3.519 \(\int \frac{1}{(c+a^2 c x^2)^{5/2} \tan ^{-1}(a x)} \, dx\)

Optimal. Leaf size=87 \[ \frac{3 \sqrt{a^2 x^2+1} \text{CosIntegral}\left (\tan ^{-1}(a x)\right )}{4 a c^2 \sqrt{a^2 c x^2+c}}+\frac{\sqrt{a^2 x^2+1} \text{CosIntegral}\left (3 \tan ^{-1}(a x)\right )}{4 a c^2 \sqrt{a^2 c x^2+c}} \]

[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])

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Rubi [A]  time = 0.132726, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {4905, 4904, 3312, 3302} \[ \frac{3 \sqrt{a^2 x^2+1} \text{CosIntegral}\left (\tan ^{-1}(a x)\right )}{4 a c^2 \sqrt{a^2 c x^2+c}}+\frac{\sqrt{a^2 x^2+1} \text{CosIntegral}\left (3 \tan ^{-1}(a x)\right )}{4 a c^2 \sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[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 4905

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])

Rule 4904

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 3312

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 3302

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]

Rubi steps

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

Mathematica [A]  time = 0.0431014, size = 50, normalized size = 0.57 \[ \frac{\left (a^2 x^2+1\right )^{5/2} \left (3 \text{CosIntegral}\left (\tan ^{-1}(a x)\right )+\text{CosIntegral}\left (3 \tan ^{-1}(a x)\right )\right )}{4 a \left (c \left (a^2 x^2+1\right )\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + a^2*x^2)^(5/2)*(3*CosIntegral[ArcTan[a*x]] + CosIntegral[3*ArcTan[a*x]]))/(4*a*(c*(1 + a^2*x^2))^(5/2))

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Maple [C]  time = 0.283, size = 179, normalized size = 2.1 \begin{align*}{\frac{-{\frac{i}{2}}{\it csgn} \left ( \arctan \left ( ax \right ) \right ){\it csgn} \left ( i\arctan \left ( ax \right ) \right ) \pi }{a{c}^{3}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}+{\frac{{\frac{i}{2}}{\it csgn} \left ( i\arctan \left ( ax \right ) \right ) \pi }{a{c}^{3}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}+{\frac{{\it Ci} \left ( 3\,\arctan \left ( ax \right ) \right ) }{4\,a{c}^{3}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}+{\frac{3\,{\it Ci} \left ( \arctan \left ( ax \right ) \right ) }{4\,a{c}^{3}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*csgn(arctan(a*x))*csgn(I*arctan(a*x))*Pi/(a^2*x^2+1)^(1/2)*(c*(a*x-I)*(a*x+I))^(1/2)/a/c^3+1/2*I*csgn(I
*arctan(a*x))*Pi/(a^2*x^2+1)^(1/2)*(c*(a*x-I)*(a*x+I))^(1/2)/a/c^3+1/4*Ci(3*arctan(a*x))/(a^2*x^2+1)^(1/2)*(c*
(a*x-I)*(a*x+I))^(1/2)/a/c^3+3/4*Ci(arctan(a*x))/(a^2*x^2+1)^(1/2)*(c*(a*x-I)*(a*x+I))^(1/2)/a/c^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a^{2} c x^{2} + c\right )}^{\frac{5}{2}} \arctan \left (a x\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{a^{2} c x^{2} + c}}{{\left (a^{6} c^{3} x^{6} + 3 \, a^{4} c^{3} x^{4} + 3 \, a^{2} c^{3} x^{2} + c^{3}\right )} \arctan \left (a x\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac{5}{2}} \operatorname{atan}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a^{2} c x^{2} + c\right )}^{\frac{5}{2}} \arctan \left (a x\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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