\(\int \frac {a+b \arctan (c x^2)}{x^4} \, dx\) [72]

Optimal result

Integrand size = 14, antiderivative size = 159 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x^4} \, dx=-\frac {2 b c}{3 x}-\frac {a+b \arctan \left (c x^2\right )}{3 x^3}+\frac {b c^{3/2} \arctan \left (1-\sqrt {2} \sqrt {c} x\right )}{3 \sqrt {2}}-\frac {b c^{3/2} \arctan \left (1+\sqrt {2} \sqrt {c} x\right )}{3 \sqrt {2}}-\frac {b c^{3/2} \log \left (1-\sqrt {2} \sqrt {c} x+c x^2\right )}{6 \sqrt {2}}+\frac {b c^{3/2} \log \left (1+\sqrt {2} \sqrt {c} x+c x^2\right )}{6 \sqrt {2}} \]

[Out]

-2/3*b*c/x+1/3*(-a-b*arctan(c*x^2))/x^3-1/6*b*c^(3/2)*arctan(-1+x*2^(1/2)*c^(1/2))*2^(1/2)-1/6*b*c^(3/2)*arcta
n(1+x*2^(1/2)*c^(1/2))*2^(1/2)-1/12*b*c^(3/2)*ln(1+c*x^2-x*2^(1/2)*c^(1/2))*2^(1/2)+1/12*b*c^(3/2)*ln(1+c*x^2+
x*2^(1/2)*c^(1/2))*2^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {4946, 331, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {a+b \arctan \left (c x^2\right )}{x^4} \, dx=-\frac {a+b \arctan \left (c x^2\right )}{3 x^3}+\frac {b c^{3/2} \arctan \left (1-\sqrt {2} \sqrt {c} x\right )}{3 \sqrt {2}}-\frac {b c^{3/2} \arctan \left (\sqrt {2} \sqrt {c} x+1\right )}{3 \sqrt {2}}-\frac {b c^{3/2} \log \left (c x^2-\sqrt {2} \sqrt {c} x+1\right )}{6 \sqrt {2}}+\frac {b c^{3/2} \log \left (c x^2+\sqrt {2} \sqrt {c} x+1\right )}{6 \sqrt {2}}-\frac {2 b c}{3 x} \]

[In]

Int[(a + b*ArcTan[c*x^2])/x^4,x]

[Out]

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

Rule 210

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

Rule 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 331

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] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.11 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x^4} \, dx=-\frac {a}{3 x^3}-\frac {2 b c}{3 x}-\frac {b \arctan \left (c x^2\right )}{3 x^3}-\frac {b c^{3/2} \arctan \left (\frac {-\sqrt {2}+2 \sqrt {c} x}{\sqrt {2}}\right )}{3 \sqrt {2}}-\frac {b c^{3/2} \arctan \left (\frac {\sqrt {2}+2 \sqrt {c} x}{\sqrt {2}}\right )}{3 \sqrt {2}}-\frac {b c^{3/2} \log \left (1-\sqrt {2} \sqrt {c} x+c x^2\right )}{6 \sqrt {2}}+\frac {b c^{3/2} \log \left (1+\sqrt {2} \sqrt {c} x+c x^2\right )}{6 \sqrt {2}} \]

[In]

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

[Out]

-1/3*a/x^3 - (2*b*c)/(3*x) - (b*ArcTan[c*x^2])/(3*x^3) - (b*c^(3/2)*ArcTan[(-Sqrt[2] + 2*Sqrt[c]*x)/Sqrt[2]])/
(3*Sqrt[2]) - (b*c^(3/2)*ArcTan[(Sqrt[2] + 2*Sqrt[c]*x)/Sqrt[2]])/(3*Sqrt[2]) - (b*c^(3/2)*Log[1 - Sqrt[2]*Sqr
t[c]*x + c*x^2])/(6*Sqrt[2]) + (b*c^(3/2)*Log[1 + Sqrt[2]*Sqrt[c]*x + c*x^2])/(6*Sqrt[2])

