Integrand size = 23, antiderivative size = 143 \[ \int \frac {x^3 (a+b \arctan (c x))}{\left (d+e x^2\right )^{5/2}} \, dx=\frac {b c x}{3 \left (c^2 d-e\right ) e \sqrt {d+e x^2}}+\frac {d (a+b \arctan (c x))}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \arctan (c x)}{e^2 \sqrt {d+e x^2}}+\frac {b c \left (2 c^2 d-3 e\right ) \arctan \left (\frac {\sqrt {c^2 d-e} x}{\sqrt {d+e x^2}}\right )}{3 \left (c^2 d-e\right )^{3/2} e^2} \]
1/3*d*(a+b*arctan(c*x))/e^2/(e*x^2+d)^(3/2)+1/3*b*c*(2*c^2*d-3*e)*arctan(x *(c^2*d-e)^(1/2)/(e*x^2+d)^(1/2))/(c^2*d-e)^(3/2)/e^2+1/3*b*c*x/(c^2*d-e)/ e/(e*x^2+d)^(1/2)+(-a-b*arctan(c*x))/e^2/(e*x^2+d)^(1/2)
Result contains complex when optimal does not.
Time = 0.49 (sec) , antiderivative size = 326, normalized size of antiderivative = 2.28 \[ \int \frac {x^3 (a+b \arctan (c x))}{\left (d+e x^2\right )^{5/2}} \, dx=\frac {2 \sqrt {c^2 d-e} \left (b c e x \left (d+e x^2\right )-a \left (c^2 d-e\right ) \left (2 d+3 e x^2\right )\right )-2 b \left (c^2 d-e\right )^{3/2} \left (2 d+3 e x^2\right ) \arctan (c x)-i b c \left (2 c^2 d-3 e\right ) \left (d+e x^2\right )^{3/2} \log \left (-\frac {12 i \sqrt {c^2 d-e} e^2 \left (c d-i e x+\sqrt {c^2 d-e} \sqrt {d+e x^2}\right )}{b \left (2 c^2 d-3 e\right ) (i+c x)}\right )+i b c \left (2 c^2 d-3 e\right ) \left (d+e x^2\right )^{3/2} \log \left (\frac {12 i \sqrt {c^2 d-e} e^2 \left (c d+i e x+\sqrt {c^2 d-e} \sqrt {d+e x^2}\right )}{b \left (2 c^2 d-3 e\right ) (-i+c x)}\right )}{6 \left (c^2 d-e\right )^{3/2} e^2 \left (d+e x^2\right )^{3/2}} \]
(2*Sqrt[c^2*d - e]*(b*c*e*x*(d + e*x^2) - a*(c^2*d - e)*(2*d + 3*e*x^2)) - 2*b*(c^2*d - e)^(3/2)*(2*d + 3*e*x^2)*ArcTan[c*x] - I*b*c*(2*c^2*d - 3*e) *(d + e*x^2)^(3/2)*Log[((-12*I)*Sqrt[c^2*d - e]*e^2*(c*d - I*e*x + Sqrt[c^ 2*d - e]*Sqrt[d + e*x^2]))/(b*(2*c^2*d - 3*e)*(I + c*x))] + I*b*c*(2*c^2*d - 3*e)*(d + e*x^2)^(3/2)*Log[((12*I)*Sqrt[c^2*d - e]*e^2*(c*d + I*e*x + S qrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*(2*c^2*d - 3*e)*(-I + c*x))])/(6*(c^2* d - e)^(3/2)*e^2*(d + e*x^2)^(3/2))
Time = 0.37 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {5511, 27, 402, 27, 291, 216}
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.
