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

Optimal. Leaf size=151 \[ \frac{x \tan ^{-1}(a x)^{5/2}}{2 c^2 \left (a^2 x^2+1\right )}+\frac{5 \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (a^2 x^2+1\right )}-\frac{15 x \sqrt{\tan ^{-1}(a x)}}{32 c^2 \left (a^2 x^2+1\right )}+\frac{15 \sqrt{\pi } S\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{128 a c^2}+\frac{\tan ^{-1}(a x)^{7/2}}{7 a c^2}-\frac{5 \tan ^{-1}(a x)^{3/2}}{16 a c^2} \]

[Out]

(-15*x*Sqrt[ArcTan[a*x]])/(32*c^2*(1 + a^2*x^2)) - (5*ArcTan[a*x]^(3/2))/(16*a*c^2) + (5*ArcTan[a*x]^(3/2))/(8
*a*c^2*(1 + a^2*x^2)) + (x*ArcTan[a*x]^(5/2))/(2*c^2*(1 + a^2*x^2)) + ArcTan[a*x]^(7/2)/(7*a*c^2) + (15*Sqrt[P
i]*FresnelS[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(128*a*c^2)

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Rubi [A]  time = 0.183928, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4892, 4930, 4970, 4406, 12, 3305, 3351} \[ \frac{x \tan ^{-1}(a x)^{5/2}}{2 c^2 \left (a^2 x^2+1\right )}+\frac{5 \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (a^2 x^2+1\right )}-\frac{15 x \sqrt{\tan ^{-1}(a x)}}{32 c^2 \left (a^2 x^2+1\right )}+\frac{15 \sqrt{\pi } S\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{128 a c^2}+\frac{\tan ^{-1}(a x)^{7/2}}{7 a c^2}-\frac{5 \tan ^{-1}(a x)^{3/2}}{16 a c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*x*Sqrt[ArcTan[a*x]])/(32*c^2*(1 + a^2*x^2)) - (5*ArcTan[a*x]^(3/2))/(16*a*c^2) + (5*ArcTan[a*x]^(3/2))/(8
*a*c^2*(1 + a^2*x^2)) + (x*ArcTan[a*x]^(5/2))/(2*c^2*(1 + a^2*x^2)) + ArcTan[a*x]^(7/2)/(7*a*c^2) + (15*Sqrt[P
i]*FresnelS[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(128*a*c^2)

Rule 4892

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

Rule 4930

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

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

Rubi steps

\begin{align*} \int \frac{\tan ^{-1}(a x)^{5/2}}{\left (c+a^2 c x^2\right )^2} \, dx &=\frac{x \tan ^{-1}(a x)^{5/2}}{2 c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^{7/2}}{7 a c^2}-\frac{1}{4} (5 a) \int \frac{x \tan ^{-1}(a x)^{3/2}}{\left (c+a^2 c x^2\right )^2} \, dx\\ &=\frac{5 \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (1+a^2 x^2\right )}+\frac{x \tan ^{-1}(a x)^{5/2}}{2 c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^{7/2}}{7 a c^2}-\frac{15}{16} \int \frac{\sqrt{\tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^2} \, dx\\ &=-\frac{15 x \sqrt{\tan ^{-1}(a x)}}{32 c^2 \left (1+a^2 x^2\right )}-\frac{5 \tan ^{-1}(a x)^{3/2}}{16 a c^2}+\frac{5 \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (1+a^2 x^2\right )}+\frac{x \tan ^{-1}(a x)^{5/2}}{2 c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^{7/2}}{7 a c^2}+\frac{1}{64} (15 a) \int \frac{x}{\left (c+a^2 c x^2\right )^2 \sqrt{\tan ^{-1}(a x)}} \, dx\\ &=-\frac{15 x \sqrt{\tan ^{-1}(a x)}}{32 c^2 \left (1+a^2 x^2\right )}-\frac{5 \tan ^{-1}(a x)^{3/2}}{16 a c^2}+\frac{5 \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (1+a^2 x^2\right )}+\frac{x \tan ^{-1}(a x)^{5/2}}{2 c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^{7/2}}{7 a c^2}+\frac{15 \operatorname{Subst}\left (\int \frac{\cos (x) \sin (x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{64 a c^2}\\ &=-\frac{15 x \sqrt{\tan ^{-1}(a x)}}{32 c^2 \left (1+a^2 x^2\right )}-\frac{5 \tan ^{-1}(a x)^{3/2}}{16 a c^2}+\frac{5 \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (1+a^2 x^2\right )}+\frac{x \tan ^{-1}(a x)^{5/2}}{2 c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^{7/2}}{7 a c^2}+\frac{15 \operatorname{Subst}\left (\int \frac{\sin (2 x)}{2 \sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{64 a c^2}\\ &=-\frac{15 x \sqrt{\tan ^{-1}(a x)}}{32 c^2 \left (1+a^2 x^2\right )}-\frac{5 \tan ^{-1}(a x)^{3/2}}{16 a c^2}+\frac{5 \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (1+a^2 x^2\right )}+\frac{x \tan ^{-1}(a x)^{5/2}}{2 c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^{7/2}}{7 a c^2}+\frac{15 \operatorname{Subst}\left (\int \frac{\sin (2 x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{128 a c^2}\\ &=-\frac{15 x \sqrt{\tan ^{-1}(a x)}}{32 c^2 \left (1+a^2 x^2\right )}-\frac{5 \tan ^{-1}(a x)^{3/2}}{16 a c^2}+\frac{5 \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (1+a^2 x^2\right )}+\frac{x \tan ^{-1}(a x)^{5/2}}{2 c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^{7/2}}{7 a c^2}+\frac{15 \operatorname{Subst}\left (\int \sin \left (2 x^2\right ) \, dx,x,\sqrt{\tan ^{-1}(a x)}\right )}{64 a c^2}\\ &=-\frac{15 x \sqrt{\tan ^{-1}(a x)}}{32 c^2 \left (1+a^2 x^2\right )}-\frac{5 \tan ^{-1}(a x)^{3/2}}{16 a c^2}+\frac{5 \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (1+a^2 x^2\right )}+\frac{x \tan ^{-1}(a x)^{5/2}}{2 c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^{7/2}}{7 a c^2}+\frac{15 \sqrt{\pi } S\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{128 a c^2}\\ \end{align*}

