Integrand size = 24, antiderivative size = 305 \[ \int \frac {x^3 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=-\frac {2}{a^4 c \sqrt {c+a^2 c x^2}}-\frac {2 x \arctan (a x)}{a^3 c \sqrt {c+a^2 c x^2}}+\frac {\arctan (a x)^2}{a^4 c \sqrt {c+a^2 c x^2}}+\frac {\sqrt {c+a^2 c x^2} \arctan (a x)^2}{a^4 c^2}+\frac {4 i \sqrt {1+a^2 x^2} \arctan (a x) \arctan \left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a^4 c \sqrt {c+a^2 c x^2}}-\frac {2 i \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a^4 c \sqrt {c+a^2 c x^2}}+\frac {2 i \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a^4 c \sqrt {c+a^2 c x^2}} \]
-2/a^4/c/(a^2*c*x^2+c)^(1/2)-2*x*arctan(a*x)/a^3/c/(a^2*c*x^2+c)^(1/2)+arc tan(a*x)^2/a^4/c/(a^2*c*x^2+c)^(1/2)+4*I*arctan(a*x)*arctan((1+I*a*x)^(1/2 )/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/a^4/c/(a^2*c*x^2+c)^(1/2)-2*I*polylog (2,-I*(1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/a^4/c/(a^2*c*x^2+ c)^(1/2)+2*I*polylog(2,I*(1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2 )/a^4/c/(a^2*c*x^2+c)^(1/2)+arctan(a*x)^2*(a^2*c*x^2+c)^(1/2)/a^4/c^2
Time = 0.75 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.69 \[ \int \frac {x^3 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {c \left (1+a^2 x^2\right )} \left (-2+3 \arctan (a x)^2-2 \cos (2 \arctan (a x))+\arctan (a x)^2 \cos (2 \arctan (a x))-\frac {4 \arctan (a x) \log \left (1-i e^{i \arctan (a x)}\right )}{\sqrt {1+a^2 x^2}}+\frac {4 \arctan (a x) \log \left (1+i e^{i \arctan (a x)}\right )}{\sqrt {1+a^2 x^2}}-\frac {4 i \operatorname {PolyLog}\left (2,-i e^{i \arctan (a x)}\right )}{\sqrt {1+a^2 x^2}}+\frac {4 i \operatorname {PolyLog}\left (2,i e^{i \arctan (a x)}\right )}{\sqrt {1+a^2 x^2}}-2 \arctan (a x) \sin (2 \arctan (a x))\right )}{2 a^4 c^2} \]
(Sqrt[c*(1 + a^2*x^2)]*(-2 + 3*ArcTan[a*x]^2 - 2*Cos[2*ArcTan[a*x]] + ArcT an[a*x]^2*Cos[2*ArcTan[a*x]] - (4*ArcTan[a*x]*Log[1 - I*E^(I*ArcTan[a*x])] )/Sqrt[1 + a^2*x^2] + (4*ArcTan[a*x]*Log[1 + I*E^(I*ArcTan[a*x])])/Sqrt[1 + a^2*x^2] - ((4*I)*PolyLog[2, (-I)*E^(I*ArcTan[a*x])])/Sqrt[1 + a^2*x^2] + ((4*I)*PolyLog[2, I*E^(I*ArcTan[a*x])])/Sqrt[1 + a^2*x^2] - 2*ArcTan[a*x ]*Sin[2*ArcTan[a*x]]))/(2*a^4*c^2)
Time = 0.90 (sec) , antiderivative size = 264, normalized size of antiderivative = 0.87, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {5499, 5465, 5425, 5421, 5429}
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.
| \(\displaystyle \int \frac {x^3 \arctan (a x)^2}{\left (a^2 c x^2+c\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 5499 |
| \(\displaystyle \frac {\int \frac {x \arctan (a x)^2}{\sqrt {a^2 c x^2+c}}dx}{a^2 c}-\frac {\int \frac {x \arctan (a x)^2}{\left (a^2 c x^2+c\right )^{3/2}}dx}{a^2}\) |
| \(\Big \downarrow \) 5465 |
| \(\displaystyle \frac {\frac {\arctan (a x)^2 \sqrt {a^2 c x^2+c}}{a^2 c}-\frac {2 \int \frac {\arctan (a x)}{\sqrt {a^2 c x^2+c}}dx}{a}}{a^2 c}-\frac {\frac {2 \int \frac {\arctan (a x)}{\left (a^2 c x^2+c\right )^{3/2}}dx}{a}-\frac {\arctan (a x)^2}{a^2 c \sqrt {a^2 c x^2+c}}}{a^2}\) |
| \(\Big \downarrow \) 5425 |
| \(\displaystyle \frac {\frac {\arctan (a x)^2 \sqrt {a^2 c x^2+c}}{a^2 c}-\frac {2 \sqrt {a^2 x^2+1} \int \frac {\arctan (a x)}{\sqrt {a^2 x^2+1}}dx}{a \sqrt {a^2 c x^2+c}}}{a^2 c}-\frac {\frac {2 \int \frac {\arctan (a x)}{\left (a^2 c x^2+c\right )^{3/2}}dx}{a}-\frac {\arctan (a x)^2}{a^2 c \sqrt {a^2 c x^2+c}}}{a^2}\) |
| \(\Big \downarrow \) 5421 |
| \(\displaystyle -\frac {\frac {2 \int \frac {\arctan (a x)}{\left (a^2 c x^2+c\right )^{3/2}}dx}{a}-\frac {\arctan (a x)^2}{a^2 c \sqrt {a^2 c x^2+c}}}{a^2}+\frac {\frac {\arctan (a x)^2 \sqrt {a^2 c x^2+c}}{a^2 c}-\frac {2 \sqrt {a^2 x^2+1} \left (-\frac {2 i \arctan (a x) \arctan \left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a}+\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{a}-\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{a}\right )}{a \sqrt {a^2 c x^2+c}}}{a^2 c}\) |
| \(\Big \downarrow \) 5429 |
