Integrand size = 21, antiderivative size = 176 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right )}{x^4} \, dx=\frac {b d^2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{9 x^3}+\frac {2 b d \left (c^2 d+9 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{9 x}-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {2 d e \left (a+b \text {sech}^{-1}(c x)\right )}{x}+e^2 x \left (a+b \text {sech}^{-1}(c x)\right )+\frac {b e^2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \arcsin (c x)}{c} \]
-1/3*d^2*(a+b*arcsech(c*x))/x^3-2*d*e*(a+b*arcsech(c*x))/x+e^2*x*(a+b*arcs ech(c*x))+b*e^2*arcsin(c*x)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)/c+1/9*b*d^2*(1 /(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1/2)/x^3+2/9*b*d*(c^2*d+9*e)*( 1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1/2)/x
Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.85 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right )}{x^4} \, dx=\frac {b c d \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (d+2 c^2 d x^2+18 e x^2\right )-3 a c \left (d^2+6 d e x^2-3 e^2 x^4\right )-3 b c \left (d^2+6 d e x^2-3 e^2 x^4\right ) \text {sech}^{-1}(c x)+9 i b e^2 x^3 \log \left (-2 i c x+2 \sqrt {\frac {1-c x}{1+c x}} (1+c x)\right )}{9 c x^3} \]
(b*c*d*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(d + 2*c^2*d*x^2 + 18*e*x^2) - 3*a*c*(d^2 + 6*d*e*x^2 - 3*e^2*x^4) - 3*b*c*(d^2 + 6*d*e*x^2 - 3*e^2*x^4)* ArcSech[c*x] + (9*I)*b*e^2*x^3*Log[(-2*I)*c*x + 2*Sqrt[(1 - c*x)/(1 + c*x) ]*(1 + c*x)])/(9*c*x^3)
Time = 0.38 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.81, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6855, 27, 1588, 25, 358, 223}
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.
| \(\displaystyle \int \frac {\left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right )}{x^4} \, dx\) |
| \(\Big \downarrow \) 6855 |
| \(\displaystyle b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \int -\frac {-3 e^2 x^4+6 d e x^2+d^2}{3 x^4 \sqrt {1-c^2 x^2}}dx-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {2 d e \left (a+b \text {sech}^{-1}(c x)\right )}{x}+e^2 x \left (a+b \text {sech}^{-1}(c x)\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {1}{3} b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \int \frac {-3 e^2 x^4+6 d e x^2+d^2}{x^4 \sqrt {1-c^2 x^2}}dx-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {2 d e \left (a+b \text {sech}^{-1}(c x)\right )}{x}+e^2 x \left (a+b \text {sech}^{-1}(c x)\right )\) |
| \(\Big \downarrow \) 1588 |
| \(\displaystyle -\frac {1}{3} b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (-\frac {1}{3} \int -\frac {2 d \left (d c^2+9 e\right )-9 e^2 x^2}{x^2 \sqrt {1-c^2 x^2}}dx-\frac {d^2 \sqrt {1-c^2 x^2}}{3 x^3}\right )-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {2 d e \left (a+b \text {sech}^{-1}(c x)\right )}{x}+e^2 x \left (a+b \text {sech}^{-1}(c x)\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{3} b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (\frac {1}{3} \int \frac {2 d \left (d c^2+9 e\right )-9 e^2 x^2}{x^2 \sqrt {1-c^2 x^2}}dx-\frac {d^2 \sqrt {1-c^2 x^2}}{3 x^3}\right )-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {2 d e \left (a+b \text {sech}^{-1}(c x)\right )}{x}+e^2 x \left (a+b \text {sech}^{-1}(c x)\right )\) |
| \(\Big \downarrow \) 358 |
| \(\displaystyle -\frac {1}{3} b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (\frac {1}{3} \left (-9 e^2 \int \frac {1}{\sqrt {1-c^2 x^2}}dx-\frac {2 d \sqrt {1-c^2 x^2} \left (c^2 d+9 e\right )}{x}\right )-\frac {d^2 \sqrt {1-c^2 x^2}}{3 x^3}\right )-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {2 d e \left (a+b \text {sech}^{-1}(c x)\right )}{x}+e^2 x \left (a+b \text {sech}^{-1}(c x)\right )\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle -\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{3 x^3}-\frac {2 d e \left (a+b \text {sech}^{-1}(c x)\right )}{x}+e^2 x \left (a+b \text {sech}^{-1}(c x)\right )-\frac {1}{3} b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (\frac {1}{3} \left (-\frac {9 e^2 \arcsin (c x)}{c}-\frac {2 d \sqrt {1-c^2 x^2} \left (c^2 d+9 e\right )}{x}\right )-\frac {d^2 \sqrt {1-c^2 x^2}}{3 x^3}\right )\) |
