Integrand size = 23, antiderivative size = 374 \[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=-\frac {b e \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right ) x (f x)^{1+m} \sqrt {-1+c^2 x^2}}{c^3 f (2+m) (3+m) (4+m) (5+m) \sqrt {c^2 x^2}}-\frac {b e^2 x (f x)^{3+m} \sqrt {-1+c^2 x^2}}{c f^3 (4+m) (5+m) \sqrt {c^2 x^2}}+\frac {d^2 (f x)^{1+m} \left (a+b \sec ^{-1}(c x)\right )}{f (1+m)}+\frac {2 d e (f x)^{3+m} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (3+m)}+\frac {e^2 (f x)^{5+m} \left (a+b \sec ^{-1}(c x)\right )}{f^5 (5+m)}-\frac {b \left (c^4 d^2 (2+m) (3+m) (4+m) (5+m)+e (1+m)^2 \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right )\right ) x (f x)^{1+m} \sqrt {1-c^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},c^2 x^2\right )}{c^3 f (1+m)^2 (2+m) (3+m) (4+m) (5+m) \sqrt {c^2 x^2} \sqrt {-1+c^2 x^2}} \]
d^2*(f*x)^(1+m)*(a+b*arcsec(c*x))/f/(1+m)+2*d*e*(f*x)^(3+m)*(a+b*arcsec(c* x))/f^3/(3+m)+e^2*(f*x)^(5+m)*(a+b*arcsec(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)))*x*(f*x)^(1+ m)*hypergeom([1/2, 1/2+1/2*m],[3/2+1/2*m],c^2*x^2)*(-c^2*x^2+1)^(1/2)/c^3/ f/(1+m)^2/(2+m)/(3+m)/(4+m)/(5+m)/(c^2*x^2)^(1/2)/(c^2*x^2-1)^(1/2)-b*e*(e *(3+m)^2+2*c^2*d*(m^2+9*m+20))*x*(f*x)^(1+m)*(c^2*x^2-1)^(1/2)/c^3/f/(4+m) /(5+m)/(m^2+5*m+6)/(c^2*x^2)^(1/2)-b*e^2*x*(f*x)^(3+m)*(c^2*x^2-1)^(1/2)/c /f^3/(4+m)/(5+m)/(c^2*x^2)^(1/2)
Time = 0.47 (sec) , antiderivative size = 291, normalized size of antiderivative = 0.78 \[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=x (f x)^m \left (\frac {a d^2}{1+m}+\frac {2 a d e x^2}{3+m}+\frac {a e^2 x^4}{5+m}+\frac {b d^2 \sec ^{-1}(c x)}{1+m}+\frac {2 b d e x^2 \sec ^{-1}(c x)}{3+m}+\frac {b e^2 x^4 \sec ^{-1}(c x)}{5+m}+\frac {b c d^2 \sqrt {1-\frac {1}{c^2 x^2}} x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},c^2 x^2\right )}{(1+m)^2 \sqrt {1-c^2 x^2}}+\frac {2 b c d e \sqrt {1-\frac {1}{c^2 x^2}} x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},c^2 x^2\right )}{(3+m)^2 \sqrt {1-c^2 x^2}}+\frac {b c e^2 \sqrt {1-\frac {1}{c^2 x^2}} x^5 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5+m}{2},\frac {7+m}{2},c^2 x^2\right )}{(5+m)^2 \sqrt {1-c^2 x^2}}\right ) \]
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*ArcSec[c*x])/(1 + m) + (2*b*d*e*x^2*ArcSec[c*x])/(3 + m) + (b*e^2* x^4*ArcSec[c*x])/(5 + m) + (b*c*d^2*Sqrt[1 - 1/(c^2*x^2)]*x*Hypergeometric 2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/((1 + m)^2*Sqrt[1 - c^2*x^2]) + ( 2*b*c*d*e*Sqrt[1 - 1/(c^2*x^2)]*x^3*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/2, c^2*x^2])/((3 + m)^2*Sqrt[1 - c^2*x^2]) + (b*c*e^2*Sqrt[1 - 1/(c^2* x^2)]*x^5*Hypergeometric2F1[1/2, (5 + m)/2, (7 + m)/2, c^2*x^2])/((5 + m)^ 2*Sqrt[1 - c^2*x^2]))
Time = 0.66 (sec) , antiderivative size = 345, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {5761, 27, 1590, 363, 279, 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.
