Integrand size = 21, antiderivative size = 189 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right )}{x^3} \, dx=\frac {b c d^2 \sqrt {1-\frac {1}{c^2 x^2}}}{4 x}-\frac {b e^2 \sqrt {1-\frac {1}{c^2 x^2}} x}{2 c}-\frac {1}{4} b c^2 d^2 \csc ^{-1}(c x)-i b d e \csc ^{-1}(c x)^2-\frac {d^2 \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}+\frac {1}{2} e^2 x^2 \left (a+b \sec ^{-1}(c x)\right )+2 b d e \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )-2 b d e \csc ^{-1}(c x) \log \left (\frac {1}{x}\right )-2 d e \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )-i b d e \operatorname {PolyLog}\left (2,e^{2 i \csc ^{-1}(c x)}\right ) \]
-1/4*b*c^2*d^2*arccsc(c*x)-I*b*d*e*arccsc(c*x)^2-1/2*d^2*(a+b*arcsec(c*x)) /x^2+1/2*e^2*x^2*(a+b*arcsec(c*x))+2*b*d*e*arccsc(c*x)*ln(1-(I/c/x+(1-1/c^ 2/x^2)^(1/2))^2)-2*b*d*e*arccsc(c*x)*ln(1/x)-2*d*e*(a+b*arcsec(c*x))*ln(1/ x)-I*b*d*e*polylog(2,(I/c/x+(1-1/c^2/x^2)^(1/2))^2)+1/4*b*c*d^2*(1-1/c^2/x ^2)^(1/2)/x-1/2*b*e^2*x*(1-1/c^2/x^2)^(1/2)/c
Time = 0.65 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.03 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right )}{x^3} \, dx=\frac {1}{4} \left (-\frac {2 a d^2}{x^2}+2 a e^2 x^2-\frac {2 b d^2 \sec ^{-1}(c x)}{x^2}+\frac {2 b e^2 x \left (-\sqrt {1-\frac {1}{c^2 x^2}}+c x \sec ^{-1}(c x)\right )}{c}+\frac {b d^2 \left (-1+c^2 x^2+c^2 x^2 \sqrt {-1+c^2 x^2} \arctan \left (\sqrt {-1+c^2 x^2}\right )\right )}{c \sqrt {1-\frac {1}{c^2 x^2}} x^3}+8 a d e \log (x)+4 i b d e \left (\sec ^{-1}(c x) \left (\sec ^{-1}(c x)+2 i \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )\right )+\operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )\right )\right ) \]
((-2*a*d^2)/x^2 + 2*a*e^2*x^2 - (2*b*d^2*ArcSec[c*x])/x^2 + (2*b*e^2*x*(-S qrt[1 - 1/(c^2*x^2)] + c*x*ArcSec[c*x]))/c + (b*d^2*(-1 + c^2*x^2 + c^2*x^ 2*Sqrt[-1 + c^2*x^2]*ArcTan[Sqrt[-1 + c^2*x^2]]))/(c*Sqrt[1 - 1/(c^2*x^2)] *x^3) + 8*a*d*e*Log[x] + (4*I)*b*d*e*(ArcSec[c*x]*(ArcSec[c*x] + (2*I)*Log [1 + E^((2*I)*ArcSec[c*x])]) + PolyLog[2, -E^((2*I)*ArcSec[c*x])]))/4
Time = 0.80 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.21, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {5763, 5231, 27, 7293, 2009}
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.
