Integrand size = 23, antiderivative size = 178 \[ \int \frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{x^6} \, dx=\frac {b c \left (4 c^2 d-7 e\right ) \sqrt {d+e x^2}}{40 x^2}-\frac {b c \left (d+e x^2\right )^{3/2}}{20 x^4}-\frac {\left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{5 d x^5}-\frac {b c \left (8 c^4 d^2-20 c^2 d e+15 e^2\right ) \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{40 \sqrt {d}}+\frac {b \left (c^2 d-e\right )^{5/2} \text {arctanh}\left (\frac {c \sqrt {d+e x^2}}{\sqrt {c^2 d-e}}\right )}{5 d} \]
-1/20*b*c*(e*x^2+d)^(3/2)/x^4-1/5*(e*x^2+d)^(5/2)*(a+b*arctan(c*x))/d/x^5+ 1/5*b*(c^2*d-e)^(5/2)*arctanh(c*(e*x^2+d)^(1/2)/(c^2*d-e)^(1/2))/d-1/40*b* c*(8*c^4*d^2-20*c^2*d*e+15*e^2)*arctanh((e*x^2+d)^(1/2)/d^(1/2))/d^(1/2)+1 /40*b*c*(4*c^2*d-7*e)*(e*x^2+d)^(1/2)/x^2
Result contains complex when optimal does not.
Time = 0.44 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.88 \[ \int \frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{x^6} \, dx=\frac {-\sqrt {d+e x^2} \left (8 a \left (d+e x^2\right )^2+b c d x \left (9 e x^2+d \left (2-4 c^2 x^2\right )\right )\right )-8 b \left (d+e x^2\right )^{5/2} \arctan (c x)+b c \sqrt {d} \left (8 c^4 d^2-20 c^2 d e+15 e^2\right ) x^5 \log (x)-b c \sqrt {d} \left (8 c^4 d^2-20 c^2 d e+15 e^2\right ) x^5 \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )+4 b \left (c^2 d-e\right )^{5/2} x^5 \log \left (-\frac {20 c d \left (c d-i e x+\sqrt {c^2 d-e} \sqrt {d+e x^2}\right )}{b \left (c^2 d-e\right )^{7/2} (i+c x)}\right )+4 b \left (c^2 d-e\right )^{5/2} x^5 \log \left (-\frac {20 c d \left (c d+i e x+\sqrt {c^2 d-e} \sqrt {d+e x^2}\right )}{b \left (c^2 d-e\right )^{7/2} (-i+c x)}\right )}{40 d x^5} \]
(-(Sqrt[d + e*x^2]*(8*a*(d + e*x^2)^2 + b*c*d*x*(9*e*x^2 + d*(2 - 4*c^2*x^ 2)))) - 8*b*(d + e*x^2)^(5/2)*ArcTan[c*x] + b*c*Sqrt[d]*(8*c^4*d^2 - 20*c^ 2*d*e + 15*e^2)*x^5*Log[x] - b*c*Sqrt[d]*(8*c^4*d^2 - 20*c^2*d*e + 15*e^2) *x^5*Log[d + Sqrt[d]*Sqrt[d + e*x^2]] + 4*b*(c^2*d - e)^(5/2)*x^5*Log[(-20 *c*d*(c*d - I*e*x + Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*(c^2*d - e)^(7/2) *(I + c*x))] + 4*b*(c^2*d - e)^(5/2)*x^5*Log[(-20*c*d*(c*d + I*e*x + Sqrt[ c^2*d - e]*Sqrt[d + e*x^2]))/(b*(c^2*d - e)^(7/2)*(-I + c*x))])/(40*d*x^5)
Time = 0.43 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {5511, 27, 354, 109, 27, 166, 27, 174, 73, 221}
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.