| \(\displaystyle \int \frac {x^3 (a+b \arctan (c x))}{\left (d+e x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 5511 |
| \(\displaystyle -b c \int -\frac {3 e x^2+2 d}{3 e^2 \left (c^2 x^2+1\right ) \left (e x^2+d\right )^{3/2}}dx-\frac {a+b \arctan (c x)}{e^2 \sqrt {d+e x^2}}+\frac {d (a+b \arctan (c x))}{3 e^2 \left (d+e x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b c \int \frac {3 e x^2+2 d}{\left (c^2 x^2+1\right ) \left (e x^2+d\right )^{3/2}}dx}{3 e^2}-\frac {a+b \arctan (c x)}{e^2 \sqrt {d+e x^2}}+\frac {d (a+b \arctan (c x))}{3 e^2 \left (d+e x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {b c \left (\frac {\int \frac {d \left (2 c^2 d-3 e\right )}{\left (c^2 x^2+1\right ) \sqrt {e x^2+d}}dx}{d \left (c^2 d-e\right )}+\frac {e x}{\left (c^2 d-e\right ) \sqrt {d+e x^2}}\right )}{3 e^2}-\frac {a+b \arctan (c x)}{e^2 \sqrt {d+e x^2}}+\frac {d (a+b \arctan (c x))}{3 e^2 \left (d+e x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b c \left (\frac {\left (2 c^2 d-3 e\right ) \int \frac {1}{\left (c^2 x^2+1\right ) \sqrt {e x^2+d}}dx}{c^2 d-e}+\frac {e x}{\left (c^2 d-e\right ) \sqrt {d+e x^2}}\right )}{3 e^2}-\frac {a+b \arctan (c x)}{e^2 \sqrt {d+e x^2}}+\frac {d (a+b \arctan (c x))}{3 e^2 \left (d+e x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {b c \left (\frac {\left (2 c^2 d-3 e\right ) \int \frac {1}{1-\frac {\left (e-c^2 d\right ) x^2}{e x^2+d}}d\frac {x}{\sqrt {e x^2+d}}}{c^2 d-e}+\frac {e x}{\left (c^2 d-e\right ) \sqrt {d+e x^2}}\right )}{3 e^2}-\frac {a+b \arctan (c x)}{e^2 \sqrt {d+e x^2}}+\frac {d (a+b \arctan (c x))}{3 e^2 \left (d+e x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {a+b \arctan (c x)}{e^2 \sqrt {d+e x^2}}+\frac {d (a+b \arctan (c x))}{3 e^2 \left (d+e x^2\right )^{3/2}}+\frac {b c \left (\frac {\left (2 c^2 d-3 e\right ) \arctan \left (\frac {x \sqrt {c^2 d-e}}{\sqrt {d+e x^2}}\right )}{\left (c^2 d-e\right )^{3/2}}+\frac {e x}{\left (c^2 d-e\right ) \sqrt {d+e x^2}}\right )}{3 e^2}\) |
(d*(a + b*ArcTan[c*x]))/(3*e^2*(d + e*x^2)^(3/2)) - (a + b*ArcTan[c*x])/(e ^2*Sqrt[d + e*x^2]) + (b*c*((e*x)/((c^2*d - e)*Sqrt[d + e*x^2]) + ((2*c^2* d - 3*e)*ArcTan[(Sqrt[c^2*d - e]*x)/Sqrt[d + e*x^2]])/(c^2*d - e)^(3/2)))/ (3*e^2)
3.13.18.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x _)^2)^(q_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^2)^q, x]}, Sim p[(a + b*ArcTan[c*x]) u, x] - Simp[b*c Int[SimplifyIntegrand[u/(1 + c^2 *x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && ((IGtQ[q, 0] && !(ILtQ[(m - 1)/2, 0] && GtQ[m + 2*q + 3, 0])) || (IGtQ[(m + 1)/2, 0] && !(ILtQ[q, 0] && GtQ[m + 2*q + 3, 0])) || (ILtQ[(m + 2*q + 1)/2, 0] && !ILt Q[(m - 1)/2, 0]))
\[\int \frac {x^{3} \left (a +b \arctan \left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 415 vs. \(2 (125) = 250\).
