\(\int \frac {x \arctan (a x)^{5/2}}{(c+a^2 c x^2)^{3/2}} \, dx\) [904]

Optimal result

Integrand size = 24, antiderivative size = 161 \[ \int \frac {x \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {15 \sqrt {\arctan (a x)}}{4 a^2 c \sqrt {c+a^2 c x^2}}+\frac {5 x \arctan (a x)^{3/2}}{2 a c \sqrt {c+a^2 c x^2}}-\frac {\arctan (a x)^{5/2}}{a^2 c \sqrt {c+a^2 c x^2}}-\frac {15 \sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{4 a^2 c \sqrt {c+a^2 c x^2}} \]

[Out]

5/2*x*arctan(a*x)^(3/2)/a/c/(a^2*c*x^2+c)^(1/2)-arctan(a*x)^(5/2)/a^2/c/(a^2*c*x^2+c)^(1/2)-15/8*FresnelC(2^(1
/2)/Pi^(1/2)*arctan(a*x)^(1/2))*2^(1/2)*Pi^(1/2)*(a^2*x^2+1)^(1/2)/a^2/c/(a^2*c*x^2+c)^(1/2)+15/4*arctan(a*x)^
(1/2)/a^2/c/(a^2*c*x^2+c)^(1/2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5050, 5018, 5025, 5024, 3385, 3433} \[ \int \frac {x \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=-\frac {15 \sqrt {\frac {\pi }{2}} \sqrt {a^2 x^2+1} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{4 a^2 c \sqrt {a^2 c x^2+c}}-\frac {\arctan (a x)^{5/2}}{a^2 c \sqrt {a^2 c x^2+c}}+\frac {5 x \arctan (a x)^{3/2}}{2 a c \sqrt {a^2 c x^2+c}}+\frac {15 \sqrt {\arctan (a x)}}{4 a^2 c \sqrt {a^2 c x^2+c}} \]

[In]

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

[Out]

(15*Sqrt[ArcTan[a*x]])/(4*a^2*c*Sqrt[c + a^2*c*x^2]) + (5*x*ArcTan[a*x]^(3/2))/(2*a*c*Sqrt[c + a^2*c*x^2]) - A
rcTan[a*x]^(5/2)/(a^2*c*Sqrt[c + a^2*c*x^2]) - (15*Sqrt[Pi/2]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[2/Pi]*Sqrt[ArcTa
n[a*x]]])/(4*a^2*c*Sqrt[c + a^2*c*x^2])

Rule 3385

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 3433

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

Rule 5018

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

Rule 5024

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 5025

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[d^(q + 1/2)*(Sqrt[1
+ c^2*x^2]/Sqrt[d + e*x^2]), Int[(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] && ILtQ[2*(q + 1), 0] &&  !(IntegerQ[q] || GtQ[d, 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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\arctan (a x)^{5/2}}{a^2 c \sqrt {c+a^2 c x^2}}+\frac {5 \int \frac {\arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{2 a} \\ & = \frac {15 \sqrt {\arctan (a x)}}{4 a^2 c \sqrt {c+a^2 c x^2}}+\frac {5 x \arctan (a x)^{3/2}}{2 a c \sqrt {c+a^2 c x^2}}-\frac {\arctan (a x)^{5/2}}{a^2 c \sqrt {c+a^2 c x^2}}-\frac {15 \int \frac {1}{\left (c+a^2 c x^2\right )^{3/2} \sqrt {\arctan (a x)}} \, dx}{8 a} \\ & = \frac {15 \sqrt {\arctan (a x)}}{4 a^2 c \sqrt {c+a^2 c x^2}}+\frac {5 x \arctan (a x)^{3/2}}{2 a c \sqrt {c+a^2 c x^2}}-\frac {\arctan (a x)^{5/2}}{a^2 c \sqrt {c+a^2 c x^2}}-\frac {\left (15 \sqrt {1+a^2 x^2}\right ) \int \frac {1}{\left (1+a^2 x^2\right )^{3/2} \sqrt {\arctan (a x)}} \, dx}{8 a c \sqrt {c+a^2 c x^2}} \\ & = \frac {15 \sqrt {\arctan (a x)}}{4 a^2 c \sqrt {c+a^2 c x^2}}+\frac {5 x \arctan (a x)^{3/2}}{2 a c \sqrt {c+a^2 c x^2}}-\frac {\arctan (a x)^{5/2}}{a^2 c \sqrt {c+a^2 c x^2}}-\frac {\left (15 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \frac {\cos (x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{8 a^2 c \sqrt {c+a^2 c x^2}} \\ & = \frac {15 \sqrt {\arctan (a x)}}{4 a^2 c \sqrt {c+a^2 c x^2}}+\frac {5 x \arctan (a x)^{3/2}}{2 a c \sqrt {c+a^2 c x^2}}-\frac {\arctan (a x)^{5/2}}{a^2 c \sqrt {c+a^2 c x^2}}-\frac {\left (15 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt {\arctan (a x)}\right )}{4 a^2 c \sqrt {c+a^2 c x^2}} \\ & = \frac {15 \sqrt {\arctan (a x)}}{4 a^2 c \sqrt {c+a^2 c x^2}}+\frac {5 x \arctan (a x)^{3/2}}{2 a c \sqrt {c+a^2 c x^2}}-\frac {\arctan (a x)^{5/2}}{a^2 c \sqrt {c+a^2 c x^2}}-\frac {15 \sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{4 a^2 c \sqrt {c+a^2 c x^2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.18 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.86 \[ \int \frac {x \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {4 \arctan (a x) \left (15+10 a x \arctan (a x)-4 \arctan (a x)^2\right )+15 i \sqrt {1+a^2 x^2} \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-i \arctan (a x)\right )-15 i \sqrt {1+a^2 x^2} \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},i \arctan (a x)\right )}{16 a^2 c \sqrt {c+a^2 c x^2} \sqrt {\arctan (a x)}} \]

[In]

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

[Out]

(4*ArcTan[a*x]*(15 + 10*a*x*ArcTan[a*x] - 4*ArcTan[a*x]^2) + (15*I)*Sqrt[1 + a^2*x^2]*Sqrt[(-I)*ArcTan[a*x]]*G
amma[1/2, (-I)*ArcTan[a*x]] - (15*I)*Sqrt[1 + a^2*x^2]*Sqrt[I*ArcTan[a*x]]*Gamma[1/2, I*ArcTan[a*x]])/(16*a^2*
c*Sqrt[c + a^2*c*x^2]*Sqrt[ArcTan[a*x]])

Maple [F]

\[\int \frac {x \arctan \left (a x \right )^{\frac {5}{2}}}{\left (a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}d x\]

[In]

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

[Out]

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

Fricas [F(-2)]

Exception generated. \[ \int \frac {x \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {x \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {x \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {x \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {x \arctan \left (a x\right )^{\frac {5}{2}}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {x \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\int \frac {x\,{\mathrm {atan}\left (a\,x\right )}^{5/2}}{{\left (c\,a^2\,x^2+c\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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