Integrand size = 26, antiderivative size = 26 \[ \int \frac {1}{x \left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{5/2}} \, dx=-\frac {2}{3 a c x \left (c+a^2 c x^2\right )^{3/2} \arctan (a x)^{3/2}}+\frac {16}{3 c \left (c+a^2 c x^2\right )^{3/2} \sqrt {\arctan (a x)}}+\frac {4}{3 a^2 c x^2 \left (c+a^2 c x^2\right )^{3/2} \sqrt {\arctan (a x)}}+\frac {4 \sqrt {2 \pi } \sqrt {1+a^2 x^2} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{c^2 \sqrt {c+a^2 c x^2}}+\frac {4 \sqrt {\frac {2 \pi }{3}} \sqrt {1+a^2 x^2} \operatorname {FresnelS}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{c^2 \sqrt {c+a^2 c x^2}}+\frac {8 \text {Int}\left (\frac {1}{x^3 \left (c+a^2 c x^2\right )^{5/2} \sqrt {\arctan (a x)}},x\right )}{3 a^2}+\frac {20}{3} \text {Int}\left (\frac {1}{x \left (c+a^2 c x^2\right )^{5/2} \sqrt {\arctan (a x)}},x\right ) \]
-2/3/a/c/x/(a^2*c*x^2+c)^(3/2)/arctan(a*x)^(3/2)+4/3*FresnelS(6^(1/2)/Pi^( 1/2)*arctan(a*x)^(1/2))*6^(1/2)*Pi^(1/2)*(a^2*x^2+1)^(1/2)/c^2/(a^2*c*x^2+ c)^(1/2)+4*FresnelS(2^(1/2)/Pi^(1/2)*arctan(a*x)^(1/2))*2^(1/2)*Pi^(1/2)*( a^2*x^2+1)^(1/2)/c^2/(a^2*c*x^2+c)^(1/2)+16/3/c/(a^2*c*x^2+c)^(3/2)/arctan (a*x)^(1/2)+4/3/a^2/c/x^2/(a^2*c*x^2+c)^(3/2)/arctan(a*x)^(1/2)+8/3*Uninte grable(1/x^3/(a^2*c*x^2+c)^(5/2)/arctan(a*x)^(1/2),x)/a^2+20/3*Unintegrabl e(1/x/(a^2*c*x^2+c)^(5/2)/arctan(a*x)^(1/2),x)
Not integrable
Time = 9.93 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {1}{x \left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{5/2}} \, dx=\int \frac {1}{x \left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{5/2}} \, dx \]
Not integrable
Time = 2.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {5503, 5437, 5503, 5506, 5505, 4906, 2009, 5560}
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.
\(\displaystyle \int \frac {1}{x \arctan (a x)^{5/2} \left (a^2 c x^2+c\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 5503 |
\(\displaystyle -\frac {8}{3} a \int \frac {1}{\left (a^2 c x^2+c\right )^{5/2} \arctan (a x)^{3/2}}dx-\frac {2 \int \frac {1}{x^2 \left (a^2 c x^2+c\right )^{5/2} \arctan (a x)^{3/2}}dx}{3 a}-\frac {2}{3 a c x \arctan (a x)^{3/2} \left (a^2 c x^2+c\right )^{3/2}}\) |
\(\Big \downarrow \) 5437 |
\(\displaystyle -\frac {2 \int \frac {1}{x^2 \left (a^2 c x^2+c\right )^{5/2} \arctan (a x)^{3/2}}dx}{3 a}-\frac {8}{3} a \left (-6 a \int \frac {x}{\left (a^2 c x^2+c\right )^{5/2} \sqrt {\arctan (a x)}}dx-\frac {2}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\right )-\frac {2}{3 a c x \arctan (a x)^{3/2} \left (a^2 c x^2+c\right )^{3/2}}\) |
\(\Big \downarrow \) 5503 |
\(\displaystyle -\frac {8}{3} a \left (-6 a \int \frac {x}{\left (a^2 c x^2+c\right )^{5/2} \sqrt {\arctan (a x)}}dx-\frac {2}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\right )-\frac {2 \left (-10 a \int \frac {1}{x \left (a^2 c x^2+c\right )^{5/2} \sqrt {\arctan (a x)}}dx-\frac {4 \int \frac {1}{x^3 \left (a^2 c x^2+c\right )^{5/2} \sqrt {\arctan (a x)}}dx}{a}-\frac {2}{a c x^2 \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\right )}{3 a}-\frac {2}{3 a c x \arctan (a x)^{3/2} \left (a^2 c x^2+c\right )^{3/2}}\) |
\(\Big \downarrow \) 5506 |
\(\displaystyle -\frac {8}{3} a \left (-\frac {6 a \sqrt {a^2 x^2+1} \int \frac {x}{\left (a^2 x^2+1\right )^{5/2} \sqrt {\arctan (a x)}}dx}{c^2 \sqrt {a^2 c x^2+c}}-\frac {2}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\right )-\frac {2 \left (-10 a \int \frac {1}{x \left (a^2 c x^2+c\right )^{5/2} \sqrt {\arctan (a x)}}dx-\frac {4 \int \frac {1}{x^3 \left (a^2 c x^2+c\right )^{5/2} \sqrt {\arctan (a x)}}dx}{a}-\frac {2}{a c x^2 \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\right )}{3 a}-\frac {2}{3 a c x \arctan (a x)^{3/2} \left (a^2 c x^2+c\right )^{3/2}}\) |
\(\Big \downarrow \) 5505 |
