3.516 \(\int \frac {x^3}{(c+a^2 c x^2)^{5/2} \tan ^{-1}(a x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.28, 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, 3312, 3299} \[ \frac {3 \sqrt {a^2 x^2+1} \text {Si}\left (\tan ^{-1}(a x)\right )}{4 a^4 c^2 \sqrt {a^2 c x^2+c}}-\frac {\sqrt {a^2 x^2+1} \text {Si}\left (3 \tan ^{-1}(a x)\right )}{4 a^4 c^2 \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/((c + a^2*c*x^2)^(5/2)*ArcTan[a*x]),x]

[Out]

(3*Sqrt[1 + a^2*x^2]*SinIntegral[ArcTan[a*x]])/(4*a^4*c^2*Sqrt[c + a^2*c*x^2]) - (Sqrt[1 + a^2*x^2]*SinIntegra
l[3*ArcTan[a*x]])/(4*a^4*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 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 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^3}{\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^3}{\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 {\sin ^3(x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c^2 \sqrt {c+a^2 c x^2}}\\ &=\frac {\sqrt {1+a^2 x^2} \operatorname {Subst}\left (\int \left (\frac {3 \sin (x)}{4 x}-\frac {\sin (3 x)}{4 x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c^2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {1+a^2 x^2} \operatorname {Subst}\left (\int \frac {\sin (3 x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{4 a^4 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^4 c^2 \sqrt {c+a^2 c x^2}}\\ &=\frac {3 \sqrt {1+a^2 x^2} \text {Si}\left (\tan ^{-1}(a x)\right )}{4 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {1+a^2 x^2} \text {Si}\left (3 \tan ^{-1}(a x)\right )}{4 a^4 c^2 \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 52, normalized size = 0.60 \[ \frac {\left (a^2 x^2+1\right )^{5/2} \left (3 \text {Si}\left (\tan ^{-1}(a x)\right )-\text {Si}\left (3 \tan ^{-1}(a x)\right )\right )}{4 a^4 \left (c \left (a^2 x^2+1\right )\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/((c + a^2*c*x^2)^(5/2)*ArcTan[a*x]),x]

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a^2*c*x^2+c)^(5/2)/arctan(a*x),x, algorithm="fricas")

[Out]

integral(sqrt(a^2*c*x^2 + c)*x^3/((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(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a^2*c*x^2+c)^(5/2)/arctan(a*x),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [C]  time = 2.68, size = 125, normalized size = 1.44 \[ -\frac {\mathrm {csgn}\left (\arctan \left (a x \right )\right ) \pi \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{4 \sqrt {a^{2} x^{2}+1}\, c^{3} a^{4}}-\frac {\Si \left (3 \arctan \left (a x \right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{4 \sqrt {a^{2} x^{2}+1}\, c^{3} a^{4}}+\frac {3 \Si \left (\arctan \left (a x \right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{4 \sqrt {a^{2} x^{2}+1}\, c^{3} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a^2*c*x^2+c)^(5/2)/arctan(a*x),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \arctan \left (a x\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a^2*c*x^2+c)^(5/2)/arctan(a*x),x, algorithm="maxima")

[Out]

integrate(x^3/((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 \[ \int \frac {x^3}{\mathrm {atan}\left (a\,x\right )\,{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(atan(a*x)*(c + a^2*c*x^2)^(5/2)),x)

[Out]

int(x^3/(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 \[ \int \frac {x^{3}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}} \operatorname {atan}{\left (a x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(a**2*c*x**2+c)**(5/2)/atan(a*x),x)

[Out]

Integral(x**3/((c*(a**2*x**2 + 1))**(5/2)*atan(a*x)), x)

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