Integrand size = 21, antiderivative size = 125 \[ \int \frac {1}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^{5/2}} \, dx=-\frac {2}{3 a c^3 \left (1+a^2 x^2\right )^2 \arctan (a x)^{3/2}}+\frac {16 x}{3 c^3 \left (1+a^2 x^2\right )^2 \sqrt {\arctan (a x)}}-\frac {4 \sqrt {2 \pi } \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{3 a c^3}-\frac {8 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{3 a c^3} \]
-2/3/a/c^3/(a^2*x^2+1)^2/arctan(a*x)^(3/2)-8/3*FresnelC(2*arctan(a*x)^(1/2 )/Pi^(1/2))*Pi^(1/2)/a/c^3-4/3*FresnelC(2*2^(1/2)/Pi^(1/2)*arctan(a*x)^(1/ 2))*2^(1/2)*Pi^(1/2)/a/c^3+16/3*x/c^3/(a^2*x^2+1)^2/arctan(a*x)^(1/2)
Result contains complex when optimal does not.
Time = 0.69 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.49 \[ \int \frac {1}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^{5/2}} \, dx=\frac {2 \left (-\frac {1}{a \left (1+a^2 x^2\right )^2}+\frac {8 x \arctan (a x)}{\left (1+a^2 x^2\right )^2}-\frac {\sqrt {2} (-i \arctan (a x))^{3/2} \Gamma \left (\frac {1}{2},-2 i \arctan (a x)\right )}{a}+\frac {\sqrt {2} \arctan (a x)^2 \Gamma \left (\frac {1}{2},2 i \arctan (a x)\right )}{a \sqrt {i \arctan (a x)}}-\frac {(-i \arctan (a x))^{3/2} \Gamma \left (\frac {1}{2},-4 i \arctan (a x)\right )}{a}+\frac {\arctan (a x)^2 \Gamma \left (\frac {1}{2},4 i \arctan (a x)\right )}{a \sqrt {i \arctan (a x)}}\right )}{3 c^3 \arctan (a x)^{3/2}} \]
(2*(-(1/(a*(1 + a^2*x^2)^2)) + (8*x*ArcTan[a*x])/(1 + a^2*x^2)^2 - (Sqrt[2 ]*((-I)*ArcTan[a*x])^(3/2)*Gamma[1/2, (-2*I)*ArcTan[a*x]])/a + (Sqrt[2]*Ar cTan[a*x]^2*Gamma[1/2, (2*I)*ArcTan[a*x]])/(a*Sqrt[I*ArcTan[a*x]]) - (((-I )*ArcTan[a*x])^(3/2)*Gamma[1/2, (-4*I)*ArcTan[a*x]])/a + (ArcTan[a*x]^2*Ga mma[1/2, (4*I)*ArcTan[a*x]])/(a*Sqrt[I*ArcTan[a*x]])))/(3*c^3*ArcTan[a*x]^ (3/2))
Time = 0.99 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.53, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {5437, 27, 5503, 5439, 3042, 3793, 2009, 5505, 4906, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\arctan (a x)^{5/2} \left (a^2 c x^2+c\right )^3} \, dx\) |
\(\Big \downarrow \) 5437 |
\(\displaystyle -\frac {8}{3} a \int \frac {x}{c^3 \left (a^2 x^2+1\right )^3 \arctan (a x)^{3/2}}dx-\frac {2}{3 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {8 a \int \frac {x}{\left (a^2 x^2+1\right )^3 \arctan (a x)^{3/2}}dx}{3 c^3}-\frac {2}{3 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^{3/2}}\) |
\(\Big \downarrow \) 5503 |
\(\displaystyle -\frac {8 a \left (\frac {2 \int \frac {1}{\left (a^2 x^2+1\right )^3 \sqrt {\arctan (a x)}}dx}{a}-6 a \int \frac {x^2}{\left (a^2 x^2+1\right )^3 \sqrt {\arctan (a x)}}dx-\frac {2 x}{a \left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}\right )}{3 c^3}-\frac {2}{3 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^{3/2}}\) |
\(\Big \downarrow \) 5439 |
\(\displaystyle -\frac {8 a \left (-6 a \int \frac {x^2}{\left (a^2 x^2+1\right )^3 \sqrt {\arctan (a x)}}dx+\frac {2 \int \frac {1}{\left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}d\arctan (a x)}{a^2}-\frac {2 x}{a \left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}\right )}{3 c^3}-\frac {2}{3 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {8 a \left (-6 a \int \frac {x^2}{\left (a^2 x^2+1\right )^3 \sqrt {\arctan (a x)}}dx+\frac {2 \int \frac {\sin \left (\arctan (a x)+\frac {\pi }{2}\right )^4}{\sqrt {\arctan (a x)}}d\arctan (a x)}{a^2}-\frac {2 x}{a \left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}\right )}{3 c^3}-\frac {2}{3 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^{3/2}}\) |
\(\Big \downarrow \) 3793 |
\(\displaystyle -\frac {8 a \left (-6 a \int \frac {x^2}{\left (a^2 x^2+1\right )^3 \sqrt {\arctan (a x)}}dx+\frac {2 \int \left (\frac {\cos (2 \arctan (a x))}{2 \sqrt {\arctan (a x)}}+\frac {\cos (4 \arctan (a x))}{8 \sqrt {\arctan (a x)}}+\frac {3}{8 \sqrt {\arctan (a x)}}\right )d\arctan (a x)}{a^2}-\frac {2 x}{a \left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}\right )}{3 c^3}-\frac {2}{3 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^{3/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {8 a \left (-6 a \int \frac {x^2}{\left (a^2 x^2+1\right )^3 \sqrt {\arctan (a x)}}dx+\frac {2 \left (\frac {1}{8} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )+\frac {1}{2} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )+\frac {3}{4} \sqrt {\arctan (a x)}\right )}{a^2}-\frac {2 x}{a \left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}\right )}{3 c^3}-\frac {2}{3 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^{3/2}}\) |
\(\Big \downarrow \) 5505 |
\(\displaystyle -\frac {8 a \left (-\frac {6 \int \frac {a^2 x^2}{\left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}d\arctan (a x)}{a^2}+\frac {2 \left (\frac {1}{8} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )+\frac {1}{2} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )+\frac {3}{4} \sqrt {\arctan (a x)}\right )}{a^2}-\frac {2 x}{a \left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}\right )}{3 c^3}-\frac {2}{3 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^{3/2}}\) |
