3.1071 \(\int \frac{1}{(c+a^2 c x^2)^3 \tan ^{-1}(a x)^{5/2}} \, dx\)

Optimal. Leaf size=125 \[ \frac{16 x}{3 c^3 \left (a^2 x^2+1\right )^2 \sqrt{\tan ^{-1}(a x)}}-\frac{2}{3 a c^3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^{3/2}}-\frac{4 \sqrt{2 \pi } \text{FresnelC}\left (2 \sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{3 a c^3}-\frac{8 \sqrt{\pi } \text{FresnelC}\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{3 a c^3} \]

[Out]

-2/(3*a*c^3*(1 + a^2*x^2)^2*ArcTan[a*x]^(3/2)) + (16*x)/(3*c^3*(1 + a^2*x^2)^2*Sqrt[ArcTan[a*x]]) - (4*Sqrt[2*
Pi]*FresnelC[2*Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/(3*a*c^3) - (8*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]]
)/(3*a*c^3)

________________________________________________________________________________________

Rubi [A]  time = 0.296887, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {4902, 4968, 4970, 4406, 3304, 3352, 4904, 3312} \[ \frac{16 x}{3 c^3 \left (a^2 x^2+1\right )^2 \sqrt{\tan ^{-1}(a x)}}-\frac{2}{3 a c^3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^{3/2}}-\frac{4 \sqrt{2 \pi } \text{FresnelC}\left (2 \sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{3 a c^3}-\frac{8 \sqrt{\pi } \text{FresnelC}\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{3 a c^3} \]

Antiderivative was successfully verified.

[In]

Int[1/((c + a^2*c*x^2)^3*ArcTan[a*x]^(5/2)),x]

[Out]

-2/(3*a*c^3*(1 + a^2*x^2)^2*ArcTan[a*x]^(3/2)) + (16*x)/(3*c^3*(1 + a^2*x^2)^2*Sqrt[ArcTan[a*x]]) - (4*Sqrt[2*
Pi]*FresnelC[2*Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/(3*a*c^3) - (8*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]]
)/(3*a*c^3)

Rule 4902

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 4968

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

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 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])
)

Rubi steps

\begin{align*} \int \frac{1}{\left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x)^{5/2}} \, dx &=-\frac{2}{3 a c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^{3/2}}-\frac{1}{3} (8 a) \int \frac{x}{\left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x)^{3/2}} \, dx\\ &=-\frac{2}{3 a c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^{3/2}}+\frac{16 x}{3 c^3 \left (1+a^2 x^2\right )^2 \sqrt{\tan ^{-1}(a x)}}-\frac{16}{3} \int \frac{1}{\left (c+a^2 c x^2\right )^3 \sqrt{\tan ^{-1}(a x)}} \, dx+\left (16 a^2\right ) \int \frac{x^2}{\left (c+a^2 c x^2\right )^3 \sqrt{\tan ^{-1}(a x)}} \, dx\\ &=-\frac{2}{3 a c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^{3/2}}+\frac{16 x}{3 c^3 \left (1+a^2 x^2\right )^2 \sqrt{\tan ^{-1}(a x)}}-\frac{16 \operatorname{Subst}\left (\int \frac{\cos ^4(x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{3 a c^3}+\frac{16 \operatorname{Subst}\left (\int \frac{\cos ^2(x) \sin ^2(x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{a c^3}\\ &=-\frac{2}{3 a c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^{3/2}}+\frac{16 x}{3 c^3 \left (1+a^2 x^2\right )^2 \sqrt{\tan ^{-1}(a x)}}-\frac{16 \operatorname{Subst}\left (\int \left (\frac{3}{8 \sqrt{x}}+\frac{\cos (2 x)}{2 \sqrt{x}}+\frac{\cos (4 x)}{8 \sqrt{x}}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{3 a c^3}+\frac{16 \operatorname{Subst}\left (\int \left (\frac{1}{8 \sqrt{x}}-\frac{\cos (4 x)}{8 \sqrt{x}}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a c^3}\\ &=-\frac{2}{3 a c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^{3/2}}+\frac{16 x}{3 c^3 \left (1+a^2 x^2\right )^2 \sqrt{\tan ^{-1}(a x)}}-\frac{2 \operatorname{Subst}\left (\int \frac{\cos (4 x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{3 a c^3}-\frac{2 \operatorname{Subst}\left (\int \frac{\cos (4 x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{a c^3}-\frac{8 \operatorname{Subst}\left (\int \frac{\cos (2 x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{3 a c^3}\\ &=-\frac{2}{3 a c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^{3/2}}+\frac{16 x}{3 c^3 \left (1+a^2 x^2\right )^2 \sqrt{\tan ^{-1}(a x)}}-\frac{4 \operatorname{Subst}\left (\int \cos \left (4 x^2\right ) \, dx,x,\sqrt{\tan ^{-1}(a x)}\right )}{3 a c^3}-\frac{4 \operatorname{Subst}\left (\int \cos \left (4 x^2\right ) \, dx,x,\sqrt{\tan ^{-1}(a x)}\right )}{a c^3}-\frac{16 \operatorname{Subst}\left (\int \cos \left (2 x^2\right ) \, dx,x,\sqrt{\tan ^{-1}(a x)}\right )}{3 a c^3}\\ &=-\frac{2}{3 a c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^{3/2}}+\frac{16 x}{3 c^3 \left (1+a^2 x^2\right )^2 \sqrt{\tan ^{-1}(a x)}}-\frac{4 \sqrt{2 \pi } C\left (2 \sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{3 a c^3}-\frac{8 \sqrt{\pi } C\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{3 a c^3}\\ \end{align*}