\(\displaystyle -\frac {8}{3} a \left (-\frac {6 \sqrt {a^2 x^2+1} \int \frac {a x}{\left (a^2 x^2+1\right )^{3/2} \sqrt {\arctan (a x)}}d\arctan (a x)}{a c^2 \sqrt {a^2 c x^2+c}}-\frac {2}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\right )-\frac {2 \left (-10 a \int \frac {1}{x \left (a^2 c x^2+c\right )^{5/2} \sqrt {\arctan (a x)}}dx-\frac {4 \int \frac {1}{x^3 \left (a^2 c x^2+c\right )^{5/2} \sqrt {\arctan (a x)}}dx}{a}-\frac {2}{a c x^2 \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\right )}{3 a}-\frac {2}{3 a c x \arctan (a x)^{3/2} \left (a^2 c x^2+c\right )^{3/2}}\) |
\(\Big \downarrow \) 4906 |
\(\displaystyle -\frac {8}{3} a \left (-\frac {6 \sqrt {a^2 x^2+1} \int \left (\frac {a x}{4 \sqrt {a^2 x^2+1} \sqrt {\arctan (a x)}}+\frac {\sin (3 \arctan (a x))}{4 \sqrt {\arctan (a x)}}\right )d\arctan (a x)}{a c^2 \sqrt {a^2 c x^2+c}}-\frac {2}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\right )-\frac {2 \left (-10 a \int \frac {1}{x \left (a^2 c x^2+c\right )^{5/2} \sqrt {\arctan (a x)}}dx-\frac {4 \int \frac {1}{x^3 \left (a^2 c x^2+c\right )^{5/2} \sqrt {\arctan (a x)}}dx}{a}-\frac {2}{a c x^2 \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\right )}{3 a}-\frac {2}{3 a c x \arctan (a x)^{3/2} \left (a^2 c x^2+c\right )^{3/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \left (-10 a \int \frac {1}{x \left (a^2 c x^2+c\right )^{5/2} \sqrt {\arctan (a x)}}dx-\frac {4 \int \frac {1}{x^3 \left (a^2 c x^2+c\right )^{5/2} \sqrt {\arctan (a x)}}dx}{a}-\frac {2}{a c x^2 \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\right )}{3 a}-\frac {8}{3} a \left (-\frac {6 \sqrt {a^2 x^2+1} \left (\frac {1}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )+\frac {1}{2} \sqrt {\frac {\pi }{6}} \operatorname {FresnelS}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )\right )}{a c^2 \sqrt {a^2 c x^2+c}}-\frac {2}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\right )-\frac {2}{3 a c x \arctan (a x)^{3/2} \left (a^2 c x^2+c\right )^{3/2}}\) |
\(\Big \downarrow \) 5560 |
\(\displaystyle -\frac {2 \left (-10 a \int \frac {1}{x \left (a^2 c x^2+c\right )^{5/2} \sqrt {\arctan (a x)}}dx-\frac {4 \int \frac {1}{x^3 \left (a^2 c x^2+c\right )^{5/2} \sqrt {\arctan (a x)}}dx}{a}-\frac {2}{a c x^2 \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\right )}{3 a}-\frac {8}{3} a \left (-\frac {6 \sqrt {a^2 x^2+1} \left (\frac {1}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )+\frac {1}{2} \sqrt {\frac {\pi }{6}} \operatorname {FresnelS}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )\right )}{a c^2 \sqrt {a^2 c x^2+c}}-\frac {2}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\right )-\frac {2}{3 a c x \arctan (a x)^{3/2} \left (a^2 c x^2+c\right )^{3/2}}\) |
3.12.7.3.1 Defintions of rubi rules used
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_.)*(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[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Unintegrab le[u*(a + b*ArcTan[c*x])^p, x] /; FreeQ[{a, b, c, p}, x] && (EqQ[u, 1] || M atchQ[u, ((d_.) + (e_.)*x)^(q_.) /; FreeQ[{d, e, q}, x]] || MatchQ[u, ((f_. )*x)^(m_.)*((d_.) + (e_.)*x)^(q_.) /; FreeQ[{d, e, f, m, q}, x]] || MatchQ[ u, ((d_.) + (e_.)*x^2)^(q_.) /; FreeQ[{d, e, q}, x]] || MatchQ[u, ((f_.)*x) ^(m_.)*((d_.) + (e_.)*x^2)^(q_.) /; FreeQ[{d, e, f, m, q}, x]])
Not integrable
Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85
\[\int \frac {1}{x \left (a^{2} c \,x^{2}+c \right )^{\frac {5}{2}} \arctan \left (a x \right )^{\frac {5}{2}}}d x\]
Exception generated. \[ \int \frac {1}{x \left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
Timed out. \[ \int \frac {1}{x \left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{5/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {1}{x \left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]
Exception generated. \[ \int \frac {1}{x \left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{5/2}} \, dx=\text {Exception raised: TypeError} \]
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
Not integrable
Time = 0.36 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x \left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{5/2}} \, dx=\int \frac {1}{x\,{\mathrm {atan}\left (a\,x\right )}^{5/2}\,{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]