Integrand size = 21, antiderivative size = 57 \[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{3/2}} \, dx=-\frac {2}{a c^2 \left (1+a^2 x^2\right ) \sqrt {\arctan (a x)}}-\frac {2 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{a c^2} \] Output:
-2/a/c^2/(a^2*x^2+1)/arctan(a*x)^(1/2)-2*Pi^(1/2)*FresnelS(2*arctan(a*x)^( 1/2)/Pi^(1/2))/a/c^2
Time = 0.30 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{3/2}} \, dx=\frac {-\frac {2}{\left (1+a^2 x^2\right ) \sqrt {\arctan (a x)}}-2 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{a c^2} \] Input:
Integrate[1/((c + a^2*c*x^2)^2*ArcTan[a*x]^(3/2)),x]
Output:
(-2/((1 + a^2*x^2)*Sqrt[ArcTan[a*x]]) - 2*Sqrt[Pi]*FresnelS[(2*Sqrt[ArcTan [a*x]])/Sqrt[Pi]])/(a*c^2)
Time = 0.46 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {5437, 27, 5505, 4906, 27, 3042, 3786, 3832}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\arctan (a x)^{3/2} \left (a^2 c x^2+c\right )^2} \, dx\) |
\(\Big \downarrow \) 5437 |
\(\displaystyle -4 a \int \frac {x}{c^2 \left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}dx-\frac {2}{a c^2 \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {4 a \int \frac {x}{\left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}dx}{c^2}-\frac {2}{a c^2 \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}\) |
\(\Big \downarrow \) 5505 |
\(\displaystyle -\frac {4 \int \frac {a x}{\left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}d\arctan (a x)}{a c^2}-\frac {2}{a c^2 \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}\) |
\(\Big \downarrow \) 4906 |
\(\displaystyle -\frac {4 \int \frac {\sin (2 \arctan (a x))}{2 \sqrt {\arctan (a x)}}d\arctan (a x)}{a c^2}-\frac {2}{a c^2 \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {2 \int \frac {\sin (2 \arctan (a x))}{\sqrt {\arctan (a x)}}d\arctan (a x)}{a c^2}-\frac {2}{a c^2 \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 \int \frac {\sin (2 \arctan (a x))}{\sqrt {\arctan (a x)}}d\arctan (a x)}{a c^2}-\frac {2}{a c^2 \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}\) |
\(\Big \downarrow \) 3786 |
\(\displaystyle -\frac {4 \int \sin (2 \arctan (a x))d\sqrt {\arctan (a x)}}{a c^2}-\frac {2}{a c^2 \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}\) |
\(\Big \downarrow \) 3832 |
\(\displaystyle -\frac {2}{a c^2 \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}-\frac {2 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{a c^2}\) |
Input:
Int[1/((c + a^2*c*x^2)^2*ArcTan[a*x]^(3/2)),x]
Output:
-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)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[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]
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]
Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b _.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(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] && IG tQ[p, 0]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_)^2)^(q_), x_S ymbol] :> Simp[(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])^(p + 1)/(b*c*d*(p + 1))), x] - Simp[2*c*((q + 1)/(b*(p + 1))) Int[x*(d + e*x^2)^q*(a + b*Arc Tan[c*x])^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && LtQ[q, -1] && LtQ[p, -1]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 2)^(q_), x_Symbol] :> Simp[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])
Time = 0.16 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.82
method | result | size |
default | \(-\frac {2 \sqrt {\arctan \left (a x \right )}\, \sqrt {\pi }\, \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right )+\cos \left (2 \arctan \left (a x \right )\right )+1}{a \,c^{2} \sqrt {\arctan \left (a x \right )}}\) | \(47\) |
Input:
int(1/(a^2*c*x^2+c)^2/arctan(a*x)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/a/c^2*(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)
Exception generated. \[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/(a^2*c*x^2+c)^2/arctan(a*x)^(3/2),x, algorithm="fricas")
Output:
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{3/2}} \, dx=\frac {\int \frac {1}{a^{4} x^{4} \operatorname {atan}^{\frac {3}{2}}{\left (a x \right )} + 2 a^{2} x^{2} \operatorname {atan}^{\frac {3}{2}}{\left (a x \right )} + \operatorname {atan}^{\frac {3}{2}}{\left (a x \right )}}\, dx}{c^{2}} \] Input:
integrate(1/(a**2*c*x**2+c)**2/atan(a*x)**(3/2),x)
Output:
Integral(1/(a**4*x**4*atan(a*x)**(3/2) + 2*a**2*x**2*atan(a*x)**(3/2) + at an(a*x)**(3/2)), x)/c**2
Exception generated. \[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{3/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(1/(a^2*c*x^2+c)^2/arctan(a*x)^(3/2),x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati ve exponent.
\[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{3/2}} \, dx=\int { \frac {1}{{\left (a^{2} c x^{2} + c\right )}^{2} \arctan \left (a x\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(a^2*c*x^2+c)^2/arctan(a*x)^(3/2),x, algorithm="giac")
Output:
integrate(1/((a^2*c*x^2 + c)^2*arctan(a*x)^(3/2)), x)
Timed out. \[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{3/2}} \, dx=\int \frac {1}{{\mathrm {atan}\left (a\,x\right )}^{3/2}\,{\left (c\,a^2\,x^2+c\right )}^2} \,d x \] Input:
int(1/(atan(a*x)^(3/2)*(c + a^2*c*x^2)^2),x)
Output:
int(1/(atan(a*x)^(3/2)*(c + a^2*c*x^2)^2), x)
\[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{3/2}} \, dx=\frac {-\frac {16 \sqrt {\mathit {atan} \left (a x \right )}\, \mathit {atan} \left (a x \right )^{2} a^{2} x^{2}}{3}-\frac {16 \sqrt {\mathit {atan} \left (a x \right )}\, \mathit {atan} \left (a x \right )^{2}}{3}-8 \sqrt {\mathit {atan} \left (a x \right )}\, \mathit {atan} \left (a x \right ) a x +16 \mathit {atan} \left (a x \right ) \left (\int \frac {\sqrt {\mathit {atan} \left (a x \right )}}{a^{4} x^{4}+2 a^{2} x^{2}+1}d x \right ) a^{3} x^{2}+16 \mathit {atan} \left (a x \right ) \left (\int \frac {\sqrt {\mathit {atan} \left (a x \right )}}{a^{4} x^{4}+2 a^{2} x^{2}+1}d x \right ) a -2 \sqrt {\mathit {atan} \left (a x \right )}}{\mathit {atan} \left (a x \right ) a \,c^{2} \left (a^{2} x^{2}+1\right )} \] Input:
int(1/(a^2*c*x^2+c)^2/atan(a*x)^(3/2),x)
Output:
(2*( - 8*sqrt(atan(a*x))*atan(a*x)**2*a**2*x**2 - 8*sqrt(atan(a*x))*atan(a *x)**2 - 12*sqrt(atan(a*x))*atan(a*x)*a*x + 24*atan(a*x)*int(sqrt(atan(a*x ))/(a**4*x**4 + 2*a**2*x**2 + 1),x)*a**3*x**2 + 24*atan(a*x)*int(sqrt(atan (a*x))/(a**4*x**4 + 2*a**2*x**2 + 1),x)*a - 3*sqrt(atan(a*x))))/(3*atan(a* x)*a*c**2*(a**2*x**2 + 1))