\(\displaystyle \int \left (d+e x^2\right )^2 (f x)^m \left (a+b \sec ^{-1}(c x)\right ) \, dx\) |
\(\Big \downarrow \) 5761 |
\(\displaystyle -\frac {b c x \int \frac {(f x)^m \left (e^2 (m+1) (m+3) x^4+2 d e (m+1) (m+5) x^2+d^2 (m+3) (m+5)\right )}{\left (m^3+9 m^2+23 m+15\right ) \sqrt {c^2 x^2-1}}dx}{\sqrt {c^2 x^2}}+\frac {d^2 (f x)^{m+1} \left (a+b \sec ^{-1}(c x)\right )}{f (m+1)}+\frac {2 d e (f x)^{m+3} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (m+3)}+\frac {e^2 (f x)^{m+5} \left (a+b \sec ^{-1}(c x)\right )}{f^5 (m+5)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {b c x \int \frac {(f x)^m \left (e^2 (m+1) (m+3) x^4+2 d e (m+1) (m+5) x^2+d^2 (m+3) (m+5)\right )}{\sqrt {c^2 x^2-1}}dx}{\left (m^3+9 m^2+23 m+15\right ) \sqrt {c^2 x^2}}+\frac {d^2 (f x)^{m+1} \left (a+b \sec ^{-1}(c x)\right )}{f (m+1)}+\frac {2 d e (f x)^{m+3} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (m+3)}+\frac {e^2 (f x)^{m+5} \left (a+b \sec ^{-1}(c x)\right )}{f^5 (m+5)}\) |
\(\Big \downarrow \) 1590 |
\(\displaystyle -\frac {b c x \left (\frac {\int \frac {(f x)^m \left (c^2 (m+3) (m+4) (m+5) d^2+e (m+1) \left (2 d \left (m^2+9 m+20\right ) c^2+e (m+3)^2\right ) x^2\right )}{\sqrt {c^2 x^2-1}}dx}{c^2 (m+4)}+\frac {e^2 (m+1) (m+3) \sqrt {c^2 x^2-1} (f x)^{m+3}}{c^2 f^3 (m+4)}\right )}{\left (m^3+9 m^2+23 m+15\right ) \sqrt {c^2 x^2}}+\frac {d^2 (f x)^{m+1} \left (a+b \sec ^{-1}(c x)\right )}{f (m+1)}+\frac {2 d e (f x)^{m+3} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (m+3)}+\frac {e^2 (f x)^{m+5} \left (a+b \sec ^{-1}(c x)\right )}{f^5 (m+5)}\) |
\(\Big \downarrow \) 363 |
\(\displaystyle -\frac {b c x \left (\frac {\frac {\left (c^4 d^2 (m+3) (m+4) (m+5)+\frac {e (m+1)^2 \left (2 c^2 d \left (m^2+9 m+20\right )+e (m+3)^2\right )}{m+2}\right ) \int \frac {(f x)^m}{\sqrt {c^2 x^2-1}}dx}{c^2}+\frac {e (m+1) \sqrt {c^2 x^2-1} (f x)^{m+1} \left (2 c^2 d \left (m^2+9 m+20\right )+e (m+3)^2\right )}{c^2 f (m+2)}}{c^2 (m+4)}+\frac {e^2 (m+1) (m+3) \sqrt {c^2 x^2-1} (f x)^{m+3}}{c^2 f^3 (m+4)}\right )}{\left (m^3+9 m^2+23 m+15\right ) \sqrt {c^2 x^2}}+\frac {d^2 (f x)^{m+1} \left (a+b \sec ^{-1}(c x)\right )}{f (m+1)}+\frac {2 d e (f x)^{m+3} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (m+3)}+\frac {e^2 (f x)^{m+5} \left (a+b \sec ^{-1}(c x)\right )}{f^5 (m+5)}\) |
\(\Big \downarrow \) 279 |
\(\displaystyle -\frac {b c x \left (\frac {\frac {\sqrt {1-c^2 x^2} \left (c^4 d^2 (m+3) (m+4) (m+5)+\frac {e (m+1)^2 \left (2 c^2 d \left (m^2+9 m+20\right )+e (m+3)^2\right )}{m+2}\right ) \int \frac {(f x)^m}{\sqrt {1-c^2 x^2}}dx}{c^2 \sqrt {c^2 x^2-1}}+\frac {e (m+1) \sqrt {c^2 x^2-1} (f x)^{m+1} \left (2 c^2 d \left (m^2+9 m+20\right )+e (m+3)^2\right )}{c^2 f (m+2)}}{c^2 (m+4)}+\frac {e^2 (m+1) (m+3) \sqrt {c^2 x^2-1} (f x)^{m+3}}{c^2 f^3 (m+4)}\right )}{\left (m^3+9 m^2+23 m+15\right ) \sqrt {c^2 x^2}}+\frac {d^2 (f x)^{m+1} \left (a+b \sec ^{-1}(c x)\right )}{f (m+1)}+\frac {2 d e (f x)^{m+3} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (m+3)}+\frac {e^2 (f x)^{m+5} \left (a+b \sec ^{-1}(c x)\right )}{f^5 (m+5)}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {d^2 (f