Integrand size = 27, antiderivative size = 211 \[ \int x^{9/2} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \, dx=\frac {60 d^2 \sqrt {x} \sqrt {d+e x^2}}{847 (-e)^{5/2}}+\frac {36 d x^{5/2} \sqrt {d+e x^2}}{847 (-e)^{3/2}}+\frac {4 x^{9/2} \sqrt {d+e x^2}}{121 \sqrt {-e}}+\frac {2}{11} x^{11/2} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )+\frac {30 d^{11/4} \sqrt {-e} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{847 e^{13/4} \sqrt {d+e x^2}} \]
2/11*x^(11/2)*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))+36/847*d*x^(5/2)*(e*x^2 +d)^(1/2)/(-e)^(3/2)+4/121*x^(9/2)*(e*x^2+d)^(1/2)/(-e)^(1/2)+60/847*d^2*x ^(1/2)*(e*x^2+d)^(1/2)/(-e)^(5/2)+30/847*d^(11/4)*(cos(2*arctan(e^(1/4)*x^ (1/2)/d^(1/4)))^2)^(1/2)/cos(2*arctan(e^(1/4)*x^(1/2)/d^(1/4)))*EllipticF( sin(2*arctan(e^(1/4)*x^(1/2)/d^(1/4))),1/2*2^(1/2))*(-e)^(1/2)*(d^(1/2)+x* e^(1/2))*((e*x^2+d)/(d^(1/2)+x*e^(1/2))^2)^(1/2)/e^(13/4)/(e*x^2+d)^(1/2)
Result contains complex when optimal does not.
Time = 11.41 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.81 \[ \int x^{9/2} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \, dx=\frac {4 \sqrt {x} \sqrt {d+e x^2} \left (15 d^2-9 d e x^2+7 e^2 x^4\right )}{847 (-e)^{5/2}}+\frac {2}{11} x^{11/2} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {60 i d^3 \sqrt {1+\frac {d}{e x^2}} x \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {i \sqrt {d}}{\sqrt {e}}}}{\sqrt {x}}\right ),-1\right )}{847 \sqrt {\frac {i \sqrt {d}}{\sqrt {e}}} (-e)^{5/2} \sqrt {d+e x^2}} \]
(4*Sqrt[x]*Sqrt[d + e*x^2]*(15*d^2 - 9*d*e*x^2 + 7*e^2*x^4))/(847*(-e)^(5/ 2)) + (2*x^(11/2)*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]])/11 - (((60*I)/847) *d^3*Sqrt[1 + d/(e*x^2)]*x*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[d])/Sqrt[e]]/S qrt[x]], -1])/(Sqrt[(I*Sqrt[d])/Sqrt[e]]*(-e)^(5/2)*Sqrt[d + e*x^2])
Time = 0.34 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5674, 262, 262, 262, 266, 761}
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.
| \(\displaystyle \int x^{9/2} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \, dx\) |
| \(\Big \downarrow \) 5674 |
| \(\displaystyle \frac {2}{11} x^{11/2} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {2}{11} \sqrt {-e} \int \frac {x^{11/2}}{\sqrt {e x^2+d}}dx\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {2}{11} x^{11/2} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {2}{11} \sqrt {-e} \left (\frac {2 x^{9/2} \sqrt {d+e x^2}}{11 e}-\frac {9 d \int \frac {x^{7/2}}{\sqrt {e x^2+d}}dx}{11 e}\right )\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {2}{11} x^{11/2} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {2}{11} \sqrt {-e} \left (\frac {2 x^{9/2} \sqrt {d+e x^2}}{11 e}-\frac {9 d \left (\frac {2 x^{5/2} \sqrt {d+e x^2}}{7 e}-\frac {5 d \int \frac {x^{3/2}}{\sqrt {e x^2+d}}dx}{7 e}\right )}{11 e}\right )\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {2}{11} x^{11/2} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {2}{11} \sqrt {-e} \left (\frac {2 x^{9/2} \sqrt {d+e x^2}}{11 e}-\frac {9 d \left (\frac {2 x^{5/2} \sqrt {d+e x^2}}{7 e}-\frac {5 d \left (\frac {2 \sqrt {x} \sqrt {d+e x^2}}{3 e}-\frac {d \int \frac {1}{\sqrt {x} \sqrt {e x^2+d}}dx}{3 