Integrand size = 16, antiderivative size = 112 \[ \int \left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=-\frac {b e x \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{6 c^2}+d x \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{3} e x^3 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {b \left (6 c^2 d+e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \arcsin (c x)}{6 c^3} \] Output:
-1/6*b*e*x*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1/2)/c^2+d*x*(a+b *arcsech(c*x))+1/3*e*x^3*(a+b*arcsech(c*x))+1/6*b*(6*c^2*d+e)*(1/(c*x+1))^ (1/2)*(c*x+1)^(1/2)*arcsin(c*x)/c^3
Result contains complex when optimal does not.
Time = 0.25 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.69 \[ \int \left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=a d x+\frac {1}{3} a e x^3+b e \sqrt {\frac {1-c x}{1+c x}} \left (-\frac {x}{6 c^2}-\frac {x^2}{6 c}\right )+b d x \text {sech}^{-1}(c x)+\frac {1}{3} b e x^3 \text {sech}^{-1}(c x)+\frac {2 b d \sqrt {\frac {1-c x}{1+c x}} \sqrt {1-c^2 x^2} \arctan \left (\frac {\sqrt {1-c^2 x^2}}{1-c x}\right )}{c-c^2 x}+\frac {i b e \log \left (-2 i c x+2 \sqrt {\frac {1-c x}{1+c x}} (1+c x)\right )}{6 c^3} \] Input:
Integrate[(d + e*x^2)*(a + b*ArcSech[c*x]),x]
Output:
a*d*x + (a*e*x^3)/3 + b*e*Sqrt[(1 - c*x)/(1 + c*x)]*(-1/6*x/c^2 - x^2/(6*c )) + b*d*x*ArcSech[c*x] + (b*e*x^3*ArcSech[c*x])/3 + (2*b*d*Sqrt[(1 - c*x) /(1 + c*x)]*Sqrt[1 - c^2*x^2]*ArcTan[Sqrt[1 - c^2*x^2]/(1 - c*x)])/(c - c^ 2*x) + ((I/6)*b*e*Log[(-2*I)*c*x + 2*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)]) /c^3
Time = 0.27 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6845, 27, 299, 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.
\(\displaystyle \int \left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right ) \, dx\) |
\(\Big \downarrow \) 6845 |
\(\displaystyle b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \int \frac {e x^2+3 d}{3 \sqrt {1-c^2 x^2}}dx+d x \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{3} e x^3 \left (a+b \text {sech}^{-1}(c x)\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \int \frac {e x^2+3 d}{\sqrt {1-c^2 x^2}}dx+d x \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{3} e x^3 \left (a+b \text {sech}^{-1}(c x)\right )\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {1}{3} b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (\frac {\left (6 c^2 d+e\right ) \int \frac {1}{\sqrt {1-c^2 x^2}}dx}{2 c^2}-\frac {e x \sqrt {1-c^2 x^2}}{2 c^2}\right )+d x \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{3} e x^3 \left (a+b \text {sech}^{-1}(c x)\right )\) |
\(\Big \downarrow \) 223 |
\(\displaystyle d x \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{3} e x^3 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{3} b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (\frac {\arcsin (c x) \left (6 c^2 d+e\right )}{2 c^3}-\frac {e x \sqrt {1-c^2 x^2}}{2 c^2}\right )\) |
Input:
Int[(d + e*x^2)*(a + b*ArcSech[c*x]),x]
Output:
d*x*(a + b*ArcSech[c*x]) + (e*x^3*(a + b*ArcSech[c*x]))/3 + (b*Sqrt[(1 + c *x)^(-1)]*Sqrt[1 + c*x]*(-1/2*(e*x*Sqrt[1 - c^2*x^2])/c^2 + ((6*c^2*d + e) *ArcSin[c*x])/(2*c^3)))/3
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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_)^2)^(p_.), x_Sym bol] :> With[{u = IntHide[(d + e*x^2)^p, x]}, Simp[(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]] /; FreeQ[{a, b, c, d, e}, x] && ( IGtQ[p, 0] || ILtQ[p + 1/2, 0])
Time = 0.32 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.07
method | result | size |
parts | \(a \left (\frac {1}{3} x^{3} e +d x \right )+\frac {b \left (\frac {c \,\operatorname {arcsech}\left (c x \right ) x^{3} e}{3}+\operatorname {arcsech}\left (c x \right ) d c x +\frac {\sqrt {-\frac {c x -1}{c x}}\, x \sqrt {\frac {c x +1}{c x}}\, \left (6 \arcsin \left (c x \right ) c^{2} d -\sqrt {-c^{2} x^{2}+1}\, e c x +\arcsin \left (c x \right ) e \right )}{6 c \sqrt {-c^{2} x^{2}+1}}\right )}{c}\) | \(120\) |
derivativedivides | \(\frac {\frac {a \left (d \,c^{3} x +\frac {1}{3} c^{3} x^{3} e \right )}{c^{2}}+\frac {b \left (\operatorname {arcsech}\left (c x \right ) d \,c^{3} x +\frac {\operatorname {arcsech}\left (c x \right ) c^{3} x^{3} e}{3}+\frac {\sqrt {-\frac {c x -1}{c x}}\, c x \sqrt {\frac {c x +1}{c x}}\, \left (6 \arcsin \left (c x \right ) c^{2} d -\sqrt {-c^{2} x^{2}+1}\, e c x +\arcsin \left (c x \right ) e \right )}{6 \sqrt {-c^{2} x^{2}+1}}\right )}{c^{2}}}{c}\) | \(135\) |
default | \(\frac {\frac {a \left (d \,c^{3} x +\frac {1}{3} c^{3} x^{3} e \right )}{c^{2}}+\frac {b \left (\operatorname {arcsech}\left (c x \right ) d \,c^{3} x +\frac {\operatorname {arcsech}\left (c x \right ) c^{3} x^{3} e}{3}+\frac {\sqrt {-\frac {c x -1}{c x}}\, c x \sqrt {\frac {c x +1}{c x}}\, \left (6 \arcsin \left (c x \right ) c^{2} d -\sqrt {-c^{2} x^{2}+1}\, e c x +\arcsin \left (c x \right ) e \right )}{6 \sqrt {-c^{2} x^{2}+1}}\right )}{c^{2}}}{c}\) | \(135\) |
Input:
int((e*x^2+d)*(a+b*arcsech(c*x)),x,method=_RETURNVERBOSE)
Output:
a*(1/3*x^3*e+d*x)+b/c*(1/3*c*arcsech(c*x)*x^3*e+arcsech(c*x)*d*c*x+1/6/c*( -(c*x-1)/c/x)^(1/2)*x*((c*x+1)/c/x)^(1/2)*(6*arcsin(c*x)*c^2*d-(-c^2*x^2+1 )^(1/2)*e*c*x+arcsin(c*x)*e)/(-c^2*x^2+1)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (64) = 128\).