Mathematica [C]  time = 0.697245, size = 186, normalized size = 1.49 \[ \frac{2 \left (\frac{\sqrt{2} \tan ^{-1}(a x)^2 \text{Gamma}\left (\frac{1}{2},2 i \tan ^{-1}(a x)\right )}{a \sqrt{i \tan ^{-1}(a x)}}+\frac{\tan ^{-1}(a x)^2 \text{Gamma}\left (\frac{1}{2},4 i \tan ^{-1}(a x)\right )}{a \sqrt{i \tan ^{-1}(a x)}}-\frac{\sqrt{2} \left (-i \tan ^{-1}(a x)\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-2 i \tan ^{-1}(a x)\right )}{a}-\frac{\left (-i \tan ^{-1}(a x)\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-4 i \tan ^{-1}(a x)\right )}{a}-\frac{1}{a \left (a^2 x^2+1\right )^2}+\frac{8 x \tan ^{-1}(a x)}{\left (a^2 x^2+1\right )^2}\right )}{3 c^3 \tan ^{-1}(a x)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((c + a^2*c*x^2)^3*ArcTan[a*x]^(5/2)),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.121, size = 113, normalized size = 0.9 \begin{align*}{\frac{1}{12\,a{c}^{3}} \left ( -16\,\sqrt{2}\sqrt{\pi }{\it FresnelC} \left ( 2\,{\frac{\sqrt{2}\sqrt{\arctan \left ( ax \right ) }}{\sqrt{\pi }}} \right ) \left ( \arctan \left ( ax \right ) \right ) ^{3/2}-32\,\sqrt{\pi }{\it FresnelC} \left ( 2\,{\frac{\sqrt{\arctan \left ( ax \right ) }}{\sqrt{\pi }}} \right ) \left ( \arctan \left ( ax \right ) \right ) ^{3/2}+16\,\sin \left ( 2\,\arctan \left ( ax \right ) \right ) \arctan \left ( ax \right ) +8\,\sin \left ( 4\,\arctan \left ( ax \right ) \right ) \arctan \left ( ax \right ) -4\,\cos \left ( 2\,\arctan \left ( ax \right ) \right ) -\cos \left ( 4\,\arctan \left ( ax \right ) \right ) -3 \right ) \left ( \arctan \left ( ax \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a^2*c*x^2+c)^3/arctan(a*x)^(5/2),x)

[Out]

1/12/a/c^3*(-16*2^(1/2)*Pi^(1/2)*FresnelC(2*2^(1/2)/Pi^(1/2)*arctan(a*x)^(1/2))*arctan(a*x)^(3/2)-32*Pi^(1/2)*
FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))*arctan(a*x)^(3/2)+16*sin(2*arctan(a*x))*arctan(a*x)+8*sin(4*arctan(a*x)
)*arctan(a*x)-4*cos(2*arctan(a*x))-cos(4*arctan(a*x))-3)/arctan(a*x)^(3/2)

________________________________________________________________________________________

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

[Out]

Exception raised: RuntimeError

________________________________________________________________________________________

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

[Out]

Exception raised: UnboundLocalError

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a^{2} c x^{2} + c\right )}^{3} \arctan \left (a x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a^2*c*x^2+c)^3/arctan(a*x)^(5/2),x, algorithm="giac")

[Out]

integrate(1/((a^2*c*x^2 + c)^3*arctan(a*x)^(5/2)), x)