Integrand size = 23, antiderivative size = 137 \[ \int \frac {x^3 (a+b \arctan (c x))}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {d (a+b \arctan (c x))}{e^2 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} (a+b \arctan (c x))}{e^2}-\frac {b \left (2 c^2 d-e\right ) \arctan \left (\frac {\sqrt {c^2 d-e} x}{\sqrt {d+e x^2}}\right )}{c \sqrt {c^2 d-e} e^2}-\frac {b \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{c e^{3/2}} \]
-b*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/c/e^(3/2)-b*(2*c^2*d-e)*arctan(x*(c^ 2*d-e)^(1/2)/(e*x^2+d)^(1/2))/c/e^2/(c^2*d-e)^(1/2)+d*(a+b*arctan(c*x))/e^ 2/(e*x^2+d)^(1/2)+(a+b*arctan(c*x))*(e*x^2+d)^(1/2)/e^2
Result contains complex when optimal does not.
Time = 0.53 (sec) , antiderivative size = 321, normalized size of antiderivative = 2.34 \[ \int \frac {x^3 (a+b \arctan (c x))}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {\frac {2 a \left (2 d+e x^2\right )}{\sqrt {d+e x^2}}+\frac {2 b \left (2 d+e x^2\right ) \arctan (c x)}{\sqrt {d+e x^2}}-\frac {i b \left (2 c^2 d-e\right ) \log \left (\frac {4 c^2 e^2 \left (-i c d+e x-i \sqrt {c^2 d-e} \sqrt {d+e x^2}\right )}{b \sqrt {c^2 d-e} \left (2 c^2 d-e\right ) (-i+c x)}\right )}{c \sqrt {c^2 d-e}}+\frac {i b \left (2 c^2 d-e\right ) \log \left (\frac {4 c^2 e^2 \left (i c d+e x+i \sqrt {c^2 d-e} \sqrt {d+e x^2}\right )}{b \sqrt {c^2 d-e} \left (2 c^2 d-e\right ) (i+c x)}\right )}{c \sqrt {c^2 d-e}}-\frac {2 b \sqrt {e} \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )}{c}}{2 e^2} \]
((2*a*(2*d + e*x^2))/Sqrt[d + e*x^2] + (2*b*(2*d + e*x^2)*ArcTan[c*x])/Sqr t[d + e*x^2] - (I*b*(2*c^2*d - e)*Log[(4*c^2*e^2*((-I)*c*d + e*x - I*Sqrt[ c^2*d - e]*Sqrt[d + e*x^2]))/(b*Sqrt[c^2*d - e]*(2*c^2*d - e)*(-I + c*x))] )/(c*Sqrt[c^2*d - e]) + (I*b*(2*c^2*d - e)*Log[(4*c^2*e^2*(I*c*d + e*x + I *Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*Sqrt[c^2*d - e]*(2*c^2*d - e)*(I + c *x))])/(c*Sqrt[c^2*d - e]) - (2*b*Sqrt[e]*Log[e*x + Sqrt[e]*Sqrt[d + e*x^2 ]])/c)/(2*e^2)
Time = 0.36 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {5511, 27, 398, 224, 219, 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.
\(\displaystyle \int \frac {x^3 (a+b \arctan (c x))}{\left (d+e x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 5511 |
\(\displaystyle -b c \int \frac {e x^2+2 d}{e^2 \left (c^2 x^2+1\right ) \sqrt {e x^2+d}}dx+\frac {\sqrt {d+e x^2} (a+b \arctan (c x))}{e^2}+\frac {d (a+b \arctan (c x))}{e^2 \sqrt {d+e x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {b c \int \frac {e x^2+2 d}{\left (c^2 x^2+1\right ) \sqrt {e x^2+d}}dx}{e^2}+\frac {\sqrt {d+e x^2} (a+b \arctan (c x))}{e^2}+\frac {d (a+b \arctan (c x))}{e^2 \sqrt {d+e x^2}}\) |
\(\Big \downarrow \) 398 |
\(\displaystyle -\frac {b c \left (\frac {e \int \frac {1}{\sqrt {e x^2+d}}dx}{c^2}+\left (2 d-\frac {e}{c^2}\right ) \int \frac {1}{\left (c^2 x^2+1\right ) \sqrt {e x^2+d}}dx\right )}{e^2}+\frac {\sqrt {d+e x^2} (a+b \arctan (c x))}{e^2}+\frac {d (a+b \arctan (c x))}{e^2 \sqrt {d+e x^2}}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle -\frac {b c \left (\left (2 d-\frac {e}{c^2}\right ) \int \frac {1}{\left (c^2 x^2+1\right ) \sqrt {e x^2+d}}dx+\frac {e \int \frac {1}{1-\frac {e x^2}{e x^2+d}}d\frac {x}{\sqrt {e x^2+d}}}{c^2}\right )}{e^2}+\frac {\sqrt {d+e x^2} (a+b \arctan (c x))}{e^2}+\frac {d (a+b \arctan (c x))}{e^2 \sqrt {d+e x^2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {b c \left (\left (2 d-\frac {e}{c^2}\right ) \int \frac {1}{\left (c^2 x^2+1\right ) \sqrt {e x^2+d}}dx+\frac {\sqrt {e} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{c^2}\right )}{e^2}+\frac {\sqrt {d+e x^2} (a+b \arctan (c x))}{e^2}+\frac {d (a+b \arctan (c x))}{e^2 \sqrt {d+e x^2}}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle -\frac {b c \left (\left (2 d-\frac {e}{c^2}\right ) \int \frac {1}{1-\frac {\left (e-c^2 d\right ) x^2}{e x^2+d}}d\frac {x}{\sqrt {e x^2+d}}+\frac {\sqrt {e} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{c^2}\right )}{e^2}+\frac {\sqrt {d+e x^2} (a+b \arctan (c x))}{e^2}+\frac {d (a+b \arctan (c x))}{e^2 \sqrt {d+e x^2}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\sqrt {d+e x^2} (a+b \arctan (c x))}{e^2}+\frac {d (a+b \arctan (c x))}{e^2 \sqrt {d+e x^2}}-\frac {b c \left (\frac {\left (2 d-\frac {e}{c^2}\right ) \arctan \left (\frac {x \sqrt {c^2 d-e}}{\sqrt {d+e x^2}}\right )}{\sqrt {c^2 d-e}}+\frac {\sqrt {e} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{c^2}\right )}{e^2}\) |
(d*(a + b*ArcTan[c*x]))/(e^2*Sqrt[d + e*x^2]) + (Sqrt[d + e*x^2]*(a + b*Ar cTan[c*x]))/e^2 - (b*c*(((2*d - e/c^2)*ArcTan[(Sqrt[c^2*d - e]*x)/Sqrt[d + e*x^2]])/Sqrt[c^2*d - e] + (Sqrt[e]*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]) /c^2))/e^2
3.13.9.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 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[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) , x_Symbol] :> Simp[f/b Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ b Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} , x]
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]))
\[\int \frac {x^{3} \left (a +b \arctan \left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 303 vs. \(2 (121) = 242\).
Time = 0.55 (sec) , antiderivative size = 1291, normalized size of antiderivative = 9.42 \[ \int \frac {x^3 (a+b \arctan (c x))}{\left (d+e x^2\right )^{3/2}} \, dx=\text {Too large to display} \]
[1/4*(2*(b*c^2*d^2 - b*d*e + (b*c^2*d*e - b*e^2)*x^2)*sqrt(e)*log(-2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) + (2*b*c^2*d^2 - b*d*e + (2*b*c^2*d*e - b*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)*sqr t(e*x^2 + d) + d^2)/(c^4*x^4 + 2*c^2*x^2 + 1)) + 4*(2*a*c^3*d^2 - 2*a*c*d* e + (a*c^3*d*e - a*c*e^2)*x^2 + (2*b*c^3*d^2 - 2*b*c*d*e + (b*c^3*d*e - b* c*e^2)*x^2)*arctan(c*x))*sqrt(e*x^2 + d))/(c^3*d^2*e^2 - c*d*e^3 + (c^3*d* e^3 - c*e^4)*x^2), -1/2*((2*b*c^2*d^2 - b*d*e + (2*b*c^2*d*e - b*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)) - (b*c^2*d^2 - b*d*e + (b*c^2*d*e - b*e^2)*x^2)*sqrt(e)*log(-2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e )*x - d) - 2*(2*a*c^3*d^2 - 2*a*c*d*e + (a*c^3*d*e - a*c*e^2)*x^2 + (2*b*c ^3*d^2 - 2*b*c*d*e + (b*c^3*d*e - b*c*e^2)*x^2)*arctan(c*x))*sqrt(e*x^2 + d))/(c^3*d^2*e^2 - c*d*e^3 + (c^3*d*e^3 - c*e^4)*x^2), 1/4*(4*(b*c^2*d^2 - b*d*e + (b*c^2*d*e - b*e^2)*x^2)*sqrt(-e)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) + (2*b*c^2*d^2 - b*d*e + (2*b*c^2*d*e - b*e^2)*x^2)*sqrt(-c^2*d + e)*l og(((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^3*d^2 - 2*a*c*d*e + (a*c^3*d*e - a*c*e^2)*x^2 + (2*b*c^3*d^2 - 2*b*c*d*e + (b*c^3*d*e - b*c*e^2)*x^2)*arctan(c*x))*sqrt...
\[ \int \frac {x^3 (a+b \arctan (c x))}{\left (d+e x^2\right )^{3/2}} \, dx=\int \frac {x^{3} \left (a + b \operatorname {atan}{\left (c x \right )}\right )}{\left (d + e x^{2}\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {x^3 (a+b \arctan (c x))}{\left (d+e x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {x^3 (a+b \arctan (c x))}{\left (d+e x^2\right )^{3/2}} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )} x^{3}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^3 (a+b \arctan (c x))}{\left (d+e x^2\right )^{3/2}} \, dx=\int \frac {x^3\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}{{\left (e\,x^2+d\right )}^{3/2}} \,d x \]