + (d*(b*e - a*f) + f*n*q*(b*c - a*d) + b*
d*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && GtQ[q, 0] && NeQ[n*(p + q + 1) + 1
, 0]

Rule 5096

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^q, x]}, Dist[a + b*ArcTan[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/(1 + c^
2*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && ((IGtQ[q, 0] &&  !(ILtQ[(m - 1)/2, 0] && GtQ[m +
2*q + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[q, 0] && GtQ[m + 2*q + 3, 0])) || (ILtQ[(m + 2*q + 1)/2, 0] &&
!ILtQ[(m - 1)/2, 0]))

Rubi steps \begin{align*} \text {integral}& = -\frac {d \sqrt {d+e x^2} (a+b \arctan (c x))}{e^2}+\frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{3 e^2}-(b c) \int \frac {\left (-2 d+e x^2\right ) \sqrt {d+e x^2}}{3 e^2 \left (1+c^2 x^2\right )} \, dx \\ & = -\frac {d \sqrt {d+e x^2} (a+b \arctan (c x))}{e^2}+\frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{3 e^2}-\frac {(b c) \int \frac {\left (-2 d+e x^2\right ) \sqrt {d+e x^2}}{1+c^2 x^2} \, dx}{3 e^2} \\ & = -\frac {b x \sqrt {d+e x^2}}{6 c e}-\frac {d \sqrt {d+e x^2} (a+b \arctan (c x))}{e^2}+\frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{3 e^2}-\frac {b \int \frac {-d \left (4 c^2 d+e\right )-e \left (3 c^2 d+2 e\right ) x^2}{\left (1+c^2 x^2\right ) \sqrt {d+e x^2}} \, dx}{6 c e^2} \\ & = -\frac {b x \sqrt {d+e x^2}}{6 c e}-\frac {d \sqrt {d+e x^2} (a+b \arctan (c x))}{e^2}+\frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{3 e^2}+\frac {\left (b \left (c^2 d-e\right ) \left (2 c^2 d+e\right )\right ) \int \frac {1}{\left (1+c^2 x^2\right ) \sqrt {d+e x^2}} \, dx}{3 c^3 e^2}+\frac {\left (b \left (3 c^2 d+2 e\right )\right ) \int \frac {1}{\sqrt {d+e x^2}} \, dx}{6 c^3 e} \\ & = -\frac {b x \sqrt {d+e x^2}}{6 c e}-\frac {d \sqrt {d+e x^2} (a+b \arctan (c x))}{e^2}+\frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{3 e^2}+\frac {\left (b \left (c^2 d-e\right ) \left (2 c^2 d+e\right )\right ) \text {Subst}\left (\int \frac {1}{1-\left (-c^2 d+e\right ) x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{3 c^3 e^2}+\frac {\left (b \left (3 c^2 d+2 e\right )\right ) \text {Subst}\left (\int \frac {1}{1-e x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{6 c^3 e} \\ & = -\frac {b x \sqrt {d+e x^2}}{6 c e}-\frac {d \sqrt {d+e x^2} (a+b \arctan (c x))}{e^2}+\frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{3 e^2}+\frac {b \sqrt {c^2 d-e} \left (2 c^2 d+e\right ) \arctan \left (\frac {\sqrt {c^2 d-e} x}{\sqrt {d+e x^2}}\right )}{3 c^3 e^2}+\frac {b \left (3 c^2 d+2 e\right ) \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{6 c^3 e^{3/2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.43 (sec) , antiderivative size = 377, normalized size of antiderivative = 2.14 \[ \int \frac {x^3 (a+b \arctan (c x))}{\sqrt {d+e x^2}} \, dx=\frac {-\frac {\sqrt {d+e x^2} \left (b e x+a c \left (4 d-2 e x^2\right )\right )}{c}+2 b \left (-2 d+e x^2\right ) \sqrt {d+e x^2} \arctan (c x)-\frac {i b \left (2 c^4 d^2-c^2 d e-e^2\right ) \log \left (\frac {12 i c^4 e^2 \left (c d-i e x+\sqrt {c^2 d-e} \sqrt {d+e x^2}\right )}{b \sqrt {c^2 d-e} \left (-2 c^4 d^2+c^2 d e+e^2\right ) (i+c x)}\right )}{c^3 \sqrt {c^2 d-e}}+\frac {i b \left (2 c^4 d^2-c^2 d e-e^2\right ) \log \left (-\frac {12 i c^4 e^2 \left (c d+i e x+\sqrt {c^2 d-e} \sqrt {d+e x^2}\right )}{b \sqrt {c^2 d-e} \left (-2 c^4 d^2+c^2 d e+e^2\right ) (-i+c x)}\right )}{c^3 \sqrt {c^2 d-e}}+\frac {b \sqrt {e} \left (3 c^2 d+2 e\right ) \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )}{c^3}}{6 e^2} \]

