Integrand size = 23, antiderivative size = 372 \[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=-\frac {b e \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right ) (f x)^{1+m} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{c^4 f (2+m) (3+m) (4+m) (5+m)}-\frac {b e^2 (f x)^{3+m} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{c^2 f^3 (4+m) (5+m)}+\frac {d^2 (f x)^{1+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \text {sech}^{-1}(c x)\right )}{f^5 (5+m)}+\frac {b \left (c^4 d^2 (2+m) (3+m) (4+m) (5+m)+e (1+m)^2 \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right )\right ) (f x)^{1+m} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},c^2 x^2\right )}{c^4 f (1+m)^2 (2+m) (3+m) (4+m) (5+m)} \]
d^2*(f*x)^(1+m)*(a+b*arcsech(c*x))/f/(1+m)+2*d*e*(f*x)^(3+m)*(a+b*arcsech( c*x))/f^3/(3+m)+e^2*(f*x)^(5+m)*(a+b*arcsech(c*x))/f^5/(5+m)+b*(c^4*d^2*(2 +m)*(3+m)*(4+m)*(5+m)+e*(1+m)^2*(e*(3+m)^2+2*c^2*d*(m^2+9*m+20)))*(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^4/f/(1+m)^2/(2+m)/(3+m)/(4+m)/(5+m)-b*e*(e*(3+m)^2+2*c^2*d*(m^ 2+9*m+20))*(f*x)^(1+m)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1/2)/ c^4/f/(4+m)/(5+m)/(m^2+5*m+6)-b*e^2*(f*x)^(3+m)*(1/(c*x+1))^(1/2)*(c*x+1)^ (1/2)*(-c^2*x^2+1)^(1/2)/c^2/f^3/(4+m)/(5+m)
Time = 0.80 (sec) , antiderivative size = 322, normalized size of antiderivative = 0.87 \[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=x (f x)^m \left (\frac {a d^2}{1+m}+\frac {2 a d e x^2}{3+m}+\frac {a e^2 x^4}{5+m}+\frac {b d^2 \text {sech}^{-1}(c x)}{1+m}+\frac {2 b d e x^2 \text {sech}^{-1}(c x)}{3+m}+\frac {b e^2 x^4 \text {sech}^{-1}(c x)}{5+m}-\frac {b d^2 \sqrt {\frac {1-c x}{1+c x}} \sqrt {1-c^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},c^2 x^2\right )}{(1+m)^2 (-1+c x)}-\frac {2 b d e x^2 \sqrt {\frac {1-c x}{1+c x}} \sqrt {1-c^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},c^2 x^2\right )}{(3+m)^2 (-1+c x)}-\frac {b e^2 x^4 \sqrt {\frac {1-c x}{1+c x}} \sqrt {1-c^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5+m}{2},\frac {7+m}{2},c^2 x^2\right )}{(5+m)^2 (-1+c x)}\right ) \]
x*(f*x)^m*((a*d^2)/(1 + m) + (2*a*d*e*x^2)/(3 + m) + (a*e^2*x^4)/(5 + m) + (b*d^2*ArcSech[c*x])/(1 + m) + (2*b*d*e*x^2*ArcSech[c*x])/(3 + m) + (b*e^ 2*x^4*ArcSech[c*x])/(5 + m) - (b*d^2*Sqrt[(1 - c*x)/(1 + c*x)]*Sqrt[1 - c^ 2*x^2]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/((1 + m)^2*( -1 + c*x)) - (2*b*d*e*x^2*Sqrt[(1 - c*x)/(1 + c*x)]*Sqrt[1 - c^2*x^2]*Hype rgeometric2F1[1/2, (3 + m)/2, (5 + m)/2, c^2*x^2])/((3 + m)^2*(-1 + c*x)) - (b*e^2*x^4*Sqrt[(1 - c*x)/(1 + c*x)]*Sqrt[1 - c^2*x^2]*Hypergeometric2F1 [1/2, (5 + m)/2, (7 + m)/2, c^2*x^2])/((5 + m)^2*(-1 + c*x)))
Time = 0.67 (sec) , antiderivative size = 328, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {6855, 27, 1590, 25, 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.
