3.7.61 \(\int \frac {1}{(c+a^2 c x^2)^{3/2} \text {ArcTan}(a x)^3} \, dx\) [661]

Optimal. Leaf size=101 \[ -\frac {1}{2 a c \sqrt {c+a^2 c x^2} \text {ArcTan}(a x)^2}+\frac {x}{2 c \sqrt {c+a^2 c x^2} \text {ArcTan}(a x)}-\frac {\sqrt {1+a^2 x^2} \text {CosIntegral}(\text {ArcTan}(a x))}{2 a c \sqrt {c+a^2 c x^2}} \]

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {5022, 5062, 5025, 5024, 3383} \begin {gather*} -\frac {\sqrt {a^2 x^2+1} \text {CosIntegral}(\text {ArcTan}(a x))}{2 a c \sqrt {a^2 c x^2+c}}+\frac {x}{2 c \text {ArcTan}(a x) \sqrt {a^2 c x^2+c}}-\frac {1}{2 a c \text {ArcTan}(a x)^2 \sqrt {a^2 c x^2+c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((c + a^2*c*x^2)^(3/2)*ArcTan[a*x]^3),x]

[Out]

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

Rule 3383

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 5022

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 5024

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 5025

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 5062

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[
(f*x)^m*(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])^(p + 1)/(b*c*d*(p + 1))), x] - Dist[f*(m/(b*c*(p + 1))), Int[
(f*x)^(m - 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[e
, c^2*d] && EqQ[m + 2*q + 2, 0] && LtQ[p, -1]

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 65, normalized size = 0.64 \begin {gather*} \frac {-1+a x \text {ArcTan}(a x)-\sqrt {1+a^2 x^2} \text {ArcTan}(a x)^2 \text {CosIntegral}(\text {ArcTan}(a x))}{2 a c \sqrt {c+a^2 c x^2} \text {ArcTan}(a x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((c + a^2*c*x^2)^(3/2)*ArcTan[a*x]^3),x]

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.35, size = 292, normalized size = 2.89

method	result	size
default	\(\frac {\left (\arctan \left (a x \right )^{2} \expIntegral \left (1, i \arctan \left (a x \right )\right ) a^{2} x^{2}+\arctan \left (a x \right ) \sqrt {a^{2} x^{2}+1}\, a x +i \sqrt {a^{2} x^{2}+1}\, a x +\expIntegral \left (1, i \arctan \left (a x \right )\right ) \arctan \left (a x \right )^{2}+i \arctan \left (a x \right ) \sqrt {a^{2} x^{2}+1}-\sqrt {a^{2} x^{2}+1}\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{4 \left (a^{2} x^{2}+1\right )^{\frac {3}{2}} \arctan \left (a x \right )^{2} a \,c^{2}}+\frac {\left (\arctan \left (a x \right )^{2} \expIntegral \left (1, -i \arctan \left (a x \right )\right ) a^{2} x^{2}+\arctan \left (a x \right ) \sqrt {a^{2} x^{2}+1}\, a x +\expIntegral \left (1, -i \arctan \left (a x \right )\right ) \arctan \left (a x \right )^{2}-i \sqrt {a^{2} x^{2}+1}\, a x -i \arctan \left (a x \right ) \sqrt {a^{2} x^{2}+1}-\sqrt {a^{2} x^{2}+1}\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{4 \left (a^{2} x^{2}+1\right )^{\frac {3}{2}} \arctan \left (a x \right )^{2} a \,c^{2}}\)	\(292\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a^2*c*x^2+c)^(3/2)/arctan(a*x)^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a^2*c*x^2+c)^(3/2)/arctan(a*x)^3,x, algorithm="maxima")

[Out]

integrate(1/((a^2*c*x^2 + c)^(3/2)*arctan(a*x)^3), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a^2*c*x^2+c)^(3/2)/arctan(a*x)^3,x, algorithm="fricas")

[Out]

integral(sqrt(a^2*c*x^2 + c)/((a^4*c^2*x^4 + 2*a^2*c^2*x^2 + c^2)*arctan(a*x)^3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {3}{2}} \operatorname {atan}^{3}{\left (a x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a^2*c*x^2+c)^(3/2)/arctan(a*x)^3,x, algorithm="giac")

[Out]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\mathrm {atan}\left (a\,x\right )}^3\,{\left (c\,a^2\,x^2+c\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(atan(a*x)^3*(c + a^2*c*x^2)^(3/2)),x)

[Out]

int(1/(atan(a*x)^3*(c + a^2*c*x^2)^(3/2)), x)

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