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.72

method	result	size
default	\(-\frac {a}{3 x^{3}}+b \left (-\frac {\arctan \left (c \,x^{2}\right )}{3 x^{3}}+\frac {2 c \left (-\frac {\sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}{x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-1\right )\right )}{8 \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-\frac {1}{x}\right )}{3}\right )\)	\(115\)
parts	\(-\frac {a}{3 x^{3}}+b \left (-\frac {\arctan \left (c \,x^{2}\right )}{3 x^{3}}+\frac {2 c \left (-\frac {\sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}{x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-1\right )\right )}{8 \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-\frac {1}{x}\right )}{3}\right )\)	\(115\)

[In]

int((a+b*arctan(c*x^2))/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/3*a/x^3+b*(-1/3/x^3*arctan(c*x^2)+2/3*c*(-1/8/(1/c^2)^(1/4)*2^(1/2)*(ln((x^2-(1/c^2)^(1/4)*x*2^(1/2)+(1/c^2
)^(1/2))/(x^2+(1/c^2)^(1/4)*x*2^(1/2)+(1/c^2)^(1/2)))+2*arctan(2^(1/2)/(1/c^2)^(1/4)*x+1)+2*arctan(2^(1/2)/(1/
c^2)^(1/4)*x-1))-1/x))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.07 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x^4} \, dx=-\frac {\left (-b^{4} c^{6}\right )^{\frac {1}{4}} x^{3} \log \left (b^{3} c^{5} x + \left (-b^{4} c^{6}\right )^{\frac {3}{4}}\right ) - i \, \left (-b^{4} c^{6}\right )^{\frac {1}{4}} x^{3} \log \left (b^{3} c^{5} x + i \, \left (-b^{4} c^{6}\right )^{\frac {3}{4}}\right ) + i \, \left (-b^{4} c^{6}\right )^{\frac {1}{4}} x^{3} \log \left (b^{3} c^{5} x - i \, \left (-b^{4} c^{6}\right )^{\frac {3}{4}}\right ) - \left (-b^{4} c^{6}\right )^{\frac {1}{4}} x^{3} \log \left (b^{3} c^{5} x - \left (-b^{4} c^{6}\right )^{\frac {3}{4}}\right ) + 4 \, b c x^{2} + 2 \, b \arctan \left (c x^{2}\right ) + 2 \, a}{6 \, x^{3}} \]

[In]

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

[Out]

-1/6*((-b^4*c^6)^(1/4)*x^3*log(b^3*c^5*x + (-b^4*c^6)^(3/4)) - I*(-b^4*c^6)^(1/4)*x^3*log(b^3*c^5*x + I*(-b^4*
c^6)^(3/4)) + I*(-b^4*c^6)^(1/4)*x^3*log(b^3*c^5*x - I*(-b^4*c^6)^(3/4)) - (-b^4*c^6)^(1/4)*x^3*log(b^3*c^5*x
- (-b^4*c^6)^(3/4)) + 4*b*c*x^2 + 2*b*arctan(c*x^2) + 2*a)/x^3