-1/3*(d^2*(a + b*ArcSech[c*x]))/x^3 - (2*d*e*(a + b*ArcSech[c*x]))/x + e^2 *x*(a + b*ArcSech[c*x]) - (b*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*(-1/3*(d^2 *Sqrt[1 - c^2*x^2])/x^3 + ((-2*d*(c^2*d + 9*e)*Sqrt[1 - c^2*x^2])/x - (9*e ^2*ArcSin[c*x])/c)/3))/3
3.2.3.3.1 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_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
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] + S imp[d/e^2 Int[(e*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e , m, p}, x] && NeQ[b*c - a*d, 0] && EqQ[Simplify[m + 2*p + 3], 0] && NeQ[m, -1]
Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c _.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, f*x, x], R = PolynomialRemainder[(a + b*x^2 + c*x^4)^p, f*x, x]}, Simp[R*(f*x)^(m + 1)*((d + e*x^2)^(q + 1)/(d*f*(m + 1))), x] + Simp[1/(d*f ^2*(m + 1)) Int[(f*x)^(m + 2)*(d + e*x^2)^q*ExpandToSum[d*f*(m + 1)*(Qx/x ) - e*R*(m + 2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q}, x] && Ne Q[b^2 - 4*a*c, 0] && IGtQ[p, 0] && LtQ[m, -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 = 0.62 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.14
| method | result | size |
| parts | \(a \left (e^{2} x -\frac {2 d e}{x}-\frac {d^{2}}{3 x^{3}}\right )+b \,c^{3} \left (\frac {\operatorname {arcsech}\left (c x \right ) e^{2} x}{c^{3}}-\frac {2 \,\operatorname {arcsech}\left (c x \right ) d e}{c^{3} x}-\frac {\operatorname {arcsech}\left (c x \right ) d^{2}}{3 x^{3} c^{3}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \left (2 \sqrt {-c^{2} x^{2}+1}\, c^{6} d^{2} x^{2}+\sqrt {-c^{2} x^{2}+1}\, c^{4} d^{2}+18 \sqrt {-c^{2} x^{2}+1}\, c^{4} d e \,x^{2}+9 \arcsin \left (c x \right ) e^{2} c^{3} x^{3}\right )}{9 c^{6} x^{2} \sqrt {-c^{2} x^{2}+1}}\right )\) | \(201\) |
| derivativedivides | \(c^{3} \left (\frac {a \left (e^{2} c x -\frac {c \,d^{2}}{3 x^{3}}-\frac {2 c d e}{x}\right )}{c^{4}}+\frac {b \left (\operatorname {arcsech}\left (c x \right ) e^{2} c x -\frac {\operatorname {arcsech}\left (c x \right ) c \,d^{2}}{3 x^{3}}-\frac {2 \,\operatorname {arcsech}\left (c x \right ) c d e}{x}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \left (2 \sqrt {-c^{2} x^{2}+1}\, c^{6} d^{2} x^{2}+\sqrt {-c^{2} x^{2}+1}\, c^{4} d^{2}+18 \sqrt {-c^{2} x^{2}+1}\, c^{4} d e \,x^{2}+9 \arcsin \left (c x \right ) e^{2} c^{3} x^{3}\right )}{9 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}\right )}{c^{4}}\right )\) | \(205\) |
| default | \(c^{3} \left (\frac {a \left (e^{2} c x -\frac {c \,d^{2}}{3 x^{3}}-\frac {2 c d e}{x}\right )}{c^{4}}+\frac {b \left (\operatorname {arcsech}\left (c x \right ) e^{2} c x -\frac {\operatorname {arcsech}\left (c x \right ) c \,d^{2}}{3 x^{3}}-\frac {2 \,\operatorname {arcsech}\left (c x \right ) c d e}{x}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \left (2 \sqrt {-c^{2} x^{2}+1}\, c^{6} d^{2} x^{2}+\sqrt {-c^{2} x^{2}+1}\, c^{4} d^{2}+18 \sqrt {-c^{2} x^{2}+1}\, c^{4} d e \,x^{2}+9 \arcsin \left (c x \right ) e^{2} c^{3} x^{3}\right )}{9 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}\right )}{c^{4}}\right )\) | \(205\) |
a*(e^2*x-2*d*e/x-1/3*d^2/x^3)+b*c^3*(1/c^3*arcsech(c*x)*e^2*x-2/c^3*arcsec h(c*x)*d*e/x-1/3*arcsech(c*x)*d^2/x^3/c^3+1/9/c^6*(-(c*x-1)/c/x)^(1/2)/x^2 *((c*x+1)/c/x)^(1/2)*(2*(-c^2*x^2+1)^(1/2)*c^6*d^2*x^2+(-c^2*x^2+1)^(1/2)* c^4*d^2+18*(-c^2*x^2+1)^(1/2)*c^4*d*e*x^2+9*arcsin(c*x)*e^2*c^3*x^3)/(-c^2 *x^2+1)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (106) = 212\).