Time = 0.60 (sec) , antiderivative size = 863, normalized size of antiderivative = 6.03 \[ \int \frac {x^3 (a+b \arctan (c x))}{\left (d+e x^2\right )^{5/2}} \, dx=\left [-\frac {{\left (2 \, b c^{3} d^{3} - 3 \, b c d^{2} e + {\left (2 \, b c^{3} d e^{2} - 3 \, b c e^{3}\right )} x^{4} + 2 \, {\left (2 \, b c^{3} d^{2} e - 3 \, b c d e^{2}\right )} x^{2}\right )} \sqrt {-c^{2} d + e} \log \left (\frac {{\left (c^{4} d^{2} - 8 \, c^{2} d e + 8 \, e^{2}\right )} x^{4} - 2 \, {\left (3 \, c^{2} d^{2} - 4 \, d e\right )} x^{2} - 4 \, {\left ({\left (c^{2} d - 2 \, e\right )} x^{3} - d x\right )} \sqrt {-c^{2} d + e} \sqrt {e x^{2} + d} + d^{2}}{c^{4} x^{4} + 2 \, c^{2} x^{2} + 1}\right ) + 4 \, {\left (2 \, a c^{4} d^{3} - 4 \, a c^{2} d^{2} e + 2 \, a d e^{2} - {\left (b c^{3} d e^{2} - b c e^{3}\right )} x^{3} + 3 \, {\left (a c^{4} d^{2} e - 2 \, a c^{2} d e^{2} + a e^{3}\right )} x^{2} - {\left (b c^{3} d^{2} e - b c d e^{2}\right )} x + {\left (2 \, b c^{4} d^{3} - 4 \, b c^{2} d^{2} e + 2 \, b d e^{2} + 3 \, {\left (b c^{4} d^{2} e - 2 \, b c^{2} d e^{2} + b e^{3}\right )} x^{2}\right )} \arctan \left (c x\right )\right )} \sqrt {e x^{2} + d}}{12 \, {\left (c^{4} d^{4} e^{2} - 2 \, c^{2} d^{3} e^{3} + d^{2} e^{4} + {\left (c^{4} d^{2} e^{4} - 2 \, c^{2} d e^{5} + e^{6}\right )} x^{4} + 2 \, {\left (c^{4} d^{3} e^{3} - 2 \, c^{2} d^{2} e^{4} + d e^{5}\right )} x^{2}\right )}}, \frac {{\left (2 \, b c^{3} d^{3} - 3 \, b c d^{2} e + {\left (2 \, b c^{3} d e^{2} - 3 \, b c e^{3}\right )} x^{4} + 2 \, {\left (2 \, b c^{3} d^{2} e - 3 \, b c d e^{2}\right )} x^{2}\right )} \sqrt {c^{2} d - e} \arctan \left (\frac {\sqrt {c^{2} d - e} {\left ({\left (c^{2} d - 2 \, e\right )} x^{2} - d\right )} \sqrt {e x^{2} + d}}{2 \, {\left ({\left (c^{2} d e - e^{2}\right )} x^{3} + {\left (c^{2} d^{2} - d e\right )} x\right )}}\right ) - 2 \, {\left (2 \, a c^{4} d^{3} - 4 \, a c^{2} d^{2} e + 2 \, a d e^{2} - {\left (b c^{3} d e^{2} - b c e^{3}\right )} x^{3} + 3 \, {\left (a c^{4} d^{2} e - 2 \, a c^{2} d e^{2} + a e^{3}\right )} x^{2} - {\left (b c^{3} d^{2} e - b c d e^{2}\right )} x + {\left (2 \, b c^{4} d^{3} - 4 \, b c^{2} d^{2} e + 2 \, b d e^{2} + 3 \, {\left (b c^{4} d^{2} e - 2 \, b c^{2} d e^{2} + b e^{3}\right )} x^{2}\right )} \arctan \left (c x\right )\right )} \sqrt {e x^{2} + d}}{6 \, {\left (c^{4} d^{4} e^{2} - 2 \, c^{2} d^{3} e^{3} + d^{2} e^{4} + {\left (c^{4} d^{2} e^{4} - 2 \, c^{2} d e^{5} + e^{6}\right )} x^{4} + 2 \, {\left (c^{4} d^{3} e^{3} - 2 \, c^{2} d^{2} e^{4} + d e^{5}\right )} x^{2}\right )}}\right ] \]