\(\displaystyle \int \left (d+e x^2\right )^2 (f x)^m \left (a+b \text {sech}^{-1}(c x)\right ) \, dx\) |
\(\Big \downarrow \) 6855 |
\(\displaystyle b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \int \frac {(f x)^m \left (e^2 (m+1) (m+3) x^4+2 d e (m+1) (m+5) x^2+d^2 (m+3) (m+5)\right )}{\left (m^3+9 m^2+23 m+15\right ) \sqrt {1-c^2 x^2}}dx+\frac {d^2 (f x)^{m+1} \left (a+b \text {sech}^{-1}(c x)\right )}{f (m+1)}+\frac {2 d e (f x)^{m+3} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (m+3)}+\frac {e^2 (f x)^{m+5} \left (a+b \text {sech}^{-1}(c x)\right )}{f^5 (m+5)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \int \frac {(f x)^m \left (e^2 (m+1) (m+3) x^4+2 d e (m+1) (m+5) x^2+d^2 (m+3) (m+5)\right )}{\sqrt {1-c^2 x^2}}dx}{m^3+9 m^2+23 m+15}+\frac {d^2 (f x)^{m+1} \left (a+b \text {sech}^{-1}(c x)\right )}{f (m+1)}+\frac {2 d e (f x)^{m+3} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (m+3)}+\frac {e^2 (f x)^{m+5} \left (a+b \text {sech}^{-1}(c x)\right )}{f^5 (m+5)}\) |
\(\Big \downarrow \) 1590 |
\(\displaystyle \frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (-\frac {\int -\frac {(f x)^m \left (c^2 (m+3) (m+4) (m+5) d^2+e (m+1) \left (2 d \left (m^2+9 m+20\right ) c^2+e (m+3)^2\right ) x^2\right )}{\sqrt {1-c^2 x^2}}dx}{c^2 (m+4)}-\frac {e^2 (m+1) (m+3) \sqrt {1-c^2 x^2} (f x)^{m+3}}{c^2 f^3 (m+4)}\right )}{m^3+9 m^2+23 m+15}+\frac {d^2 (f x)^{m+1} \left (a+b \text {sech}^{-1}(c x)\right )}{f (m+1)}+\frac {2 d e (f x)^{m+3} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (m+3)}+\frac {e^2 (f x)^{m+5} \left (a+b \text {sech}^{-1}(c x)\right )}{f^5 (m+5)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (\frac {\int \frac {(f x)^m \left (c^2 (m+3) (m+4) (m+5) d^2+e (m+1) \left (2 d \left (m^2+9 m+20\right ) c^2+e (m+3)^2\right ) x^2\right )}{\sqrt {1-c^2 x^2}}dx}{c^2 (m+4)}-\frac {e^2 (m+1) (m+3) \sqrt {1-c^2 x^2} (f x)^{m+3}}{c^2 f^3 (m+4)}\right )}{m^3+9 m^2+23 m+15}+\frac {d^2 (f x)^{m+1} \left (a+b \text {sech}^{-1}(c x)\right )}{f (m+1)}+\frac {2 d e (f x)^{m+3} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (m+3)}+\frac {e^2 (f x)^{m+5} \left (a+b \text {sech}^{-1}(c x)\right )}{f^5 (m+5)}\) |
\(\Big \downarrow \) 363 |
\(\displaystyle \frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (\frac {\frac {\left (c^4 d^2 (m+3) (m+4) (m+5)+\frac {e (m+1)^2 \left (2 c^2 d \left (m^2+9 m+20\right )+e (m+3)^2\right )}{m+2}\right ) \int \frac {(f x)^m}{\sqrt {1-c^2 x^2}}dx}{c^2}-\frac {e (m+1) \sqrt {1-c^2 x^2} (f x)^{m+1} \left (2 c^2 d \left (m^2+9 m+20\right )+e (m+3)^2\right )}{c^2 f (m+2)}}{c^2 (m+4)}-\frac {e^2 (m+1) (m+3) \sqrt {1-c^2 x^2} (f x)^{m+3}}{c^2 f^3 (m+4)}\right )}{m^3+9 m^2+23 m+15}+\frac {d^2 (f x)^{m+1} \left (a+b \text {sech}^{-1}(c x)\right )}{f (m+1)}+\frac {2 d e (f x)^{m+3} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (m+3)}+\frac {e^2 (f x)^{m+5} \left (a+b \text {sech}^{-1}(c x)\right )}{f^5 (m+5)}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {d^2 (f x)^{m+1} \left (a+b \text {sech}^{-1}(c x)\right )}{f (m+1)}+\frac {2 d e (f x)^{m+3} \left (a+b \text {sech}^{-1}(c x)\right )}{f^3 (m+3)}+\frac {e^2 (f x)^{m+5} \left (a+b \text {sech}^{-1}(c x)\right )}{f^5 (m+5)}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (\frac {\frac {(f x)^{m+1} \left (c^4 d^2 (m+3) (m+4) (m+5)+\frac {e (m+1)^2 \left (2 c^2 d \left (m^2+9 m+20\right )+e (m+3)^2\right )}{m+2}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},c^2 x^2\right )}{c^2 f (m+1)}-\frac {e (m+1) \sqrt {1-c^2 x^2} (f x)^{m+1} \left (2 c^2 d \left (m^2+9 m+20\right )+e (m+3)^2\right )}{c^2 f (m+2)}}{c^2 (m+4)}-\frac {e^2 (m+1) (m+3) \sqrt {1-c^2 x^2} (f x)^{m+3}}{c^2 f^3 (m+4)}\right )}{m^3+9 m^2+23 m+15}\) |
