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

Optimal. Leaf size=115 \[ -\frac{3 \sqrt{a^2 x^2+1} \text{Si}\left (\tan ^{-1}(a x)\right )}{4 a c^2 \sqrt{a^2 c x^2+c}}-\frac{3 \sqrt{a^2 x^2+1} \text{Si}\left (3 \tan ^{-1}(a x)\right )}{4 a c^2 \sqrt{a^2 c x^2+c}}-\frac{1}{a c \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)} \]

[Out]

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

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Rubi [A]  time = 0.244665, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {4902, 4971, 4970, 4406, 3299} \[ -\frac{3 \sqrt{a^2 x^2+1} \text{Si}\left (\tan ^{-1}(a x)\right )}{4 a c^2 \sqrt{a^2 c x^2+c}}-\frac{3 \sqrt{a^2 x^2+1} \text{Si}\left (3 \tan ^{-1}(a x)\right )}{4 a c^2 \sqrt{a^2 c x^2+c}}-\frac{1}{a c \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4902

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 4971

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[(d^(q +
1/2)*Sqrt[1 + c^2*x^2])/Sqrt[d + e*x^2], Int[x^m*(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] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] &&  !(IntegerQ[q] || GtQ[d, 0])

Rule 4970

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])

Rule 4406

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 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rubi steps

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

Mathematica [A]  time = 0.186117, size = 61, normalized size = 0.53 \[ \frac{-3 \left (a^2 x^2+1\right )^{3/2} \left (\text{Si}\left (\tan ^{-1}(a x)\right )+\text{Si}\left (3 \tan ^{-1}(a x)\right )\right )-\frac{4}{\tan ^{-1}(a x)}}{4 a c \left (a^2 c x^2+c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.341, size = 586, normalized size = 5.1 \begin{align*} \text{result too large to display} \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)^2,x)

[Out]

1/8*I*(3*arctan(a*x)*Ei(1,3*I*arctan(a*x))*x^4*a^4-(a^2*x^2+1)^(1/2)*x^3*a^3+6*arctan(a*x)*Ei(1,3*I*arctan(a*x
))*x^2*a^2-3*I*(a^2*x^2+1)^(1/2)*x^2*a^2+3*(a^2*x^2+1)^(1/2)*x*a+3*Ei(1,3*I*arctan(a*x))*arctan(a*x)+I*(a^2*x^
2+1)^(1/2))/(a^2*x^2+1)^(1/2)*(c*(a*x-I)*(a*x+I))^(1/2)/(a^4*x^4+2*a^2*x^2+1)/arctan(a*x)/a/c^3-1/8*I*(3*arcta
n(a*x)*Ei(1,-3*I*arctan(a*x))*x^4*a^4+6*arctan(a*x)*Ei(1,-3*I*arctan(a*x))*x^2*a^2-(a^2*x^2+1)^(1/2)*x^3*a^3+3
*I*(a^2*x^2+1)^(1/2)*x^2*a^2+3*Ei(1,-3*I*arctan(a*x))*arctan(a*x)+3*(a^2*x^2+1)^(1/2)*x*a-I*(a^2*x^2+1)^(1/2))
/(a^2*x^2+1)^(1/2)*(c*(a*x-I)*(a*x+I))^(1/2)/(a^4*x^4+2*a^2*x^2+1)/arctan(a*x)/a/c^3+3/8*I*(arctan(a*x)*Ei(1,I
*arctan(a*x))*x^2*a^2+Ei(1,I*arctan(a*x))*arctan(a*x)+(a^2*x^2+1)^(1/2)*x*a+I*(a^2*x^2+1)^(1/2))/(a^2*x^2+1)^(
3/2)*(c*(a*x-I)*(a*x+I))^(1/2)/arctan(a*x)/a/c^3-3/8*I*(arctan(a*x)*Ei(1,-I*arctan(a*x))*x^2*a^2+Ei(1,-I*arcta
n(a*x))*arctan(a*x)+(a^2*x^2+1)^(1/2)*x*a-I*(a^2*x^2+1)^(1/2))/(a^2*x^2+1)^(3/2)*(c*(a*x-I)*(a*x+I))^(1/2)/arc
tan(a*x)/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 )^{2}}\,{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)^2,x, algorithm="maxima")

[Out]

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

[Out]

Integral(1/((c*(a**2*x**2 + 1))**(5/2)*atan(a*x)**2), 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 )^{2}}\,{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)^2,x, algorithm="giac")

[Out]

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