x)^{m+1} \left (a+b \sec ^{-1}(c x)\right )}{f (m+1)}+\frac {2 d e (f x)^{m+3} \left (a+b \sec ^{-1}(c x)\right )}{f^3 (m+3)}+\frac {e^2 (f x)^{m+5} \left (a+b \sec ^{-1}(c x)\right )}{f^5 (m+5)}-\frac {b c x \left (\frac {e^2 (m+1) (m+3) \sqrt {c^2 x^2-1} (f x)^{m+3}}{c^2 f^3 (m+4)}+\frac {\frac {e (m+1) \sqrt {c^2 x^2-1} (f x)^{m+1} \left (2 c^2 d \left (m^2+9 m+20\right )+e (m+3)^2\right )}{c^2 f (m+2)}+\frac {\sqrt {1-c^2 x^2} (f x)^{m+1} \left (c^4 d^2 (m+3) (m+4) (m+5)+\frac {e (m+1)^2 \left (2 c^2 d \left (m^2+9 m+20\right )+e (m+3)^2\right )}{m+2}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},c^2 x^2\right )}{c^2 f (m+1) \sqrt {c^2 x^2-1}}}{c^2 (m+4)}\right )}{\left (m^3+9 m^2+23 m+15\right ) \sqrt {c^2 x^2}}\) |
(d^2*(f*x)^(1 + m)*(a + b*ArcSec[c*x]))/(f*(1 + m)) + (2*d*e*(f*x)^(3 + m) *(a + b*ArcSec[c*x]))/(f^3*(3 + m)) + (e^2*(f*x)^(5 + m)*(a + b*ArcSec[c*x ]))/(f^5*(5 + m)) - (b*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)*Sqrt[1 - c^2*x^2]*Hypergeometric2F1[1/2, (1 + m) /2, (3 + m)/2, c^2*x^2])/(c^2*f*(1 + m)*Sqrt[-1 + c^2*x^2]))/(c^2*(4 + m)) ))/((15 + 23*m + 9*m^2 + m^3)*Sqrt[c^2*x^2])
3.2.62.3.1 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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(c*x)^m* (1 + b*(x^2/a))^p, x], 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_.) + ArcSec[(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]}, Sim p[(a + b*ArcSec[c*x]) u, x] - Simp[b*c*(x/Sqrt[c^2*x^2]) Int[SimplifyIn tegrand[u/(x*Sqrt[c^2*x^2 - 1]), x], x], x]] /; FreeQ[{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]))
\[\int \left (f x \right )^{m} \left (e \,x^{2}+d \right )^{2} \left (a +b \,\operatorname {arcsec}\left (c x \right )\right )d x\]
\[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int { {\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]
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)*arcsec(c*x))*(f*x)^m, x)
\[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int \left (f x\right )^{m} \left (a + b \operatorname {asec}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{2}\, dx \]
\[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int { {\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]
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)) + (((b*e^2*f^m*m^2 + 4*b*e^2*f^m*m + 3*b*e^2*f^m)*x^5 + 2* (b*d*e*f^m*m^2 + 6*b*d*e*f^m*m + 5*b*d*e*f^m)*x^3 + (b*d^2*f^m*m^2 + 8*b*d ^2*f^m*m + 15*b*d^2*f^m)*x)*x^m*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)) - (m^3 + 9*m^2 + 23*m + 15)*integrate(-(b*d^2*f^m*m^2 + 8*b*d^2*f^m*m + (b*e^2*f ^m*m^2 + 4*b*e^2*f^m*m + 3*b*e^2*f^m)*x^4 + 15*b*d^2*f^m + 2*(b*d*e*f^m*m^ 2 + 6*b*d*e*f^m*m + 5*b*d*e*f^m)*x^2)*sqrt(c*x + 1)*sqrt(c*x - 1)*x^m/(m^3 - (c^2*m^3 + 9*c^2*m^2 + 23*c^2*m + 15*c^2)*x^2 + 9*m^2 + 23*m + 15), x)) /(m^3 + 9*m^2 + 23*m + 15)
\[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int { {\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]
Timed out. \[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int {\left (f\,x\right )}^m\,{\left (e\,x^2+d\right )}^2\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right ) \,d x \]