\(\Big \downarrow \) 4906 |
\(\displaystyle -\frac {8 a \left (-\frac {6 \int \left (\frac {1}{8 \sqrt {\arctan (a x)}}-\frac {\cos (4 \arctan (a x))}{8 \sqrt {\arctan (a x)}}\right )d\arctan (a x)}{a^2}+\frac {2 \left (\frac {1}{8} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )+\frac {1}{2} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )+\frac {3}{4} \sqrt {\arctan (a x)}\right )}{a^2}-\frac {2 x}{a \left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}\right )}{3 c^3}-\frac {2}{3 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^{3/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {8 a \left (-\frac {6 \left (\frac {1}{4} \sqrt {\arctan (a x)}-\frac {1}{8} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )\right )}{a^2}+\frac {2 \left (\frac {1}{8} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )+\frac {1}{2} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )+\frac {3}{4} \sqrt {\arctan (a x)}\right )}{a^2}-\frac {2 x}{a \left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}\right )}{3 c^3}-\frac {2}{3 a c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^{3/2}}\) |
-2/(3*a*c^3*(1 + a^2*x^2)^2*ArcTan[a*x]^(3/2)) - (8*a*((-2*x)/(a*(1 + a^2* x^2)^2*Sqrt[ArcTan[a*x]]) - (6*(Sqrt[ArcTan[a*x]]/4 - (Sqrt[Pi/2]*FresnelC [2*Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/8))/a^2 + (2*((3*Sqrt[ArcTan[a*x]])/4 + (Sqrt[Pi/2]*FresnelC[2*Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/8 + (Sqrt[Pi]*Fresne lC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/2))/a^2))/(3*c^3)
3.11.71.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In t[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]))
Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b _.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(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] && IG tQ[p, 0]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_)^2)^(q_), x_S ymbol] :> Simp[(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])^(p + 1)/(b*c*d*(p + 1))), x] - Simp[2*c*((q + 1)/(b*(p + 1))) Int[x*(d + e*x^2)^q*(a + b*Arc Tan[c*x])^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && LtQ[q, -1] && LtQ[p, -1]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_), x_ Symbol] :> Simp[d^q/c Subst[Int[(a + b*x)^p/Cos[x]^(2*(q + 1)), x], x, Ar cTan[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])
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] + (-Simp[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] - Simp[m/(b*c*(p + 1)) Int[x^(m - 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p + 1), x], x]) /; F reeQ[{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]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 2)^(q_), x_Symbol] :> Simp[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])
Time = 0.18 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.90
method | result | size |
default | \(\frac {-16 \sqrt {2}\, \sqrt {\pi }\, \operatorname {FresnelC}\left (\frac {2 \sqrt {2}\, \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right ) \arctan \left (a x \right )^{\frac {3}{2}}-32 \sqrt {\pi }\, \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right ) \arctan \left (a x \right )^{\frac {3}{2}}+16 \sin \left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )+8 \sin \left (4 \arctan \left (a x \right )\right ) \arctan \left (a x \right )-4 \cos \left (2 \arctan \left (a x \right )\right )-\cos \left (4 \arctan \left (a x \right )\right )-3}{12 a \,c^{3} \arctan \left (a x \right )^{\frac {3}{2}}}\) | \(113\) |
1/12/a/c^3*(-16*2^(1/2)*Pi^(1/2)*FresnelC(2*2^(1/2)/Pi^(1/2)*arctan(a*x)^( 1/2))*arctan(a*x)^(3/2)-32*Pi^(1/2)*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2)) *arctan(a*x)^(3/2)+16*sin(2*arctan(a*x))*arctan(a*x)+8*sin(4*arctan(a*x))* arctan(a*x)-4*cos(2*arctan(a*x))-cos(4*arctan(a*x))-3)/arctan(a*x)^(3/2)
Exception generated. \[ \int \frac {1}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \frac {1}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^{5/2}} \, dx=\frac {\int \frac {1}{a^{6} x^{6} \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )} + 3 a^{4} x^{4} \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )} + \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )}}\, dx}{c^{3}} \]
Integral(1/(a**6*x**6*atan(a*x)**(5/2) + 3*a**4*x**4*atan(a*x)**(5/2) + 3* a**2*x**2*atan(a*x)**(5/2) + atan(a*x)**(5/2)), x)/c**3
Exception generated. \[ \int \frac {1}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]
Timed out. \[ \int \frac {1}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^{5/2}} \, dx=\int \frac {1}{{\mathrm {atan}\left (a\,x\right )}^{5/2}\,{\left (c\,a^2\,x^2+c\right )}^3} \,d x \]