Integrand size = 19, antiderivative size = 152 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right )}{x^6} \, dx=\frac {2 b c^3 \left (12 c^2 d+25 e\right ) \sqrt {-1+c^2 x^2}}{225 \sqrt {c^2 x^2}}+\frac {b c d \sqrt {-1+c^2 x^2}}{25 x^4 \sqrt {c^2 x^2}}+\frac {b c \left (12 c^2 d+25 e\right ) \sqrt {-1+c^2 x^2}}{225 x^2 \sqrt {c^2 x^2}}-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac {e \left (a+b \sec ^{-1}(c x)\right )}{3 x^3} \]
-1/5*d*(a+b*arcsec(c*x))/x^5-1/3*e*(a+b*arcsec(c*x))/x^3+2/225*b*c^3*(12*c ^2*d+25*e)*(c^2*x^2-1)^(1/2)/(c^2*x^2)^(1/2)+1/25*b*c*d*(c^2*x^2-1)^(1/2)/ x^4/(c^2*x^2)^(1/2)+1/225*b*c*(12*c^2*d+25*e)*(c^2*x^2-1)^(1/2)/x^2/(c^2*x ^2)^(1/2)
Time = 0.10 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.62 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right )}{x^6} \, dx=\frac {-15 a \left (3 d+5 e x^2\right )+b c \sqrt {1-\frac {1}{c^2 x^2}} x \left (25 e x^2 \left (1+2 c^2 x^2\right )+3 d \left (3+4 c^2 x^2+8 c^4 x^4\right )\right )-15 b \left (3 d+5 e x^2\right ) \sec ^{-1}(c x)}{225 x^5} \]
(-15*a*(3*d + 5*e*x^2) + b*c*Sqrt[1 - 1/(c^2*x^2)]*x*(25*e*x^2*(1 + 2*c^2* x^2) + 3*d*(3 + 4*c^2*x^2 + 8*c^4*x^4)) - 15*b*(3*d + 5*e*x^2)*ArcSec[c*x] )/(225*x^5)
Time = 0.30 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.86, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {5761, 27, 359, 245, 242}
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.
| \(\displaystyle \int \frac {\left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right )}{x^6} \, dx\) |
| \(\Big \downarrow \) 5761 |
| \(\displaystyle -\frac {b c x \int -\frac {5 e x^2+3 d}{15 x^6 \sqrt {c^2 x^2-1}}dx}{\sqrt {c^2 x^2}}-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac {e \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b c x \int \frac {5 e x^2+3 d}{x^6 \sqrt {c^2 x^2-1}}dx}{15 \sqrt {c^2 x^2}}-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac {e \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}\) |
| \(\Big \downarrow \) 359 |
| \(\displaystyle \frac {b c x \left (\frac {1}{5} \left (12 c^2 d+25 e\right ) \int \frac {1}{x^4 \sqrt {c^2 x^2-1}}dx+\frac {3 d \sqrt {c^2 x^2-1}}{5 x^5}\right )}{15 \sqrt {c^2 x^2}}-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac {e \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}\) |
| \(\Big \downarrow \) 245 |
| \(\displaystyle \frac {b c x \left (\frac {1}{5} \left (12 c^2 d+25 e\right ) \left (\frac {2}{3} c^2 \int \frac {1}{x^2 \sqrt {c^2 x^2-1}}dx+\frac {\sqrt {c^2 x^2-1}}{3 x^3}\right )+\frac {3 d \sqrt {c^2 x^2-1}}{5 x^5}\right )}{15 \sqrt {c^2 x^2}}-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac {e \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}\) |
| \(\Big \downarrow \) 242 |
| \(\displaystyle -\frac {d \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac {e \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}+\frac {b c x \left (\frac {1}{5} \left (\frac {2 c^2 \sqrt {c^2 x^2-1}}{3 x}+\frac {\sqrt {c^2 x^2-1}}{3 x^3}\right ) \left (12 c^2 d+25 e\right )+\frac {3 d \sqrt {c^2 x^2-1}}{5 x^5}\right )}{15 \sqrt {c^2 x^2}}\) |
(b*c*x*((3*d*Sqrt[-1 + c^2*x^2])/(5*x^5) + ((12*c^2*d + 25*e)*(Sqrt[-1 + c ^2*x^2]/(3*x^3) + (2*c^2*Sqrt[-1 + c^2*x^2])/(3*x)))/5))/(15*Sqrt[c^2*x^2] ) - (d*(a + b*ArcSec[c*x]))/(5*x^5) - (e*(a + b*ArcSec[c*x]))/(3*x^3)
