Integrand size = 24, antiderivative size = 280 \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{3/2}} \, dx=-\frac {2 x}{a c \left (c+a^2 c x^2\right )^{3/2} \sqrt {\arctan (a x)}}+\frac {3 \sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{a^2 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {2 \pi } \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{a^2 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {\frac {\pi }{6}} \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{a^2 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {\frac {2 \pi }{3}} \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{a^2 c^2 \sqrt {c+a^2 c x^2}} \]
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)
Result contains complex when optimal does not.
Time = 0.46 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.07 \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{3/2}} \, dx=-\frac {i \left (-8 i a x+\left (1+a^2 x^2\right )^{3/2} \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-i \arctan (a x)\right )-\left (1+a^2 x^2\right )^{3/2} \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},i \arctan (a x)\right )+\sqrt {3+3 a^2 x^2} \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-3 i \arctan (a x)\right )+a^2 x^2 \sqrt {3+3 a^2 x^2} \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-3 i \arctan (a x)\right )-\sqrt {3+3 a^2 x^2} \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},3 i \arctan (a x)\right )-a^2 x^2 \sqrt {3+3 a^2 x^2} \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},3 i \arctan (a x)\right )\right )}{4 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt {c+a^2 c x^2} \sqrt {\arctan (a x)}} \]
((-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]]*Gam ma[1/2, (-3*I)*ArcTan[a*x]] - Sqrt[3 + 3*a^2*x^2]*Sqrt[I*ArcTan[a*x]]*Gamm a[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]])
Time = 1.53 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.83, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5503, 5440, 5439, 3042, 3793, 2009, 5506, 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.
| \(\displaystyle \int \frac {x}{\arctan (a x)^{3/2} \left (a^2 c x^2+c\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 5503 |
| \(\displaystyle \frac {2 \int \frac {1}{\left (a^2 c x^2+c\right )^{5/2} \sqrt {\arctan (a x)}}dx}{a}-4 a \int \frac {x^2}{\left (a^2 c x^2+c\right )^{5/2} \sqrt {\arctan (a x)}}dx-\frac {2 x}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\) |
| \(\Big \downarrow \) 5440 |
| \(\displaystyle \frac {2 \sqrt {a^2 x^2+1} \int \frac {1}{\left (a^2 x^2+1\right )^{5/2} \sqrt {\arctan (a x)}}dx}{a c^2 \sqrt {a^2 c x^2+c}}-4 a \int \frac {x^2}{\left (a^2 c x^2+c\right )^{5/2} \sqrt {\arctan (a x)}}dx-\frac {2 x}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\) |
| \(\Big \downarrow \) 5439 |
| \(\displaystyle \frac {2 \sqrt {a^2 x^2+1} \int \frac {1}{\left (a^2 x^2+1\right )^{3/2} \sqrt {\arctan (a x)}}d\arctan (a x)}{a^2 c^2 \sqrt {a^2 c x^2+c}}-4 a \int \frac {x^2}{\left (a^2 c x^2+c\right )^{5/2} \sqrt {\arctan (a x)}}dx-\frac {2 x}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \sqrt {a^2 x^2+1} \int \frac {\sin \left (\arctan (a x)+\frac {\pi }{2}\right )^3}{\sqrt {\arctan (a x)}}d\arctan (a x)}{a^2 c^2 \sqrt {a^2 c x^2+c}}-4 a \int \frac {x^2}{\left (a^2 c x^2+c\right )^{5/2} \sqrt {\arctan (a x)}}dx-\frac {2 x}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\) |
| \(\Big \downarrow \) 3793 |
| \(\displaystyle \frac {2 \sqrt {a^2 x^2+1} \int \left (\frac {\cos (3 \arctan (a x))}{4 \sqrt {\arctan (a x)}}+\frac {3}{4 \sqrt {a^2 x^2+1} \sqrt {\arctan (a x)}}\right )d\arctan (a x)}{a^2 c^2 \sqrt {a^2 c x^2+c}}-4 a \int \frac {x^2}{\left (a^2 c x^2+c\right )^{5/2} \sqrt {\arctan (a x)}}dx-\frac {2 x}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -4 a \int \frac {x^2}{\left (a^2 c x^2+c\right )^{5/2} \sqrt {\arctan (a x)}}dx+\frac {2 \sqrt {a^2 x^2+1} \left (\frac {3}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )+\frac {1}{2} \sqrt {\frac {\pi }{6}} \operatorname {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )\right )}{a^2 c^2 \sqrt {a^2 c x^2+c}}-\frac {2 x}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\) |
