Integrand size = 19, antiderivative size = 126 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right )}{x^4} \, dx=\frac {b d \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{9 x^3}+\frac {b \left (2 c^2 d+9 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{9 x}-\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {e \left (a+b \text {sech}^{-1}(c x)\right )}{x} \]
-1/3*d*(a+b*arcsech(c*x))/x^3-e*(a+b*arcsech(c*x))/x+1/9*b*d*(1/(c*x+1))^( 1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1/2)/x^3+1/9*b*(2*c^2*d+9*e)*(1/(c*x+1))^ (1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1/2)/x
Time = 0.14 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.60 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right )}{x^4} \, dx=\frac {-3 a \left (d+3 e x^2\right )+b \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (d+2 c^2 d x^2+9 e x^2\right )-3 b \left (d+3 e x^2\right ) \text {sech}^{-1}(c x)}{9 x^3} \]
(-3*a*(d + 3*e*x^2) + b*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(d + 2*c^2*d*x ^2 + 9*e*x^2) - 3*b*(d + 3*e*x^2)*ArcSech[c*x])/(9*x^3)
Time = 0.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.87, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {6855, 27, 359, 242}
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.
\(\displaystyle \int \frac {\left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right )}{x^4} \, dx\) |
\(\Big \downarrow \) 6855 |
\(\displaystyle b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \int -\frac {3 e x^2+d}{3 x^4 \sqrt {1-c^2 x^2}}dx-\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {e \left (a+b \text {sech}^{-1}(c x)\right )}{x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{3} b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \int \frac {3 e x^2+d}{x^4 \sqrt {1-c^2 x^2}}dx-\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {e \left (a+b \text {sech}^{-1}(c x)\right )}{x}\) |
\(\Big \downarrow \) 359 |
\(\displaystyle -\frac {1}{3} b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (\frac {1}{3} \left (2 c^2 d+9 e\right ) \int \frac {1}{x^2 \sqrt {1-c^2 x^2}}dx-\frac {d \sqrt {1-c^2 x^2}}{3 x^3}\right )-\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {e \left (a+b \text {sech}^{-1}(c x)\right )}{x}\) |
\(\Big \downarrow \) 242 |
\(\displaystyle -\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {e \left (a+b \text {sech}^{-1}(c x)\right )}{x}-\frac {1}{3} b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (-\frac {\sqrt {1-c^2 x^2} \left (2 c^2 d+9 e\right )}{3 x}-\frac {d \sqrt {1-c^2 x^2}}{3 x^3}\right )\) |
-1/3*(b*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*(-1/3*(d*Sqrt[1 - c^2*x^2])/x^3 - ((2*c^2*d + 9*e)*Sqrt[1 - c^2*x^2])/(3*x))) - (d*(a + b*ArcSech[c*x]))/ (3*x^3) - (e*(a + b*ArcSech[c*x]))/x
3.1.92.3.1 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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, p}, x ] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
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]))
Time = 0.35 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.87
method | result | size |
parts | \(a \left (-\frac {e}{x}-\frac {d}{3 x^{3}}\right )+b \,c^{3} \left (-\frac {\operatorname {arcsech}\left (c x \right ) e}{c^{3} x}-\frac {\operatorname {arcsech}\left (c x \right ) d}{3 x^{3} c^{3}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \left (2 c^{4} d \,x^{2}+9 e \,c^{2} x^{2}+c^{2} d \right )}{9 c^{4} x^{2}}\right )\) | \(110\) |
derivativedivides | \(c^{3} \left (\frac {a \left (-\frac {e}{c x}-\frac {d}{3 c \,x^{3}}\right )}{c^{2}}+\frac {b \left (-\frac {\operatorname {arcsech}\left (c x \right ) e}{c x}-\frac {\operatorname {arcsech}\left (c x \right ) d}{3 c \,x^{3}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \left (2 c^{4} d \,x^{2}+9 e \,c^{2} x^{2}+c^{2} d \right )}{9 c^{2} x^{2}}\right )}{c^{2}}\right )\) | \(123\) |
default | \(c^{3} \left (\frac {a \left (-\frac {e}{c x}-\frac {d}{3 c \,x^{3}}\right )}{c^{2}}+\frac {b \left (-\frac {\operatorname {arcsech}\left (c x \right ) e}{c x}-\frac {\operatorname {arcsech}\left (c x \right ) d}{3 c \,x^{3}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \left (2 c^{4} d \,x^{2}+9 e \,c^{2} x^{2}+c^{2} d \right )}{9 c^{2} x^{2}}\right )}{c^{2}}\right )\) | \(123\) |
a*(-e/x-1/3*d/x^3)+b*c^3*(-1/c^3*arcsech(c*x)*e/x-1/3*arcsech(c*x)*d/x^3/c ^3+1/9/c^4*(-(c*x-1)/c/x)^(1/2)/x^2*((c*x+1)/c/x)^(1/2)*(2*c^4*d*x^2+9*c^2 *e*x^2+c^2*d))
Time = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.84 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right )}{x^4} \, dx=-\frac {9 \, a e x^{2} + 3 \, a d + 3 \, {\left (3 \, b e x^{2} + b d\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) - {\left (b c d x + {\left (2 \, b c^{3} d + 9 \, b c e\right )} x^{3}\right )} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}}}{9 \, x^{3}} \]
-1/9*(9*a*e*x^2 + 3*a*d + 3*(3*b*e*x^2 + b*d)*log((c*x*sqrt(-(c^2*x^2 - 1) /(c^2*x^2)) + 1)/(c*x)) - (b*c*d*x + (2*b*c^3*d + 9*b*c*e)*x^3)*sqrt(-(c^2 *x^2 - 1)/(c^2*x^2)))/x^3
\[ \int \frac {\left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right )}{x^4} \, dx=\int \frac {\left (a + b \operatorname {asech}{\left (c x \right )}\right ) \left (d + e x^{2}\right )}{x^{4}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.72 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right )}{x^4} \, dx={\left (c \sqrt {\frac {1}{c^{2} x^{2}} - 1} - \frac {\operatorname {arsech}\left (c x\right )}{x}\right )} b e + \frac {1}{9} \, b d {\left (\frac {c^{4} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {3}{2}} + 3 \, c^{4} \sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c} - \frac {3 \, \operatorname {arsech}\left (c x\right )}{x^{3}}\right )} - \frac {a e}{x} - \frac {a d}{3 \, x^{3}} \]
(c*sqrt(1/(c^2*x^2) - 1) - arcsech(c*x)/x)*b*e + 1/9*b*d*((c^4*(1/(c^2*x^2 ) - 1)^(3/2) + 3*c^4*sqrt(1/(c^2*x^2) - 1))/c - 3*arcsech(c*x)/x^3) - a*e/ x - 1/3*a*d/x^3
\[ \int \frac {\left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right )}{x^4} \, dx=\int { \frac {{\left (e x^{2} + d\right )} {\left (b \operatorname {arsech}\left (c x\right ) + a\right )}}{x^{4}} \,d x } \]
Timed out. \[ \int \frac {\left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right )}{x^4} \, dx=\int \frac {\left (e\,x^2+d\right )\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}{x^4} \,d x \]