\(\int \frac {x^3}{(c+a^2 c x^2)^{5/2} \arctan (a x)^2} \, dx\) [589]

Optimal result

Integrand size = 24, antiderivative size = 118 \[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^2} \, dx=-\frac {x^3}{a c \left (c+a^2 c x^2\right )^{3/2} \arctan (a x)}+\frac {3 \sqrt {1+a^2 x^2} \operatorname {CosIntegral}(\arctan (a x))}{4 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {3 \sqrt {1+a^2 x^2} \operatorname {CosIntegral}(3 \arctan (a x))}{4 a^4 c^2 \sqrt {c+a^2 c x^2}} \]

[Out]

-x^3/a/c/(a^2*c*x^2+c)^(3/2)/arctan(a*x)+3/4*Ci(arctan(a*x))*(a^2*x^2+1)^(1/2)/a^4/c^2/(a^2*c*x^2+c)^(1/2)-3/4
*Ci(3*arctan(a*x))*(a^2*x^2+1)^(1/2)/a^4/c^2/(a^2*c*x^2+c)^(1/2)

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {5062, 5091, 5090, 4491, 3383} \[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^2} \, dx=-\frac {x^3}{a c \arctan (a x) \left (a^2 c x^2+c\right )^{3/2}}+\frac {3 \sqrt {a^2 x^2+1} \operatorname {CosIntegral}(\arctan (a x))}{4 a^4 c^2 \sqrt {a^2 c x^2+c}}-\frac {3 \sqrt {a^2 x^2+1} \operatorname {CosIntegral}(3 \arctan (a x))}{4 a^4 c^2 \sqrt {a^2 c x^2+c}} \]

[In]

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

[Out]

-(x^3/(a*c*(c + a^2*c*x^2)^(3/2)*ArcTan[a*x])) + (3*Sqrt[1 + a^2*x^2]*CosIntegral[ArcTan[a*x]])/(4*a^4*c^2*Sqr
t[c + a^2*c*x^2]) - (3*Sqrt[1 + a^2*x^2]*CosIntegral[3*ArcTan[a*x]])/(4*a^4*c^2*Sqrt[c + a^2*c*x^2])

Rule 3383

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

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 5062

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

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.69 \[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^2} \, dx=\frac {-\frac {4 a^3 c x^3}{\left (1+a^2 x^2\right ) \arctan (a x)}+3 c \sqrt {1+a^2 x^2} (\operatorname {CosIntegral}(\arctan (a x))-\operatorname {CosIntegral}(3 \arctan (a x)))}{4 a^4 c^3 \sqrt {c+a^2 c x^2}} \]

[In]

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

[Out]

((-4*a^3*c*x^3)/((1 + a^2*x^2)*ArcTan[a*x]) + 3*c*Sqrt[1 + a^2*x^2]*(CosIntegral[ArcTan[a*x]] - CosIntegral[3*
ArcTan[a*x]]))/(4*a^4*c^3*Sqrt[c + a^2*c*x^2])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 15.63 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.62

method	result	size
default	\(-\frac {\left (3 \arctan \left (a x \right ) \operatorname {Ei}_{1}\left (-i \arctan \left (a x \right )\right ) a^{4} x^{4}+3 \arctan \left (a x \right ) \operatorname {Ei}_{1}\left (i \arctan \left (a x \right )\right ) a^{4} x^{4}-3 \arctan \left (a x \right ) \operatorname {Ei}_{1}\left (3 i \arctan \left (a x \right )\right ) a^{4} x^{4}-3 \arctan \left (a x \right ) \operatorname {Ei}_{1}\left (-3 i \arctan \left (a x \right )\right ) a^{4} x^{4}+8 \sqrt {a^{2} x^{2}+1}\, a^{3} x^{3}+6 \arctan \left (a x \right ) \operatorname {Ei}_{1}\left (-i \arctan \left (a x \right )\right ) a^{2} x^{2}+6 \arctan \left (a x \right ) \operatorname {Ei}_{1}\left (i \arctan \left (a x \right )\right ) a^{2} x^{2}-6 \arctan \left (a x \right ) \operatorname {Ei}_{1}\left (3 i \arctan \left (a x \right )\right ) a^{2} x^{2}-6 \arctan \left (a x \right ) \operatorname {Ei}_{1}\left (-3 i \arctan \left (a x \right )\right ) a^{2} x^{2}+3 \,\operatorname {Ei}_{1}\left (-i \arctan \left (a x \right )\right ) \arctan \left (a x \right )+3 \,\operatorname {Ei}_{1}\left (i \arctan \left (a x \right )\right ) \arctan \left (a x \right )-3 \,\operatorname {Ei}_{1}\left (3 i \arctan \left (a x \right )\right ) \arctan \left (a x \right )-3 \,\operatorname {Ei}_{1}\left (-3 i \arctan \left (a x \right )\right ) \arctan \left (a x \right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{8 \sqrt {a^{2} x^{2}+1}\, \arctan \left (a x \right ) a^{4} c^{3} \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right )}\)	\(309\)

[In]

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

[Out]

-1/8*(3*arctan(a*x)*Ei(1,-I*arctan(a*x))*a^4*x^4+3*arctan(a*x)*Ei(1,I*arctan(a*x))*a^4*x^4-3*arctan(a*x)*Ei(1,
3*I*arctan(a*x))*a^4*x^4-3*arctan(a*x)*Ei(1,-3*I*arctan(a*x))*a^4*x^4+8*(a^2*x^2+1)^(1/2)*a^3*x^3+6*arctan(a*x
)*Ei(1,-I*arctan(a*x))*a^2*x^2+6*arctan(a*x)*Ei(1,I*arctan(a*x))*a^2*x^2-6*arctan(a*x)*Ei(1,3*I*arctan(a*x))*a
^2*x^2-6*arctan(a*x)*Ei(1,-3*I*arctan(a*x))*a^2*x^2+3*Ei(1,-I*arctan(a*x))*arctan(a*x)+3*Ei(1,I*arctan(a*x))*a
rctan(a*x)-3*Ei(1,3*I*arctan(a*x))*arctan(a*x)-3*Ei(1,-3*I*arctan(a*x))*arctan(a*x))/(a^2*x^2+1)^(1/2)*(c*(a*x
-I)*(I+a*x))^(1/2)/arctan(a*x)/a^4/c^3/(a^4*x^4+2*a^2*x^2+1)

Fricas [F]

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

[In]

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

[Out]

integral(sqrt(a^2*c*x^2 + c)*x^3/((a^6*c^3*x^6 + 3*a^4*c^3*x^4 + 3*a^2*c^3*x^2 + c^3)*arctan(a*x)^2), x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

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

[In]

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

[In]

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

[Out]

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