\(\int \frac {a+b \arctan (c x)}{x^4 (d+e x^2)^{3/2}} \, dx\) [1216]

Optimal result

Integrand size = 23, antiderivative size = 249 \[ \int \frac {a+b \arctan (c x)}{x^4 \left (d+e x^2\right )^{3/2}} \, dx=-\frac {b c \sqrt {d+e x^2}}{6 d^2 x^2}-\frac {a+b \arctan (c x)}{3 d x^3 \sqrt {d+e x^2}}+\frac {4 e (a+b \arctan (c x))}{3 d^2 x \sqrt {d+e x^2}}+\frac {8 e^2 x (a+b \arctan (c x))}{3 d^3 \sqrt {d+e x^2}}+\frac {b c e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{6 d^{5/2}}+\frac {b c \left (c^2 d+4 e\right ) \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 d^{5/2}}-\frac {b \left (c^4 d^2+4 c^2 d e-8 e^2\right ) \text {arctanh}\left (\frac {c \sqrt {d+e x^2}}{\sqrt {c^2 d-e}}\right )}{3 d^3 \sqrt {c^2 d-e}} \]

[Out]

1/6*b*c*e*arctanh((e*x^2+d)^(1/2)/d^(1/2))/d^(5/2)+1/3*b*c*(c^2*d+4*e)*arctanh((e*x^2+d)^(1/2)/d^(1/2))/d^(5/2
)-1/3*b*(c^4*d^2+4*c^2*d*e-8*e^2)*arctanh(c*(e*x^2+d)^(1/2)/(c^2*d-e)^(1/2))/d^3/(c^2*d-e)^(1/2)+1/3*(-a-b*arc
tan(c*x))/d/x^3/(e*x^2+d)^(1/2)+4/3*e*(a+b*arctan(c*x))/d^2/x/(e*x^2+d)^(1/2)+8/3*e^2*x*(a+b*arctan(c*x))/d^3/
(e*x^2+d)^(1/2)-1/6*b*c*(e*x^2+d)^(1/2)/d^2/x^2

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {277, 197, 5096, 12, 6857, 272, 44, 65, 214, 455} \[ \int \frac {a+b \arctan (c x)}{x^4 \left (d+e x^2\right )^{3/2}} \, dx=\frac {8 e^2 x (a+b \arctan (c x))}{3 d^3 \sqrt {d+e x^2}}+\frac {4 e (a+b \arctan (c x))}{3 d^2 x \sqrt {d+e x^2}}-\frac {a+b \arctan (c x)}{3 d x^3 \sqrt {d+e x^2}}+\frac {b c \left (c^2 d+4 e\right ) \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 d^{5/2}}-\frac {b \left (c^4 d^2+4 c^2 d e-8 e^2\right ) \text {arctanh}\left (\frac {c \sqrt {d+e x^2}}{\sqrt {c^2 d-e}}\right )}{3 d^3 \sqrt {c^2 d-e}}+\frac {b c e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{6 d^{5/2}}-\frac {b c \sqrt {d+e x^2}}{6 d^2 x^2} \]

[In]

Int[(a + b*ArcTan[c*x])/(x^4*(d + e*x^2)^(3/2)),x]

[Out]

-1/6*(b*c*Sqrt[d + e*x^2])/(d^2*x^2) - (a + b*ArcTan[c*x])/(3*d*x^3*Sqrt[d + e*x^2]) + (4*e*(a + b*ArcTan[c*x]
))/(3*d^2*x*Sqrt[d + e*x^2]) + (8*e^2*x*(a + b*ArcTan[c*x]))/(3*d^3*Sqrt[d + e*x^2]) + (b*c*e*ArcTanh[Sqrt[d +
e*x^2]/Sqrt[d]])/(6*d^(5/2)) + (b*c*(c^2*d + 4*e)*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(3*d^(5/2)) - (b*(c^4*d^2
+ 4*c^2*d*e - 8*e^2)*ArcTanh[(c*Sqrt[d + e*x^2])/Sqrt[c^2*d - e]])/(3*d^3*Sqrt[c^2*d - e])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 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 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 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 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 277

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^(p + 1)/(a*(m + 1))), x]
- Dist[b*((m + n*(p + 1) + 1)/(a*(m + 1))), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 455