Time = 0.31 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.52 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right )}{x^4} \, dx=\frac {9 \, a c e^{2} x^{4} - 18 \, b e^{2} x^{3} \arctan \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{c x}\right ) - 18 \, a c d e x^{2} + 3 \, {\left (b c d^{2} + 6 \, b c d e - 3 \, b c e^{2}\right )} x^{3} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{x}\right ) - 3 \, a c d^{2} + 3 \, {\left (3 \, b c e^{2} x^{4} - 6 \, b c d e x^{2} - b c d^{2} + {\left (b c d^{2} + 6 \, b c d e - 3 \, b c e^{2}\right )} x^{3}\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) + {\left (b c^{2} d^{2} x + 2 \, {\left (b c^{4} d^{2} + 9 \, b c^{2} d e\right )} x^{3}\right )} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}}}{9 \, c x^{3}} \]
1/9*(9*a*c*e^2*x^4 - 18*b*e^2*x^3*arctan((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2 )) - 1)/(c*x)) - 18*a*c*d*e*x^2 + 3*(b*c*d^2 + 6*b*c*d*e - 3*b*c*e^2)*x^3* log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) - 1)/x) - 3*a*c*d^2 + 3*(3*b*c*e^2 *x^4 - 6*b*c*d*e*x^2 - b*c*d^2 + (b*c*d^2 + 6*b*c*d*e - 3*b*c*e^2)*x^3)*lo g((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) + (b*c^2*d^2*x + 2*(b*c^ 4*d^2 + 9*b*c^2*d*e)*x^3)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))/(c*x^3)
\[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right )}{x^4} \, dx=\int \frac {\left (a + b \operatorname {asech}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{2}}{x^{4}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.76 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right )}{x^4} \, dx=2 \, {\left (c \sqrt {\frac {1}{c^{2} x^{2}} - 1} - \frac {\operatorname {arsech}\left (c x\right )}{x}\right )} b d e + a e^{2} x + \frac {1}{9} \, b d^{2} {\left (\frac {c^{4} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {3}{2}} + 3 \, c^{4} \sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c} - \frac {3 \, \operatorname {arsech}\left (c x\right )}{x^{3}}\right )} + \frac {{\left (c x \operatorname {arsech}\left (c x\right ) - \arctan \left (\sqrt {\frac {1}{c^{2} x^{2}} - 1}\right )\right )} b e^{2}}{c} - \frac {2 \, a d e}{x} - \frac {a d^{2}}{3 \, x^{3}} \]
2*(c*sqrt(1/(c^2*x^2) - 1) - arcsech(c*x)/x)*b*d*e + a*e^2*x + 1/9*b*d^2*( (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) + (c*x*arcsech(c*x) - arctan(sqrt(1/(c^2*x^2) - 1)))*b*e^2/c - 2 *a*d*e/x - 1/3*a*d^2/x^3
\[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right )}{x^4} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arsech}\left (c x\right ) + a\right )}}{x^{4}} \,d x } \]
Timed out. \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right )}{x^4} \, dx=\int \frac {{\left (e\,x^2+d\right )}^2\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}{x^4} \,d x \]