3.1.74.3.1 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[(c*x)^ (m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, p}, x ] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + 2*(p + 1) + 1)/(a*(m + 1))) Int[x^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, m, p}, x] && ILtQ[Si mplify[(m + 1)/2 + p + 1], 0] && NeQ[m, -1]
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] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
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]))
Time = 0.31 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.84
| method | result | size |
| parts | \(a \left (-\frac {d}{5 x^{5}}-\frac {e}{3 x^{3}}\right )+b \,c^{5} \left (-\frac {\operatorname {arcsec}\left (c x \right ) d}{5 x^{5} c^{5}}-\frac {\operatorname {arcsec}\left (c x \right ) e}{3 c^{5} x^{3}}+\frac {\left (c^{2} x^{2}-1\right ) \left (24 c^{6} d \,x^{4}+50 c^{4} e \,x^{4}+12 c^{4} d \,x^{2}+25 c^{2} e \,x^{2}+9 c^{2} d \right )}{225 c^{8} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x^{6}}\right )\) | \(127\) |
| derivativedivides | \(c^{5} \left (\frac {a \left (-\frac {d}{5 c^{3} x^{5}}-\frac {e}{3 c^{3} x^{3}}\right )}{c^{2}}+\frac {b \left (-\frac {\operatorname {arcsec}\left (c x \right ) d}{5 c^{3} x^{5}}-\frac {\operatorname {arcsec}\left (c x \right ) e}{3 c^{3} x^{3}}+\frac {\left (c^{2} x^{2}-1\right ) \left (24 c^{6} d \,x^{4}+50 c^{4} e \,x^{4}+12 c^{4} d \,x^{2}+25 c^{2} e \,x^{2}+9 c^{2} d \right )}{225 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{6} x^{6}}\right )}{c^{2}}\right )\) | \(140\) |
| default | \(c^{5} \left (\frac {a \left (-\frac {d}{5 c^{3} x^{5}}-\frac {e}{3 c^{3} x^{3}}\right )}{c^{2}}+\frac {b \left (-\frac {\operatorname {arcsec}\left (c x \right ) d}{5 c^{3} x^{5}}-\frac {\operatorname {arcsec}\left (c x \right ) e}{3 c^{3} x^{3}}+\frac {\left (c^{2} x^{2}-1\right ) \left (24 c^{6} d \,x^{4}+50 c^{4} e \,x^{4}+12 c^{4} d \,x^{2}+25 c^{2} e \,x^{2}+9 c^{2} d \right )}{225 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{6} x^{6}}\right )}{c^{2}}\right )\) | \(140\) |
a*(-1/5*d/x^5-1/3*e/x^3)+b*c^5*(-1/5*arcsec(c*x)*d/x^5/c^5-1/3/c^5*arcsec( c*x)*e/x^3+1/225/c^8*(c^2*x^2-1)*(24*c^6*d*x^4+50*c^4*e*x^4+12*c^4*d*x^2+2 5*c^2*e*x^2+9*c^2*d)/((c^2*x^2-1)/c^2/x^2)^(1/2)/x^6)
Time = 0.28 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.59 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right )}{x^6} \, dx=-\frac {75 \, a e x^{2} + 45 \, a d + 15 \, {\left (5 \, b e x^{2} + 3 \, b d\right )} \operatorname {arcsec}\left (c x\right ) - {\left (2 \, {\left (12 \, b c^{4} d + 25 \, b c^{2} e\right )} x^{4} + {\left (12 \, b c^{2} d + 25 \, b e\right )} x^{2} + 9 \, b d\right )} \sqrt {c^{2} x^{2} - 1}}{225 \, x^{5}} \]
-1/225*(75*a*e*x^2 + 45*a*d + 15*(5*b*e*x^2 + 3*b*d)*arcsec(c*x) - (2*(12* b*c^4*d + 25*b*c^2*e)*x^4 + (12*b*c^2*d + 25*b*e)*x^2 + 9*b*d)*sqrt(c^2*x^ 2 - 1))/x^5
Time = 4.70 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.84 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right )}{x^6} \, dx=- \frac {a d}{5 x^{5}} - \frac {a e}{3 x^{3}} - \frac {b d \operatorname {asec}{\left (c x \right )}}{5 x^{5}} - \frac {b e \operatorname {asec}{\left (c x \right )}}{3 x^{3}} + \frac {b d \left (\begin {cases} \frac {8 c^{5} \sqrt {c^{2} x^{2} - 1}}{15 x} + \frac {4 c^{3} \sqrt {c^{2} x^{2} - 1}}{15 x^{3}} + \frac {c \sqrt {c^{2} x^{2} - 1}}{5 