\(\int \frac {x}{(c+a^2 c x^2)^{5/2} \arctan (a x)^{3/2}} \, dx\) [1036]

Optimal result

Integrand size = 24, antiderivative size = 280 \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{3/2}} \, dx=-\frac {2 x}{a c \left (c+a^2 c x^2\right )^{3/2} \sqrt {\arctan (a x)}}+\frac {3 \sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{a^2 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {2 \pi } \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{a^2 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {\frac {\pi }{6}} \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{a^2 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {\frac {2 \pi }{3}} \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{a^2 c^2 \sqrt {c+a^2 c x^2}} \]

[Out]

1/2*FresnelC(6^(1/2)/Pi^(1/2)*arctan(a*x)^(1/2))*6^(1/2)*Pi^(1/2)*(a^2*x^2+1)^(1/2)/a^2/c^2/(a^2*c*x^2+c)^(1/2
)+1/2*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^2/(a^2*c*x^2+c)^(1
/2)-2*x/a/c/(a^2*c*x^2+c)^(3/2)/arctan(a*x)^(1/2)

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5088, 5091, 5090, 4491, 3385, 3433, 5025, 5024, 3393} \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{3/2}} \, dx=-\frac {\sqrt {2 \pi } \sqrt {a^2 x^2+1} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{a^2 c^2 \sqrt {a^2 c x^2+c}}+\frac {3 \sqrt {\frac {\pi }{2}} \sqrt {a^2 x^2+1} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{a^2 c^2 \sqrt {a^2 c x^2+c}}+\frac {\sqrt {\frac {2 \pi }{3}} \sqrt {a^2 x^2+1} \operatorname {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{a^2 c^2 \sqrt {a^2 c x^2+c}}+\frac {\sqrt {\frac {\pi }{6}} \sqrt {a^2 x^2+1} \operatorname {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{a^2 c^2 \sqrt {a^2 c x^2+c}}-\frac {2 x}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}} \]

[In]

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

[Out]

(-2*x)/(a*c*(c + a^2*c*x^2)^(3/2)*Sqrt[ArcTan[a*x]]) + (3*Sqrt[Pi/2]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[2/Pi]*Sqr
t[ArcTan[a*x]]])/(a^2*c^2*Sqrt[c + a^2*c*x^2]) - (Sqrt[2*Pi]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[2/Pi]*Sqrt[ArcTan
[a*x]]])/(a^2*c^2*Sqrt[c + a^2*c*x^2]) + (Sqrt[Pi/6]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[6/Pi]*Sqrt[ArcTan[a*x]]])
/(a^2*c^2*Sqrt[c + a^2*c*x^2]) + (Sqrt[(2*Pi)/3]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[6/Pi]*Sqrt[ArcTan[a*x]]])/(a^
2*c^2*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 3393

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])
)

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 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 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 5088

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 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])

