Integrand size = 20, antiderivative size = 201 \[ \int x^m \left (c+a^2 c x^2\right )^2 \arctan (a x) \, dx=\frac {c^2 x^{1+m} \arctan (a x)}{1+m}+\frac {2 a^2 c^2 x^{3+m} \arctan (a x)}{3+m}+\frac {a^4 c^2 x^{5+m} \arctan (a x)}{5+m}-\frac {a c^2 x^{2+m} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-a^2 x^2\right )}{2+3 m+m^2}-\frac {2 a^3 c^2 x^{4+m} \operatorname {Hypergeometric2F1}\left (1,\frac {4+m}{2},\frac {6+m}{2},-a^2 x^2\right )}{12+7 m+m^2}-\frac {a^5 c^2 x^{6+m} \operatorname {Hypergeometric2F1}\left (1,\frac {6+m}{2},\frac {8+m}{2},-a^2 x^2\right )}{(5+m) (6+m)} \]
c^2*x^(1+m)*arctan(a*x)/(1+m)+2*a^2*c^2*x^(3+m)*arctan(a*x)/(3+m)+a^4*c^2* x^(5+m)*arctan(a*x)/(5+m)-a*c^2*x^(2+m)*hypergeom([1, 1+1/2*m],[2+1/2*m],- a^2*x^2)/(m^2+3*m+2)-2*a^3*c^2*x^(4+m)*hypergeom([1, 2+1/2*m],[3+1/2*m],-a ^2*x^2)/(m^2+7*m+12)-a^5*c^2*x^(6+m)*hypergeom([1, 3+1/2*m],[4+1/2*m],-a^2 *x^2)/(5+m)/(6+m)
Time = 0.10 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.87 \[ \int x^m \left (c+a^2 c x^2\right )^2 \arctan (a x) \, dx=c^2 x^{1+m} \left (\frac {\arctan (a x)}{1+m}+\frac {2 a^2 x^2 \arctan (a x)}{3+m}+\frac {a^4 x^4 \arctan (a x)}{5+m}-\frac {a x \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-a^2 x^2\right )}{2+3 m+m^2}-\frac {2 a^3 x^3 \operatorname {Hypergeometric2F1}\left (1,\frac {4+m}{2},\frac {6+m}{2},-a^2 x^2\right )}{12+7 m+m^2}-\frac {a^5 x^5 \operatorname {Hypergeometric2F1}\left (1,\frac {6+m}{2},\frac {8+m}{2},-a^2 x^2\right )}{(5+m) (6+m)}\right ) \]
c^2*x^(1 + m)*(ArcTan[a*x]/(1 + m) + (2*a^2*x^2*ArcTan[a*x])/(3 + m) + (a^ 4*x^4*ArcTan[a*x])/(5 + m) - (a*x*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/ 2, -(a^2*x^2)])/(2 + 3*m + m^2) - (2*a^3*x^3*Hypergeometric2F1[1, (4 + m)/ 2, (6 + m)/2, -(a^2*x^2)])/(12 + 7*m + m^2) - (a^5*x^5*Hypergeometric2F1[1 , (6 + m)/2, (8 + m)/2, -(a^2*x^2)])/((5 + m)*(6 + m)))
Time = 0.38 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {5483, 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.
| \(\displaystyle \int x^m \arctan (a x) \left (a^2 c x^2+c\right )^2 \, dx\) |
| \(\Big \downarrow \) 5483 |
| \(\displaystyle \int \left (a^4 c^2 x^{m+4} \arctan (a x)+2 a^2 c^2 x^{m+2} \arctan (a x)+c^2 x^m \arctan (a x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^4 c^2 x^{m+5} \arctan (a x)}{m+5}+\frac {2 a^2 c^2 x^{m+3} \arctan (a x)}{m+3}-\frac {a c^2 x^{m+2} \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-a^2 x^2\right )}{m^2+3 m+2}-\frac {a^5 c^2 x^{m+6} \operatorname {Hypergeometric2F1}\left (1,\frac {m+6}{2},\frac {m+8}{2},-a^2 x^2\right )}{(m+5) (m+6)}-\frac {2 a^3 c^2 x^{m+4} \operatorname {Hypergeometric2F1}\left (1,\frac {m+4}{2},\frac {m+6}{2},-a^2 x^2\right )}{m^2+7 m+12}+\frac {c^2 x^{m+1} \arctan (a x)}{m+1}\) |
(c^2*x^(1 + m)*ArcTan[a*x])/(1 + m) + (2*a^2*c^2*x^(3 + m)*ArcTan[a*x])/(3 + m) + (a^4*c^2*x^(5 + m)*ArcTan[a*x])/(5 + m) - (a*c^2*x^(2 + m)*Hyperge ometric2F1[1, (2 + m)/2, (4 + m)/2, -(a^2*x^2)])/(2 + 3*m + m^2) - (2*a^3* c^2*x^(4 + m)*Hypergeometric2F1[1, (4 + m)/2, (6 + m)/2, -(a^2*x^2)])/(12 + 7*m + m^2) - (a^5*c^2*x^(6 + m)*Hypergeometric2F1[1, (6 + m)/2, (8 + m)/ 2, -(a^2*x^2)])/((5 + m)*(6 + m))
3.3.49.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_)*((d_) + (e_. )*(x_)^2)^(q_), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*ArcTan[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2* d] && IGtQ[p, 0] && IGtQ[q, 1] && (EqQ[p, 1] || IntegerQ[m])
Result contains higher order function than in optimal. Order 9 vs. order 5.