x^{5}} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\\frac {8 i c^{5} \sqrt {- c^{2} x^{2} + 1}}{15 x} + \frac {4 i c^{3} \sqrt {- c^{2} x^{2} + 1}}{15 x^{3}} + \frac {i c \sqrt {- c^{2} x^{2} + 1}}{5 x^{5}} & \text {otherwise} \end {cases}\right )}{5 c} + \frac {b e \left (\begin {cases} \frac {2 c^{3} \sqrt {c^{2} x^{2} - 1}}{3 x} + \frac {c \sqrt {c^{2} x^{2} - 1}}{3 x^{3}} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\\frac {2 i c^{3} \sqrt {- c^{2} x^{2} + 1}}{3 x} + \frac {i c \sqrt {- c^{2} x^{2} + 1}}{3 x^{3}} & \text {otherwise} \end {cases}\right )}{3 c} \]
-a*d/(5*x**5) - a*e/(3*x**3) - b*d*asec(c*x)/(5*x**5) - b*e*asec(c*x)/(3*x **3) + b*d*Piecewise((8*c**5*sqrt(c**2*x**2 - 1)/(15*x) + 4*c**3*sqrt(c**2 *x**2 - 1)/(15*x**3) + c*sqrt(c**2*x**2 - 1)/(5*x**5), Abs(c**2*x**2) > 1) , (8*I*c**5*sqrt(-c**2*x**2 + 1)/(15*x) + 4*I*c**3*sqrt(-c**2*x**2 + 1)/(1 5*x**3) + I*c*sqrt(-c**2*x**2 + 1)/(5*x**5), True))/(5*c) + b*e*Piecewise( (2*c**3*sqrt(c**2*x**2 - 1)/(3*x) + c*sqrt(c**2*x**2 - 1)/(3*x**3), Abs(c* *2*x**2) > 1), (2*I*c**3*sqrt(-c**2*x**2 + 1)/(3*x) + I*c*sqrt(-c**2*x**2 + 1)/(3*x**3), True))/(3*c)
Time = 0.20 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.90 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right )}{x^6} \, dx=\frac {1}{75} \, b d {\left (\frac {3 \, c^{6} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {5}{2}} - 10 \, c^{6} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} + 15 \, c^{6} \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{c} - \frac {15 \, \operatorname {arcsec}\left (c x\right )}{x^{5}}\right )} - \frac {1}{9} \, b e {\left (\frac {c^{4} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} - 3 \, c^{4} \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{c} + \frac {3 \, \operatorname {arcsec}\left (c x\right )}{x^{3}}\right )} - \frac {a e}{3 \, x^{3}} - \frac {a d}{5 \, x^{5}} \]
1/75*b*d*((3*c^6*(-1/(c^2*x^2) + 1)^(5/2) - 10*c^6*(-1/(c^2*x^2) + 1)^(3/2 ) + 15*c^6*sqrt(-1/(c^2*x^2) + 1))/c - 15*arcsec(c*x)/x^5) - 1/9*b*e*((c^4 *(-1/(c^2*x^2) + 1)^(3/2) - 3*c^4*sqrt(-1/(c^2*x^2) + 1))/c + 3*arcsec(c*x )/x^3) - 1/3*a*e/x^3 - 1/5*a*d/x^5
Time = 0.29 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.04 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right )}{x^6} \, dx=\frac {1}{225} \, {\left (24 \, b c^{4} d \sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 50 \, b c^{2} e \sqrt {-\frac {1}{c^{2} x^{2}} + 1} + \frac {12 \, b c^{2} d \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{x^{2}} + \frac {25 \, b e \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{x^{2}} - \frac {75 \, b e \arccos \left (\frac {1}{c x}\right )}{c x^{3}} + \frac {9 \, b d \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{x^{4}} - \frac {75 \, a e}{c x^{3}} - \frac {45 \, b d \arccos \left (\frac {1}{c x}\right )}{c x^{5}} - \frac {45 \, a d}{c x^{5}}\right )} c \]
1/225*(24*b*c^4*d*sqrt(-1/(c^2*x^2) + 1) + 50*b*c^2*e*sqrt(-1/(c^2*x^2) + 1) + 12*b*c^2*d*sqrt(-1/(c^2*x^2) + 1)/x^2 + 25*b*e*sqrt(-1/(c^2*x^2) + 1) /x^2 - 75*b*e*arccos(1/(c*x))/(c*x^3) + 9*b*d*sqrt(-1/(c^2*x^2) + 1)/x^4 - 75*a*e/(c*x^3) - 45*b*d*arccos(1/(c*x))/(c*x^5) - 45*a*d/(c*x^5))*c
Timed out. \[ \int \frac {\left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right )}{x^6} \, dx=\int \frac {\left (e\,x^2+d\right )\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}{x^6} \,d x \]