\(\int \frac {1}{(c+a^2 c x^2)^2 \arctan (a x)^{3/2}} \, dx\) [993]

Optimal result

Integrand size = 21, antiderivative size = 57 \[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{3/2}} \, dx=-\frac {2}{a c^2 \left (1+a^2 x^2\right ) \sqrt {\arctan (a x)}}-\frac {2 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{a c^2} \]

[Out]

-2*FresnelS(2*arctan(a*x)^(1/2)/Pi^(1/2))*Pi^(1/2)/a/c^2-2/a/c^2/(a^2*x^2+1)/arctan(a*x)^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5022, 5090, 4491, 12, 3386, 3432} \[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{3/2}} \, dx=-\frac {2}{a c^2 \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}-\frac {2 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{a c^2} \]

[In]

Int[1/((c + a^2*c*x^2)^2*ArcTan[a*x]^(3/2)),x]

[Out]

-2/(a*c^2*(1 + a^2*x^2)*Sqrt[ArcTan[a*x]]) - (2*Sqrt[Pi]*FresnelS[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(a*c^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 3386

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Sin[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 3432

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

Rule 4491

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 5022

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

Rule 5090

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

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{3/2}} \, dx=\frac {-\frac {2}{\left (1+a^2 x^2\right ) \sqrt {\arctan (a x)}}-2 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{a c^2} \]

[In]

Integrate[1/((c + a^2*c*x^2)^2*ArcTan[a*x]^(3/2)),x]

[Out]

(-2/((1 + a^2*x^2)*Sqrt[ArcTan[a*x]]) - 2*Sqrt[Pi]*FresnelS[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(a*c^2)

Maple [A] (verified)

Time = 2.06 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.82

method	result	size
default	\(-\frac {2 \sqrt {\arctan \left (a x \right )}\, \sqrt {\pi }\, \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right )+\cos \left (2 \arctan \left (a x \right )\right )+1}{c^{2} a \sqrt {\arctan \left (a x \right )}}\)	\(47\)

[In]

int(1/(a^2*c*x^2+c)^2/arctan(a*x)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/c^2/a*(2*arctan(a*x)^(1/2)*Pi^(1/2)*FresnelS(2*arctan(a*x)^(1/2)/Pi^(1/2))+cos(2*arctan(a*x))+1)/arctan(a*x
)^(1/2)

Fricas [F(-2)]

Exception generated. \[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{3/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/(a^2*c*x^2+c)^2/arctan(a*x)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{3/2}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(1/(a^2*c*x^2+c)^2/arctan(a*x)^(3/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(1/(a^2*c*x^2+c)^2/arctan(a*x)^(3/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{3/2}} \, dx=\int \frac {1}{{\mathrm {atan}\left (a\,x\right )}^{3/2}\,{\left (c\,a^2\,x^2+c\right )}^2} \,d x \]

[In]

int(1/(atan(a*x)^(3/2)*(c + a^2*c*x^2)^2),x)

[Out]

int(1/(atan(a*x)^(3/2)*(c + a^2*c*x^2)^2), x)