Time = 57.86 (sec) , antiderivative size = 376, normalized size of antiderivative = 1.87
| method | result | size |
| meijerg | \(\frac {a^{-m -1} c^{2} \left (-\frac {4 x^{m} a^{m} \left (a^{4} m^{2} x^{4}+2 a^{4} m \,x^{4}-a^{2} m^{2} x^{2}-4 a^{2} m \,x^{2}+m^{2}+6 m +8\right )}{\left (5+m \right ) m \left (2+m \right ) \left (4+m \right )}+\frac {8 x^{6+m} a^{6+m} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{\left (10+2 m \right ) \sqrt {a^{2} x^{2}}}+\frac {2 x^{m} a^{m} \operatorname {LerchPhi}\left (-a^{2} x^{2}, 1, \frac {m}{2}\right )}{5+m}\right )}{4}+\frac {a^{-m -1} c^{2} \left (-\frac {4 x^{m} a^{m} \left (a^{2} m \,x^{2}-m -2\right )}{\left (3+m \right ) m \left (2+m \right )}+\frac {8 x^{4+m} a^{4+m} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{\left (6+2 m \right ) \sqrt {a^{2} x^{2}}}+\frac {2 x^{m} a^{m} \left (-m -4\right ) \operatorname {LerchPhi}\left (-a^{2} x^{2}, 1, \frac {m}{2}\right )}{\left (4+m \right ) \left (3+m \right )}\right )}{2}+\frac {a^{-m -1} c^{2} \left (\frac {4 x^{m} a^{m} \left (-m -2\right )}{\left (2+m \right ) \left (1+m \right ) m}+\frac {8 x^{2+m} a^{2+m} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{\left (2+2 m \right ) \sqrt {a^{2} x^{2}}}+\frac {2 x^{m} a^{m} \operatorname {LerchPhi}\left (-a^{2} x^{2}, 1, \frac {m}{2}\right )}{1+m}\right )}{4}\) | \(376\) |
1/4*a^(-m-1)*c^2*(-4*x^m*a^m*(a^4*m^2*x^4+2*a^4*m*x^4-a^2*m^2*x^2-4*a^2*m* x^2+m^2+6*m+8)/(5+m)/m/(2+m)/(4+m)+8*x^(6+m)*a^(6+m)/(10+2*m)/(a^2*x^2)^(1 /2)*arctan((a^2*x^2)^(1/2))+2*x^m*a^m/(5+m)*LerchPhi(-a^2*x^2,1,1/2*m))+1/ 2*a^(-m-1)*c^2*(-4*x^m*a^m*(a^2*m*x^2-m-2)/(3+m)/m/(2+m)+8*x^(4+m)*a^(4+m) /(6+2*m)/(a^2*x^2)^(1/2)*arctan((a^2*x^2)^(1/2))+2/(4+m)*x^m*a^m*(-m-4)/(3 +m)*LerchPhi(-a^2*x^2,1,1/2*m))+1/4*a^(-m-1)*c^2*(4/(2+m)*x^m*a^m*(-m-2)/( 1+m)/m+8*x^(2+m)*a^(2+m)/(2+2*m)/(a^2*x^2)^(1/2)*arctan((a^2*x^2)^(1/2))+2 *x^m*a^m/(1+m)*LerchPhi(-a^2*x^2,1,1/2*m))
\[ \int x^m \left (c+a^2 c x^2\right )^2 \arctan (a x) \, dx=\int { {\left (a^{2} c x^{2} + c\right )}^{2} x^{m} \arctan \left (a x\right ) \,d x } \]
\[ \int x^m \left (c+a^2 c x^2\right )^2 \arctan (a x) \, dx=c^{2} \left (\int x^{m} \operatorname {atan}{\left (a x \right )}\, dx + \int 2 a^{2} x^{2} x^{m} \operatorname {atan}{\left (a x \right )}\, dx + \int a^{4} x^{4} x^{m} \operatorname {atan}{\left (a x \right )}\, dx\right ) \]
c**2*(Integral(x**m*atan(a*x), x) + Integral(2*a**2*x**2*x**m*atan(a*x), x ) + Integral(a**4*x**4*x**m*atan(a*x), x))
\[ \int x^m \left (c+a^2 c x^2\right )^2 \arctan (a x) \, dx=\int { {\left (a^{2} c x^{2} + c\right )}^{2} x^{m} \arctan \left (a x\right ) \,d x } \]
(((a^4*c^2*m^2 + 4*a^4*c^2*m + 3*a^4*c^2)*x^5 + 2*(a^2*c^2*m^2 + 6*a^2*c^2 *m + 5*a^2*c^2)*x^3 + (c^2*m^2 + 8*c^2*m + 15*c^2)*x)*x^m*arctan(a*x) - (m ^3 + 9*m^2 + 23*m + 15)*integrate(((a^5*c^2*m^2 + 4*a^5*c^2*m + 3*a^5*c^2) *x^5 + 2*(a^3*c^2*m^2 + 6*a^3*c^2*m + 5*a^3*c^2)*x^3 + (a*c^2*m^2 + 8*a*c^ 2*m + 15*a*c^2)*x)*x^m/(m^3 + (a^2*m^3 + 9*a^2*m^2 + 23*a^2*m + 15*a^2)*x^ 2 + 9*m^2 + 23*m + 15), x))/(m^3 + 9*m^2 + 23*m + 15)
\[ \int x^m \left (c+a^2 c x^2\right )^2 \arctan (a x) \, dx=\int { {\left (a^{2} c x^{2} + c\right )}^{2} x^{m} \arctan \left (a x\right ) \,d x } \]
Timed out. \[ \int x^m \left (c+a^2 c x^2\right )^2 \arctan (a x) \, dx=\int x^m\,\mathrm {atan}\left (a\,x\right )\,{\left (c\,a^2\,x^2+c\right )}^2 \,d x \]