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

Optimal. Leaf size=91 \[ \frac {x \sqrt {\tan ^{-1}(a x)}}{c \sqrt {a^2 c x^2+c}}-\frac {\sqrt {\frac {\pi }{2}} \sqrt {a^2 x^2+1} S\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{a c \sqrt {a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4905, 4904, 3296, 3305, 3351} \[ \frac {x \sqrt {\tan ^{-1}(a x)}}{c \sqrt {a^2 c x^2+c}}-\frac {\sqrt {\frac {\pi }{2}} \sqrt {a^2 x^2+1} S\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{a c \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3305

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Sin[(f*x^2)/d], x], x,
Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 4904

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 4905

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])

Rubi steps

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

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Mathematica [C]  time = 0.15, size = 94, normalized size = 1.03 \[ \frac {\left (a^2 x^2+1\right )^{3/2} \left (\sqrt {-i \tan ^{-1}(a x)} \Gamma \left (\frac {3}{2},-i \tan ^{-1}(a x)\right )+\sqrt {i \tan ^{-1}(a x)} \Gamma \left (\frac {3}{2},i \tan ^{-1}(a x)\right )\right )}{2 a \left (c \left (a^2 x^2+1\right )\right )^{3/2} \sqrt {\tan ^{-1}(a x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1 + a^2*x^2)^(3/2)*(Sqrt[(-I)*ArcTan[a*x]]*Gamma[3/2, (-I)*ArcTan[a*x]] + Sqrt[I*ArcTan[a*x]]*Gamma[3/2, I*A
rcTan[a*x]]))/(2*a*(c*(1 + a^2*x^2))^(3/2)*Sqrt[ArcTan[a*x]])

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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

[Out]

sage0*x

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maple [F]  time = 1.54, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\arctan \left (a x \right )}}{\left (a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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