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

Optimal. Leaf size=131 \[ \frac {\sqrt {\frac {\pi }{2}} \sqrt {a^2 x^2+1} C\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{2 a^3 c^2 \sqrt {a^2 c x^2+c}}-\frac {\sqrt {\frac {\pi }{6}} \sqrt {a^2 x^2+1} C\left (\sqrt {\frac {6}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{2 a^3 c^2 \sqrt {a^2 c x^2+c}} \]

[Out]

-1/12*FresnelC(6^(1/2)/Pi^(1/2)*arctan(a*x)^(1/2))*6^(1/2)*Pi^(1/2)*(a^2*x^2+1)^(1/2)/a^3/c^2/(a^2*c*x^2+c)^(1
/2)+1/4*FresnelC(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^3/c^2/(a^2*c*x^2+c)^
(1/2)

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Rubi [A]  time = 0.30, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {4971, 4970, 4406, 3304, 3352} \[ \frac {\sqrt {\frac {\pi }{2}} \sqrt {a^2 x^2+1} \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{2 a^3 c^2 \sqrt {a^2 c x^2+c}}-\frac {\sqrt {\frac {\pi }{6}} \sqrt {a^2 x^2+1} \text {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{2 a^3 c^2 \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Pi/2]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/(2*a^3*c^2*Sqrt[c + a^2*c*x^2]) - (Sqrt[
Pi/6]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[6/Pi]*Sqrt[ArcTan[a*x]]])/(2*a^3*c^2*Sqrt[c + a^2*c*x^2])

Rule 3304

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Cos[(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 3352

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

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} \sqrt {\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} \sqrt {\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)}{\sqrt {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 \sqrt {x}}-\frac {\cos (3 x)}{4 \sqrt {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)}{\sqrt {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)}{\sqrt {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 \cos \left (x^2\right ) \, dx,x,\sqrt {\tan ^{-1}(a x)}\right )}{2 a^3 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {1+a^2 x^2} \operatorname {Subst}\left (\int \cos \left (3 x^2\right ) \, dx,x,\sqrt {\tan ^{-1}(a x)}\right )}{2 a^3 c^2 \sqrt {c+a^2 c x^2}}\\ &=\frac {\sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} C\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{2 a^3 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {\frac {\pi }{6}} \sqrt {1+a^2 x^2} C\left (\sqrt {\frac {6}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{2 a^3 c^2 \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.20, size = 159, normalized size = 1.21 \[ -\frac {i \sqrt {a^2 x^2+1} \left (3 \sqrt {-i \tan ^{-1}(a x)} \Gamma \left (\frac {1}{2},-i \tan ^{-1}(a x)\right )-3 \sqrt {i \tan ^{-1}(a x)} \Gamma \left (\frac {1}{2},i \tan ^{-1}(a x)\right )+\sqrt {3} \left (\sqrt {i \tan ^{-1}(a x)} \Gamma \left (\frac {1}{2},3 i \tan ^{-1}(a x)\right )-\sqrt {-i \tan ^{-1}(a x)} \Gamma \left (\frac {1}{2},-3 i \tan ^{-1}(a x)\right )\right )\right )}{24 a^3 c^2 \sqrt {c \left (a^2 x^2+1\right )} \sqrt {\tan ^{-1}(a x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-1/24*I)*Sqrt[1 + a^2*x^2]*(3*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-I)*ArcTan[a*x]] - 3*Sqrt[I*ArcTan[a*x]]*Ga
mma[1/2, I*ArcTan[a*x]] + Sqrt[3]*(-(Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-3*I)*ArcTan[a*x]]) + Sqrt[I*ArcTan[a*
x]]*Gamma[1/2, (3*I)*ArcTan[a*x]])))/(a^3*c^2*Sqrt[c*(1 + a^2*x^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(x^2/(a^2*c*x^2+c)^(5/2)/arctan(a*x)^(1/2),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(x^2/(a^2*c*x^2+c)^(5/2)/arctan(a*x)^(1/2),x, algorithm="giac")

[Out]

sage0*x

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maple [F]  time = 9.92, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\left (a^{2} c \,x^{2}+c \right )^{\frac {5}{2}} \sqrt {\arctan \left (a x \right )}}\, dx \]

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)^(1/2),x)

[Out]

int(x^2/(a^2*c*x^2+c)^(5/2)/arctan(a*x)^(1/2),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(x^2/(a^2*c*x^2+c)^(5/2)/arctan(a*x)^(1/2),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 \[ \int \frac {x^2}{\sqrt {\mathrm {atan}\left (a\,x\right )}\,{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)**(1/2),x)

[Out]

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

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