Integrand size = 21, antiderivative size = 206 \[ \int (f x)^m \left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=-\frac {b e (f x)^{1+m} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{c^2 f (2+m) (3+m)}+\frac {d (f x)^{1+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f (1+m)}+\frac {e (f x)^{3+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (3+m)}+\frac {b \left (e (1+m)^2+c^2 d (2+m) (3+m)\right ) (f x)^{1+m} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},c^2 x^2\right )}{c^2 f (1+m)^2 (2+m) (3+m)} \]
d*(f*x)^(1+m)*(a+b*arcsech(c*x))/f/(1+m)+e*(f*x)^(3+m)*(a+b*arcsech(c*x))/ f^3/(3+m)+b*(e*(1+m)^2+c^2*d*(2+m)*(3+m))*(f*x)^(1+m)*hypergeom([1/2, 1/2+ 1/2*m],[3/2+1/2*m],c^2*x^2)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)/c^2/f/(1+m)^2/ (2+m)/(3+m)-b*e*(f*x)^(1+m)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^( 1/2)/c^2/f/(2+m)/(3+m)
Time = 0.53 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.92 \[ \int (f x)^m \left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=x (f x)^m \left (-\frac {b d \sqrt {\frac {1-c x}{1+c x}} \sqrt {1-c^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},c^2 x^2\right )}{(1+m)^2 (-1+c x)}+\frac {\frac {(3+m) \left (d (3+m)+e (1+m) x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right )}{1+m}-\frac {b e x^2 \sqrt {\frac {1-c x}{1+c x}} \sqrt {1-c^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},c^2 x^2\right )}{-1+c x}}{(3+m)^2}\right ) \]
x*(f*x)^m*(-((b*d*Sqrt[(1 - c*x)/(1 + c*x)]*Sqrt[1 - c^2*x^2]*Hypergeometr ic2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/((1 + m)^2*(-1 + c*x))) + (((3 + m)*(d*(3 + m) + e*(1 + m)*x^2)*(a + b*ArcSech[c*x]))/(1 + m) - (b*e*x^2* Sqrt[(1 - c*x)/(1 + c*x)]*Sqrt[1 - c^2*x^2]*Hypergeometric2F1[1/2, (3 + m) /2, (5 + m)/2, c^2*x^2])/(-1 + c*x))/(3 + m)^2)
Time = 0.39 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.89, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {6855, 27, 363, 278}
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.
| \(\displaystyle \int \left (d+e x^2\right ) (f x)^m \left (a+b \text {sech}^{-1}(c x)\right ) \, dx\) |
| \(\Big \downarrow \) 6855 |
| \(\displaystyle b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \int \frac {(f x)^m \left (e (m+1) x^2+d (m+3)\right )}{\left (m^2+4 m+3\right ) \sqrt {1-c^2 x^2}}dx+\frac {d (f x)^{m+1} \left (a+b \text {sech}^{-1}(c x)\right )}{f (m+1)}+\frac {e (f x)^{m+3} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (m+3)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \int \frac {(f x)^m \left (e (m+1) x^2+d (m+3)\right )}{\sqrt {1-c^2 x^2}}dx}{m^2+4 m+3}+\frac {d (f x)^{m+1} \left (a+b \text {sech}^{-1}(c x)\right )}{f (m+1)}+\frac {e (f x)^{m+3} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (m+3)}\) |
| \(\Big \downarrow \) 363 |
| \(\displaystyle \frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (\left (\frac {e (m+1)^2}{c^2 (m+2)}+d (m+3)\right ) \int \frac {(f x)^m}{\sqrt {1-c^2 x^2}}dx-\frac {e (m+1) \sqrt {1-c^2 x^2} (f x)^{m+1}}{c^2 f (m+2)}\right )}{m^2+4 m+3}+\frac {d (f x)^{m+1} \left (a+b \text {sech}^{-1}(c x)\right )}{f (m+1)}+\frac {e (f x)^{m+3} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (m+3)}\) |
| \(\Big \downarrow \) 278 |
| \(\displaystyle \frac {d (f x)^{m+1} \left (a+b \text {sech}^{-1}(c x)\right )}{f (m+1)}+\frac {e (f x)^{m+3} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (m+3)}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (\frac {(f x)^{m+1} \left (\frac {e (m+1)^2}{c^2 (m+2)}+d (m+3)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},c^2 x^2\right )}{f (m+1)}-\frac {e (m+1) \sqrt {1-c^2 x^2} (f x)^{m+1}}{c^2 f (m+2)}\right )}{m^2+4 m+3}\) |
(d*(f*x)^(1 + m)*(a + b*ArcSech[c*x]))/(f*(1 + m)) + (e*(f*x)^(3 + m)*(a + b*ArcSech[c*x]))/(f^3*(3 + m)) + (b*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*(- ((e*(1 + m)*(f*x)^(1 + m)*Sqrt[1 - c^2*x^2])/(c^2*f*(2 + m))) + (((e*(1 + m)^2)/(c^2*(2 + m)) + d*(3 + m))*(f*x)^(1 + m)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/(f*(1 + m))))/(3 + 4*m + m^2)
3.2.79.3.1 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[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
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]))
\[\int \left (f x \right )^{m} \left (e \,x^{2}+d \right ) \left (a +b \,\operatorname {arcsech}\left (c x \right )\right )d x\]
\[ \int (f x)^m \left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=\int { {\left (e x^{2} + d\right )} {\left (b \operatorname {arsech}\left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]
\[ \int (f x)^m \left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=\int \left (f x\right )^{m} \left (a + b \operatorname {asech}{\left (c x \right )}\right ) \left (d + e x^{2}\right )\, dx \]
\[ \int (f x)^m \left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=\int { {\left (e x^{2} + d\right )} {\left (b \operatorname {arsech}\left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]
a*e*f^m*x^3*x^m/(m + 3) + (f*x)^(m + 1)*a*d/(f*(m + 1)) + ((b*e*f^m*(m + 1 )*x^3*x^m + b*d*f^m*(m + 3)*x*x^m)*log(sqrt(c*x + 1)*sqrt(-c*x + 1) + 1) - (b*e*f^m*(m + 1)*x^3*x^m + b*d*f^m*(m + 3)*x*x^m)*log(x))/(m^2 + 4*m + 3) - integrate((b*c^2*e*f^m*(m + 3)*x^2*log(c) - (e*f^m*(m + 3)*log(c) - e*f ^m)*b)*x^2*x^m/(c^2*(m + 3)*x^2 - m - 3), x) - integrate((b*c^2*d*f^m*(m + 1)*x^2*log(c) - (d*f^m*(m + 1)*log(c) - d*f^m)*b)*x^m/(c^2*(m + 1)*x^2 - m - 1), x) + integrate((b*c^2*e*f^m*(m + 1)*x^4*x^m + b*c^2*d*f^m*(m + 3)* x^2*x^m)/((m^2 + 4*m + 3)*c^2*x^2 + ((m^2 + 4*m + 3)*c^2*x^2 - m^2 - 4*m - 3)*sqrt(c*x + 1)*sqrt(-c*x + 1) - m^2 - 4*m - 3), x)
\[ \int (f x)^m \left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=\int { {\left (e x^{2} + d\right )} {\left (b \operatorname {arsech}\left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]
Timed out. \[ \int (f x)^m \left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=\int {\left (f\,x\right )}^m\,\left (e\,x^2+d\right )\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right ) \,d x \]