Integrand size = 21, antiderivative size = 110 \[ \int \frac {x (a+b \arctan (c x))}{\left (d+e x^2\right )^{5/2}} \, dx=-\frac {b c x}{3 d \left (c^2 d-e\right ) \sqrt {d+e x^2}}-\frac {a+b \arctan (c x)}{3 e \left (d+e x^2\right )^{3/2}}+\frac {b c^3 \arctan \left (\frac {\sqrt {c^2 d-e} x}{\sqrt {d+e x^2}}\right )}{3 \left (c^2 d-e\right )^{3/2} e} \]
1/3*(-a-b*arctan(c*x))/e/(e*x^2+d)^(3/2)+1/3*b*c^3*arctan(x*(c^2*d-e)^(1/2 )/(e*x^2+d)^(1/2))/(c^2*d-e)^(3/2)/e-1/3*b*c*x/d/(c^2*d-e)/(e*x^2+d)^(1/2)
Result contains complex when optimal does not.
Time = 0.57 (sec) , antiderivative size = 259, normalized size of antiderivative = 2.35 \[ \int \frac {x (a+b \arctan (c x))}{\left (d+e x^2\right )^{5/2}} \, dx=\frac {1}{6} \left (-\frac {2 a}{e \left (d+e x^2\right )^{3/2}}-\frac {2 b c x}{\left (c^2 d^2-d e\right ) \sqrt {d+e x^2}}-\frac {2 b \arctan (c x)}{e \left (d+e x^2\right )^{3/2}}-\frac {i b c^3 \log \left (-\frac {12 i \sqrt {c^2 d-e} e \left (c d-i e x+\sqrt {c^2 d-e} \sqrt {d+e x^2}\right )}{b c^2 (i+c x)}\right )}{\left (c^2 d-e\right )^{3/2} e}+\frac {i b c^3 \log \left (\frac {12 i \sqrt {c^2 d-e} e \left (c d+i e x+\sqrt {c^2 d-e} \sqrt {d+e x^2}\right )}{b c^2 (-i+c x)}\right )}{\left (c^2 d-e\right )^{3/2} e}\right ) \]
((-2*a)/(e*(d + e*x^2)^(3/2)) - (2*b*c*x)/((c^2*d^2 - d*e)*Sqrt[d + e*x^2] ) - (2*b*ArcTan[c*x])/(e*(d + e*x^2)^(3/2)) - (I*b*c^3*Log[((-12*I)*Sqrt[c ^2*d - e]*e*(c*d - I*e*x + Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*c^2*(I + c *x))])/((c^2*d - e)^(3/2)*e) + (I*b*c^3*Log[((12*I)*Sqrt[c^2*d - e]*e*(c*d + I*e*x + Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*c^2*(-I + c*x))])/((c^2*d - e)^(3/2)*e))/6
Time = 0.28 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {5509, 296, 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.
\(\displaystyle \int \frac {x (a+b \arctan (c x))}{\left (d+e x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 5509 |
\(\displaystyle \frac {b c \int \frac {1}{\left (c^2 x^2+1\right ) \left (e x^2+d\right )^{3/2}}dx}{3 e}-\frac {a+b \arctan (c x)}{3 e \left (d+e x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 296 |
\(\displaystyle \frac {b c \left (\frac {c^2 \int \frac {1}{\left (c^2 x^2+1\right ) \sqrt {e x^2+d}}dx}{c^2 d-e}-\frac {e x}{d \left (c^2 d-e\right ) \sqrt {d+e x^2}}\right )}{3 e}-\frac {a+b \arctan (c x)}{3 e \left (d+e x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {b c \left (\frac {c^2 \int \frac {1}{1-\frac {\left (e-c^2 d\right ) x^2}{e x^2+d}}d\frac {x}{\sqrt {e x^2+d}}}{c^2 d-e}-\frac {e x}{d \left (c^2 d-e\right ) \sqrt {d+e x^2}}\right )}{3 e}-\frac {a+b \arctan (c x)}{3 e \left (d+e x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {b c \left (\frac {c^2 \arctan \left (\frac {x \sqrt {c^2 d-e}}{\sqrt {d+e x^2}}\right )}{\left (c^2 d-e\right )^{3/2}}-\frac {e x}{d \left (c^2 d-e\right ) \sqrt {d+e x^2}}\right )}{3 e}-\frac {a+b \arctan (c x)}{3 e \left (d+e x^2\right )^{3/2}}\) |
-1/3*(a + b*ArcTan[c*x])/(e*(d + e*x^2)^(3/2)) + (b*c*(-((e*x)/(d*(c^2*d - e)*Sqrt[d + e*x^2])) + (c^2*ArcTan[(Sqrt[c^2*d - e]*x)/Sqrt[d + e*x^2]])/ (c^2*d - e)^(3/2)))/(3*e)
3.13.20.3.1 Defintions of rubi rules used
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_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[(b*c + 2*(p + 1)*(b*c - a*d))/(2*a*(p + 1)*(b*c - a*d)) Int[ (a + b*x^2)^(p + 1)*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, q}, x] && N eQ[b*c - a*d, 0] && EqQ[2*(p + q + 2) + 1, 0] && (LtQ[p, -1] || !LtQ[q, -1 ]) && NeQ[p, -1]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*(x_)*((d_.) + (e_.)*(x_)^2)^(q_.), x _Symbol] :> Simp[(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])/(2*e*(q + 1))), x ] - Simp[b*(c/(2*e*(q + 1))) Int[(d + e*x^2)^(q + 1)/(1 + c^2*x^2), x], x ] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[q, -1]
\[\int \frac {x \left (a +b \arctan \left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 323 vs. \(2 (94) = 188\).