Time = 0.15 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.87 \[ \int \left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=\frac {2 \, a c^{3} e x^{3} - b c^{2} e x^{2} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 6 \, a c^{3} d x - 2 \, {\left (6 \, b c^{2} d + b e\right )} \arctan \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{c x}\right ) - 2 \, {\left (3 \, b c^{3} d + b c^{3} e\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{x}\right ) + 2 \, {\left (b c^{3} e x^{3} + 3 \, b c^{3} d x - 3 \, b c^{3} d - b c^{3} e\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )}{6 \, c^{3}} \] Input:
integrate((e*x^2+d)*(a+b*arcsech(c*x)),x, algorithm="fricas")
Output:
1/6*(2*a*c^3*e*x^3 - b*c^2*e*x^2*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 6*a*c^3* d*x - 2*(6*b*c^2*d + b*e)*arctan((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) - 1)/ (c*x)) - 2*(3*b*c^3*d + b*c^3*e)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) - 1)/x) + 2*(b*c^3*e*x^3 + 3*b*c^3*d*x - 3*b*c^3*d - b*c^3*e)*log((c*x*sqrt (-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)))/c^3
\[ \int \left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=\int \left (a + b \operatorname {asech}{\left (c x \right )}\right ) \left (d + e x^{2}\right )\, dx \] Input:
integrate((e*x**2+d)*(a+b*asech(c*x)),x)
Output:
Integral((a + b*asech(c*x))*(d + e*x**2), x)
Time = 0.11 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.96 \[ \int \left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=\frac {1}{3} \, a e x^{3} + \frac {1}{6} \, {\left (2 \, x^{3} \operatorname {arsech}\left (c x\right ) - \frac {\frac {\sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c^{2} {\left (\frac {1}{c^{2} x^{2}} - 1\right )} + c^{2}} + \frac {\arctan \left (\sqrt {\frac {1}{c^{2} x^{2}} - 1}\right )}{c^{2}}}{c}\right )} b e + a d x + \frac {{\left (c x \operatorname {arsech}\left (c x\right ) - \arctan \left (\sqrt {\frac {1}{c^{2} x^{2}} - 1}\right )\right )} b d}{c} \] Input:
integrate((e*x^2+d)*(a+b*arcsech(c*x)),x, algorithm="maxima")
Output:
1/3*a*e*x^3 + 1/6*(2*x^3*arcsech(c*x) - (sqrt(1/(c^2*x^2) - 1)/(c^2*(1/(c^ 2*x^2) - 1) + c^2) + arctan(sqrt(1/(c^2*x^2) - 1))/c^2)/c)*b*e + a*d*x + ( c*x*arcsech(c*x) - arctan(sqrt(1/(c^2*x^2) - 1)))*b*d/c
\[ \int \left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=\int { {\left (e x^{2} + d\right )} {\left (b \operatorname {arsech}\left (c x\right ) + a\right )} \,d x } \] Input:
integrate((e*x^2+d)*(a+b*arcsech(c*x)),x, algorithm="giac")
Output:
integrate((e*x^2 + d)*(b*arcsech(c*x) + a), x)
Timed out. \[ \int \left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=\int \left (e\,x^2+d\right )\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right ) \,d x \] Input:
int((d + e*x^2)*(a + b*acosh(1/(c*x))),x)
Output:
int((d + e*x^2)*(a + b*acosh(1/(c*x))), x)
\[ \int \left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=\left (\int \mathit {asech} \left (c x \right )d x \right ) b d +\left (\int \mathit {asech} \left (c x \right ) x^{2}d x \right ) b e +a d x +\frac {a e \,x^{3}}{3} \] Input:
int((e*x^2+d)*(a+b*asech(c*x)),x)
Output:
(3*int(asech(c*x),x)*b*d + 3*int(asech(c*x)*x**2,x)*b*e + 3*a*d*x + a*e*x* *3)/3