Mathematica [A]  time = 0.266976, size = 85, normalized size = 0.56 \[ \frac{105 \sqrt{\pi } S\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )+2 \sqrt{\tan ^{-1}(a x)} \left (64 \tan ^{-1}(a x)^3+7 \left (16 \tan ^{-1}(a x)^2-15\right ) \sin \left (2 \tan ^{-1}(a x)\right )+140 \tan ^{-1}(a x) \cos \left (2 \tan ^{-1}(a x)\right )\right )}{896 a c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(105*Sqrt[Pi]*FresnelS[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]] + 2*Sqrt[ArcTan[a*x]]*(64*ArcTan[a*x]^3 + 140*ArcTan[a*
x]*Cos[2*ArcTan[a*x]] + 7*(-15 + 16*ArcTan[a*x]^2)*Sin[2*ArcTan[a*x]]))/(896*a*c^2)

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Maple [A]  time = 0.109, size = 102, normalized size = 0.7 \begin{align*}{\frac{1}{7\,a{c}^{2}} \left ( \arctan \left ( ax \right ) \right ) ^{{\frac{7}{2}}}}+{\frac{\sin \left ( 2\,\arctan \left ( ax \right ) \right ) }{4\,a{c}^{2}} \left ( \arctan \left ( ax \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{5\,\cos \left ( 2\,\arctan \left ( ax \right ) \right ) }{16\,a{c}^{2}} \left ( \arctan \left ( ax \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{15\,\sin \left ( 2\,\arctan \left ( ax \right ) \right ) }{64\,a{c}^{2}}\sqrt{\arctan \left ( ax \right ) }}+{\frac{15\,\sqrt{\pi }}{128\,a{c}^{2}}{\it FresnelS} \left ( 2\,{\frac{\sqrt{\arctan \left ( ax \right ) }}{\sqrt{\pi }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*arctan(a*x)^(7/2)/a/c^2+1/4/a/c^2*arctan(a*x)^(5/2)*sin(2*arctan(a*x))+5/16/a/c^2*arctan(a*x)^(3/2)*cos(2*
arctan(a*x))-15/64/a/c^2*arctan(a*x)^(1/2)*sin(2*arctan(a*x))+15/128*FresnelS(2*arctan(a*x)^(1/2)/Pi^(1/2))*Pi
^(1/2)/a/c^2

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

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\operatorname{atan}^{\frac{5}{2}}{\left (a x \right )}}{a^{4} x^{4} + 2 a^{2} x^{2} + 1}\, dx}{c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atan(a*x)**(5/2)/(a**4*x**4 + 2*a**2*x**2 + 1), x)/c**2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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