| \(\Big \downarrow \) 5506 |
| \(\displaystyle -\frac {4 a \sqrt {a^2 x^2+1} \int \frac {x^2}{\left (a^2 x^2+1\right )^{5/2} \sqrt {\arctan (a x)}}dx}{c^2 \sqrt {a^2 c x^2+c}}+\frac {2 \sqrt {a^2 x^2+1} \left (\frac {3}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )+\frac {1}{2} \sqrt {\frac {\pi }{6}} \operatorname {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )\right )}{a^2 c^2 \sqrt {a^2 c x^2+c}}-\frac {2 x}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\) |
| \(\Big \downarrow \) 5505 |
| \(\displaystyle -\frac {4 \sqrt {a^2 x^2+1} \int \frac {a^2 x^2}{\left (a^2 x^2+1\right )^{3/2} \sqrt {\arctan (a x)}}d\arctan (a x)}{a^2 c^2 \sqrt {a^2 c x^2+c}}+\frac {2 \sqrt {a^2 x^2+1} \left (\frac {3}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )+\frac {1}{2} \sqrt {\frac {\pi }{6}} \operatorname {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )\right )}{a^2 c^2 \sqrt {a^2 c x^2+c}}-\frac {2 x}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\) |
| \(\Big \downarrow \) 4906 |
| \(\displaystyle -\frac {4 \sqrt {a^2 x^2+1} \int \left (\frac {1}{4 \sqrt {a^2 x^2+1} \sqrt {\arctan (a x)}}-\frac {\cos (3 \arctan (a x))}{4 \sqrt {\arctan (a x)}}\right )d\arctan (a x)}{a^2 c^2 \sqrt {a^2 c x^2+c}}+\frac {2 \sqrt {a^2 x^2+1} \left (\frac {3}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )+\frac {1}{2} \sqrt {\frac {\pi }{6}} \operatorname {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )\right )}{a^2 c^2 \sqrt {a^2 c x^2+c}}-\frac {2 x}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {4 \sqrt {a^2 x^2+1} \left (\frac {1}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )-\frac {1}{2} \sqrt {\frac {\pi }{6}} \operatorname {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )\right )}{a^2 c^2 \sqrt {a^2 c x^2+c}}+\frac {2 \sqrt {a^2 x^2+1} \left (\frac {3}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )+\frac {1}{2} \sqrt {\frac {\pi }{6}} \operatorname {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )\right )}{a^2 c^2 \sqrt {a^2 c x^2+c}}-\frac {2 x}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\) |
(-2*x)/(a*c*(c + a^2*c*x^2)^(3/2)*Sqrt[ArcTan[a*x]]) - (4*Sqrt[1 + a^2*x^2 ]*((Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/2 - (Sqrt[Pi/6]*Fre snelC[Sqrt[6/Pi]*Sqrt[ArcTan[a*x]]])/2))/(a^2*c^2*Sqrt[c + a^2*c*x^2]) + ( 2*Sqrt[1 + a^2*x^2]*((3*Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]]) /2 + (Sqrt[Pi/6]*FresnelC[Sqrt[6/Pi]*Sqrt[ArcTan[a*x]]])/2))/(a^2*c^2*Sqrt [c + a^2*c*x^2])
3.11.36.3.1 Defintions of rubi rules used
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_ 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_.)*((d_) + (e_.)*(x_)^2)^(q_), x_ Symbol] :> Simp[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])
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])
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 2)^(q_), x_Symbol] :> Simp[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] && !(I ntegerQ[q] || GtQ[d, 0])
\[\int \frac {x}{\left (a^{2} c \,x^{2}+c \right )^{\frac {5}{2}} \arctan \left (a x \right )^{\frac {3}{2}}}d x\]
Exception generated. \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{3/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \frac {x}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{3/2}} \, dx=\int \frac {x}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}} \operatorname {atan}^{\frac {3}{2}}{\left (a x \right )}}\, dx \]
Exception generated. \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{3/2}} \, dx=\text {Exception raised: RuntimeError} \]
Exception generated. \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{3/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{3/2}} \, dx=\int \frac {x}{{\mathrm {atan}\left (a\,x\right )}^{3/2}\,{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]