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 20.04 (sec) , antiderivative size = 529, normalized size of antiderivative = 3.33 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x^4} \, dx=\begin {cases} - \frac {a}{3 x^{3}} & \text {for}\: c = 0 \\- \frac {a - \infty i b}{3 x^{3}} & \text {for}\: c = - \frac {i}{x^{2}} \\- \frac {a + \infty i b}{3 x^{3}} & \text {for}\: c = \frac {i}{x^{2}} \\- \frac {2 a x^{4}}{6 x^{7} + \frac {6 x^{3}}{c^{2}}} - \frac {2 a}{6 c^{2} x^{7} + 6 x^{3}} + \frac {2 b c^{3} x^{7} \left (- \frac {1}{c^{2}}\right )^{\frac {3}{4}} \log {\left (x - \sqrt [4]{- \frac {1}{c^{2}}} \right )}}{6 x^{7} + \frac {6 x^{3}}{c^{2}}} - \frac {b c^{3} x^{7} \left (- \frac {1}{c^{2}}\right )^{\frac {3}{4}} \log {\left (x^{2} + \sqrt {- \frac {1}{c^{2}}} \right )}}{6 x^{7} + \frac {6 x^{3}}{c^{2}}} + \frac {2 b c^{3} x^{7} \left (- \frac {1}{c^{2}}\right )^{\frac {3}{4}} \operatorname {atan}{\left (\frac {x}{\sqrt [4]{- \frac {1}{c^{2}}}} \right )}}{6 x^{7} + \frac {6 x^{3}}{c^{2}}} + \frac {2 b c^{2} x^{7} \sqrt [4]{- \frac {1}{c^{2}}} \operatorname {atan}{\left (c x^{2} \right )}}{6 x^{7} + \frac {6 x^{3}}{c^{2}}} - \frac {4 b c x^{6}}{6 x^{7} + \frac {6 x^{3}}{c^{2}}} + \frac {2 b c x^{3} \left (- \frac {1}{c^{2}}\right )^{\frac {3}{4}} \log {\left (x - \sqrt [4]{- \frac {1}{c^{2}}} \right )}}{6 x^{7} + \frac {6 x^{3}}{c^{2}}} - \frac {b c x^{3} \left (- \frac {1}{c^{2}}\right )^{\frac {3}{4}} \log {\left (x^{2} + \sqrt {- \frac {1}{c^{2}}} \right )}}{6 x^{7} + \frac {6 x^{3}}{c^{2}}} + \frac {2 b c x^{3} \left (- \frac {1}{c^{2}}\right )^{\frac {3}{4}} \operatorname {atan}{\left (\frac {x}{\sqrt [4]{- \frac {1}{c^{2}}}} \right )}}{6 x^{7} + \frac {6 x^{3}}{c^{2}}} - \frac {2 b x^{4} \operatorname {atan}{\left (c x^{2} \right )}}{6 x^{7} + \frac {6 x^{3}}{c^{2}}} + \frac {2 b x^{3} \sqrt [4]{- \frac {1}{c^{2}}} \operatorname {atan}{\left (c x^{2} \right )}}{6 x^{7} + \frac {6 x^{3}}{c^{2}}} - \frac {4 b x^{2}}{6 c x^{7} + \frac {6 x^{3}}{c}} - \frac {2 b \operatorname {atan}{\left (c x^{2} \right )}}{6 c^{2} x^{7} + 6 x^{3}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-a/(3*x**3), Eq(c, 0)), (-(a - oo*I*b)/(3*x**3), Eq(c, -I/x**2)), (-(a + oo*I*b)/(3*x**3), Eq(c, I/
x**2)), (-2*a*x**4/(6*x**7 + 6*x**3/c**2) - 2*a/(6*c**2*x**7 + 6*x**3) + 2*b*c**3*x**7*(-1/c**2)**(3/4)*log(x
- (-1/c**2)**(1/4))/(6*x**7 + 6*x**3/c**2) - b*c**3*x**7*(-1/c**2)**(3/4)*log(x**2 + sqrt(-1/c**2))/(6*x**7 +
6*x**3/c**2) + 2*b*c**3*x**7*(-1/c**2)**(3/4)*atan(x/(-1/c**2)**(1/4))/(6*x**7 + 6*x**3/c**2) + 2*b*c**2*x**7*
(-1/c**2)**(1/4)*atan(c*x**2)/(6*x**7 + 6*x**3/c**2) - 4*b*c*x**6/(6*x**7 + 6*x**3/c**2) + 2*b*c*x**3*(-1/c**2
)**(3/4)*log(x - (-1/c**2)**(1/4))/(6*x**7 + 6*x**3/c**2) - b*c*x**3*(-1/c**2)**(3/4)*log(x**2 + sqrt(-1/c**2)
)/(6*x**7 + 6*x**3/c**2) + 2*b*c*x**3*(-1/c**2)**(3/4)*atan(x/(-1/c**2)**(1/4))/(6*x**7 + 6*x**3/c**2) - 2*b*x
**4*atan(c*x**2)/(6*x**7 + 6*x**3/c**2) + 2*b*x**3*(-1/c**2)**(1/4)*atan(c*x**2)/(6*x**7 + 6*x**3/c**2) - 4*b*
x**2/(6*c*x**7 + 6*x**3/c) - 2*b*atan(c*x**2)/(6*c**2*x**7 + 6*x**3), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.89 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x^4} \, dx=-\frac {1}{12} \, {\left ({\left (c^{2} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x + \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right )}{c^{\frac {3}{2}}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x - \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right )}{c^{\frac {3}{2}}} - \frac {\sqrt {2} \log \left (c x^{2} + \sqrt {2} \sqrt {c} x + 1\right )}{c^{\frac {3}{2}}} + \frac {\sqrt {2} \log \left (c x^{2} - \sqrt {2} \sqrt {c} x + 1\right )}{c^{\frac {3}{2}}}\right )} + \frac {8}{x}\right )} c + \frac {4 \, \arctan \left (c x^{2}\right )}{x^{3}}\right )} b - \frac {a}{3 \, x^{3}} \]