Rule 5091

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 x}{a c \left (c+a^2 c x^2\right )^{3/2} \sqrt {\arctan (a x)}}+\frac {2 \int \frac {1}{\left (c+a^2 c x^2\right )^{5/2} \sqrt {\arctan (a x)}} \, dx}{a}-(4 a) \int \frac {x^2}{\left (c+a^2 c x^2\right )^{5/2} \sqrt {\arctan (a x)}} \, dx \\ & = -\frac {2 x}{a c \left (c+a^2 c x^2\right )^{3/2} \sqrt {\arctan (a x)}}+\frac {\left (2 \sqrt {1+a^2 x^2}\right ) \int \frac {1}{\left (1+a^2 x^2\right )^{5/2} \sqrt {\arctan (a x)}} \, dx}{a c^2 \sqrt {c+a^2 c x^2}}-\frac {\left (4 a \sqrt {1+a^2 x^2}\right ) \int \frac {x^2}{\left (1+a^2 x^2\right )^{5/2} \sqrt {\arctan (a x)}} \, dx}{c^2 \sqrt {c+a^2 c x^2}} \\ & = -\frac {2 x}{a c \left (c+a^2 c x^2\right )^{3/2} \sqrt {\arctan (a x)}}+\frac {\left (2 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \frac {\cos ^3(x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{a^2 c^2 \sqrt {c+a^2 c x^2}}-\frac {\left (4 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \frac {\cos (x) \sin ^2(x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{a^2 c^2 \sqrt {c+a^2 c x^2}} \\ & = -\frac {2 x}{a c \left (c+a^2 c x^2\right )^{3/2} \sqrt {\arctan (a x)}}+\frac {\left (2 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \left (\frac {3 \cos (x)}{4 \sqrt {x}}+\frac {\cos (3 x)}{4 \sqrt {x}}\right ) \, dx,x,\arctan (a x)\right )}{a^2 c^2 \sqrt {c+a^2 c x^2}}-\frac {\left (4 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \left (\frac {\cos (x)}{4 \sqrt {x}}-\frac {\cos (3 x)}{4 \sqrt {x}}\right ) \, dx,x,\arctan (a x)\right )}{a^2 c^2 \sqrt {c+a^2 c x^2}} \\ & = -\frac {2 x}{a c \left (c+a^2 c x^2\right )^{3/2} \sqrt {\arctan (a x)}}+\frac {\sqrt {1+a^2 x^2} \text {Subst}\left (\int \frac {\cos (3 x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{2 a^2 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {1+a^2 x^2} \text {Subst}\left (\int \frac {\cos (x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{a^2 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \text {Subst}\left (\int \frac {\cos (3 x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{a^2 c^2 \sqrt {c+a^2 c x^2}}+\frac {\left (3 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \frac {\cos (x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{2 a^2 c^2 \sqrt {c+a^2 c x^2}} \\ & = -\frac {2 x}{a c \left (c+a^2 c x^2\right )^{3/2} \sqrt {\arctan (a x)}}+\frac {\sqrt {1+a^2 x^2} \text {Subst}\left (\int \cos \left (3 x^2\right ) \, dx,x,\sqrt {\arctan (a x)}\right )}{a^2 c^2 \sqrt {c+a^2 c x^2}}-\frac {\left (2 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt {\arctan (a x)}\right )}{a^2 c^2 \sqrt {c+a^2 c x^2}}+\frac {\left (2 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \cos \left (3 x^2\right ) \, dx,x,\sqrt {\arctan (a x)}\right )}{a^2 c^2 \sqrt {c+a^2 c x^2}}+\frac {\left (3 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt {\arctan (a x)}\right )}{a^2 c^2 \sqrt {c+a^2 c x^2}} \\ & = -\frac {2 x}{a c \left (c+a^2 c x^2\right )^{3/2} \sqrt {\arctan (a x)}}+\frac {3 \sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{a^2 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {2 \pi } \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{a^2 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {\frac {\pi }{6}} \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{a^2 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {\frac {2 \pi }{3}} \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{a^2 c^2 \sqrt {c+a^2 c x^2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.46 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.07 \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{3/2}} \, dx=-\frac {i \left (-8 i a x+\left (1+a^2 x^2\right )^{3/2} \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-i \arctan (a x)\right )-\left (1+a^2 x^2\right )^{3/2} \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},i \arctan (a x)\right )+\sqrt {3+3 a^2 x^2} \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-3 i \arctan (a x)\right )+a^2 x^2 \sqrt {3+3 a^2 x^2} \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-3 i \arctan (a x)\right )-\sqrt {3+3 a^2 x^2} \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},3 i \arctan (a x)\right )-a^2 x^2 \sqrt {3+3 a^2 x^2} \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},3 i \arctan (a x)\right )\right )}{4 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt {c+a^2 c x^2} \sqrt {\arctan (a x)}} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F(-2)]

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

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {x}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{3/2}} \, dx=\int \frac {x}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}} \operatorname {atan}^{\frac {3}{2}}{\left (a x \right )}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [F(-2)]

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

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

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

[In]

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

[Out]

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