Integrand size = 23, antiderivative size = 177 \[ \int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^2}-\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \arctan \left (\frac {\sqrt {e} \sqrt {1-c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{c e^{3/2}}-\frac {2 b \sqrt {d} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{e^2} \]
-b*arctan(e^(1/2)*(-c^2*x^2+1)^(1/2)/c/(e*x^2+d)^(1/2))*(1/(c*x+1))^(1/2)* (c*x+1)^(1/2)/c/e^(3/2)-2*b*arctanh((e*x^2+d)^(1/2)/d^(1/2)/(-c^2*x^2+1)^( 1/2))*d^(1/2)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)/e^2+d*(a+b*arcsech(c*x))/e^2 /(e*x^2+d)^(1/2)+(a+b*arcsech(c*x))*(e*x^2+d)^(1/2)/e^2
Time = 23.17 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.41 \[ \int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {\left (2 d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}+\frac {b \sqrt {\frac {1-c x}{1+c x}} \sqrt {1-c^2 x^2} \left (\sqrt {-c^2} \sqrt {-c^2 d-e} \sqrt {e} \sqrt {\frac {c^2 \left (d+e x^2\right )}{c^2 d+e}} \arcsin \left (\frac {c \sqrt {e} \sqrt {1-c^2 x^2}}{\sqrt {-c^2} \sqrt {-c^2 d-e}}\right )+2 c^3 \sqrt {d} \sqrt {-d-e x^2} \arctan \left (\frac {\sqrt {d} \sqrt {1-c^2 x^2}}{\sqrt {-d-e x^2}}\right )\right )}{c^3 e^2 (-1+c x) \sqrt {d+e x^2}} \]
((2*d + e*x^2)*(a + b*ArcSech[c*x]))/(e^2*Sqrt[d + e*x^2]) + (b*Sqrt[(1 - c*x)/(1 + c*x)]*Sqrt[1 - c^2*x^2]*(Sqrt[-c^2]*Sqrt[-(c^2*d) - e]*Sqrt[e]*S qrt[(c^2*(d + e*x^2))/(c^2*d + e)]*ArcSin[(c*Sqrt[e]*Sqrt[1 - c^2*x^2])/(S qrt[-c^2]*Sqrt[-(c^2*d) - e])] + 2*c^3*Sqrt[d]*Sqrt[-d - e*x^2]*ArcTan[(Sq rt[d]*Sqrt[1 - c^2*x^2])/Sqrt[-d - e*x^2]]))/(c^3*e^2*(-1 + c*x)*Sqrt[d + e*x^2])
Time = 0.45 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.91, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {6855, 27, 435, 175, 66, 104, 218, 220}
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.
| \(\displaystyle \int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 6855 |
| \(\displaystyle b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \int \frac {e x^2+2 d}{e^2 x \sqrt {1-c^2 x^2} \sqrt {e x^2+d}}dx+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^2}+\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \int \frac {e x^2+2 d}{x \sqrt {1-c^2 x^2} \sqrt {e x^2+d}}dx}{e^2}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^2}+\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}\) |
| \(\Big \downarrow \) 435 |
| \(\displaystyle \frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \int \frac {e x^2+2 d}{x^2 \sqrt {1-c^2 x^2} \sqrt {e x^2+d}}dx^2}{2 e^2}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^2}+\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}\) |
| \(\Big \downarrow \) 175 |
| \(\displaystyle \frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (e \int \frac {1}{\sqrt {1-c^2 x^2} \sqrt {e x^2+d}}dx^2+2 d \int \frac {1}{x^2 \sqrt {1-c^2 x^2} \sqrt {e x^2+d}}dx^2\right )}{2 e^2}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^2}+\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (2 d \int \frac {1}{x^2 \sqrt {1-c^2 x^2} \sqrt {e x^2+d}}dx^2+2 e \int \frac {1}{-e x^4-c^2}d\frac {\sqrt {1-c^2 x^2}}{\sqrt {e x^2+d}}\right )}{2 e^2}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^2}+\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (4 d \int \frac {1}{x^4-d}d\frac {\sqrt {e x^2+d}}{\sqrt {1-c^2 x^2}}+2 e \int \frac {1}{-e x^4-c^2}d\frac {\sqrt {1-c^2 x^2}}{\sqrt {e x^2+d}}\right )}{2 e^2}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^2}+\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (4 d \int \frac {1}{x^4-d}d\frac {\sqrt {e x^2+d}}{\sqrt {1-c^2 x^2}}-\frac {2 \sqrt {e} \arctan \left (\frac {\sqrt {e} \sqrt {1-c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{c}\right )}{2 e^2}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^2}+\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle \frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^2}+\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (-\frac {2 \sqrt {e} \arctan \left (\frac {\sqrt {e} \sqrt {1-c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{c}-4 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )\right )}{2 e^2}\) |
