\(\int \frac {\sqrt {d+e x^2} (a+b \arctan (c x))}{x^4} \, dx\) [1180]

Optimal result

Integrand size = 23, antiderivative size = 137 \[ \int \frac {\sqrt {d+e x^2} (a+b \arctan (c x))}{x^4} \, dx=-\frac {b c \sqrt {d+e x^2}}{6 x^2}-\frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{3 d x^3}+\frac {b c \left (2 c^2 d-3 e\right ) \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{6 \sqrt {d}}-\frac {b \left (c^2 d-e\right )^{3/2} \text {arctanh}\left (\frac {c \sqrt {d+e x^2}}{\sqrt {c^2 d-e}}\right )}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {270, 5096, 12, 457, 100, 162, 65, 214} \[ \int \frac {\sqrt {d+e x^2} (a+b \arctan (c x))}{x^4} \, dx=-\frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{3 d x^3}-\frac {b \left (c^2 d-e\right )^{3/2} \text {arctanh}\left (\frac {c \sqrt {d+e x^2}}{\sqrt {c^2 d-e}}\right )}{3 d}+\frac {b c \left (2 c^2 d-3 e\right ) \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{6 \sqrt {d}}-\frac {b c \sqrt {d+e x^2}}{6 x^2} \]

[In]

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

[Out]

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

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 100

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((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 - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

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 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && 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*} \text {integral}& = -\frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{3 d x^3}-(b c) \int \frac {\left (d+e x^2\right )^{3/2}}{3 x^3 \left (-d-c^2 d x^2\right )} \, dx \\ & = -\frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{3 d x^3}-\frac {1}{3} (b c) \int \frac {\left (d+e x^2\right )^{3/2}}{x^3 \left (-d-c^2 d x^2\right )} \, dx \\ & = -\frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{3 d x^3}-\frac {1}{6} (b c) \text {Subst}\left (\int \frac {(d+e x)^{3/2}}{x^2 \left (-d-c^2 d x\right )} \, dx,x,x^2\right ) \\ & = -\frac {b c \sqrt {d+e x^2}}{6 x^2}-\frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{3 d x^3}-\frac {(b c) \text {Subst}\left (\int \frac {-\frac {1}{2} d^2 \left (2 c^2 d-3 e\right )-\frac {1}{2} d \left (c^2 d-2 e\right ) e x}{x \left (-d-c^2 d x\right ) \sqrt {d+e x}} \, dx,x,x^2\right )}{6 d} \\ & = -\frac {b c \sqrt {d+e x^2}}{6 x^2}-\frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{3 d x^3}-\frac {1}{12} \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 )-\frac {1}{6} \left (b c \left (c^2 d-e\right )^2\right ) \text {Subst}\left (\int \frac {1}{\left (-d-c^2 d x\right ) \sqrt {d+e x}} \, dx,x,x^2\right ) \\ & = -\frac {b c \sqrt {d+e x^2}}{6 x^2}-\frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{3 d x^3}-\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 e}-\frac {\left (b c \left (c^2 d-e\right )^2\right ) \text {Subst}\left (\int \frac {1}{-d+\frac {c^2 d^2}{e}-\frac {c^2 d x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{3 e} \\ & = -\frac {b c \sqrt {d+e x^2}}{6 x^2}-\frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{3 d x^3}+\frac {b c \left (2 c^2 d-3 e\right ) \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{6 \sqrt {d}}-\frac {b \left (c^2 d-e\right )^{3/2} \text {arctanh}\left (\frac {c \sqrt {d+e x^2}}{\sqrt {c^2 d-e}}\right )}{3 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.49 (sec) , antiderivative size = 288, normalized size of antiderivative = 2.10 \[ \int \frac {\sqrt {d+e x^2} (a+b \arctan (c x))}{x^4} \, dx=-\frac {\sqrt {d+e x^2} \left (b c d x+2 a \left (d+e x^2\right )\right )+2 b \left (d+e x^2\right )^{3/2} \arctan (c x)+b c \sqrt {d} \left (2 c^2 d-3 e\right ) x^3 \log (x)-b c \sqrt {d} \left (2 c^2 d-3 e\right ) x^3 \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )+b \left (c^2 d-e\right )^{3/2} x^3 \log \left (\frac {12 c d \left (c d-i e x+\sqrt {c^2 d-e} \sqrt {d+e x^2}\right )}{b \left (c^2 d-e\right )^{5/2} (i+c x)}\right )+b \left (c^2 d-e\right )^{3/2} x^3 \log \left (\frac {12 c d \left (c d+i e x+\sqrt {c^2 d-e} \sqrt {d+e x^2}\right )}{b \left (c^2 d-e\right )^{5/2} (-i+c x)}\right )}{6 d x^3} \]

[In]

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

[Out]

