\(\int \frac {x^2 \arctan (a x)^{5/2}}{(c+a^2 c x^2)^2} \, dx\) [866]

Optimal result

Integrand size = 24, antiderivative size = 157 \[ \int \frac {x^2 \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {15 x \sqrt {\arctan (a x)}}{32 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {5 \arctan (a x)^{3/2}}{16 a^3 c^2}-\frac {5 \arctan (a x)^{3/2}}{8 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {x \arctan (a x)^{5/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^{7/2}}{7 a^3 c^2}-\frac {15 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{128 a^3 c^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5056, 5050, 5012, 5090, 4491, 12, 3386, 3432} \[ \int \frac {x^2 \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^2} \, dx=-\frac {15 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{128 a^3 c^2}+\frac {\arctan (a x)^{7/2}}{7 a^3 c^2}+\frac {5 \arctan (a x)^{3/2}}{16 a^3 c^2}-\frac {x \arctan (a x)^{5/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}+\frac {15 x \sqrt {\arctan (a x)}}{32 a^2 c^2 \left (a^2 x^2+1\right )}-\frac {5 \arctan (a x)^{3/2}}{8 a^3 c^2 \left (a^2 x^2+1\right )} \]

[In]

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

[Out]

(15*x*Sqrt[ArcTan[a*x]])/(32*a^2*c^2*(1 + a^2*x^2)) + (5*ArcTan[a*x]^(3/2))/(16*a^3*c^2) - (5*ArcTan[a*x]^(3/2
))/(8*a^3*c^2*(1 + a^2*x^2)) - (x*ArcTan[a*x]^(5/2))/(2*a^2*c^2*(1 + a^2*x^2)) + ArcTan[a*x]^(7/2)/(7*a^3*c^2)
- (15*Sqrt[Pi]*FresnelS[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(128*a^3*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 3386

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 3432

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

Rule 4491

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 5012

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 5050

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 5056

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^2)/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> Simp[(a + b*ArcTan
[c*x])^(p + 1)/(2*b*c^3*d^2*(p + 1)), x] + (Dist[b*(p/(2*c)), Int[x*((a + b*ArcTan[c*x])^(p - 1)/(d + e*x^2)^2
), x], x] - Simp[x*((a + b*ArcTan[c*x])^p/(2*c^2*d*(d + e*x^2))), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c
^2*d] && GtQ[p, 0]

Rule 5090

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

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.71 \[ \int \frac {x^2 \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {4 \sqrt {\arctan (a x)} \left (105 a x+70 \left (-1+a^2 x^2\right ) \arctan (a x)-112 a x \arctan (a x)^2+32 \left (1+a^2 x^2\right ) \arctan (a x)^3\right )-105 \sqrt {\pi } \left (1+a^2 x^2\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{896 a^3 c^2 \left (1+a^2 x^2\right )} \]

[In]

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

[Out]

(4*Sqrt[ArcTan[a*x]]*(105*a*x + 70*(-1 + a^2*x^2)*ArcTan[a*x] - 112*a*x*ArcTan[a*x]^2 + 32*(1 + a^2*x^2)*ArcTa
n[a*x]^3) - 105*Sqrt[Pi]*(1 + a^2*x^2)*FresnelS[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(896*a^3*c^2*(1 + a^2*x^2))

Maple [A] (verified)

Time = 25.51 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.59

method	result	size
default	\(\frac {128 \arctan \left (a x \right )^{\frac {7}{2}} \sqrt {\pi }-224 \arctan \left (a x \right )^{\frac {5}{2}} \sqrt {\pi }\, \sin \left (2 \arctan \left (a x \right )\right )-280 \arctan \left (a x \right )^{\frac {3}{2}} \sqrt {\pi }\, \cos \left (2 \arctan \left (a x \right )\right )+210 \sqrt {\arctan \left (a x \right )}\, \sqrt {\pi }\, \sin \left (2 \arctan \left (a x \right )\right )-105 \pi \,\operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right )}{896 c^{2} a^{3} \sqrt {\pi }}\)	\(93\)

[In]

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

[Out]

1/896/c^2/a^3*(128*arctan(a*x)^(7/2)*Pi^(1/2)-224*arctan(a*x)^(5/2)*Pi^(1/2)*sin(2*arctan(a*x))-280*arctan(a*x
)^(3/2)*Pi^(1/2)*cos(2*arctan(a*x))+210*arctan(a*x)^(1/2)*Pi^(1/2)*sin(2*arctan(a*x))-105*Pi*FresnelS(2*arctan
(a*x)^(1/2)/Pi^(1/2)))/Pi^(1/2)

Fricas [F(-2)]

Exception generated. \[ \int \frac {x^2 \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x^2*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)

Sympy [F]

\[ \int \frac {x^2 \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {\int \frac {x^{2} \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )}}{a^{4} x^{4} + 2 a^{2} x^{2} + 1}\, dx}{c^{2}} \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^2 \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [F]

\[ \int \frac {x^2 \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\int { \frac {x^{2} \arctan \left (a x\right )^{\frac {5}{2}}}{{\left (a^{2} c x^{2} + c\right )}^{2}} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\int \frac {x^2\,{\mathrm {atan}\left (a\,x\right )}^{5/2}}{{\left (c\,a^2\,x^2+c\right )}^2} \,d x \]

[In]

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

[Out]

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