(d*(a + b*ArcSech[c*x]))/(e^2*Sqrt[d + e*x^2]) + (Sqrt[d + e*x^2]*(a + b*A rcSech[c*x]))/e^2 + (b*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*((-2*Sqrt[e]*Arc Tan[(Sqrt[e]*Sqrt[1 - c^2*x^2])/(c*Sqrt[d + e*x^2])])/c - 4*Sqrt[d]*ArcTan h[Sqrt[d + e*x^2]/(Sqrt[d]*Sqrt[1 - c^2*x^2])]))/(2*e^2)
3.2.60.3.1 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[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && !GtQ[c - a*(d/b), 0]
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[q Subst[Int[x^(q*(m + 1) - 1) /(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] ] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_ )))/((a_.) + (b_.)*(x_)), x_] :> Simp[h/b Int[(c + d*x)^n*(e + f*x)^p, x] , x] + Simp[(b*g - a*h)/b Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[(x_)^(m_.)*((a_.) + (b_.)*(x_)^2)^(p_.)*((c_.) + (d_.)*(x_)^2)^(q_.)*(( e_.) + (f_.)*(x_)^2)^(r_.), x_Symbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2) *(a + b*x)^p*(c + d*x)^q*(e + f*x)^r, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, f, p, q, r}, x] && IntegerQ[(m - 1)/2]
Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*( x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Si mp[(a + b*ArcSech[c*x]) u, x] + Simp[b*Sqrt[1 + c*x]*Sqrt[1/(1 + c*x)] Int[SimplifyIntegrand[u/(x*Sqrt[1 - c*x]*Sqrt[1 + c*x]), x], x], x]] /; Fre eQ[{a, b, c, d, e, f, m, p}, x] && ((IGtQ[p, 0] && !(ILtQ[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) || (IGtQ[(m + 1)/2, 0] && !(ILtQ[p, 0] && GtQ[m + 2 *p + 3, 0])) || (ILtQ[(m + 2*p + 1)/2, 0] && !ILtQ[(m - 1)/2, 0]))
\[\int \frac {x^{3} \left (a +b \,\operatorname {arcsech}\left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 310 vs. \(2 (117) = 234\).
Time = 0.36 (sec) , antiderivative size = 1311, normalized size of antiderivative = 7.41 \[ \int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\text {Too large to display} \]
[-1/4*((b*e*x^2 + b*d)*sqrt(-e)*log(8*c^4*e^2*x^4 + c^4*d^2 - 6*c^2*d*e + 8*(c^4*d*e - c^2*e^2)*x^2 - 4*(2*c^4*e*x^3 + (c^4*d - c^2*e)*x)*sqrt(e*x^2 + d)*sqrt(-e)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + e^2) - 4*(b*c*e*x^2 + 2*b* c*d)*sqrt(e*x^2 + d)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) - 2*(b*c*e*x^2 + b*c*d)*sqrt(d)*log(((c^4*d^2 - 6*c^2*d*e + e^2)*x^4 - 8*(c ^2*d^2 - d*e)*x^2 + 4*((c^3*d - c*e)*x^3 - 2*c*d*x)*sqrt(e*x^2 + d)*sqrt(d )*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 8*d^2)/x^4) - 4*(a*c*e*x^2 + 2*a*c*d)*s qrt(e*x^2 + d))/(c*e^3*x^2 + c*d*e^2), -1/2*((b*e*x^2 + b*d)*sqrt(e)*arcta n(1/2*(2*c^2*e*x^3 + (c^2*d - e)*x)*sqrt(e*x^2 + d)*sqrt(e)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2))/(c^2*e^2*x^4 + (c^2*d*e - e^2)*x^2 - d*e)) - 2*(b*c*e*x^2 + 2*b*c*d)*sqrt(e*x^2 + d)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/( c*x)) - (b*c*e*x^2 + b*c*d)*sqrt(d)*log(((c^4*d^2 - 6*c^2*d*e + e^2)*x^4 - 8*(c^2*d^2 - d*e)*x^2 + 4*((c^3*d - c*e)*x^3 - 2*c*d*x)*sqrt(e*x^2 + d)*s qrt(d)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 8*d^2)/x^4) - 2*(a*c*e*x^2 + 2*a*c *d)*sqrt(e*x^2 + d))/(c*e^3*x^2 + c*d*e^2), -1/4*(4*(b*c*e*x^2 + b*c*d)*sq rt(-d)*arctan(-1/2*((c^3*d - c*e)*x^3 - 2*c*d*x)*sqrt(e*x^2 + d)*sqrt(-d)* sqrt(-(c^2*x^2 - 1)/(c^2*x^2))/(c^2*d*e*x^4 + (c^2*d^2 - d*e)*x^2 - d^2)) + (b*e*x^2 + b*d)*sqrt(-e)*log(8*c^4*e^2*x^4 + c^4*d^2 - 6*c^2*d*e + 8*(c^ 4*d*e - c^2*e^2)*x^2 - 4*(2*c^4*e*x^3 + (c^4*d - c^2*e)*x)*sqrt(e*x^2 + d) *sqrt(-e)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + e^2) - 4*(b*c*e*x^2 + 2*b*c*...
\[ \int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\int \frac {x^{3} \left (a + b \operatorname {asech}{\left (c x \right )}\right )}{\left (d + e x^{2}\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\int { \frac {{\left (b \operatorname {arsech}\left (c x\right ) + a\right )} x^{3}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\int \frac {x^3\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}{{\left (e\,x^2+d\right )}^{3/2}} \,d x \]