| \(\displaystyle \int \frac {\left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right )}{x^3} \, dx\) |
| \(\Big \downarrow \) 5763 |
| \(\displaystyle -\int \left (\frac {d}{x^2}+e\right )^2 x^3 \left (a+b \arccos \left (\frac {1}{c x}\right )\right )d\frac {1}{x}\) |
| \(\Big \downarrow \) 5231 |
| \(\displaystyle -\frac {b \int -\frac {-\frac {d^2}{x^2}-4 e \log \left (\frac {1}{x}\right ) d+e^2 x^2}{2 \sqrt {1-\frac {1}{c^2 x^2}}}d\frac {1}{x}}{c}-\frac {d^2 \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{2 x^2}-2 d e \log \left (\frac {1}{x}\right ) \left (a+b \arccos \left (\frac {1}{c x}\right )\right )+\frac {1}{2} e^2 x^2 \left (a+b \arccos \left (\frac {1}{c x}\right )\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b \int \frac {-\frac {d^2}{x^2}-4 e \log \left (\frac {1}{x}\right ) d+e^2 x^2}{\sqrt {1-\frac {1}{c^2 x^2}}}d\frac {1}{x}}{2 c}-\frac {d^2 \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{2 x^2}-2 d e \log \left (\frac {1}{x}\right ) \left (a+b \arccos \left (\frac {1}{c x}\right )\right )+\frac {1}{2} e^2 x^2 \left (a+b \arccos \left (\frac {1}{c x}\right )\right )\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {b \int \left (-\frac {d^2}{\sqrt {1-\frac {1}{c^2 x^2}} x^2}-\frac {4 e \log \left (\frac {1}{x}\right ) d}{\sqrt {1-\frac {1}{c^2 x^2}}}+\frac {e^2 x^2}{\sqrt {1-\frac {1}{c^2 x^2}}}\right )d\frac {1}{x}}{2 c}-\frac {d^2 \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{2 x^2}-2 d e \log \left (\frac {1}{x}\right ) \left (a+b \arccos \left (\frac {1}{c x}\right )\right )+\frac {1}{2} e^2 x^2 \left (a+b \arccos \left (\frac {1}{c x}\right )\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {d^2 \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{2 x^2}-2 d e \log \left (\frac {1}{x}\right ) \left (a+b \arccos \left (\frac {1}{c x}\right )\right )+\frac {1}{2} e^2 x^2 \left (a+b \arccos \left (\frac {1}{c x}\right )\right )+\frac {b \left (-\frac {1}{2} c^3 d^2 \arcsin \left (\frac {1}{c x}\right )-2 i c d e \operatorname {PolyLog}\left (2,e^{2 i \arcsin \left (\frac {1}{c x}\right )}\right )-2 i c d e \arcsin \left (\frac {1}{c x}\right )^2+4 c d e \arcsin \left (\frac {1}{c x}\right ) \log \left (1-e^{2 i \arcsin \left (\frac {1}{c x}\right )}\right )-4 c d e \log \left (\frac {1}{x}\right ) \arcsin \left (\frac {1}{c x}\right )+\frac {c^2 d^2 \sqrt {1-\frac {1}{c^2 x^2}}}{2 x}-e^2 x \sqrt {1-\frac {1}{c^2 x^2}}\right )}{2 c}\) |
-1/2*(d^2*(a + b*ArcCos[1/(c*x)]))/x^2 + (e^2*x^2*(a + b*ArcCos[1/(c*x)])) /2 - 2*d*e*(a + b*ArcCos[1/(c*x)])*Log[x^(-1)] + (b*((c^2*d^2*Sqrt[1 - 1/( c^2*x^2)])/(2*x) - e^2*Sqrt[1 - 1/(c^2*x^2)]*x - (c^3*d^2*ArcSin[1/(c*x)]) /2 - (2*I)*c*d*e*ArcSin[1/(c*x)]^2 + 4*c*d*e*ArcSin[1/(c*x)]*Log[1 - E^((2 *I)*ArcSin[1/(c*x)])] - 4*c*d*e*ArcSin[1/(c*x)]*Log[x^(-1)] - (2*I)*c*d*e* PolyLog[2, E^((2*I)*ArcSin[1/(c*x)])]))/(2*c)
3.1.90.3.1 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[((a_.) + ArcCos[(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]}, Simp [(a + b*ArcCos[c*x]) u, x] + Simp[b*c Int[SimplifyIntegrand[u/Sqrt[1 - c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[c^2*d + e, 0] && IntegerQ[p] && (GtQ[p, 0] || (IGtQ[(m - 1)/2, 0] && LeQ[m + p, 0]))
Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_.) + (e_.)*(x_) ^2)^(p_.), x_Symbol] :> -Subst[Int[(e + d*x^2)^p*((a + b*ArcCos[x/c])^n/x^( m + 2*(p + 1))), x], x, 1/x] /; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[n, 0] && IntegerQ[m] && IntegerQ[p]
Time = 2.42 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.14
| method | result | size |
| parts | \(a \left (\frac {e^{2} x^{2}}{2}-\frac {d^{2}}{2 x^{2}}+2 d e \ln \left (x \right )\right )+i b d e \operatorname {arcsec}\left (c x \right )^{2}+\frac {b c \,d^{2} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}{4 x}+\frac {b \,c^{2} d^{2} \operatorname {arcsec}\left (c x \right )}{4}-\frac {b \,\operatorname {arcsec}\left (c x \right ) d^{2}}{2 x^{2}}+\frac {b \,e^{2} \operatorname {arcsec}\left (c x \right ) x^{2}}{2}-\frac {b \,e^{2} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x}{2 c}-\frac {i b \,e^{2}}{2 c^{2}}-2 b d e \,\operatorname {arcsec}\left (c x \right ) \ln \left (1+{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )+i b d e \operatorname {polylog}\left (2, -{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )\) | \(216\) |
