3.13.8 \(\int \frac {a+b \text {ArcTan}(c x)}{x^4 \sqrt {d+e x^2}} \, dx\) [1208]

Optimal. Leaf size=179 \[ -\frac {b c \sqrt {d+e x^2}}{6 d x^2}-\frac {\sqrt {d+e x^2} (a+b \text {ArcTan}(c x))}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} (a+b \text {ArcTan}(c x))}{3 d^2 x}+\frac {b c \left (2 c^2 d+3 e\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{6 d^{3/2}}-\frac {b \sqrt {c^2 d-e} \left (c^2 d+2 e\right ) \tanh ^{-1}\left (\frac {c \sqrt {d+e x^2}}{\sqrt {c^2 d-e}}\right )}{3 d^2} \]

[Out]

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

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Rubi [A]
time = 0.19, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {277, 270, 5096, 12, 587, 154, 162, 65, 214} \begin {gather*} \frac {2 e \sqrt {d+e x^2} (a+b \text {ArcTan}(c x))}{3 d^2 x}-\frac {\sqrt {d+e x^2} (a+b \text {ArcTan}(c x))}{3 d x^3}+\frac {b c \left (2 c^2 d+3 e\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{6 d^{3/2}}-\frac {b \sqrt {c^2 d-e} \left (c^2 d+2 e\right ) \tanh ^{-1}\left (\frac {c \sqrt {d+e x^2}}{\sqrt {c^2 d-e}}\right )}{3 d^2}-\frac {b c \sqrt {d+e x^2}}{6 d x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(b*c*Sqrt[d + e*x^2])/(d*x^2) - (Sqrt[d + e*x^2]*(a + b*ArcTan[c*x]))/(3*d*x^3) + (2*e*Sqrt[d + e*x^2]*(a
+ b*ArcTan[c*x]))/(3*d^2*x) + (b*c*(2*c^2*d + 3*e)*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(6*d^(3/2)) - (b*Sqrt[c^
2*d - e]*(c^2*d + 2*e)*ArcTanh[(c*Sqrt[d + e*x^2])/Sqrt[c^2*d - e]])/(3*d^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 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 154

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && ILtQ[m, -1] && GtQ[n, 0]

Rule 162

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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 270

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

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 587

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^(r_.), x
_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q*(e + f*x)^r, x], x, x^n],
x] /; FreeQ[{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[Simplify[(m + 1)/n]]

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*} \int \frac {a+b \tan ^{-1}(c x)}{x^4 \sqrt {d+e x^2}} \, dx &=-\frac {\sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 x}-(b c) \int \frac {\sqrt {d+e x^2} \left (-d+2 e x^2\right )}{3 d^2 x^3 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {\sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 x}-\frac {(b c) \int \frac {\sqrt {d+e x^2} \left (-d+2 e x^2\right )}{x^3 \left (1+c^2 x^2\right )} \, dx}{3 d^2}\\ &=-\frac {\sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 x}-\frac {(b c) \text {Subst}\left (\int \frac {\sqrt {d+e x} (-d+2 e x)}{x^2 \left (1+c^2 x\right )} \, dx,x,x^2\right )}{6 d^2}\\ &=-\frac {b c \sqrt {d+e x^2}}{6 d x^2}-\frac {\sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 x}-\frac {(b c) \text {Subst}\left (\int \frac {\frac {1}{2} d \left (2 c^2 d+3 e\right )+\frac {1}{2} e \left (c^2 d+4 e\right ) x}{x \left (1+c^2 x\right ) \sqrt {d+e x}} \, dx,x,x^2\right )}{6 d^2}\\ &=-\frac {b c \sqrt {d+e x^2}}{6 d x^2}-\frac {\sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 x}+\frac {\left (b c \left (c^2 d-e\right ) \left (c^2 d+2 e\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^2}-\frac {\left (b c \left (2 c^2 d+3 e\right )\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^2\right )}{12 d}\\ &=-\frac {b c \sqrt {d+e x^2}}{6 d x^2}-\frac {\sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 x}+\frac {\left (b c \left (c^2 d-e\right ) \left (c^2 d+2 e\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^2 e}-\frac {\left (b c \left (2 c^2 d+3 e\right )\right ) \text {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{6 d e}\\ &=-\frac {b c \sqrt {d+e x^2}}{6 d x^2}-\frac {\sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 x}+\frac {b c \left (2 c^2 d+3 e\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{6 d^{3/2}}-\frac {b \sqrt {c^2 d-e} \left (c^2 d+2 e\right ) \tanh ^{-1}\left (\frac {c \sqrt {d+e x^2}}{\sqrt {c^2 d-e}}\right )}{3 d^2}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.38, size = 372, normalized size = 2.08 \begin {gather*} -\frac {\frac {\sqrt {d+e x^2} \left (b c d x+2 a \left (d-2 e x^2\right )\right )}{x^3}+\frac {2 b \left (d-2 e x^2\right ) \sqrt {d+e x^2} \text {ArcTan}(c x)}{x^3}+b c \sqrt {d} \left (2 c^2 d+3 e\right ) \log (x)-b c \sqrt {d} \left (2 c^2 d+3 e\right ) \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )+\frac {b \left (c^4 d^2+c^2 d e-2 e^2\right ) \log \left (\frac {12 c d^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 (c^4 d^2+c^2 d e-2 e^2\right ) (i+c x)}\right )}{\sqrt {c^2 d-e}}+\frac {b \left (c^4 d^2+c^2 d e-2 e^2\right ) \log \left (\frac {12 c d^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 (c^4 d^2+c^2 d e-2 e^2\right ) (-i+c x)}\right )}{\sqrt {c^2 d-e}}}{6 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.14, size = 0, normalized size = 0.00 \[\int \frac {a +b \arctan \left (c x \right )}{x^{4} \sqrt {e \,x^{2}+d}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x))/x^4/(e*x^2+d)^(1/2),x)

