\(\int \frac {x^2 \arctan (a x)^{3/2}}{(c+a^2 c x^2)^2} \, dx\) [785]

Optimal result

Integrand size = 24, antiderivative size = 127 \[ \int \frac {x^2 \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {3 \sqrt {\arctan (a x)}}{16 a^3 c^2}-\frac {3 \sqrt {\arctan (a x)}}{8 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {x \arctan (a x)^{3/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^{5/2}}{5 a^3 c^2}+\frac {3 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{32 a^3 c^2} \]

[Out]

-1/2*x*arctan(a*x)^(3/2)/a^2/c^2/(a^2*x^2+1)+1/5*arctan(a*x)^(5/2)/a^3/c^2+3/32*FresnelC(2*arctan(a*x)^(1/2)/P
i^(1/2))*Pi^(1/2)/a^3/c^2+3/16*arctan(a*x)^(1/2)/a^3/c^2-3/8*arctan(a*x)^(1/2)/a^3/c^2/(a^2*x^2+1)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5056, 5050, 5024, 3393, 3385, 3433} \[ \int \frac {x^2 \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {3 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{32 a^3 c^2}+\frac {\arctan (a x)^{5/2}}{5 a^3 c^2}+\frac {3 \sqrt {\arctan (a x)}}{16 a^3 c^2}-\frac {x \arctan (a x)^{3/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}-\frac {3 \sqrt {\arctan (a x)}}{8 a^3 c^2 \left (a^2 x^2+1\right )} \]

[In]

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

[Out]

(3*Sqrt[ArcTan[a*x]])/(16*a^3*c^2) - (3*Sqrt[ArcTan[a*x]])/(8*a^3*c^2*(1 + a^2*x^2)) - (x*ArcTan[a*x]^(3/2))/(
2*a^2*c^2*(1 + a^2*x^2)) + ArcTan[a*x]^(5/2)/(5*a^3*c^2) + (3*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]
])/(32*a^3*c^2)

Rule 3385

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Cos[f*(x^2/d)],
x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 3433

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 5024

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[d^q/c, Subst[Int[(a
+ b*x)^p/Cos[x]^(2*(q + 1)), x], x, ArcTan[c*x]], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && ILtQ
[2*(q + 1), 0] && (IntegerQ[q] || GtQ[d, 0])

Rule 5050

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[(d + e*x^2)^(
q + 1)*((a + b*ArcTan[c*x])^p/(2*e*(q + 1))), x] - Dist[b*(p/(2*c*(q + 1))), Int[(d + e*x^2)^q*(a + b*ArcTan[c
*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[e, c^2*d] && GtQ[p, 0] && NeQ[q, -1]