Time = 0.57 (sec) , antiderivative size = 679, normalized size of antiderivative = 6.17 \[ \int \frac {x (a+b \arctan (c x))}{\left (d+e x^2\right )^{5/2}} \, dx=\left [\frac {{\left (b c^{3} d e^{2} x^{4} + 2 \, b c^{3} d^{2} e x^{2} + b c^{3} d^{3}\right )} \sqrt {-c^{2} d + e} \log \left (\frac {{\left (c^{4} d^{2} - 8 \, c^{2} d e + 8 \, e^{2}\right )} x^{4} - 2 \, {\left (3 \, c^{2} d^{2} - 4 \, d e\right )} x^{2} + 4 \, {\left ({\left (c^{2} d - 2 \, e\right )} x^{3} - d x\right )} \sqrt {-c^{2} d + e} \sqrt {e x^{2} + d} + d^{2}}{c^{4} x^{4} + 2 \, c^{2} x^{2} + 1}\right ) - 4 \, {\left (a c^{4} d^{3} - 2 \, a c^{2} d^{2} e + a d e^{2} + {\left (b c^{3} d e^{2} - b c e^{3}\right )} x^{3} + {\left (b c^{3} d^{2} e - b c d e^{2}\right )} x + {\left (b c^{4} d^{3} - 2 \, b c^{2} d^{2} e + b d e^{2}\right )} \arctan \left (c x\right )\right )} \sqrt {e x^{2} + d}}{12 \, {\left (c^{4} d^{5} e - 2 \, c^{2} d^{4} e^{2} + d^{3} e^{3} + {\left (c^{4} d^{3} e^{3} - 2 \, c^{2} d^{2} e^{4} + d e^{5}\right )} x^{4} + 2 \, {\left (c^{4} d^{4} e^{2} - 2 \, c^{2} d^{3} e^{3} + d^{2} e^{4}\right )} x^{2}\right )}}, \frac {{\left (b c^{3} d e^{2} x^{4} + 2 \, b c^{3} d^{2} e x^{2} + b c^{3} d^{3}\right )} \sqrt {c^{2} d - e} \arctan \left (\frac {\sqrt {c^{2} d - e} {\left ({\left (c^{2} d - 2 \, e\right )} x^{2} - d\right )} \sqrt {e x^{2} + d}}{2 \, {\left ({\left (c^{2} d e - e^{2}\right )} x^{3} + {\left (c^{2} d^{2} - d e\right )} x\right )}}\right ) - 2 \, {\left (a c^{4} d^{3} - 2 \, a c^{2} d^{2} e + a d e^{2} + {\left (b c^{3} d e^{2} - b c e^{3}\right )} x^{3} + {\left (b c^{3} d^{2} e - b c d e^{2}\right )} x + {\left (b c^{4} d^{3} - 2 \, b c^{2} d^{2} e + b d e^{2}\right )} \arctan \left (c x\right )\right )} \sqrt {e x^{2} + d}}{6 \, {\left (c^{4} d^{5} e - 2 \, c^{2} d^{4} e^{2} + d^{3} e^{3} + {\left (c^{4} d^{3} e^{3} - 2 \, c^{2} d^{2} e^{4} + d e^{5}\right )} x^{4} + 2 \, {\left (c^{4} d^{4} e^{2} - 2 \, c^{2} d^{3} e^{3} + d^{2} e^{4}\right )} x^{2}\right )}}\right ] \]
[1/12*((b*c^3*d*e^2*x^4 + 2*b*c^3*d^2*e*x^2 + b*c^3*d^3)*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*(a*c^4*d^3 - 2*a*c^2*d^2*e + a*d*e^2 + (b*c^3*d*e^2 - b*c*e^3)*x^3 + (b*c^3*d^2*e - b*c*d*e^2)*x + (b*c^4*d^3 - 2*b*c^2*d^2*e + b*d*e^2)*arctan(c*x))*sqrt(e*x^2 + d))/(c^4*d^5*e - 2*c^2*d^4*e^2 + d^3*e^ 3 + (c^4*d^3*e^3 - 2*c^2*d^2*e^4 + d*e^5)*x^4 + 2*(c^4*d^4*e^2 - 2*c^2*d^3 *e^3 + d^2*e^4)*x^2), 1/6*((b*c^3*d*e^2*x^4 + 2*b*c^3*d^2*e*x^2 + b*c^3*d^ 3)*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*(a*c^4*d^3 - 2* a*c^2*d^2*e + a*d*e^2 + (b*c^3*d*e^2 - b*c*e^3)*x^3 + (b*c^3*d^2*e - b*c*d *e^2)*x + (b*c^4*d^3 - 2*b*c^2*d^2*e + b*d*e^2)*arctan(c*x))*sqrt(e*x^2 + d))/(c^4*d^5*e - 2*c^2*d^4*e^2 + d^3*e^3 + (c^4*d^3*e^3 - 2*c^2*d^2*e^4 + d*e^5)*x^4 + 2*(c^4*d^4*e^2 - 2*c^2*d^3*e^3 + d^2*e^4)*x^2)]
\[ \int \frac {x (a+b \arctan (c x))}{\left (d+e x^2\right )^{5/2}} \, dx=\int \frac {x \left (a + b \operatorname {atan}{\left (c x \right )}\right )}{\left (d + e x^{2}\right )^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {x (a+b \arctan (c x))}{\left (d+e x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \]
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-c^2*d>0)', see `assume?` for m ore detail
\[ \int \frac {x (a+b \arctan (c x))}{\left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )} x}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {x (a+b \arctan (c x))}{\left (d+e x^2\right )^{5/2}} \, dx=\int \frac {x\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}{{\left (e\,x^2+d\right )}^{5/2}} \,d x \]