\(\int \frac {1}{(c+a^2 c x^2)^{3/2} \arctan (a x)^{5/2}} \, dx\) [1097]

Optimal result

Integrand size = 23, antiderivative size = 126 \[ \int \frac {1}{\left (c+a^2 c x^2\right )^{3/2} \arctan (a x)^{5/2}} \, dx=-\frac {2}{3 a c \sqrt {c+a^2 c x^2} \arctan (a x)^{3/2}}+\frac {4 x}{3 c \sqrt {c+a^2 c x^2} \sqrt {\arctan (a x)}}-\frac {4 \sqrt {2 \pi } \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{3 a c \sqrt {c+a^2 c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {5022, 5062, 5025, 5024, 3385, 3433} \[ \int \frac {1}{\left (c+a^2 c x^2\right )^{3/2} \arctan (a x)^{5/2}} \, dx=-\frac {4 \sqrt {2 \pi } \sqrt {a^2 x^2+1} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{3 a c \sqrt {a^2 c x^2+c}}+\frac {4 x}{3 c \sqrt {\arctan (a x)} \sqrt {a^2 c x^2+c}}-\frac {2}{3 a c \arctan (a x)^{3/2} \sqrt {a^2 c x^2+c}} \]

[In]

Int[1/((c + a^2*c*x^2)^(3/2)*ArcTan[a*x]^(5/2)),x]

[Out]

-2/(3*a*c*Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^(3/2)) + (4*x)/(3*c*Sqrt[c + a^2*c*x^2]*Sqrt[ArcTan[a*x]]) - (4*Sqrt
[2*Pi]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/(3*a*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 5022

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

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.95 \[ \int \frac {1}{\left (c+a^2 c x^2\right )^{3/2} \arctan (a x)^{5/2}} \, dx=\frac {-2+4 a x \arctan (a x)-2 \sqrt {1+a^2 x^2} (-i \arctan (a x))^{3/2} \Gamma \left (\frac {1}{2},-i \arctan (a x)\right )-2 \sqrt {1+a^2 x^2} (i \arctan (a x))^{3/2} \Gamma \left (\frac {1}{2},i \arctan (a x)\right )}{3 a c \sqrt {c+a^2 c x^2} \arctan (a x)^{3/2}} \]

[In]

Integrate[1/((c + a^2*c*x^2)^(3/2)*ArcTan[a*x]^(5/2)),x]

[Out]

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

Maple [F]

\[\int \frac {1}{\left (a^{2} c \,x^{2}+c \right )^{\frac {3}{2}} \arctan \left (a x \right )^{\frac {5}{2}}}d x\]

[In]

int(1/(a^2*c*x^2+c)^(3/2)/arctan(a*x)^(5/2),x)

[Out]

int(1/(a^2*c*x^2+c)^(3/2)/arctan(a*x)^(5/2),x)

Fricas [F(-2)]

Exception generated. \[ \int \frac {1}{\left (c+a^2 c x^2\right )^{3/2} \arctan (a x)^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/(a^2*c*x^2+c)^(3/2)/arctan(a*x)^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\left (c+a^2 c x^2\right )^{3/2} \arctan (a x)^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(1/(a^2*c*x^2+c)^(3/2)/arctan(a*x)^(5/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(1/(a^2*c*x^2+c)^(3/2)/arctan(a*x)^(5/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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