| \(\displaystyle -\frac {\frac {2 \left (\frac {x \arctan (a x)}{c \sqrt {a^2 c x^2+c}}+\frac {1}{a c \sqrt {a^2 c x^2+c}}\right )}{a}-\frac {\arctan (a x)^2}{a^2 c \sqrt {a^2 c x^2+c}}}{a^2}+\frac {\frac {\arctan (a x)^2 \sqrt {a^2 c x^2+c}}{a^2 c}-\frac {2 \sqrt {a^2 x^2+1} \left (-\frac {2 i \arctan (a x) \arctan \left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a}+\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{a}-\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{a}\right )}{a \sqrt {a^2 c x^2+c}}}{a^2 c}\) |
-((-(ArcTan[a*x]^2/(a^2*c*Sqrt[c + a^2*c*x^2])) + (2*(1/(a*c*Sqrt[c + a^2* c*x^2]) + (x*ArcTan[a*x])/(c*Sqrt[c + a^2*c*x^2])))/a)/a^2) + ((Sqrt[c + a ^2*c*x^2]*ArcTan[a*x]^2)/(a^2*c) - (2*Sqrt[1 + a^2*x^2]*(((-2*I)*ArcTan[a* x]*ArcTan[Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/a + (I*PolyLog[2, ((-I)*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/a - (I*PolyLog[2, (I*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/a))/(a*Sqrt[c + a^2*c*x^2]))/(a^2*c)
3.4.39.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[-2*I*(a + b*ArcTan[c*x])*(ArcTan[Sqrt[1 + I*c*x]/Sqrt[1 - I*c*x]]/ (c*Sqrt[d])), x] + (Simp[I*b*(PolyLog[2, (-I)*(Sqrt[1 + I*c*x]/Sqrt[1 - I*c *x])]/(c*Sqrt[d])), x] - Simp[I*b*(PolyLog[2, I*(Sqrt[1 + I*c*x]/Sqrt[1 - I *c*x])]/(c*Sqrt[d])), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[d, 0]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_S ymbol] :> Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2] Int[(a + b*ArcTan[c*x])^ p/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] & & IGtQ[p, 0] && !GtQ[d, 0]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbo l] :> Simp[b/(c*d*Sqrt[d + e*x^2]), x] + Simp[x*((a + b*ArcTan[c*x])/(d*Sqr t[d + e*x^2])), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d]
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[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^2 )^(q_), x_Symbol] :> Simp[1/e Int[x^(m - 2)*(d + e*x^2)^(q + 1)*(a + b*Ar cTan[c*x])^p, x], x] - Simp[d/e Int[x^(m - 2)*(d + e*x^2)^q*(a + b*ArcTan [c*x])^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegersQ [p, 2*q] && LtQ[q, -1] && IGtQ[m, 1] && NeQ[p, -1]
Time = 1.20 (sec) , antiderivative size = 294, normalized size of antiderivative = 0.96
| method | result | size |
| default | \(\frac {\left (\arctan \left (a x \right )^{2}-2+2 i \arctan \left (a x \right )\right ) \left (i a x +1\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 \left (a^{2} x^{2}+1\right ) a^{4} c^{2}}-\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (i a x -1\right ) \left (\arctan \left (a x \right )^{2}-2-2 i \arctan \left (a x \right )\right )}{2 \left (a^{2} x^{2}+1\right ) a^{4} c^{2}}+\frac {\arctan \left (a x \right )^{2} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{a^{4} c^{2}}+\frac {2 \left (\arctan \left (a x \right ) \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-\arctan \left (a x \right ) \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-i \operatorname {dilog}\left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+i \operatorname {dilog}\left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{\sqrt {a^{2} x^{2}+1}\, a^{4} c^{2}}\) | \(294\) |
1/2*(arctan(a*x)^2-2+2*I*arctan(a*x))*(1+I*a*x)*(c*(a*x-I)*(I+a*x))^(1/2)/ (a^2*x^2+1)/a^4/c^2-1/2*(c*(a*x-I)*(I+a*x))^(1/2)*(I*a*x-1)*(arctan(a*x)^2 -2-2*I*arctan(a*x))/(a^2*x^2+1)/a^4/c^2+arctan(a*x)^2*(c*(a*x-I)*(I+a*x))^ (1/2)/a^4/c^2+2*(arctan(a*x)*ln(1+I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-arctan(a* x)*ln(1-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-I*dilog(1+I*(1+I*a*x)/(a^2*x^2+1)^( 1/2))+I*dilog(1-I*(1+I*a*x)/(a^2*x^2+1)^(1/2)))*(c*(a*x-I)*(I+a*x))^(1/2)/ (a^2*x^2+1)^(1/2)/a^4/c^2
\[ \int \frac {x^3 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {x^{3} \arctan \left (a x\right )^{2}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {x^3 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\int \frac {x^{3} \operatorname {atan}^{2}{\left (a x \right )}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {x^3 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {x^{3} \arctan \left (a x\right )^{2}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}}} \,d x } \]
Exception generated. \[ \int \frac {x^3 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {x^3 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\int \frac {x^3\,{\mathrm {atan}\left (a\,x\right )}^2}{{\left (c\,a^2\,x^2+c\right )}^{3/2}} \,d x \]