| derivativedivides | \(c^{2} \left (\frac {a \,x^{2} e^{2}}{2 c^{2}}+\frac {2 a d e \ln \left (c x \right )}{c^{2}}-\frac {a \,d^{2}}{2 c^{2} x^{2}}+\frac {i b d e \operatorname {arcsec}\left (c x \right )^{2}}{c^{2}}+\frac {b \,d^{2} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}{4 c x}+\frac {b \,\operatorname {arcsec}\left (c x \right ) d^{2}}{4}-\frac {b \,\operatorname {arcsec}\left (c x \right ) d^{2}}{2 c^{2} x^{2}}+\frac {b \,\operatorname {arcsec}\left (c x \right ) x^{2} e^{2}}{2 c^{2}}-\frac {b \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, e^{2} x}{2 c^{3}}-\frac {i b \,e^{2}}{2 c^{4}}-\frac {2 b \ln \left (1+{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right ) d e \,\operatorname {arcsec}\left (c x \right )}{c^{2}}+\frac {i b \operatorname {polylog}\left (2, -{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right ) d e}{c^{2}}\right )\) | \(245\) |
| default | \(c^{2} \left (\frac {a \,x^{2} e^{2}}{2 c^{2}}+\frac {2 a d e \ln \left (c x \right )}{c^{2}}-\frac {a \,d^{2}}{2 c^{2} x^{2}}+\frac {i b d e \operatorname {arcsec}\left (c x \right )^{2}}{c^{2}}+\frac {b \,d^{2} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}{4 c x}+\frac {b \,\operatorname {arcsec}\left (c x \right ) d^{2}}{4}-\frac {b \,\operatorname {arcsec}\left (c x \right ) d^{2}}{2 c^{2} x^{2}}+\frac {b \,\operatorname {arcsec}\left (c x \right ) x^{2} e^{2}}{2 c^{2}}-\frac {b \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, e^{2} x}{2 c^{3}}-\frac {i b \,e^{2}}{2 c^{4}}-\frac {2 b \ln \left (1+{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right ) d e \,\operatorname {arcsec}\left (c x \right )}{c^{2}}+\frac {i b \operatorname {polylog}\left (2, -{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right ) d e}{c^{2}}\right )\) | \(245\) |
a*(1/2*e^2*x^2-1/2*d^2/x^2+2*d*e*ln(x))+I*b*d*e*arcsec(c*x)^2+1/4*b*c*d^2/ x*((c^2*x^2-1)/c^2/x^2)^(1/2)+1/4*b*c^2*d^2*arcsec(c*x)-1/2*b*arcsec(c*x)* d^2/x^2+1/2*b*e^2*arcsec(c*x)*x^2-1/2*b/c*e^2*((c^2*x^2-1)/c^2/x^2)^(1/2)* x-1/2*I*b/c^2*e^2-2*b*d*e*arcsec(c*x)*ln(1+(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2 )+I*b*d*e*polylog(2,-(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2)
\[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right )}{x^3} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )}}{x^{3}} \,d x } \]
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))/x^3, x)
\[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right )}{x^3} \, dx=\int \frac {\left (a + b \operatorname {asec}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{2}}{x^{3}}\, dx \]
\[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right )}{x^3} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )}}{x^{3}} \,d x } \]
1/2*a*e^2*x^2 - 1/4*b*d^2*((c^4*x*sqrt(-1/(c^2*x^2) + 1)/(c^2*x^2*(1/(c^2* x^2) - 1) - 1) - c^3*arctan(c*x*sqrt(-1/(c^2*x^2) + 1)))/c + 2*arcsec(c*x) /x^2) + 2*a*d*e*log(x) - 1/2*a*d^2/x^2 - 1/4*(-2*I*b*c^2*e^2*x^2*log(c) - 4*I*b*c^2*d*e*log(-c*x + 1)*log(x) - 4*I*b*c^2*d*e*log(x)^2 - 4*I*b*c^2*d* e*dilog(c*x) - 4*I*b*c^2*d*e*dilog(-c*x) - I*b*e^2*log(c*x - 1) + I*(b*e^2 *(log(c*x + 1)/c^2 + log(c*x - 1)/c^2) + 16*b*d*e*integrate(1/2*log(x)/(c^ 2*x^3 - x), x))*c^2 + 4*c^2*integrate(1/2*(b*e^2*x^2 + 4*b*d*e*log(x))*sqr t(c*x + 1)*sqrt(c*x - 1)/(c^2*x^3 - x), x) - 2*(b*c^2*e^2*x^2 + 4*b*c^2*d* e*log(x))*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)) + (I*b*c^2*e^2*x^2 + 4*I*b*c ^2*d*e*log(x))*log(c^2*x^2) + (-4*I*b*c^2*d*e*log(x) - I*b*e^2)*log(c*x + 1) - 2*(I*b*c^2*e^2*x^2 + 4*I*b*c^2*d*e*log(c))*log(x))/c^2
\[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right )}{x^3} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )}}{x^{3}} \,d x } \]
Timed out. \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right )}{x^3} \, dx=\int \frac {{\left (e\,x^2+d\right )}^2\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}{x^3} \,d x \]