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]

-5/16*arctan(a*x)^(3/2)/a/c^2+5/8*arctan(a*x)^(3/2)/a/c^2/(a^2*x^2+1)+1/2*x*arctan(a*x)^(5/2)/c^2/(a^2*x^2+1)+
1/7*arctan(a*x)^(7/2)/a/c^2+15/128*FresnelS(2*arctan(a*x)^(1/2)/Pi^(1/2))*Pi^(1/2)/a/c^2-15/32*x*arctan(a*x)^(
1/2)/c^2/(a^2*x^2+1)

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 151, normalized size of antiderivative = 1.00, 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 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]

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 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])

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 = 0.28, 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)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

________________________________________________________________________________________

maple [A]  time = 0.50, size = 102, normalized size = 0.68 \[ \frac {\arctan \left (a x \right )^{\frac {7}{2}}}{7 a \,c^{2}}+\frac {\arctan \left (a x \right )^{\frac {5}{2}} \sin \left (2 \arctan \left (a x \right )\right )}{4 a \,c^{2}}+\frac {5 \arctan \left (a x \right )^{\frac {3}{2}} \cos \left (2 \arctan \left (a x \right )\right )}{16 a \,c^{2}}-\frac {15 \sqrt {\arctan \left (a x \right )}\, \sin \left (2 \arctan \left (a x \right )\right )}{64 a \,c^{2}}+\frac {15 \,\mathrm {S}\left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right ) \sqrt {\pi }}{128 a \,c^{2}} \]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {atan}\left (a\,x\right )}^{5/2}}{{\left (c\,a^2\,x^2+c\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\operatorname {atan}^{\frac {5}{2}}{\left (a x \right )}}{a^{4} x^{4} + 2 a^{2} x^{2} + 1}\, dx}{c^{2}} \]

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

________________________________________________________________________________________