| \(\displaystyle \int \frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{x^6} \, dx\) |
| \(\Big \downarrow \) 5511 |
| \(\displaystyle -b c \int -\frac {\left (e x^2+d\right )^{5/2}}{5 d x^5 \left (c^2 x^2+1\right )}dx-\frac {\left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{5 d x^5}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b c \int \frac {\left (e x^2+d\right )^{5/2}}{x^5 \left (c^2 x^2+1\right )}dx}{5 d}-\frac {\left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{5 d x^5}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {b c \int \frac {\left (e x^2+d\right )^{5/2}}{x^6 \left (c^2 x^2+1\right )}dx^2}{10 d}-\frac {\left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{5 d x^5}\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle \frac {b c \left (-\frac {1}{2} \int \frac {\sqrt {e x^2+d} \left (\left (c^2 d-4 e\right ) e x^2+d \left (4 c^2 d-7 e\right )\right )}{2 x^4 \left (c^2 x^2+1\right )}dx^2-\frac {d \left (d+e x^2\right )^{3/2}}{2 x^4}\right )}{10 d}-\frac {\left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{5 d x^5}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b c \left (-\frac {1}{4} \int \frac {\sqrt {e x^2+d} \left (\left (c^2 d-4 e\right ) e x^2+d \left (4 c^2 d-7 e\right )\right )}{x^4 \left (c^2 x^2+1\right )}dx^2-\frac {d \left (d+e x^2\right )^{3/2}}{2 x^4}\right )}{10 d}-\frac {\left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{5 d x^5}\) |
| \(\Big \downarrow \) 166 |
| \(\displaystyle \frac {b c \left (\frac {1}{4} \left (\frac {d \left (4 c^2 d-7 e\right ) \sqrt {d+e x^2}}{x^2}-\int -\frac {e \left (4 d^2 c^4-9 d e c^2+8 e^2\right ) x^2+d \left (8 d^2 c^4-20 d e c^2+15 e^2\right )}{2 x^2 \left (c^2 x^2+1\right ) \sqrt {e x^2+d}}dx^2\right )-\frac {d \left (d+e x^2\right )^{3/2}}{2 x^4}\right )}{10 d}-\frac {\left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{5 d x^5}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b c \left (\frac {1}{4} \left (\frac {1}{2} \int \frac {e \left (4 d^2 c^4-9 d e c^2+8 e^2\right ) x^2+d \left (8 d^2 c^4-20 d e c^2+15 e^2\right )}{x^2 \left (c^2 x^2+1\right ) \sqrt {e x^2+d}}dx^2+\frac {d \left (4 c^2 d-7 e\right ) \sqrt {d+e x^2}}{x^2}\right )-\frac {d \left (d+e x^2\right )^{3/2}}{2 x^4}\right )}{10 d}-\frac {\left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{5 d x^5}\) |
| \(\Big \downarrow \) 174 |
| \(\displaystyle \frac {b c \left (\frac {1}{4} \left (\frac {1}{2} \left (d \left (8 c^4 d^2-20 c^2 d e+15 e^2\right ) \int \frac {1}{x^2 \sqrt {e x^2+d}}dx^2-8 \left (c^2 d-e\right )^3 \int \frac {1}{\left (c^2 x^2+1\right ) \sqrt {e x^2+d}}dx^2\right )+\frac {d \left (4 c^2 d-7 e\right ) \sqrt {d+e x^2}}{x^2}\right )-\frac {d \left (d+e x^2\right )^{3/2}}{2 x^4}\right )}{10 d}-\frac {\left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{5 d x^5}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {b c \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {2 d \left (8 c^4 d^2-20 c^2 d e+15 e^2\right ) \int \frac {1}{\frac {x^4}{e}-\frac {d}{e}}d\sqrt {e x^2+d}}{e}-\frac {16 \left (c^2 d-e\right )^3 \int \frac {1}{\frac {c^2 x^4}{e}-\frac {c^2 d}{e}+1}d\sqrt {e x^2+d}}{e}\right )+\frac {d \left (4 c^2 d-7 e\right ) \sqrt {d+e x^2}}{x^2}\right )-\frac {d \left (d+e x^2\right )^{3/2}}{2 x^4}\right )}{10 d}-\frac {\left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{5 d x^5}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {b c \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {16 \left (c^2 d-e\right )^{5/2} \text {arctanh}\left (\frac {c \sqrt {d+e x^2}}{\sqrt {c^2 d-e}}\right )}{c}-2 \sqrt {d} \left (8 c^4 d^2-20 c^2 d e+15 e^2\right ) \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )\right )+\frac {d \left (4 c^2 d-7 e\right ) \sqrt {d+e x^2}}{x^2}\right )-\frac {d \left (d+e x^2\right )^{3/2}}{2 x^4}\right )}{10 d}-\frac {\left (d+e x^2\right )^{5/2} (a+b \arctan (c x))}{5 d x^5}\) |