[In]

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

[Out]

-1/12*((c^2*(2*sqrt(2)*arctan(1/2*sqrt(2)*(2*c*x + sqrt(2)*sqrt(c))/sqrt(c))/c^(3/2) + 2*sqrt(2)*arctan(1/2*sq
rt(2)*(2*c*x - sqrt(2)*sqrt(c))/sqrt(c))/c^(3/2) - sqrt(2)*log(c*x^2 + sqrt(2)*sqrt(c)*x + 1)/c^(3/2) + sqrt(2
)*log(c*x^2 - sqrt(2)*sqrt(c)*x + 1)/c^(3/2)) + 8/x)*c + 4*arctan(c*x^2)/x^3)*b - 1/3*a/x^3

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x^4} \, dx=-\frac {1}{12} \, b c^{3} {\left (\frac {2 \, \sqrt {2} \sqrt {{\left | c \right |}} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \frac {\sqrt {2}}{\sqrt {{\left | c \right |}}}\right )} \sqrt {{\left | c \right |}}\right )}{c^{2}} + \frac {2 \, \sqrt {2} \sqrt {{\left | c \right |}} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \frac {\sqrt {2}}{\sqrt {{\left | c \right |}}}\right )} \sqrt {{\left | c \right |}}\right )}{c^{2}} - \frac {\sqrt {2} \sqrt {{\left | c \right |}} \log \left (x^{2} + \frac {\sqrt {2} x}{\sqrt {{\left | c \right |}}} + \frac {1}{{\left | c \right |}}\right )}{c^{2}} + \frac {\sqrt {2} \sqrt {{\left | c \right |}} \log \left (x^{2} - \frac {\sqrt {2} x}{\sqrt {{\left | c \right |}}} + \frac {1}{{\left | c \right |}}\right )}{c^{2}}\right )} - \frac {2 \, b c x^{2} + b \arctan \left (c x^{2}\right ) + a}{3 \, x^{3}} \]

[In]

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

[Out]

-1/12*b*c^3*(2*sqrt(2)*sqrt(abs(c))*arctan(1/2*sqrt(2)*(2*x + sqrt(2)/sqrt(abs(c)))*sqrt(abs(c)))/c^2 + 2*sqrt
(2)*sqrt(abs(c))*arctan(1/2*sqrt(2)*(2*x - sqrt(2)/sqrt(abs(c)))*sqrt(abs(c)))/c^2 - sqrt(2)*sqrt(abs(c))*log(
x^2 + sqrt(2)*x/sqrt(abs(c)) + 1/abs(c))/c^2 + sqrt(2)*sqrt(abs(c))*log(x^2 - sqrt(2)*x/sqrt(abs(c)) + 1/abs(c
))/c^2) - 1/3*(2*b*c*x^2 + b*arctan(c*x^2) + a)/x^3

Mupad [B] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.40 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x^4} \, dx=-\frac {2\,b\,c\,x^2+a}{3\,x^3}-\frac {b\,\mathrm {atan}\left (c\,x^2\right )}{3\,x^3}-\frac {{\left (-1\right )}^{1/4}\,b\,c^{3/2}\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {c}\,x\right )}{3}-\frac {{\left (-1\right )}^{1/4}\,b\,c^{3/2}\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {c}\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{3} \]

[In]

int((a + b*atan(c*x^2))/x^4,x)

[Out]

- (a + 2*b*c*x^2)/(3*x^3) - (b*atan(c*x^2))/(3*x^3) - ((-1)^(1/4)*b*c^(3/2)*atan((-1)^(1/4)*c^(1/2)*x))/3 - ((
-1)^(1/4)*b*c^(3/2)*atan((-1)^(1/4)*c^(1/2)*x*1i)*1i)/3