Rule 5056

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^2)/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> Simp[(a + b*ArcTan
[c*x])^(p + 1)/(2*b*c^3*d^2*(p + 1)), x] + (Dist[b*(p/(2*c)), Int[x*((a + b*ArcTan[c*x])^(p - 1)/(d + e*x^2)^2
), x], x] - Simp[x*((a + b*ArcTan[c*x])^p/(2*c^2*d*(d + e*x^2))), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c
^2*d] && GtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {x \arctan (a x)^{3/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^{5/2}}{5 a^3 c^2}+\frac {3 \int \frac {x \sqrt {\arctan (a x)}}{\left (c+a^2 c x^2\right )^2} \, dx}{4 a} \\ & = -\frac {3 \sqrt {\arctan (a x)}}{8 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {x \arctan (a x)^{3/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^{5/2}}{5 a^3 c^2}+\frac {3 \int \frac {1}{\left (c+a^2 c x^2\right )^2 \sqrt {\arctan (a x)}} \, dx}{16 a^2} \\ & = -\frac {3 \sqrt {\arctan (a x)}}{8 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {x \arctan (a x)^{3/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^{5/2}}{5 a^3 c^2}+\frac {3 \text {Subst}\left (\int \frac {\cos ^2(x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{16 a^3 c^2} \\ & = -\frac {3 \sqrt {\arctan (a x)}}{8 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {x \arctan (a x)^{3/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^{5/2}}{5 a^3 c^2}+\frac {3 \text {Subst}\left (\int \left (\frac {1}{2 \sqrt {x}}+\frac {\cos (2 x)}{2 \sqrt {x}}\right ) \, dx,x,\arctan (a x)\right )}{16 a^3 c^2} \\ & = \frac {3 \sqrt {\arctan (a x)}}{16 a^3 c^2}-\frac {3 \sqrt {\arctan (a x)}}{8 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {x \arctan (a x)^{3/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^{5/2}}{5 a^3 c^2}+\frac {3 \text {Subst}\left (\int \frac {\cos (2 x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{32 a^3 c^2} \\ & = \frac {3 \sqrt {\arctan (a x)}}{16 a^3 c^2}-\frac {3 \sqrt {\arctan (a x)}}{8 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {x \arctan (a x)^{3/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^{5/2}}{5 a^3 c^2}+\frac {3 \text {Subst}\left (\int \cos \left (2 x^2\right ) \, dx,x,\sqrt {\arctan (a x)}\right )}{16 a^3 c^2} \\ & = \frac {3 \sqrt {\arctan (a x)}}{16 a^3 c^2}-\frac {3 \sqrt {\arctan (a x)}}{8 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {x \arctan (a x)^{3/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^{5/2}}{5 a^3 c^2}+\frac {3 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{32 a^3 c^2} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.48 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.47 \[ \int \frac {x^2 \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {\frac {16 \sqrt {\arctan (a x)} \left (15 \left (-1+a^2 x^2\right )-40 a x \arctan (a x)+16 \left (1+a^2 x^2\right ) \arctan (a x)^2\right )}{1+a^2 x^2}+60 \left (-2 \sqrt {\arctan (a x)}+\sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )\right )+\frac {15 \left (8 \arctan (a x)-i \sqrt {2} \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-2 i \arctan (a x)\right )+i \sqrt {2} \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},2 i \arctan (a x)\right )\right )}{\sqrt {\arctan (a x)}}}{1280 a^3 c^2} \]

[In]

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

[Out]

((16*Sqrt[ArcTan[a*x]]*(15*(-1 + a^2*x^2) - 40*a*x*ArcTan[a*x] + 16*(1 + a^2*x^2)*ArcTan[a*x]^2))/(1 + a^2*x^2
) + 60*(-2*Sqrt[ArcTan[a*x]] + Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]]) + (15*(8*ArcTan[a*x] - I*Sqr
t[2]*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-2*I)*ArcTan[a*x]] + I*Sqrt[2]*Sqrt[I*ArcTan[a*x]]*Gamma[1/2, (2*I)*Ar
cTan[a*x]]))/Sqrt[ArcTan[a*x]])/(1280*a^3*c^2)

Maple [A] (verified)

Time = 7.45 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.59

method	result	size
default	\(\frac {32 \arctan \left (a x \right )^{\frac {5}{2}} \sqrt {\pi }-40 \arctan \left (a x \right )^{\frac {3}{2}} \sin \left (2 \arctan \left (a x \right )\right ) \sqrt {\pi }+15 \pi \,\operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right )-30 \sqrt {\arctan \left (a x \right )}\, \sqrt {\pi }\, \cos \left (2 \arctan \left (a x \right )\right )}{160 c^{2} a^{3} \sqrt {\pi }}\)	\(75\)

[In]

int(x^2*arctan(a*x)^(3/2)/(a^2*c*x^2+c)^2,x,method=_RETURNVERBOSE)

[Out]

1/160/c^2/a^3/Pi^(1/2)*(32*arctan(a*x)^(5/2)*Pi^(1/2)-40*arctan(a*x)^(3/2)*sin(2*arctan(a*x))*Pi^(1/2)+15*Pi*F
resnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))-30*arctan(a*x)^(1/2)*Pi^(1/2)*cos(2*arctan(a*x)))

Fricas [F(-2)]

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

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

Sympy [F]

\[ \int \frac {x^2 \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {\int \frac {x^{2} \operatorname {atan}^{\frac {3}{2}}{\left (a x \right )}}{a^{4} x^{4} + 2 a^{2} x^{2} + 1}\, dx}{c^{2}} \]

[In]

integrate(x**2*atan(a*x)**(3/2)/(a**2*c*x**2+c)**2,x)

[Out]

Integral(x**2*atan(a*x)**(3/2)/(a**4*x**4 + 2*a**2*x**2 + 1), x)/c**2

Maxima [F(-2)]

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

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

int((x^2*atan(a*x)^(3/2))/(c + a^2*c*x^2)^2,x)

[Out]

int((x^2*atan(a*x)^(3/2))/(c + a^2*c*x^2)^2, x)