\(\int \frac {1}{(c+a^2 c x^2)^{5/2} \arctan (a x)} \, dx\) [519]

Optimal result

Integrand size = 21, antiderivative size = 87 \[ \int \frac {1}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)} \, dx=\frac {3 \sqrt {1+a^2 x^2} \operatorname {CosIntegral}(\arctan (a x))}{4 a c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \operatorname {CosIntegral}(3 \arctan (a x))}{4 a c^2 \sqrt {c+a^2 c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {5025, 5024, 3393, 3383} \[ \int \frac {1}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)} \, dx=\frac {3 \sqrt {a^2 x^2+1} \operatorname {CosIntegral}(\arctan (a x))}{4 a c^2 \sqrt {a^2 c x^2+c}}+\frac {\sqrt {a^2 x^2+1} \operatorname {CosIntegral}(3 \arctan (a x))}{4 a c^2 \sqrt {a^2 c x^2+c}} \]

[In]

Int[1/((c + a^2*c*x^2)^(5/2)*ArcTan[a*x]),x]

[Out]

(3*Sqrt[1 + a^2*x^2]*CosIntegral[ArcTan[a*x]])/(4*a*c^2*Sqrt[c + a^2*c*x^2]) + (Sqrt[1 + a^2*x^2]*CosIntegral[
3*ArcTan[a*x]])/(4*a*c^2*Sqrt[c + a^2*c*x^2])

Rule 3383

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - 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 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 5025

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[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])

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.61 \[ \int \frac {1}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)} \, dx=\frac {\left (1+a^2 x^2\right )^{3/2} (3 \operatorname {CosIntegral}(\arctan (a x))+\operatorname {CosIntegral}(3 \arctan (a x)))}{4 a c \left (c \left (1+a^2 x^2\right )\right )^{3/2}} \]

[In]

Integrate[1/((c + a^2*c*x^2)^(5/2)*ArcTan[a*x]),x]

[Out]

((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)
)

Maple [C] (warning: unable to verify)

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	\(-\frac {i \operatorname {csgn}\left (\arctan \left (a x \right )\right ) \operatorname {csgn}\left (i \arctan \left (a x \right )\right ) \pi \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 \sqrt {a^{2} x^{2}+1}\, c^{3} a}+\frac {i \operatorname {csgn}\left (i \arctan \left (a x \right )\right ) \pi \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 \sqrt {a^{2} x^{2}+1}\, c^{3} a}+\frac {\operatorname {Ci}\left (3 \arctan \left (a x \right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{4 \sqrt {a^{2} x^{2}+1}\, c^{3} a}+\frac {3 \,\operatorname {Ci}\left (\arctan \left (a x \right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{4 \sqrt {a^{2} x^{2}+1}\, c^{3} a}\)	\(179\)

[In]

int(1/(a^2*c*x^2+c)^(5/2)/arctan(a*x),x,method=_RETURNVERBOSE)

[Out]

-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

Fricas [F]

\[ \int \frac {1}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)} \, dx=\int { \frac {1}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \arctan \left (a x\right )} \,d x } \]

[In]

integrate(1/(a^2*c*x^2+c)^(5/2)/arctan(a*x),x, algorithm="fricas")

[Out]

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)

Sympy [F]

\[ \int \frac {1}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)} \, dx=\int \frac {1}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}} \operatorname {atan}{\left (a x \right )}}\, dx \]

[In]

integrate(1/(a**2*c*x**2+c)**(5/2)/atan(a*x),x)

[Out]

Integral(1/((c*(a**2*x**2 + 1))**(5/2)*atan(a*x)), x)

Maxima [F]

\[ \int \frac {1}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)} \, dx=\int { \frac {1}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \arctan \left (a x\right )} \,d x } \]

[In]

integrate(1/(a^2*c*x^2+c)^(5/2)/arctan(a*x),x, algorithm="maxima")

[Out]

integrate(1/((a^2*c*x^2 + c)^(5/2)*arctan(a*x)), x)

Giac [F]

\[ \int \frac {1}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)} \, dx=\int { \frac {1}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \arctan \left (a x\right )} \,d x } \]

[In]

integrate(1/(a^2*c*x^2+c)^(5/2)/arctan(a*x),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)} \, dx=\int \frac {1}{\mathrm {atan}\left (a\,x\right )\,{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]

[In]

int(1/(atan(a*x)*(c + a^2*c*x^2)^(5/2)),x)

[Out]

int(1/(atan(a*x)*(c + a^2*c*x^2)^(5/2)), x)