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

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {a+b \arctan (c x)}{3 d x^3 \sqrt {d+e x^2}}+\frac {4 e (a+b \arctan (c x))}{3 d^2 x \sqrt {d+e x^2}}+\frac {8 e^2 x (a+b \arctan (c x))}{3 d^3 \sqrt {d+e x^2}}-(b c) \int \frac {-d^2+4 d e x^2+8 e^2 x^4}{3 d^3 x^3 \left (1+c^2 x^2\right ) \sqrt {d+e x^2}} \, dx \\ & = -\frac {a+b \arctan (c x)}{3 d x^3 \sqrt {d+e x^2}}+\frac {4 e (a+b \arctan (c x))}{3 d^2 x \sqrt {d+e x^2}}+\frac {8 e^2 x (a+b \arctan (c x))}{3 d^3 \sqrt {d+e x^2}}-\frac {(b c) \int \frac {-d^2+4 d e x^2+8 e^2 x^4}{x^3 \left (1+c^2 x^2\right ) \sqrt {d+e x^2}} \, dx}{3 d^3} \\ & = -\frac {a+b \arctan (c x)}{3 d x^3 \sqrt {d+e x^2}}+\frac {4 e (a+b \arctan (c x))}{3 d^2 x \sqrt {d+e x^2}}+\frac {8 e^2 x (a+b \arctan (c x))}{3 d^3 \sqrt {d+e x^2}}-\frac {(b c) \int \left (-\frac {d^2}{x^3 \sqrt {d+e x^2}}+\frac {d \left (c^2 d+4 e\right )}{x \sqrt {d+e x^2}}-\frac {\left (c^4 d^2+4 c^2 d e-8 e^2\right ) x}{\left (1+c^2 x^2\right ) \sqrt {d+e x^2}}\right ) \, dx}{3 d^3} \\ & = -\frac {a+b \arctan (c x)}{3 d x^3 \sqrt {d+e x^2}}+\frac {4 e (a+b \arctan (c x))}{3 d^2 x \sqrt {d+e x^2}}+\frac {8 e^2 x (a+b \arctan (c x))}{3 d^3 \sqrt {d+e x^2}}+\frac {(b c) \int \frac {1}{x^3 \sqrt {d+e x^2}} \, dx}{3 d}-\frac {\left (b c \left (c^2 d+4 e\right )\right ) \int \frac {1}{x \sqrt {d+e x^2}} \, dx}{3 d^2}+\frac {\left (b c \left (c^4 d^2+4 c^2 d e-8 e^2\right )\right ) \int \frac {x}{\left (1+c^2 x^2\right ) \sqrt {d+e x^2}} \, dx}{3 d^3} \\ & = -\frac {a+b \arctan (c x)}{3 d x^3 \sqrt {d+e x^2}}+\frac {4 e (a+b \arctan (c x))}{3 d^2 x \sqrt {d+e x^2}}+\frac {8 e^2 x (a+b \arctan (c x))}{3 d^3 \sqrt {d+e x^2}}+\frac {(b c) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {d+e x}} \, dx,x,x^2\right )}{6 d}-\frac {\left (b c \left (c^2 d+4 e\right )\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^2\right )}{6 d^2}+\frac {\left (b c \left (c^4 d^2+4 c^2 d e-8 e^2\right )\right ) \text {Subst}\left (\int \frac {1}{\left (1+c^2 x\right ) \sqrt {d+e x}} \, dx,x,x^2\right )}{6 d^3} \\ & = -\frac {b c \sqrt {d+e x^2}}{6 d^2 x^2}-\frac {a+b \arctan (c x)}{3 d x^3 \sqrt {d+e x^2}}+\frac {4 e (a+b \arctan (c x))}{3 d^2 x \sqrt {d+e x^2}}+\frac {8 e^2 x (a+b \arctan (c x))}{3 d^3 \sqrt {d+e x^2}}-\frac {(b c e) \text {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^2\right )}{12 d^2}-\frac {\left (b c \left (c^2 d+4 e\right )\right ) \text {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{3 d^2 e}+\frac {\left (b c \left (c^4 d^2+4 c^2 d e-8 e^2\right )\right ) \text {Subst}\left (\int \frac {1}{1-\frac {c^2 d}{e}+\frac {c^2 x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{3 d^3 e} \\ & = -\frac {b c \sqrt {d+e x^2}}{6 d^2 x^2}-\frac {a+b \arctan (c x)}{3 d x^3 \sqrt {d+e x^2}}+\frac {4 e (a+b \arctan (c x))}{3 d^2 x \sqrt {d+e x^2}}+\frac {8 e^2 x (a+b \arctan (c x))}{3 d^3 \sqrt {d+e x^2}}+\frac {b c \left (c^2 d+4 e\right ) \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 d^{5/2}}-\frac {b \left (c^4 d^2+4 c^2 d e-8 e^2\right ) \text {arctanh}\left (\frac {c \sqrt {d+e x^2}}{\sqrt {c^2 d-e}}\right )}{3 d^3 \sqrt {c^2 d-e}}-\frac {(b c) \text {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{6 d^2} \\ & = -\frac {b c \sqrt {d+e x^2}}{6 d^2 x^2}-\frac {a+b \arctan (c x)}{3 d x^3 \sqrt {d+e x^2}}+\frac {4 e (a+b \arctan (c x))}{3 d^2 x \sqrt {d+e x^2}}+\frac {8 e^2 x (a+b \arctan (c x))}{3 d^3 \sqrt {d+e x^2}}+\frac {b c e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{6 d^{5/2}}+\frac {b c \left (c^2 d+4 e\right ) \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 d^{5/2}}-\frac {b \left (c^4 d^2+4 c^2 d e-8 e^2\right ) \text {arctanh}\left (\frac {c \sqrt {d+e x^2}}{\sqrt {c^2 d-e}}\right )}{3 d^3 \sqrt {c^2 d-e}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.56 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.63 \[ \int \frac {a+b \arctan (c x)}{x^4 \left (d+e x^2\right )^{3/2}} \, dx=-\frac {\frac {b c d x \left (d+e x^2\right )+2 a \left (d^2-4 d e x^2-8 e^2 x^4\right )}{x^3 \sqrt {d+e x^2}}+\frac {2 b \left (d^2-4 d e x^2-8 e^2 x^4\right ) \arctan (c x)}{x^3 \sqrt {d+e x^2}}+b c \sqrt {d} \left (2 c^2 d+9 e\right ) \log (x)-b c \sqrt {d} \left (2 c^2 d+9 e\right ) \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )+\frac {b \left (c^4 d^2+4 c^2 d e-8 e^2\right ) \log \left (\frac {12 c d^3 \left (c d-i e x+\sqrt {c^2 d-e} \sqrt {d+e x^2}\right )}{b \sqrt {c^2 d-e} \left (c^4 d^2+4 c^2 d e-8 e^2\right ) (i+c x)}\right )}{\sqrt {c^2 d-e}}+\frac {b \left (c^4 d^2+4 c^2 d e-8 e^2\right ) \log \left (\frac {12 c d^3 \left (c d+i e x+\sqrt {c^2 d-e} \sqrt {d+e x^2}\right )}{b \sqrt {c^2 d-e} \left (c^4 d^2+4 c^2 d e-8 e^2\right ) (-i+c x)}\right )}{\sqrt {c^2 d-e}}}{6 d^3} \]

