3.1034 \(\int \frac{x^3}{(c+a^2 c x^2)^{5/2} \tan ^{-1}(a x)^{3/2}} \, dx\)

Optimal. Leaf size=160 \[ \frac{3 \sqrt{\frac{\pi }{2}} \sqrt{a^2 x^2+1} \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{a^4 c^2 \sqrt{a^2 c x^2+c}}-\frac{\sqrt{\frac{3 \pi }{2}} \sqrt{a^2 x^2+1} \text{FresnelC}\left (\sqrt{\frac{6}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{a^4 c^2 \sqrt{a^2 c x^2+c}}-\frac{2 x^3}{a c \left (a^2 c x^2+c\right )^{3/2} \sqrt{\tan ^{-1}(a x)}} \]

[Out]

(-2*x^3)/(a*c*(c + a^2*c*x^2)^(3/2)*Sqrt[ArcTan[a*x]]) + (3*Sqrt[Pi/2]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[2/Pi]*S
qrt[ArcTan[a*x]]])/(a^4*c^2*Sqrt[c + a^2*c*x^2]) - (Sqrt[(3*Pi)/2]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[6/Pi]*Sqrt[
ArcTan[a*x]]])/(a^4*c^2*Sqrt[c + a^2*c*x^2])

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Rubi [A]  time = 0.433501, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {4942, 4971, 4970, 4406, 3304, 3352} \[ \frac{3 \sqrt{\frac{\pi }{2}} \sqrt{a^2 x^2+1} \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{a^4 c^2 \sqrt{a^2 c x^2+c}}-\frac{\sqrt{\frac{3 \pi }{2}} \sqrt{a^2 x^2+1} \text{FresnelC}\left (\sqrt{\frac{6}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{a^4 c^2 \sqrt{a^2 c x^2+c}}-\frac{2 x^3}{a c \left (a^2 c x^2+c\right )^{3/2} \sqrt{\tan ^{-1}(a x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x^3)/(a*c*(c + a^2*c*x^2)^(3/2)*Sqrt[ArcTan[a*x]]) + (3*Sqrt[Pi/2]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[2/Pi]*S
qrt[ArcTan[a*x]]])/(a^4*c^2*Sqrt[c + a^2*c*x^2]) - (Sqrt[(3*Pi)/2]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[6/Pi]*Sqrt[
ArcTan[a*x]]])/(a^4*c^2*Sqrt[c + a^2*c*x^2])

Rule 4942

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]

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])

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 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 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]

Rubi steps

\begin{align*} \int \frac{x^3}{\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^{3/2}} \, dx &=-\frac{2 x^3}{a c \left (c+a^2 c x^2\right )^{3/2} \sqrt{\tan ^{-1}(a x)}}+\frac{6 \int \frac{x^2}{\left (c+a^2 c x^2\right )^{5/2} \sqrt{\tan ^{-1}(a x)}} \, dx}{a}\\ &=-\frac{2 x^3}{a c \left (c+a^2 c x^2\right )^{3/2} \sqrt{\tan ^{-1}(a x)}}+\frac{\left (6 \sqrt{1+a^2 x^2}\right ) \int \frac{x^2}{\left (1+a^2 x^2\right )^{5/2} \sqrt{\tan ^{-1}(a x)}} \, dx}{a c^2 \sqrt{c+a^2 c x^2}}\\ &=-\frac{2 x^3}{a c \left (c+a^2 c x^2\right )^{3/2} \sqrt{\tan ^{-1}(a x)}}+\frac{\left (6 \sqrt{1+a^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\cos (x) \sin ^2(x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c^2 \sqrt{c+a^2 c x^2}}\\ &=-\frac{2 x^3}{a c \left (c+a^2 c x^2\right )^{3/2} \sqrt{\tan ^{-1}(a x)}}+\frac{\left (6 \sqrt{1+a^2 x^2}\right ) \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^4 c^2 \sqrt{c+a^2 c x^2}}\\ &=-\frac{2 x^3}{a c \left (c+a^2 c x^2\right )^{3/2} \sqrt{\tan ^{-1}(a x)}}+\frac{\left (3 \sqrt{1+a^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\cos (x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{2 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{\cos (3 x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{2 a^4 c^2 \sqrt{c+a^2 c x^2}}\\ &=-\frac{2 x^3}{a c \left (c+a^2 c x^2\right )^{3/2} \sqrt{\tan ^{-1}(a x)}}+\frac{\left (3 \sqrt{1+a^2 x^2}\right ) \operatorname{Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt{\tan ^{-1}(a x)}\right )}{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 \cos \left (3 x^2\right ) \, dx,x,\sqrt{\tan ^{-1}(a x)}\right )}{a^4 c^2 \sqrt{c+a^2 c x^2}}\\ &=-\frac{2 x^3}{a c \left (c+a^2 c x^2\right )^{3/2} \sqrt{\tan ^{-1}(a x)}}+\frac{3 \sqrt{\frac{\pi }{2}} \sqrt{1+a^2 x^2} C\left (\sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{a^4 c^2 \sqrt{c+a^2 c x^2}}-\frac{\sqrt{\frac{3 \pi }{2}} \sqrt{1+a^2 x^2} C\left (\sqrt{\frac{6}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{a^4 c^2 \sqrt{c+a^2 c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.517108, size = 182, normalized size = 1.14 \[ \frac{-\frac{8 a^3 c x^3}{a^2 x^2+1}-i c \sqrt{a^2 x^2+1} \left (3 \sqrt{-i \tan ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},-i \tan ^{-1}(a x)\right )-3 \sqrt{i \tan ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},i \tan ^{-1}(a x)\right )+\sqrt{3} \left (\sqrt{i \tan ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},3 i \tan ^{-1}(a x)\right )-\sqrt{-i \tan ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},-3 i \tan ^{-1}(a x)\right )\right )\right )}{4 a^4 c^3 \sqrt{a^2 c x^2+c} \sqrt{\tan ^{-1}(a x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-8*a^3*c*x^3)/(1 + a^2*x^2) - I*c*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]]*Gamma[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]])))/(4*a^4*c^3*Sqrt[c + a^2*c*x^2]*Sqrt[ArcTan[a*x]]
)

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Maple [F]  time = 4.371, size = 0, normalized size = 0. \begin{align*} \int{{x}^{3} \left ({a}^{2}c{x}^{2}+c \right ) ^{-{\frac{5}{2}}} \left ( \arctan \left ( ax \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

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)^(3/2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

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)^(3/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{{\left (a^{2} c x^{2} + c\right )}^{\frac{5}{2}} \arctan \left (a x\right )^{\frac{3}{2}}}\,{d x} \end{align*}

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)^(3/2),x, algorithm="giac")

[Out]

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