Integrand size = 21, antiderivative size = 546 \[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^2} \, dx=-\frac {e \left (a+b \sec ^{-1}(c x)\right )}{2 d^2 \left (e+\frac {d}{x^2}\right )}+\frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b d^2}-\frac {b \sqrt {e} \arctan \left (\frac {\sqrt {c^2 d+e}}{c \sqrt {e} \sqrt {1-\frac {1}{c^2 x^2}} x}\right )}{2 d^2 \sqrt {c^2 d+e}}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d^2}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d^2}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 d^2}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 d^2}+\frac {i b \operatorname {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d^2}+\frac {i b \operatorname {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d^2}+\frac {i b \operatorname {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 d^2}+\frac {i b \operatorname {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 d^2} \] Output:
-1/2*e*(a+b*arcsec(c*x))/d^2/(e+d/x^2)+1/2*I*(a+b*arcsec(c*x))^2/b/d^2-1/2 *b*e^(1/2)*arctan((c^2*d+e)^(1/2)/c/e^(1/2)/(1-1/c^2/x^2)^(1/2)/x)/d^2/(c^ 2*d+e)^(1/2)-1/2*(a+b*arcsec(c*x))*ln(1-c*(-d)^(1/2)*(1/c/x+I*(1-1/c^2/x^2 )^(1/2))/(e^(1/2)-(c^2*d+e)^(1/2)))/d^2-1/2*(a+b*arcsec(c*x))*ln(1+c*(-d)^ (1/2)*(1/c/x+I*(1-1/c^2/x^2)^(1/2))/(e^(1/2)-(c^2*d+e)^(1/2)))/d^2-1/2*(a+ b*arcsec(c*x))*ln(1-c*(-d)^(1/2)*(1/c/x+I*(1-1/c^2/x^2)^(1/2))/(e^(1/2)+(c ^2*d+e)^(1/2)))/d^2-1/2*(a+b*arcsec(c*x))*ln(1+c*(-d)^(1/2)*(1/c/x+I*(1-1/ c^2/x^2)^(1/2))/(e^(1/2)+(c^2*d+e)^(1/2)))/d^2+1/2*I*b*polylog(2,-c*(-d)^( 1/2)*(1/c/x+I*(1-1/c^2/x^2)^(1/2))/(e^(1/2)-(c^2*d+e)^(1/2)))/d^2+1/2*I*b* polylog(2,c*(-d)^(1/2)*(1/c/x+I*(1-1/c^2/x^2)^(1/2))/(e^(1/2)-(c^2*d+e)^(1 /2)))/d^2+1/2*I*b*polylog(2,-c*(-d)^(1/2)*(1/c/x+I*(1-1/c^2/x^2)^(1/2))/(e ^(1/2)+(c^2*d+e)^(1/2)))/d^2+1/2*I*b*polylog(2,c*(-d)^(1/2)*(1/c/x+I*(1-1/ c^2/x^2)^(1/2))/(e^(1/2)+(c^2*d+e)^(1/2)))/d^2
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1190\) vs. \(2(546)=1092\).
Time = 0.73 (sec) , antiderivative size = 1190, normalized size of antiderivative = 2.18 \[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^2} \, dx =\text {Too large to display} \] Input:
Integrate[(a + b*ArcSec[c*x])/(x*(d + e*x^2)^2),x]
Output:
((2*a*d)/(d + e*x^2) + (b*Sqrt[d]*ArcSec[c*x])/(Sqrt[d] - I*Sqrt[e]*x) + ( b*Sqrt[d]*ArcSec[c*x])/(Sqrt[d] + I*Sqrt[e]*x) + (2*I)*b*ArcSec[c*x]^2 + 2 *b*ArcSin[1/(c*x)] - (8*I)*b*ArcSin[Sqrt[1 - (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt [2]]*ArcTan[(((-I)*c*Sqrt[d] + Sqrt[e])*Tan[ArcSec[c*x]/2])/Sqrt[c^2*d + e ]] - (8*I)*b*ArcSin[Sqrt[1 + (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*ArcTan[((I* c*Sqrt[d] + Sqrt[e])*Tan[ArcSec[c*x]/2])/Sqrt[c^2*d + e]] - 2*b*ArcSec[c*x ]*Log[1 + (I*(Sqrt[e] - Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] - 4*b*ArcSin[Sqrt[1 + (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*Log[1 + (I*(Sqrt[e] - Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] - 2*b*ArcSec[c*x]*Log[ 1 + (I*(-Sqrt[e] + Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] - 4*b* ArcSin[Sqrt[1 - (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*Log[1 + (I*(-Sqrt[e] + S qrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] - 2*b*ArcSec[c*x]*Log[1 - (I*(Sqrt[e] + Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] + 4*b*ArcSi n[Sqrt[1 - (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*Log[1 - (I*(Sqrt[e] + Sqrt[c^ 2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] - 2*b*ArcSec[c*x]*Log[1 + (I*(Sq rt[e] + Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] + 4*b*ArcSin[Sqrt [1 + (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*Log[1 + (I*(Sqrt[e] + Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] + 4*a*Log[x] - (b*Sqrt[e]*Log[(2*Sqrt[ d]*Sqrt[e]*(Sqrt[e] + c*(I*c*Sqrt[d] - Sqrt[-(c^2*d) - e]*Sqrt[1 - 1/(c^2* x^2)])*x))/(Sqrt[-(c^2*d) - e]*(Sqrt[d] - I*Sqrt[e]*x))])/Sqrt[-(c^2*d)...
Time = 1.58 (sec) , antiderivative size = 602, normalized size of antiderivative = 1.10, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5763, 5233, 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.