-1/5*((d + e*x^2)^(5/2)*(a + b*ArcTan[c*x]))/(d*x^5) + (b*c*(-1/2*(d*(d + e*x^2)^(3/2))/x^4 + ((d*(4*c^2*d - 7*e)*Sqrt[d + e*x^2])/x^2 + (-2*Sqrt[d] *(8*c^4*d^2 - 20*c^2*d*e + 15*e^2)*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]] + (16* (c^2*d - e)^(5/2)*ArcTanh[(c*Sqrt[d + e*x^2])/Sqrt[c^2*d - e]])/c)/2)/4))/ (10*d)
3.12.92.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f *x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] - Simp[1/(b*(b*e - a*f)*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b* c*(f*g - e*h)*(m + 1) + (b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h )*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, p}, x] && ILtQ[m, -1] && GtQ[n, 0]
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* ((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d) Int[(e + f*x)^ p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d) Int[(e + f*x)^p/(c + d *x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
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]))
\[\int \frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \arctan \left (c x \right )\right )}{x^{6}}d x\]
Time = 0.60 (sec) , antiderivative size = 1145, normalized size of antiderivative = 6.43 \[ \int \frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{x^6} \, dx=\text {Too large to display} \]
[1/80*(4*(b*c^4*d^2 - 2*b*c^2*d*e + b*e^2)*sqrt(c^2*d - e)*x^5*log((c^4*e^ 2*x^4 + 8*c^4*d^2 - 8*c^2*d*e + 2*(4*c^4*d*e - 3*c^2*e^2)*x^2 + 4*(c^3*e*x ^2 + 2*c^3*d - c*e)*sqrt(c^2*d - e)*sqrt(e*x^2 + d) + e^2)/(c^4*x^4 + 2*c^ 2*x^2 + 1)) + (8*b*c^5*d^2 - 20*b*c^3*d*e + 15*b*c*e^2)*sqrt(d)*x^5*log(-( e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) - 2*(8*a*e^2*x^4 + 2*b*c*d^2 *x + 16*a*d*e*x^2 - (4*b*c^3*d^2 - 9*b*c*d*e)*x^3 + 8*a*d^2 + 8*(b*e^2*x^4 + 2*b*d*e*x^2 + b*d^2)*arctan(c*x))*sqrt(e*x^2 + d))/(d*x^5), 1/80*(8*(b* c^4*d^2 - 2*b*c^2*d*e + b*e^2)*sqrt(-c^2*d + e)*x^5*arctan(-1/2*(c^2*e*x^2 + 2*c^2*d - e)*sqrt(-c^2*d + e)*sqrt(e*x^2 + d)/(c^3*d^2 - c*d*e + (c^3*d *e - c*e^2)*x^2)) + (8*b*c^5*d^2 - 20*b*c^3*d*e + 15*b*c*e^2)*sqrt(d)*x^5* log(-(e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) - 2*(8*a*e^2*x^4 + 2*b *c*d^2*x + 16*a*d*e*x^2 - (4*b*c^3*d^2 - 9*b*c*d*e)*x^3 + 8*a*d^2 + 8*(b*e ^2*x^4 + 2*b*d*e*x^2 + b*d^2)*arctan(c*x))*sqrt(e*x^2 + d))/(d*x^5), 1/40* ((8*b*c^5*d^2 - 20*b*c^3*d*e + 15*b*c*e^2)*sqrt(-d)*x^5*arctan(sqrt(-d)/sq rt(e*x^2 + d)) + 2*(b*c^4*d^2 - 2*b*c^2*d*e + b*e^2)*sqrt(c^2*d - e)*x^5*l og((c^4*e^2*x^4 + 8*c^4*d^2 - 8*c^2*d*e + 2*(4*c^4*d*e - 3*c^2*e^2)*x^2 + 4*(c^3*e*x^2 + 2*c^3*d - c*e)*sqrt(c^2*d - e)*sqrt(e*x^2 + d) + e^2)/(c^4* x^4 + 2*c^2*x^2 + 1)) - (8*a*e^2*x^4 + 2*b*c*d^2*x + 16*a*d*e*x^2 - (4*b*c ^3*d^2 - 9*b*c*d*e)*x^3 + 8*a*d^2 + 8*(b*e^2*x^4 + 2*b*d*e*x^2 + b*d^2)*ar ctan(c*x))*sqrt(e*x^2 + d))/(d*x^5), 1/40*(4*(b*c^4*d^2 - 2*b*c^2*d*e +...
\[ \int \frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{x^6} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{\frac {3}{2}}}{x^{6}}\, dx \]
Exception generated. \[ \int \frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{x^6} \, dx=\text {Exception raised: ValueError} \]
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
Timed out. \[ \int \frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{x^6} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\left (d+e x^2\right )^{3/2} (a+b \arctan (c x))}{x^6} \, dx=\int \frac {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^{3/2}}{x^6} \,d x \]