[In]

Integrate[(a + b*ArcTan[c*x])/(x^4*(d + e*x^2)^(3/2)),x]

[Out]

-1/6*((b*c*d*x*(d + e*x^2) + 2*a*(d^2 - 4*d*e*x^2 - 8*e^2*x^4))/(x^3*Sqrt[d + e*x^2]) + (2*b*(d^2 - 4*d*e*x^2
- 8*e^2*x^4)*ArcTan[c*x])/(x^3*Sqrt[d + e*x^2]) + b*c*Sqrt[d]*(2*c^2*d + 9*e)*Log[x] - b*c*Sqrt[d]*(2*c^2*d +
9*e)*Log[d + Sqrt[d]*Sqrt[d + e*x^2]] + (b*(c^4*d^2 + 4*c^2*d*e - 8*e^2)*Log[(12*c*d^3*(c*d - I*e*x + Sqrt[c^2
*d - e]*Sqrt[d + e*x^2]))/(b*Sqrt[c^2*d - e]*(c^4*d^2 + 4*c^2*d*e - 8*e^2)*(I + c*x))])/Sqrt[c^2*d - e] + (b*(
c^4*d^2 + 4*c^2*d*e - 8*e^2)*Log[(12*c*d^3*(c*d + I*e*x + Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*Sqrt[c^2*d - e]
*(c^4*d^2 + 4*c^2*d*e - 8*e^2)*(-I + c*x))])/Sqrt[c^2*d - e])/d^3

Maple [F]

\[\int \frac {a +b \arctan \left (c x \right )}{x^{4} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}d x\]

[In]

int((a+b*arctan(c*x))/x^4/(e*x^2+d)^(3/2),x)

[Out]

int((a+b*arctan(c*x))/x^4/(e*x^2+d)^(3/2),x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 452 vs. \(2 (209) = 418\).