-1/6*(Sqrt[d + e*x^2]*(b*c*d*x + 2*a*(d + e*x^2)) + 2*b*(d + e*x^2)^(3/2)*ArcTan[c*x] + b*c*Sqrt[d]*(2*c^2*d -
3*e)*x^3*Log[x] - b*c*Sqrt[d]*(2*c^2*d - 3*e)*x^3*Log[d + Sqrt[d]*Sqrt[d + e*x^2]] + b*(c^2*d - e)^(3/2)*x^3*
Log[(12*c*d*(c*d - I*e*x + Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*(c^2*d - e)^(5/2)*(I + c*x))] + b*(c^2*d - e)^
(3/2)*x^3*Log[(12*c*d*(c*d + I*e*x + Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*(c^2*d - e)^(5/2)*(-I + c*x))])/(d*x
^3)

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 858, normalized size of antiderivative = 6.26 \[ \int \frac {\sqrt {d+e x^2} (a+b \arctan (c x))}{x^4} \, dx=\left [-\frac {{\left (b c^{2} d - b e\right )} \sqrt {c^{2} d - e} x^{3} \log \left (\frac {c^{4} e^{2} x^{4} + 8 \, c^{4} d^{2} - 8 \, c^{2} d e + 2 \, {\left (4 \, c^{4} d e - 3 \, c^{2} e^{2}\right )} x^{2} + 4 \, {\left (c^{3} e x^{2} + 2 \, c^{3} d - c e\right )} \sqrt {c^{2} d - e} \sqrt {e x^{2} + d} + e^{2}}{c^{4} x^{4} + 2 \, c^{2} x^{2} + 1}\right ) + {\left (2 \, b c^{3} d - 3 \, b c e\right )} \sqrt {d} x^{3} \log \left (-\frac {e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) + 2 \, {\left (b c d x + 2 \, a e x^{2} + 2 \, a d + 2 \, {\left (b e x^{2} + b d\right )} \arctan \left (c x\right )\right )} \sqrt {e x^{2} + d}}{12 \, d x^{3}}, -\frac {2 \, {\left (b c^{2} d - b e\right )} \sqrt {-c^{2} d + e} x^{3} \arctan \left (-\frac {{\left (c^{2} e x^{2} + 2 \, c^{2} d - e\right )} \sqrt {-c^{2} d + e} \sqrt {e x^{2} + d}}{2 \, {\left (c^{3} d^{2} - c d e + {\left (c^{3} d e - c e^{2}\right )} x^{2}\right )}}\right ) + {\left (2 \, b c^{3} d - 3 \, b c e\right )} \sqrt {d} x^{3} \log \left (-\frac {e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) + 2 \, {\left (b c d x + 2 \, a e x^{2} + 2 \, a d + 2 \, {\left (b e x^{2} + b d\right )} \arctan \left (c x\right )\right )} \sqrt {e x^{2} + d}}{12 \, d x^{3}}, -\frac {2 \, {\left (2 \, b c^{3} d - 3 \, b c e\right )} \sqrt {-d} x^{3} \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x^{2} + d}}\right ) + {\left (b c^{2} d - b e\right )} \sqrt {c^{2} d - e} x^{3} \log \left (\frac {c^{4} e^{2} x^{4} + 8 \, c^{4} d^{2} - 8 \, c^{2} d e + 2 \, {\left (4 \, c^{4} d e - 3 \, c^{2} e^{2}\right )} x^{2} + 4 \, {\left (c^{3} e x^{2} + 2 \, c^{3} d - c e\right )} \sqrt {c^{2} d - e} \sqrt {e x^{2} + d} + e^{2}}{c^{4} x^{4} + 2 \, c^{2} x^{2} + 1}\right ) + 2 \, {\left (b c d x + 2 \, a e x^{2} + 2 \, a d + 2 \, {\left (b e x^{2} + b d\right )} \arctan \left (c x\right )\right )} \sqrt {e x^{2} + d}}{12 \, d x^{3}}, -\frac {{\left (b c^{2} d - b e\right )} \sqrt {-c^{2} d + e} x^{3} \arctan \left (-\frac {{\left (c^{2} e x^{2} + 2 \, c^{2} d - e\right )} \sqrt {-c^{2} d + e} \sqrt {e x^{2} + d}}{2 \, {\left (c^{3} d^{2} - c d e + {\left (c^{3} d e - c e^{2}\right )} x^{2}\right )}}\right ) + {\left (2 \, b c^{3} d - 3 \, b c e\right )} \sqrt {-d} x^{3} \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x^{2} + d}}\right ) + {\left (b c d x + 2 \, a e x^{2} + 2 \, a d + 2 \, {\left (b e x^{2} + b d\right )} \arctan \left (c x\right )\right )} \sqrt {e x^{2} + d}}{6 \, d x^{3}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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