(d^2*(f*x)^(1 + m)*(a + b*ArcSech[c*x]))/(f*(1 + m)) + (2*d*e*(f*x)^(3 + m )*(a + b*ArcSech[c*x]))/(f^3*(3 + m)) + (e^2*(f*x)^(5 + m)*(a + b*ArcSech[ c*x]))/(f^5*(5 + m)) + (b*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*(-((e^2*(1 + m)*(3 + m)*(f*x)^(3 + m)*Sqrt[1 - c^2*x^2])/(c^2*f^3*(4 + m))) + (-((e*(1 + m)*(e*(3 + m)^2 + 2*c^2*d*(20 + 9*m + m^2))*(f*x)^(1 + m)*Sqrt[1 - c^2*x ^2])/(c^2*f*(2 + m))) + ((c^4*d^2*(3 + m)*(4 + m)*(5 + m) + (e*(1 + m)^2*( e*(3 + m)^2 + 2*c^2*d*(20 + 9*m + m^2)))/(2 + m))*(f*x)^(1 + m)*Hypergeome tric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/(c^2*f*(1 + m)))/(c^2*(4 + m) )))/(15 + 23*m + 9*m^2 + m^3)
3.2.78.3.1 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[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[c^p*(f*x)^(m + 4*p - 1)*((d + e*x^2)^ (q + 1)/(e*f^(4*p - 1)*(m + 4*p + 2*q + 1))), x] + Simp[1/(e*(m + 4*p + 2*q + 1)) Int[(f*x)^m*(d + e*x^2)^q*ExpandToSum[e*(m + 4*p + 2*q + 1)*((a + b*x^2 + c*x^4)^p - c^p*x^(4*p)) - d*c^p*(m + 4*p - 1)*x^(4*p - 2), x], x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && !IntegerQ[q] && NeQ[m + 4*p + 2*q + 1, 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]))
\[\int \left (f x \right )^{m} \left (e \,x^{2}+d \right )^{2} \left (a +b \,\operatorname {arcsech}\left (c x \right )\right )d x\]
\[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=\int { {\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arsech}\left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]
integral((a*e^2*x^4 + 2*a*d*e*x^2 + a*d^2 + (b*e^2*x^4 + 2*b*d*e*x^2 + b*d ^2)*arcsech(c*x))*(f*x)^m, x)
\[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=\int \left (f x\right )^{m} \left (a + b \operatorname {asech}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{2}\, dx \]
\[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=\int { {\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arsech}\left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]
a*e^2*f^m*x^5*x^m/(m + 5) + 2*a*d*e*f^m*x^3*x^m/(m + 3) + (f*x)^(m + 1)*a* d^2/(f*(m + 1)) + (((m^2 + 4*m + 3)*b*e^2*f^m*x^5*x^m + 2*(m^2 + 6*m + 5)* b*d*e*f^m*x^3*x^m + (m^2 + 8*m + 15)*b*d^2*f^m*x*x^m)*log(sqrt(c*x + 1)*sq rt(-c*x + 1) + 1) - ((m^2 + 4*m + 3)*b*e^2*f^m*x^5*x^m + 2*(m^2 + 6*m + 5) *b*d*e*f^m*x^3*x^m + (m^2 + 8*m + 15)*b*d^2*f^m*x*x^m)*log(x))/(m^3 + 9*m^ 2 + 23*m + 15) - integrate((b*c^2*e^2*f^m*(m + 5)*x^2*log(c) - (e^2*f^m*(m + 5)*log(c) - e^2*f^m)*b)*x^4*x^m/(c^2*(m + 5)*x^2 - m - 5), x) - integra te(2*(b*c^2*d*e*f^m*(m + 3)*x^2*log(c) - (d*e*f^m*(m + 3)*log(c) - d*e*f^m )*b)*x^2*x^m/(c^2*(m + 3)*x^2 - m - 3), x) - integrate((b*c^2*d^2*f^m*(m + 1)*x^2*log(c) - (d^2*f^m*(m + 1)*log(c) - d^2*f^m)*b)*x^m/(c^2*(m + 1)*x^ 2 - m - 1), x) + integrate(((m^2 + 4*m + 3)*b*c^2*e^2*f^m*x^6*x^m + 2*(m^2 + 6*m + 5)*b*c^2*d*e*f^m*x^4*x^m + (m^2 + 8*m + 15)*b*c^2*d^2*f^m*x^2*x^m )/((m^3 + 9*m^2 + 23*m + 15)*c^2*x^2 - m^3 + ((m^3 + 9*m^2 + 23*m + 15)*c^ 2*x^2 - m^3 - 9*m^2 - 23*m - 15)*sqrt(c*x + 1)*sqrt(-c*x + 1) - 9*m^2 - 23 *m - 15), x)
\[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=\int { {\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arsech}\left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]
Timed out. \[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=\int {\left (f\,x\right )}^m\,{\left (e\,x^2+d\right )}^2\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right ) \,d x \]