Integrand size = 24, antiderivative size = 161 \[ \int \frac {x \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {15 \sqrt {\arctan (a x)}}{4 a^2 c \sqrt {c+a^2 c x^2}}+\frac {5 x \arctan (a x)^{3/2}}{2 a c \sqrt {c+a^2 c x^2}}-\frac {\arctan (a x)^{5/2}}{a^2 c \sqrt {c+a^2 c x^2}}-\frac {15 \sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{4 a^2 c \sqrt {c+a^2 c x^2}} \]
5/2*x*arctan(a*x)^(3/2)/a/c/(a^2*c*x^2+c)^(1/2)-arctan(a*x)^(5/2)/a^2/c/(a ^2*c*x^2+c)^(1/2)-15/8*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/(a^2*c*x^2+c)^(1/2)+15/4*arctan(a*x)^(1 /2)/a^2/c/(a^2*c*x^2+c)^(1/2)
Result contains complex when optimal does not.
Time = 0.18 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.86 \[ \int \frac {x \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {4 \arctan (a x) \left (15+10 a x \arctan (a x)-4 \arctan (a x)^2\right )+15 i \sqrt {1+a^2 x^2} \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-i \arctan (a x)\right )-15 i \sqrt {1+a^2 x^2} \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},i \arctan (a x)\right )}{16 a^2 c \sqrt {c+a^2 c x^2} \sqrt {\arctan (a x)}} \]
(4*ArcTan[a*x]*(15 + 10*a*x*ArcTan[a*x] - 4*ArcTan[a*x]^2) + (15*I)*Sqrt[1 + a^2*x^2]*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-I)*ArcTan[a*x]] - (15*I)*S qrt[1 + a^2*x^2]*Sqrt[I*ArcTan[a*x]]*Gamma[1/2, I*ArcTan[a*x]])/(16*a^2*c* Sqrt[c + a^2*c*x^2]*Sqrt[ArcTan[a*x]])
Time = 0.71 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {5465, 5433, 5440, 5439, 3042, 3785, 3833}
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.
\(\displaystyle \int \frac {x \arctan (a x)^{5/2}}{\left (a^2 c x^2+c\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 5465 |
\(\displaystyle \frac {5 \int \frac {\arctan (a x)^{3/2}}{\left (a^2 c x^2+c\right )^{3/2}}dx}{2 a}-\frac {\arctan (a x)^{5/2}}{a^2 c \sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 5433 |
\(\displaystyle \frac {5 \left (-\frac {3}{4} \int \frac {1}{\left (a^2 c x^2+c\right )^{3/2} \sqrt {\arctan (a x)}}dx+\frac {x \arctan (a x)^{3/2}}{c \sqrt {a^2 c x^2+c}}+\frac {3 \sqrt {\arctan (a x)}}{2 a c \sqrt {a^2 c x^2+c}}\right )}{2 a}-\frac {\arctan (a x)^{5/2}}{a^2 c \sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 5440 |
\(\displaystyle \frac {5 \left (-\frac {3 \sqrt {a^2 x^2+1} \int \frac {1}{\left (a^2 x^2+1\right )^{3/2} \sqrt {\arctan (a x)}}dx}{4 c \sqrt {a^2 c x^2+c}}+\frac {x \arctan (a x)^{3/2}}{c \sqrt {a^2 c x^2+c}}+\frac {3 \sqrt {\arctan (a x)}}{2 a c \sqrt {a^2 c x^2+c}}\right )}{2 a}-\frac {\arctan (a x)^{5/2}}{a^2 c \sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 5439 |
\(\displaystyle \frac {5 \left (-\frac {3 \sqrt {a^2 x^2+1} \int \frac {1}{\sqrt {a^2 x^2+1} \sqrt {\arctan (a x)}}d\arctan (a x)}{4 a c \sqrt {a^2 c x^2+c}}+\frac {x \arctan (a x)^{3/2}}{c \sqrt {a^2 c x^2+c}}+\frac {3 \sqrt {\arctan (a x)}}{2 a c \sqrt {a^2 c x^2+c}}\right )}{2 a}-\frac {\arctan (a x)^{5/2}}{a^2 c \sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5 \left (-\frac {3 \sqrt {a^2 x^2+1} \int \frac {\sin \left (\arctan (a x)+\frac {\pi }{2}\right )}{\sqrt {\arctan (a x)}}d\arctan (a x)}{4 a c \sqrt {a^2 c x^2+c}}+\frac {x \arctan (a x)^{3/2}}{c \sqrt {a^2 c x^2+c}}+\frac {3 \sqrt {\arctan (a x)}}{2 a c \sqrt {a^2 c x^2+c}}\right )}{2 a}-\frac {\arctan (a x)^{5/2}}{a^2 c \sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 3785 |
\(\displaystyle \frac {5 \left (-\frac {3 \sqrt {a^2 x^2+1} \int \frac {1}{\sqrt {a^2 x^2+1}}d\sqrt {\arctan (a x)}}{2 a c \sqrt {a^2 c x^2+c}}+\frac {x \arctan (a x)^{3/2}}{c \sqrt {a^2 c x^2+c}}+\frac {3 \sqrt {\arctan (a x)}}{2 a c \sqrt {a^2 c x^2+c}}\right )}{2 a}-\frac {\arctan (a x)^{5/2}}{a^2 c \sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 3833 |
\(\displaystyle \frac {5 \left (-\frac {3 \sqrt {\frac {\pi }{2}} \sqrt {a^2 x^2+1} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{2 a c \sqrt {a^2 c x^2+c}}+\frac {x \arctan (a x)^{3/2}}{c \sqrt {a^2 c x^2+c}}+\frac {3 \sqrt {\arctan (a x)}}{2 a c \sqrt {a^2 c x^2+c}}\right )}{2 a}-\frac {\arctan (a x)^{5/2}}{a^2 c \sqrt {a^2 c x^2+c}}\) |
-(ArcTan[a*x]^(5/2)/(a^2*c*Sqrt[c + a^2*c*x^2])) + (5*((3*Sqrt[ArcTan[a*x] ])/(2*a*c*Sqrt[c + a^2*c*x^2]) + (x*ArcTan[a*x]^(3/2))/(c*Sqrt[c + a^2*c*x ^2]) - (3*Sqrt[Pi/2]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[2/Pi]*Sqrt[ArcTan[a*x ]]])/(2*a*c*Sqrt[c + a^2*c*x^2])))/(2*a)
3.10.4.3.1 Defintions of rubi rules used
Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> S imp[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]
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]
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[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q_ .), x_Symbol] :> Simp[(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])^p/(2*e*(q + 1))), x] - Simp[b*(p/(2*c*(q + 1))) Int[(d + e*x^2)^q*(a + b*ArcTan[c*x]) ^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[e, c^2*d] && GtQ[p, 0] && NeQ[q, -1]
\[\int \frac {x \arctan \left (a x \right )^{\frac {5}{2}}}{\left (a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}d x\]
Exception generated. \[ \int \frac {x \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
Timed out. \[ \int \frac {x \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {x \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {x \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {x \arctan \left (a x\right )^{\frac {5}{2}}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\int \frac {x\,{\mathrm {atan}\left (a\,x\right )}^{5/2}}{{\left (c\,a^2\,x^2+c\right )}^{3/2}} \,d x \]