e}\right )}{7 e}\right )}{11 e}\right )\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {2}{11} x^{11/2} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {2}{11} \sqrt {-e} \left (\frac {2 x^{9/2} \sqrt {d+e x^2}}{11 e}-\frac {9 d \left (\frac {2 x^{5/2} \sqrt {d+e x^2}}{7 e}-\frac {5 d \left (\frac {2 \sqrt {x} \sqrt {d+e x^2}}{3 e}-\frac {2 d \int \frac {1}{\sqrt {e x^2+d}}d\sqrt {x}}{3 e}\right )}{7 e}\right )}{11 e}\right )\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {2}{11} x^{11/2} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {2}{11} \sqrt {-e} \left (\frac {2 x^{9/2} \sqrt {d+e x^2}}{11 e}-\frac {9 d \left (\frac {2 x^{5/2} \sqrt {d+e x^2}}{7 e}-\frac {5 d \left (\frac {2 \sqrt {x} \sqrt {d+e x^2}}{3 e}-\frac {d^{3/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{3 e^{5/4} \sqrt {d+e x^2}}\right )}{7 e}\right )}{11 e}\right )\) |
(2*x^(11/2)*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]])/11 - (2*Sqrt[-e]*((2*x^( 9/2)*Sqrt[d + e*x^2])/(11*e) - (9*d*((2*x^(5/2)*Sqrt[d + e*x^2])/(7*e) - ( 5*d*((2*Sqrt[x]*Sqrt[d + e*x^2])/(3*e) - (d^(3/4)*(Sqrt[d] + Sqrt[e]*x)*Sq rt[(d + e*x^2)/(Sqrt[d] + Sqrt[e]*x)^2]*EllipticF[2*ArcTan[(e^(1/4)*Sqrt[x ])/d^(1/4)], 1/2])/(3*e^(5/4)*Sqrt[d + e*x^2])))/(7*e)))/(11*e)))/11
3.1.18.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[ArcTan[((c_.)*(x_))/Sqrt[(a_.) + (b_.)*(x_)^2]]*((d_.)*(x_))^(m_.), x_S ymbol] :> Simp[(d*x)^(m + 1)*(ArcTan[(c*x)/Sqrt[a + b*x^2]]/(d*(m + 1))), x ] - Simp[c/(d*(m + 1)) Int[(d*x)^(m + 1)/Sqrt[a + b*x^2], x], x] /; FreeQ [{a, b, c, d, m}, x] && EqQ[b + c^2, 0] && NeQ[m, -1]
\[\int x^{\frac {9}{2}} \arctan \left (\frac {x \sqrt {-e}}{\sqrt {e \,x^{2}+d}}\right )d x\]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.13 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.45 \[ \int x^{9/2} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \, dx=\frac {2 \, {\left (77 \, e^{4} x^{\frac {11}{2}} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) + 30 \, d^{3} \sqrt {-e} \sqrt {e} {\rm weierstrassPInverse}\left (-\frac {4 \, d}{e}, 0, x\right ) - 2 \, {\left (7 \, e^{3} x^{4} - 9 \, d e^{2} x^{2} + 15 \, d^{2} e\right )} \sqrt {e x^{2} + d} \sqrt {-e} \sqrt {x}\right )}}{847 \, e^{4}} \]
2/847*(77*e^4*x^(11/2)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) + 30*d^3*sqrt(-e )*sqrt(e)*weierstrassPInverse(-4*d/e, 0, x) - 2*(7*e^3*x^4 - 9*d*e^2*x^2 + 15*d^2*e)*sqrt(e*x^2 + d)*sqrt(-e)*sqrt(x))/e^4
Timed out. \[ \int x^{9/2} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \, dx=\text {Timed out} \]
\[ \int x^{9/2} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \, dx=\int { x^{\frac {9}{2}} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) \,d x } \]
2/11*x^(11/2)*arctan2(sqrt(-e)*x, sqrt(e*x^2 + d)) - 2*d*sqrt(-e)*integrat e(-1/11*x*e^(1/2*log(e*x^2 + d) + 9/2*log(x))/(e^2*x^4 + d*e*x^2 - (e*x^2 + d)^2), x)
\[ \int x^{9/2} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \, dx=\int { x^{\frac {9}{2}} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) \,d x } \]
Timed out. \[ \int x^{9/2} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \, dx=\int x^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {-e}\,x}{\sqrt {e\,x^2+d}}\right ) \,d x \]