[-1/12*((2*b*c^3*d^3 - 3*b*c*d^2*e + (2*b*c^3*d*e^2 - 3*b*c*e^3)*x^4 + 2*( 2*b*c^3*d^2*e - 3*b*c*d*e^2)*x^2)*sqrt(-c^2*d + e)*log(((c^4*d^2 - 8*c^2*d *e + 8*e^2)*x^4 - 2*(3*c^2*d^2 - 4*d*e)*x^2 - 4*((c^2*d - 2*e)*x^3 - d*x)* sqrt(-c^2*d + e)*sqrt(e*x^2 + d) + d^2)/(c^4*x^4 + 2*c^2*x^2 + 1)) + 4*(2* a*c^4*d^3 - 4*a*c^2*d^2*e + 2*a*d*e^2 - (b*c^3*d*e^2 - b*c*e^3)*x^3 + 3*(a *c^4*d^2*e - 2*a*c^2*d*e^2 + a*e^3)*x^2 - (b*c^3*d^2*e - b*c*d*e^2)*x + (2 *b*c^4*d^3 - 4*b*c^2*d^2*e + 2*b*d*e^2 + 3*(b*c^4*d^2*e - 2*b*c^2*d*e^2 + b*e^3)*x^2)*arctan(c*x))*sqrt(e*x^2 + d))/(c^4*d^4*e^2 - 2*c^2*d^3*e^3 + d ^2*e^4 + (c^4*d^2*e^4 - 2*c^2*d*e^5 + e^6)*x^4 + 2*(c^4*d^3*e^3 - 2*c^2*d^ 2*e^4 + d*e^5)*x^2), 1/6*((2*b*c^3*d^3 - 3*b*c*d^2*e + (2*b*c^3*d*e^2 - 3* b*c*e^3)*x^4 + 2*(2*b*c^3*d^2*e - 3*b*c*d*e^2)*x^2)*sqrt(c^2*d - e)*arctan (1/2*sqrt(c^2*d - e)*((c^2*d - 2*e)*x^2 - d)*sqrt(e*x^2 + d)/((c^2*d*e - e ^2)*x^3 + (c^2*d^2 - d*e)*x)) - 2*(2*a*c^4*d^3 - 4*a*c^2*d^2*e + 2*a*d*e^2 - (b*c^3*d*e^2 - b*c*e^3)*x^3 + 3*(a*c^4*d^2*e - 2*a*c^2*d*e^2 + a*e^3)*x ^2 - (b*c^3*d^2*e - b*c*d*e^2)*x + (2*b*c^4*d^3 - 4*b*c^2*d^2*e + 2*b*d*e^ 2 + 3*(b*c^4*d^2*e - 2*b*c^2*d*e^2 + b*e^3)*x^2)*arctan(c*x))*sqrt(e*x^2 + d))/(c^4*d^4*e^2 - 2*c^2*d^3*e^3 + d^2*e^4 + (c^4*d^2*e^4 - 2*c^2*d*e^5 + e^6)*x^4 + 2*(c^4*d^3*e^3 - 2*c^2*d^2*e^4 + d*e^5)*x^2)]
\[ \int \frac {x^3 (a+b \arctan (c x))}{\left (d+e x^2\right )^{5/2}} \, dx=\int \frac {x^{3} \left (a + b \operatorname {atan}{\left (c x \right )}\right )}{\left (d + e x^{2}\right )^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {x^3 (a+b \arctan (c x))}{\left (d+e x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int \frac {x^3 (a+b \arctan (c x))}{\left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )} x^{3}}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^3 (a+b \arctan (c x))}{\left (d+e x^2\right )^{5/2}} \, dx=\int \frac {x^3\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}{{\left (e\,x^2+d\right )}^{5/2}} \,d x \]