[Out]

int((a+b*arctan(c*x))/x^4/(e*x^2+d)^(1/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))/x^4/(e*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

1/3*a*(2*sqrt(x^2*e + d)*e/(d^2*x) - sqrt(x^2*e + d)/(d*x^3)) + b*integrate(arctan(c*x)/(sqrt(x^2*e + d)*x^4),
x)

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Fricas [A]
time = 2.72, size = 936, normalized size = 5.23 \begin {gather*} \left [\frac {{\left (b c^{2} d x^{3} + 2 \, b x^{3} e\right )} \sqrt {c^{2} d - e} \log \left (\frac {8 \, c^{4} d^{2} - 4 \, {\left (2 \, c^{3} d + {\left (c^{3} x^{2} - c\right )} e\right )} \sqrt {c^{2} d - e} \sqrt {x^{2} e + d} + {\left (c^{4} x^{4} - 6 \, c^{2} x^{2} + 1\right )} e^{2} + 8 \, {\left (c^{4} d x^{2} - c^{2} d\right )} e}{c^{4} x^{4} + 2 \, c^{2} x^{2} + 1}\right ) + {\left (2 \, b c^{3} d x^{3} + 3 \, b c x^{3} e\right )} \sqrt {d} \log \left (-\frac {x^{2} e + 2 \, \sqrt {x^{2} e + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - 2 \, {\left (b c d x - 4 \, a x^{2} e + 2 \, a d - 2 \, {\left (2 \, b x^{2} e - b d\right )} \arctan \left (c x\right )\right )} \sqrt {x^{2} e + d}}{12 \, d^{2} x^{3}}, -\frac {2 \, {\left (b c^{2} d x^{3} + 2 \, b x^{3} e\right )} \sqrt {-c^{2} d + e} \arctan \left (-\frac {{\left (2 \, c^{2} d + {\left (c^{2} x^{2} - 1\right )} e\right )} \sqrt {-c^{2} d + e} \sqrt {x^{2} e + d}}{2 \, {\left (c^{3} d^{2} - c x^{2} e^{2} + {\left (c^{3} d x^{2} - c d\right )} e\right )}}\right ) - {\left (2 \, b c^{3} d x^{3} + 3 \, b c x^{3} e\right )} \sqrt {d} \log \left (-\frac {x^{2} e + 2 \, \sqrt {x^{2} e + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) + 2 \, {\left (b c d x - 4 \, a x^{2} e + 2 \, a d - 2 \, {\left (2 \, b x^{2} e - b d\right )} \arctan \left (c x\right )\right )} \sqrt {x^{2} e + d}}{12 \, d^{2} x^{3}}, -\frac {2 \, {\left (2 \, b c^{3} d x^{3} + 3 \, b c x^{3} e\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d}}{\sqrt {x^{2} e + d}}\right ) - {\left (b c^{2} d x^{3} + 2 \, b x^{3} e\right )} \sqrt {c^{2} d - e} \log \left (\frac {8 \, c^{4} d^{2} - 4 \, {\left (2 \, c^{3} d + {\left (c^{3} x^{2} - c\right )} e\right )} \sqrt {c^{2} d - e} \sqrt {x^{2} e + d} + {\left (c^{4} x^{4} - 6 \, c^{2} x^{2} + 1\right )} e^{2} + 8 \, {\left (c^{4} d x^{2} - c^{2} d\right )} e}{c^{4} x^{4} + 2 \, c^{2} x^{2} + 1}\right ) + 2 \, {\left (b c d x - 4 \, a x^{2} e + 2 \, a d - 2 \, {\left (2 \, b x^{2} e - b d\right )} \arctan \left (c x\right )\right )} \sqrt {x^{2} e + d}}{12 \, d^{2} x^{3}}, -\frac {{\left (b c^{2} d x^{3} + 2 \, b x^{3} e\right )} \sqrt {-c^{2} d + e} \arctan \left (-\frac {{\left (2 \, c^{2} d + {\left (c^{2} x^{2} - 1\right )} e\right )} \sqrt {-c^{2} d + e} \sqrt {x^{2} e + d}}{2 \, {\left (c^{3} d^{2} - c x^{2} e^{2} + {\left (c^{3} d x^{2} - c d\right )} e\right )}}\right ) + {\left (2 \, b c^{3} d x^{3} + 3 \, b c x^{3} e\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d}}{\sqrt {x^{2} e + d}}\right ) + {\left (b c d x - 4 \, a x^{2} e + 2 \, a d - 2 \, {\left (2 \, b x^{2} e - b d\right )} \arctan \left (c x\right )\right )} \sqrt {x^{2} e + d}}{6 \, d^{2} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))/x^4/(e*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \operatorname {atan}{\left (c x \right )}}{x^{4} \sqrt {d + e x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x))/x**4/(e*x**2+d)**(1/2),x)

[Out]

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

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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((a+b*arctan(c*x))/x^4/(e*x^2+d)^(1/2),x, algorithm="giac")

[Out]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,\mathrm {atan}\left (c\,x\right )}{x^4\,\sqrt {e\,x^2+d}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))/(x^4*(d + e*x^2)^(1/2)),x)

[Out]

int((a + b*atan(c*x))/(x^4*(d + e*x^2)^(1/2)), x)

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