Time = 0.55 (sec) , antiderivative size = 1920, normalized size of antiderivative = 7.71 \[ \int \frac {a+b \arctan (c x)}{x^4 \left (d+e x^2\right )^{3/2}} \, dx=\text {Too large to display} \]

[In]

integrate((a+b*arctan(c*x))/x^4/(e*x^2+d)^(3/2),x, algorithm="fricas")

[Out]

[-1/12*(((b*c^4*d^2*e + 4*b*c^2*d*e^2 - 8*b*e^3)*x^5 + (b*c^4*d^3 + 4*b*c^2*d^2*e - 8*b*d*e^2)*x^3)*sqrt(c^2*d
- e)*log((c^4*e^2*x^4 + 8*c^4*d^2 - 8*c^2*d*e + 2*(4*c^4*d*e - 3*c^2*e^2)*x^2 + 4*(c^3*e*x^2 + 2*c^3*d - c*e)
*sqrt(c^2*d - e)*sqrt(e*x^2 + d) + e^2)/(c^4*x^4 + 2*c^2*x^2 + 1)) - ((2*b*c^5*d^2*e + 7*b*c^3*d*e^2 - 9*b*c*e
^3)*x^5 + (2*b*c^5*d^3 + 7*b*c^3*d^2*e - 9*b*c*d*e^2)*x^3)*sqrt(d)*log(-(e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(d) + 2
*d)/x^2) + 2*(2*a*c^2*d^3 - 16*(a*c^2*d*e^2 - a*e^3)*x^4 - 2*a*d^2*e + (b*c^3*d^2*e - b*c*d*e^2)*x^3 - 8*(a*c^
2*d^2*e - a*d*e^2)*x^2 + (b*c^3*d^3 - b*c*d^2*e)*x + 2*(b*c^2*d^3 - 8*(b*c^2*d*e^2 - b*e^3)*x^4 - b*d^2*e - 4*
(b*c^2*d^2*e - b*d*e^2)*x^2)*arctan(c*x))*sqrt(e*x^2 + d))/((c^2*d^4*e - d^3*e^2)*x^5 + (c^2*d^5 - d^4*e)*x^3)
, -1/12*(2*((b*c^4*d^2*e + 4*b*c^2*d*e^2 - 8*b*e^3)*x^5 + (b*c^4*d^3 + 4*b*c^2*d^2*e - 8*b*d*e^2)*x^3)*sqrt(-c
^2*d + e)*arctan(-1/2*(c^2*e*x^2 + 2*c^2*d - e)*sqrt(-c^2*d + e)*sqrt(e*x^2 + d)/(c^3*d^2 - c*d*e + (c^3*d*e -
c*e^2)*x^2)) - ((2*b*c^5*d^2*e + 7*b*c^3*d*e^2 - 9*b*c*e^3)*x^5 + (2*b*c^5*d^3 + 7*b*c^3*d^2*e - 9*b*c*d*e^2)
*x^3)*sqrt(d)*log(-(e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) + 2*(2*a*c^2*d^3 - 16*(a*c^2*d*e^2 - a*e^3)*
x^4 - 2*a*d^2*e + (b*c^3*d^2*e - b*c*d*e^2)*x^3 - 8*(a*c^2*d^2*e - a*d*e^2)*x^2 + (b*c^3*d^3 - b*c*d^2*e)*x +
2*(b*c^2*d^3 - 8*(b*c^2*d*e^2 - b*e^3)*x^4 - b*d^2*e - 4*(b*c^2*d^2*e - b*d*e^2)*x^2)*arctan(c*x))*sqrt(e*x^2
+ d))/((c^2*d^4*e - d^3*e^2)*x^5 + (c^2*d^5 - d^4*e)*x^3), -1/12*(2*((2*b*c^5*d^2*e + 7*b*c^3*d*e^2 - 9*b*c*e^
3)*x^5 + (2*b*c^5*d^3 + 7*b*c^3*d^2*e - 9*b*c*d*e^2)*x^3)*sqrt(-d)*arctan(sqrt(-d)/sqrt(e*x^2 + d)) + ((b*c^4*
d^2*e + 4*b*c^2*d*e^2 - 8*b*e^3)*x^5 + (b*c^4*d^3 + 4*b*c^2*d^2*e - 8*b*d*e^2)*x^3)*sqrt(c^2*d - e)*log((c^4*e
^2*x^4 + 8*c^4*d^2 - 8*c^2*d*e + 2*(4*c^4*d*e - 3*c^2*e^2)*x^2 + 4*(c^3*e*x^2 + 2*c^3*d - c*e)*sqrt(c^2*d - e)
*sqrt(e*x^2 + d) + e^2)/(c^4*x^4 + 2*c^2*x^2 + 1)) + 2*(2*a*c^2*d^3 - 16*(a*c^2*d*e^2 - a*e^3)*x^4 - 2*a*d^2*e
+ (b*c^3*d^2*e - b*c*d*e^2)*x^3 - 8*(a*c^2*d^2*e - a*d*e^2)*x^2 + (b*c^3*d^3 - b*c*d^2*e)*x + 2*(b*c^2*d^3 -
8*(b*c^2*d*e^2 - b*e^3)*x^4 - b*d^2*e - 4*(b*c^2*d^2*e - b*d*e^2)*x^2)*arctan(c*x))*sqrt(e*x^2 + d))/((c^2*d^4
*e - d^3*e^2)*x^5 + (c^2*d^5 - d^4*e)*x^3), -1/6*(((b*c^4*d^2*e + 4*b*c^2*d*e^2 - 8*b*e^3)*x^5 + (b*c^4*d^3 +
4*b*c^2*d^2*e - 8*b*d*e^2)*x^3)*sqrt(-c^2*d + e)*arctan(-1/2*(c^2*e*x^2 + 2*c^2*d - e)*sqrt(-c^2*d + e)*sqrt(e
*x^2 + d)/(c^3*d^2 - c*d*e + (c^3*d*e - c*e^2)*x^2)) + ((2*b*c^5*d^2*e + 7*b*c^3*d*e^2 - 9*b*c*e^3)*x^5 + (2*b
*c^5*d^3 + 7*b*c^3*d^2*e - 9*b*c*d*e^2)*x^3)*sqrt(-d)*arctan(sqrt(-d)/sqrt(e*x^2 + d)) + (2*a*c^2*d^3 - 16*(a*
c^2*d*e^2 - a*e^3)*x^4 - 2*a*d^2*e + (b*c^3*d^2*e - b*c*d*e^2)*x^3 - 8*(a*c^2*d^2*e - a*d*e^2)*x^2 + (b*c^3*d^
3 - b*c*d^2*e)*x + 2*(b*c^2*d^3 - 8*(b*c^2*d*e^2 - b*e^3)*x^4 - b*d^2*e - 4*(b*c^2*d^2*e - b*d*e^2)*x^2)*arcta
n(c*x))*sqrt(e*x^2 + d))/((c^2*d^4*e - d^3*e^2)*x^5 + (c^2*d^5 - d^4*e)*x^3)]

