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

Optimal. Leaf size=142 \[ -\frac {\sqrt {a^2 x^2+1} \text {Si}\left (\tan ^{-1}(a x)\right )}{4 a^3 c^2 \sqrt {a^2 c x^2+c}}+\frac {3 \sqrt {a^2 x^2+1} \text {Si}\left (3 \tan ^{-1}(a x)\right )}{4 a^3 c^2 \sqrt {a^2 c x^2+c}}-\frac {1}{a^3 c^2 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}+\frac {1}{a^3 c \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)} \]

[Out]

1/a^3/c/(a^2*c*x^2+c)^(3/2)/arctan(a*x)-1/a^3/c^2/arctan(a*x)/(a^2*c*x^2+c)^(1/2)-1/4*Si(arctan(a*x))*(a^2*x^2
+1)^(1/2)/a^3/c^2/(a^2*c*x^2+c)^(1/2)+3/4*Si(3*arctan(a*x))*(a^2*x^2+1)^(1/2)/a^3/c^2/(a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.58, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4964, 4902, 4971, 4970, 3299, 4406} \[ -\frac {\sqrt {a^2 x^2+1} \text {Si}\left (\tan ^{-1}(a x)\right )}{4 a^3 c^2 \sqrt {a^2 c x^2+c}}+\frac {3 \sqrt {a^2 x^2+1} \text {Si}\left (3 \tan ^{-1}(a x)\right )}{4 a^3 c^2 \sqrt {a^2 c x^2+c}}-\frac {1}{a^3 c^2 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}+\frac {1}{a^3 c \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[1/e, Int[
x^(m - 2)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p, x], x] - Dist[d/e, Int[x^(m - 2)*(d + e*x^2)^q*(a + b*Arc
Tan[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegersQ[p, 2*q] && LtQ[q, -1] && IGtQ[m
, 1] && NeQ[p, -1]

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 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])

Rubi steps

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

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Mathematica [A]  time = 0.29, size = 99, normalized size = 0.70 \[ -\frac {\left (a^2 x^2+1\right )^{3/2} \tan ^{-1}(a x) \text {Si}\left (\tan ^{-1}(a x)\right )-3 \left (a^2 x^2+1\right )^{3/2} \tan ^{-1}(a x) \text {Si}\left (3 \tan ^{-1}(a x)\right )+4 a^2 x^2}{4 a^3 c^2 \left (a^2 x^2+1\right ) \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a^{2} c x^{2} + c} x^{2}}{{\left (a^{6} c^{3} x^{6} + 3 \, a^{4} c^{3} x^{4} + 3 \, a^{2} c^{3} x^{2} + c^{3}\right )} \arctan \left (a x\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a^2*c*x^2 + c)*x^2/((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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [C]  time = 4.13, size = 586, normalized size = 4.13 \[ -\frac {i \left (3 \Ei \left (1, 3 i \arctan \left (a x \right )\right ) \arctan \left (a x \right ) x^{4} a^{4}-\sqrt {a^{2} x^{2}+1}\, x^{3} a^{3}+6 \Ei \left (1, 3 i \arctan \left (a x \right )\right ) \arctan \left (a x \right ) x^{2} a^{2}-3 i \sqrt {a^{2} x^{2}+1}\, x^{2} a^{2}+3 \sqrt {a^{2} x^{2}+1}\, x a +3 \Ei \left (1, 3 i \arctan \left (a x \right )\right ) \arctan \left (a x \right )+i \sqrt {a^{2} x^{2}+1}\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{8 \sqrt {a^{2} x^{2}+1}\, \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right ) \arctan \left (a x \right ) c^{3} a^{3}}+\frac {i \left (3 \Ei \left (1, -3 i \arctan \left (a x \right )\right ) \arctan \left (a x \right ) x^{4} a^{4}+6 \Ei \left (1, -3 i \arctan \left (a x \right )\right ) \arctan \left (a x \right ) x^{2} a^{2}-\sqrt {a^{2} x^{2}+1}\, x^{3} a^{3}+3 i \sqrt {a^{2} x^{2}+1}\, x^{2} a^{2}+3 \Ei \left (1, -3 i \arctan \left (a x \right )\right ) \arctan \left (a x \right )+3 \sqrt {a^{2} x^{2}+1}\, x a -i \sqrt {a^{2} x^{2}+1}\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{8 \sqrt {a^{2} x^{2}+1}\, \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right ) \arctan \left (a x \right ) c^{3} a^{3}}+\frac {i \left (\Ei \left (1, i \arctan \left (a x \right )\right ) \arctan \left (a x \right ) x^{2} a^{2}+\Ei \left (1, i \arctan \left (a x \right )\right ) \arctan \left (a x \right )+\sqrt {a^{2} x^{2}+1}\, x a +i \sqrt {a^{2} x^{2}+1}\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{8 \left (a^{2} x^{2}+1\right )^{\frac {3}{2}} \arctan \left (a x \right ) c^{3} a^{3}}-\frac {i \left (\arctan \left (a x \right ) \Ei \left (1, -i \arctan \left (a x \right )\right ) x^{2} a^{2}+\Ei \left (1, -i \arctan \left (a x \right )\right ) \arctan \left (a x \right )+\sqrt {a^{2} x^{2}+1}\, x a -i \sqrt {a^{2} x^{2}+1}\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{8 \left (a^{2} x^{2}+1\right )^{\frac {3}{2}} \arctan \left (a x \right ) c^{3} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*I*(3*Ei(1,3*I*arctan(a*x))*arctan(a*x)*x^4*a^4-(a^2*x^2+1)^(1/2)*x^3*a^3+6*Ei(1,3*I*arctan(a*x))*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)*(I+a*x))^(1/2)/(a^4*x^4+2*a^2*x^2+1)/arctan(a*x)/c^3/a^3+1/8*I*(3*Ei
(1,-3*I*arctan(a*x))*arctan(a*x)*x^4*a^4+6*Ei(1,-3*I*arctan(a*x))*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)*(I+a*x))^(1/2)/(a^4*x^4+2*a^2*x^2+1)/arctan(a*x)/c^3/a^3+1/8*I*(Ei(1,I*arctan
(a*x))*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)*(I+a*x))^(1/2)/arctan(a*x)/c^3/a^3-1/8*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)*(I+a*x))^(1
/2)/arctan(a*x)/c^3/a^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \arctan \left (a x\right )^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2}{{\mathrm {atan}\left (a\,x\right )}^2\,{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}} \operatorname {atan}^{2}{\left (a x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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