Integrand size = 21, antiderivative size = 87 \[ \int \frac {1}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)} \, dx=\frac {3 \sqrt {1+a^2 x^2} \operatorname {CosIntegral}(\arctan (a x))}{4 a c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \operatorname {CosIntegral}(3 \arctan (a x))}{4 a c^2 \sqrt {c+a^2 c x^2}} \]
3/4*Ci(arctan(a*x))*(a^2*x^2+1)^(1/2)/a/c^2/(a^2*c*x^2+c)^(1/2)+1/4*Ci(3*a rctan(a*x))*(a^2*x^2+1)^(1/2)/a/c^2/(a^2*c*x^2+c)^(1/2)
Time = 0.10 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.61 \[ \int \frac {1}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)} \, dx=\frac {\left (1+a^2 x^2\right )^{3/2} (3 \operatorname {CosIntegral}(\arctan (a x))+\operatorname {CosIntegral}(3 \arctan (a x)))}{4 a c \left (c \left (1+a^2 x^2\right )\right )^{3/2}} \]
((1 + a^2*x^2)^(3/2)*(3*CosIntegral[ArcTan[a*x]] + CosIntegral[3*ArcTan[a* x]]))/(4*a*c*(c*(1 + a^2*x^2))^(3/2))
Time = 0.40 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.63, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {5440, 5439, 3042, 3793, 2009}
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.
\(\displaystyle \int \frac {1}{\arctan (a x) \left (a^2 c x^2+c\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 5440 |
\(\displaystyle \frac {\sqrt {a^2 x^2+1} \int \frac {1}{\left (a^2 x^2+1\right )^{5/2} \arctan (a x)}dx}{c^2 \sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 5439 |
\(\displaystyle \frac {\sqrt {a^2 x^2+1} \int \frac {1}{\left (a^2 x^2+1\right )^{3/2} \arctan (a x)}d\arctan (a x)}{a c^2 \sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {a^2 x^2+1} \int \frac {\sin \left (\arctan (a x)+\frac {\pi }{2}\right )^3}{\arctan (a x)}d\arctan (a x)}{a c^2 \sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 3793 |
\(\displaystyle \frac {\sqrt {a^2 x^2+1} \int \left (\frac {\cos (3 \arctan (a x))}{4 \arctan (a x)}+\frac {3}{4 \sqrt {a^2 x^2+1} \arctan (a x)}\right )d\arctan (a x)}{a c^2 \sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2 x^2+1} \left (\frac {3}{4} \operatorname {CosIntegral}(\arctan (a x))+\frac {1}{4} \operatorname {CosIntegral}(3 \arctan (a x))\right )}{a c^2 \sqrt {a^2 c x^2+c}}\) |
(Sqrt[1 + a^2*x^2]*((3*CosIntegral[ArcTan[a*x]])/4 + CosIntegral[3*ArcTan[ a*x]]/4))/(a*c^2*Sqrt[c + a^2*c*x^2])
3.6.19.3.1 Defintions of rubi rules used
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In t[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]))
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_), x_ Symbol] :> Simp[d^q/c Subst[Int[(a + b*x)^p/Cos[x]^(2*(q + 1)), x], x, Ar cTan[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])
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_), x_ Symbol] :> Simp[d^(q + 1/2)*(Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]) Int[(1 + c^2*x^2)^q*(a + b*ArcTan[c*x])^p, 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])
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 6.21 (sec) , antiderivative size = 179, normalized size of antiderivative = 2.06
method | result | size |
default | \(-\frac {i \operatorname {csgn}\left (\arctan \left (a x \right )\right ) \operatorname {csgn}\left (i \arctan \left (a x \right )\right ) \pi \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 \sqrt {a^{2} x^{2}+1}\, c^{3} a}+\frac {i \operatorname {csgn}\left (i \arctan \left (a x \right )\right ) \pi \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 \sqrt {a^{2} x^{2}+1}\, c^{3} a}+\frac {\operatorname {Ci}\left (3 \arctan \left (a x \right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{4 \sqrt {a^{2} x^{2}+1}\, c^{3} a}+\frac {3 \,\operatorname {Ci}\left (\arctan \left (a x \right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{4 \sqrt {a^{2} x^{2}+1}\, c^{3} a}\) | \(179\) |
-1/2*I*csgn(arctan(a*x))*csgn(I*arctan(a*x))*Pi*(c*(a*x-I)*(I+a*x))^(1/2)/ (a^2*x^2+1)^(1/2)/c^3/a+1/2*I*csgn(I*arctan(a*x))*Pi*(c*(a*x-I)*(I+a*x))^( 1/2)/(a^2*x^2+1)^(1/2)/c^3/a+1/4*Ci(3*arctan(a*x))*(c*(a*x-I)*(I+a*x))^(1/ 2)/(a^2*x^2+1)^(1/2)/c^3/a+3/4*Ci(arctan(a*x))*(c*(a*x-I)*(I+a*x))^(1/2)/( a^2*x^2+1)^(1/2)/c^3/a
\[ \int \frac {1}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)} \, dx=\int { \frac {1}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \arctan \left (a x\right )} \,d x } \]
integral(sqrt(a^2*c*x^2 + c)/((a^6*c^3*x^6 + 3*a^4*c^3*x^4 + 3*a^2*c^3*x^2 + c^3)*arctan(a*x)), x)
\[ \int \frac {1}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)} \, dx=\int \frac {1}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}} \operatorname {atan}{\left (a x \right )}}\, dx \]
\[ \int \frac {1}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)} \, dx=\int { \frac {1}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \arctan \left (a x\right )} \,d x } \]
\[ \int \frac {1}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)} \, dx=\int { \frac {1}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \arctan \left (a x\right )} \,d x } \]
Timed out. \[ \int \frac {1}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)} \, dx=\int \frac {1}{\mathrm {atan}\left (a\,x\right )\,{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]