Sympy [F]

\[ \int \frac {a+b \arctan (c x)}{x^4 \left (d+e x^2\right )^{3/2}} \, dx=\int \frac {a + b \operatorname {atan}{\left (c x \right )}}{x^{4} \left (d + e x^{2}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate((a+b*atan(c*x))/x**4/(e*x**2+d)**(3/2),x)

[Out]

Integral((a + b*atan(c*x))/(x**4*(d + e*x**2)**(3/2)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b \arctan (c x)}{x^4 \left (d+e x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((a+b*arctan(c*x))/x^4/(e*x^2+d)^(3/2),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]

\[ \int \frac {a+b \arctan (c x)}{x^4 \left (d+e x^2\right )^{3/2}} \, dx=\int { \frac {b \arctan \left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} x^{4}} \,d x } \]

[In]

integrate((a+b*arctan(c*x))/x^4/(e*x^2+d)^(3/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \arctan (c x)}{x^4 \left (d+e x^2\right )^{3/2}} \, dx=\int \frac {a+b\,\mathrm {atan}\left (c\,x\right )}{x^4\,{\left (e\,x^2+d\right )}^{3/2}} \,d x \]

[In]

int((a + b*atan(c*x))/(x^4*(d + e*x^2)^(3/2)),x)

[Out]

int((a + b*atan(c*x))/(x^4*(d + e*x^2)^(3/2)), x)