[In]

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

[Out]

(-((Sqrt[d + e*x^2]*(b*e*x + a*c*(4*d - 2*e*x^2)))/c) + 2*b*(-2*d + e*x^2)*Sqrt[d + e*x^2]*ArcTan[c*x] - (I*b*
(2*c^4*d^2 - c^2*d*e - e^2)*Log[((12*I)*c^4*e^2*(c*d - I*e*x + Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*Sqrt[c^2*d
- e]*(-2*c^4*d^2 + c^2*d*e + e^2)*(I + c*x))])/(c^3*Sqrt[c^2*d - e]) + (I*b*(2*c^4*d^2 - c^2*d*e - e^2)*Log[(
(-12*I)*c^4*e^2*(c*d + I*e*x + Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*Sqrt[c^2*d - e]*(-2*c^4*d^2 + c^2*d*e + e^
2)*(-I + c*x))])/(c^3*Sqrt[c^2*d - e]) + (b*Sqrt[e]*(3*c^2*d + 2*e)*Log[e*x + Sqrt[e]*Sqrt[d + e*x^2]])/c^3)/(
6*e^2)

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.63 (sec) , antiderivative size = 882, normalized size of antiderivative = 5.01 \[ \int \frac {x^3 (a+b \arctan (c x))}{\sqrt {d+e x^2}} \, dx=\left [\frac {{\left (3 \, b c^{2} d + 2 \, b e\right )} \sqrt {e} \log \left (-2 \, e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {e} x - d\right ) + {\left (2 \, b c^{2} d + b e\right )} \sqrt {-c^{2} d + e} \log \left (\frac {{\left (c^{4} d^{2} - 8 \, c^{2} d e + 8 \, e^{2}\right )} x^{4} - 2 \, {\left (3 \, c^{2} d^{2} - 4 \, d e\right )} x^{2} + 4 \, {\left ({\left (c^{2} d - 2 \, e\right )} x^{3} - d x\right )} \sqrt {-c^{2} d + e} \sqrt {e x^{2} + d} + d^{2}}{c^{4} x^{4} + 2 \, c^{2} x^{2} + 1}\right ) + 2 \, {\left (2 \, a c^{3} e x^{2} - 4 \, a c^{3} d - b c^{2} e x + 2 \, {\left (b c^{3} e x^{2} - 2 \, b c^{3} d\right )} \arctan \left (c x\right )\right )} \sqrt {e x^{2} + d}}{12 \, c^{3} e^{2}}, \frac {2 \, {\left (2 \, b c^{2} d + b e\right )} \sqrt {c^{2} d - e} \arctan \left (\frac {\sqrt {c^{2} d - e} {\left ({\left (c^{2} d - 2 \, e\right )} x^{2} - d\right )} \sqrt {e x^{2} + d}}{2 \, {\left ({\left (c^{2} d e - e^{2}\right )} x^{3} + {\left (c^{2} d^{2} - d e\right )} x\right )}}\right ) + {\left (3 \, b c^{2} d + 2 \, b e\right )} \sqrt {e} \log \left (-2 \, e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {e} x - d\right ) + 2 \, {\left (2 \, a c^{3} e x^{2} - 4 \, a c^{3} d - b c^{2} e x + 2 \, {\left (b c^{3} e x^{2} - 2 \, b c^{3} d\right )} \arctan \left (c x\right )\right )} \sqrt {e x^{2} + d}}{12 \, c^{3} e^{2}}, -\frac {2 \, {\left (3 \, b c^{2} d + 2 \, b e\right )} \sqrt {-e} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) - {\left (2 \, b c^{2} d + b e\right )} \sqrt {-c^{2} d + e} \log \left (\frac {{\left (c^{4} d^{2} - 8 \, c^{2} d e + 8 \, e^{2}\right )} x^{4} - 2 \, {\left (3 \, c^{2} d^{2} - 4 \, d e\right )} x^{2} + 4 \, {\left ({\left (c^{2} d - 2 \, e\right )} x^{3} - d x\right )} \sqrt {-c^{2} d + e} \sqrt {e x^{2} + d} + d^{2}}{c^{4} x^{4} + 2 \, c^{2} x^{2} + 1}\right ) - 2 \, {\left (2 \, a c^{3} e x^{2} - 4 \, a c^{3} d - b c^{2} e x + 2 \, {\left (b c^{3} e x^{2} - 2 \, b c^{3} d\right )} \arctan \left (c x\right )\right )} \sqrt {e x^{2} + d}}{12 \, c^{3} e^{2}}, \frac {{\left (2 \, b c^{2} d + b e\right )} \sqrt {c^{2} d - e} \arctan \left (\frac {\sqrt {c^{2} d - e} {\left ({\left (c^{2} d - 2 \, e\right )} x^{2} - d\right )} \sqrt {e x^{2} + d}}{2 \, {\left ({\left (c^{2} d e - e^{2}\right )} x^{3} + {\left (c^{2} d^{2} - d e\right )} x\right )}}\right ) - {\left (3 \, b c^{2} d + 2 \, b e\right )} \sqrt {-e} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) + {\left (2 \, a c^{3} e x^{2} - 4 \, a c^{3} d - b c^{2} e x + 2 \, {\left (b c^{3} e x^{2} - 2 \, b c^{3} d\right )} \arctan \left (c x\right )\right )} \sqrt {e x^{2} + d}}{6 \, c^{3} e^{2}}\right ] \]

