Integrand size = 23, antiderivative size = 155 \[ \int \frac {\arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=-\frac {15 x \sqrt {\arctan (a x)}}{4 c \sqrt {c+a^2 c x^2}}+\frac {5 \arctan (a x)^{3/2}}{2 a c \sqrt {c+a^2 c x^2}}+\frac {x \arctan (a x)^{5/2}}{c \sqrt {c+a^2 c x^2}}+\frac {15 \sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{4 a c \sqrt {c+a^2 c x^2}} \]
5/2*arctan(a*x)^(3/2)/a/c/(a^2*c*x^2+c)^(1/2)+x*arctan(a*x)^(5/2)/c/(a^2*c *x^2+c)^(1/2)+15/8*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)/a/c/(a^2*c*x^2+c)^(1/2)-15/4*x*arctan(a*x)^(1/2)/ c/(a^2*c*x^2+c)^(1/2)
Time = 0.10 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.63 \[ \int \frac {\arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {\arctan (a x)} \left (-15 a x+10 \arctan (a x)+4 a x \arctan (a x)^2\right )+15 \sqrt {2 \pi } \sqrt {1+a^2 x^2} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{8 a c \sqrt {c+a^2 c x^2}} \]
(2*Sqrt[ArcTan[a*x]]*(-15*a*x + 10*ArcTan[a*x] + 4*a*x*ArcTan[a*x]^2) + 15 *Sqrt[2*Pi]*Sqrt[1 + a^2*x^2]*FresnelS[Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/(8*a *c*Sqrt[c + a^2*c*x^2])
Time = 0.63 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.98, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {5433, 5440, 5439, 3042, 3777, 25, 3042, 3786, 3832}
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.
\(\displaystyle \int \frac {\arctan (a x)^{5/2}}{\left (a^2 c x^2+c\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 5433 |
\(\displaystyle -\frac {15}{4} \int \frac {\sqrt {\arctan (a x)}}{\left (a^2 c x^2+c\right )^{3/2}}dx+\frac {x \arctan (a x)^{5/2}}{c \sqrt {a^2 c x^2+c}}+\frac {5 \arctan (a x)^{3/2}}{2 a c \sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 5440 |
\(\displaystyle -\frac {15 \sqrt {a^2 x^2+1} \int \frac {\sqrt {\arctan (a x)}}{\left (a^2 x^2+1\right )^{3/2}}dx}{4 c \sqrt {a^2 c x^2+c}}+\frac {x \arctan (a x)^{5/2}}{c \sqrt {a^2 c x^2+c}}+\frac {5 \arctan (a x)^{3/2}}{2 a c \sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 5439 |
\(\displaystyle -\frac {15 \sqrt {a^2 x^2+1} \int \frac {\sqrt {\arctan (a x)}}{\sqrt {a^2 x^2+1}}d\arctan (a x)}{4 a c \sqrt {a^2 c x^2+c}}+\frac {x \arctan (a x)^{5/2}}{c \sqrt {a^2 c x^2+c}}+\frac {5 \arctan (a x)^{3/2}}{2 a c \sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {15 \sqrt {a^2 x^2+1} \int \sqrt {\arctan (a x)} \sin \left (\arctan (a x)+\frac {\pi }{2}\right )d\arctan (a x)}{4 a c \sqrt {a^2 c x^2+c}}+\frac {x \arctan (a x)^{5/2}}{c \sqrt {a^2 c x^2+c}}+\frac {5 \arctan (a x)^{3/2}}{2 a c \sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle -\frac {15 \sqrt {a^2 x^2+1} \left (\frac {1}{2} \int -\frac {a x}{\sqrt {a^2 x^2+1} \sqrt {\arctan (a x)}}d\arctan (a x)+\frac {a x \sqrt {\arctan (a x)}}{\sqrt {a^2 x^2+1}}\right )}{4 a c \sqrt {a^2 c x^2+c}}+\frac {x \arctan (a x)^{5/2}}{c \sqrt {a^2 c x^2+c}}+\frac {5 \arctan (a x)^{3/2}}{2 a c \sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {15 \sqrt {a^2 x^2+1} \left (\frac {a x \sqrt {\arctan (a