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

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

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4971, 4970, 4406, 3302} \[ \frac {\sqrt {a^2 x^2+1} \text {CosIntegral}\left (\tan ^{-1}(a x)\right )}{4 a^3 c^2 \sqrt {a^2 c x^2+c}}-\frac {\sqrt {a^2 x^2+1} \text {CosIntegral}\left (3 \tan ^{-1}(a x)\right )}{4 a^3 c^2 \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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

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Mathematica [A]  time = 0.12, size = 53, normalized size = 0.61 \[ \frac {\sqrt {c \left (a^2 x^2+1\right )} \left (\text {Ci}\left (\tan ^{-1}(a x)\right )-\text {Ci}\left (3 \tan ^{-1}(a x)\right )\right )}{4 a^3 c^3 \sqrt {a^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a^{2} c x^{2} + c} x^{2}}{{\left (a^{6} c^{3} x^{6} + 3 \, a^{4} c^{3} x^{4} + 3 \, a^{2} c^{3} x^{2} + c^{3}\right )} \arctan \left (a x\right )}, x\right ) \]

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),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)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

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),x, algorithm="giac")

[Out]

sage0*x

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maple [C]  time = 3.59, size = 84, normalized size = 0.97 \[ -\frac {\Ci \left (3 \arctan \left (a x \right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{4 \sqrt {a^{2} x^{2}+1}\, c^{3} a^{3}}+\frac {\Ci \left (\arctan \left (a x \right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{4 \sqrt {a^{2} x^{2}+1}\, c^{3} a^{3}} \]

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),x)

[Out]

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

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

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),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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),x)

[Out]

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

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