[In]

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

[Out]

[1/12*((3*b*c^2*d + 2*b*e)*sqrt(e)*log(-2*e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) + (2*b*c^2*d + b*e)*sqrt(-c
^2*d + e)*log(((c^4*d^2 - 8*c^2*d*e + 8*e^2)*x^4 - 2*(3*c^2*d^2 - 4*d*e)*x^2 + 4*((c^2*d - 2*e)*x^3 - d*x)*sqr
t(-c^2*d + e)*sqrt(e*x^2 + d) + d^2)/(c^4*x^4 + 2*c^2*x^2 + 1)) + 2*(2*a*c^3*e*x^2 - 4*a*c^3*d - b*c^2*e*x + 2
*(b*c^3*e*x^2 - 2*b*c^3*d)*arctan(c*x))*sqrt(e*x^2 + d))/(c^3*e^2), 1/12*(2*(2*b*c^2*d + b*e)*sqrt(c^2*d - e)*
arctan(1/2*sqrt(c^2*d - e)*((c^2*d - 2*e)*x^2 - d)*sqrt(e*x^2 + d)/((c^2*d*e - e^2)*x^3 + (c^2*d^2 - d*e)*x))
+ (3*b*c^2*d + 2*b*e)*sqrt(e)*log(-2*e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) + 2*(2*a*c^3*e*x^2 - 4*a*c^3*d -
b*c^2*e*x + 2*(b*c^3*e*x^2 - 2*b*c^3*d)*arctan(c*x))*sqrt(e*x^2 + d))/(c^3*e^2), -1/12*(2*(3*b*c^2*d + 2*b*e)
*sqrt(-e)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) - (2*b*c^2*d + b*e)*sqrt(-c^2*d + e)*log(((c^4*d^2 - 8*c^2*d*e +
8*e^2)*x^4 - 2*(3*c^2*d^2 - 4*d*e)*x^2 + 4*((c^2*d - 2*e)*x^3 - d*x)*sqrt(-c^2*d + e)*sqrt(e*x^2 + d) + d^2)/(
c^4*x^4 + 2*c^2*x^2 + 1)) - 2*(2*a*c^3*e*x^2 - 4*a*c^3*d - b*c^2*e*x + 2*(b*c^3*e*x^2 - 2*b*c^3*d)*arctan(c*x)
)*sqrt(e*x^2 + d))/(c^3*e^2), 1/6*((2*b*c^2*d + b*e)*sqrt(c^2*d - e)*arctan(1/2*sqrt(c^2*d - e)*((c^2*d - 2*e)
*x^2 - d)*sqrt(e*x^2 + d)/((c^2*d*e - e^2)*x^3 + (c^2*d^2 - d*e)*x)) - (3*b*c^2*d + 2*b*e)*sqrt(-e)*arctan(sqr
t(-e)*x/sqrt(e*x^2 + d)) + (2*a*c^3*e*x^2 - 4*a*c^3*d - b*c^2*e*x + 2*(b*c^3*e*x^2 - 2*b*c^3*d)*arctan(c*x))*s
qrt(e*x^2 + d))/(c^3*e^2)]

Sympy [F]

\[ \int \frac {x^3 (a+b \arctan (c x))}{\sqrt {d+e x^2}} \, dx=\int \frac {x^{3} \left (a + b \operatorname {atan}{\left (c x \right )}\right )}{\sqrt {d + e x^{2}}}\, dx \]

[In]

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

[Out]

Integral(x**3*(a + b*atan(c*x))/sqrt(d + e*x**2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^3 (a+b \arctan (c x))}{\sqrt {d+e x^2}} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 (a+b \arctan (c x))}{\sqrt {d+e x^2}} \, dx=\int \frac {x^3\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}{\sqrt {e\,x^2+d}} \,d x \]

[In]

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

[Out]

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