x)}}{\sqrt {a^2 x^2+1}}-\frac {1}{2} \int \frac {a x}{\sqrt {a^2 x^2+1} \sqrt {\arctan (a x)}}d\arctan (a x)\right )}{4 a c \sqrt {a^2 c x^2+c}}+\frac {x \arctan (a x)^{5/2}}{c \sqrt {a^2 c x^2+c}}+\frac {5 \arctan (a x)^{3/2}}{2 a c \sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {15 \sqrt {a^2 x^2+1} \left (\frac {a x \sqrt {\arctan (a x)}}{\sqrt {a^2 x^2+1}}-\frac {1}{2} \int \frac {\sin (\arctan (a x))}{\sqrt {\arctan (a x)}}d\arctan (a x)\right )}{4 a c \sqrt {a^2 c x^2+c}}+\frac {x \arctan (a x)^{5/2}}{c \sqrt {a^2 c x^2+c}}+\frac {5 \arctan (a x)^{3/2}}{2 a c \sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 3786 |
\(\displaystyle -\frac {15 \sqrt {a^2 x^2+1} \left (\frac {a x \sqrt {\arctan (a x)}}{\sqrt {a^2 x^2+1}}-\int \frac {a x}{\sqrt {a^2 x^2+1}}d\sqrt {\arctan (a x)}\right )}{4 a c \sqrt {a^2 c x^2+c}}+\frac {x \arctan (a x)^{5/2}}{c \sqrt {a^2 c x^2+c}}+\frac {5 \arctan (a x)^{3/2}}{2 a c \sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 3832 |
\(\displaystyle -\frac {15 \sqrt {a^2 x^2+1} \left (\frac {a x \sqrt {\arctan (a x)}}{\sqrt {a^2 x^2+1}}-\sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )\right )}{4 a c \sqrt {a^2 c x^2+c}}+\frac {x \arctan (a x)^{5/2}}{c \sqrt {a^2 c x^2+c}}+\frac {5 \arctan (a x)^{3/2}}{2 a c \sqrt {a^2 c x^2+c}}\) |
(5*ArcTan[a*x]^(3/2))/(2*a*c*Sqrt[c + a^2*c*x^2]) + (x*ArcTan[a*x]^(5/2))/ (c*Sqrt[c + a^2*c*x^2]) - (15*Sqrt[1 + a^2*x^2]*((a*x*Sqrt[ArcTan[a*x]])/S qrt[1 + a^2*x^2] - Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]]))/(4* a*c*Sqrt[c + a^2*c*x^2])
3.10.5.3.1 Defintions of rubi rules used
Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( -(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1)*C os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[2/d Subst[Int[Sin[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]
Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)/((d_) + (e_.)*(x_)^2)^(3/2), x_ Symbol] :> Simp[b*p*((a + b*ArcTan[c*x])^(p - 1)/(c*d*Sqrt[d + e*x^2])), x] + (Simp[x*((a + b*ArcTan[c*x])^p/(d*Sqrt[d + e*x^2])), x] - Simp[b^2*p*(p - 1) Int[(a + b*ArcTan[c*x])^(p - 2)/(d + e*x^2)^(3/2), x], x]) /; FreeQ[ {a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[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_.)*((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])
\[\int \frac {\arctan \left (a x \right )^{\frac {5}{2}}}{\left (a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}d x\]
Exception generated. \[ \int \frac {\arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \frac {\arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\int \frac {\operatorname {atan}^{\frac {5}{2}}{\left (a x \right )}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {\arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {\arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {\arctan \left (a x\right )^{\frac {5}{2}}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\int \frac {{\mathrm {atan}\left (a\,x\right )}^{5/2}}{{\left (c\,a^2\,x^2+c\right )}^{3/2}} \,d x \]