\(\displaystyle \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 5763 |
\(\displaystyle -\int \frac {a+b \arccos \left (\frac {1}{c x}\right )}{\left (\frac {d}{x^2}+e\right )^2 x^3}d\frac {1}{x}\) |
\(\Big \downarrow \) 5233 |
\(\displaystyle -\int \left (\frac {a+b \arccos \left (\frac {1}{c x}\right )}{d \left (\frac {d}{x^2}+e\right ) x}-\frac {e \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{d \left (\frac {d}{x^2}+e\right )^2 x}\right )d\frac {1}{x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\left (a+b \arccos \left (\frac {1}{c x}\right )\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \arccos \left (\frac {1}{c x}\right )}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d^2}-\frac {\left (a+b \arccos \left (\frac {1}{c x}\right )\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \arccos \left (\frac {1}{c x}\right )}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d^2}-\frac {\left (a+b \arccos \left (\frac {1}{c x}\right )\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \arccos \left (\frac {1}{c x}\right )}}{\sqrt {c^2 d+e}+\sqrt {e}}\right )}{2 d^2}-\frac {\left (a+b \arccos \left (\frac {1}{c x}\right )\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \arccos \left (\frac {1}{c x}\right )}}{\sqrt {c^2 d+e}+\sqrt {e}}\right )}{2 d^2}-\frac {e \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{2 d^2 \left (\frac {d}{x^2}+e\right )}+\frac {i \left (a+b \arccos \left (\frac {1}{c x}\right )\right )^2}{2 b d^2}+\frac {i b \operatorname {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \arccos \left (\frac {1}{c x}\right )}}{\sqrt {e}-\sqrt {d c^2+e}}\right )}{2 d^2}+\frac {i b \operatorname {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \arccos \left (\frac {1}{c x}\right )}}{\sqrt {e}-\sqrt {d c^2+e}}\right )}{2 d^2}+\frac {i b \operatorname {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \arccos \left (\frac {1}{c x}\right )}}{\sqrt {e}+\sqrt {d c^2+e}}\right )}{2 d^2}+\frac {i b \operatorname {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \arccos \left (\frac {1}{c x}\right )}}{\sqrt {e}+\sqrt {d c^2+e}}\right )}{2 d^2}-\frac {b \sqrt {e} \arctan \left (\frac {\sqrt {c^2 d+e}}{c \sqrt {e} x \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{2 d^2 \sqrt {c^2 d+e}}\) |
Input:
Int[(a + b*ArcSec[c*x])/(x*(d + e*x^2)^2),x]
Output:
-1/2*(e*(a + b*ArcCos[1/(c*x)]))/(d^2*(e + d/x^2)) + ((I/2)*(a + b*ArcCos[ 1/(c*x)])^2)/(b*d^2) - (b*Sqrt[e]*ArcTan[Sqrt[c^2*d + e]/(c*Sqrt[e]*Sqrt[1 - 1/(c^2*x^2)]*x)])/(2*d^2*Sqrt[c^2*d + e]) - ((a + b*ArcCos[1/(c*x)])*Lo g[1 - (c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] - Sqrt[c^2*d + e])])/(2* d^2) - ((a + b*ArcCos[1/(c*x)])*Log[1 + (c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)])) /(Sqrt[e] - Sqrt[c^2*d + e])])/(2*d^2) - ((a + b*ArcCos[1/(c*x)])*Log[1 - (c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] + Sqrt[c^2*d + e])])/(2*d^2) - ((a + b*ArcCos[1/(c*x)])*Log[1 + (c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt [e] + Sqrt[c^2*d + e])])/(2*d^2) + ((I/2)*b*PolyLog[2, -((c*Sqrt[-d]*E^(I* ArcCos[1/(c*x)]))/(Sqrt[e] - Sqrt[c^2*d + e]))])/d^2 + ((I/2)*b*PolyLog[2, (c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] - Sqrt[c^2*d + e])])/d^2 + (( I/2)*b*PolyLog[2, -((c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] + Sqrt[c^2 *d + e]))])/d^2 + ((I/2)*b*PolyLog[2, (c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/( Sqrt[e] + Sqrt[c^2*d + e])])/d^2
Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_.)*((d_) + (e_ .)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcCos[c*x])^n, ( f*x)^m*(d + e*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[c^2*d + e, 0] && IGtQ[n, 0] && IntegerQ[p] && IntegerQ[m]
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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 6.22 (sec) , antiderivative size = 2108, normalized size of antiderivative = 3.86
method | result | size |
parts | \(\text {Expression too large to display}\) | \(2108\) |
derivativedivides | \(\text {Expression too large to display}\) | \(2157\) |
default | \(\text {Expression too large to display}\) | \(2157\) |
Input:
int((a+b*arcsec(c*x))/x/(e*x^2+d)^2,x,method=_RETURNVERBOSE)
Output:
a/d^2*ln(x)-1/2*a/d^2*ln(e*x^2+d)+1/2*a/d/(e*x^2+d)+b*(-1/8*I*(-(e*(c^2*d+ e))^(1/2)*c^2*d+2*c^2*d*e-2*(e*(c^2*d+e))^(1/2)*e+2*e^2)*polylog(2,d*c^2*( 1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+e))^(1/2)-2*e))/d^2/e/( c^2*d+e)+1/4*(-(e*(c^2*d+e))^(1/2)*c^2*d+2*c^2*d*e-2*(e*(c^2*d+e))^(1/2)*e +2*e^2)/d^2/e/(c^2*d+e)*ln(1-d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d -2*(e*(c^2*d+e))^(1/2)-2*e))*arcsec(c*x)+1/4*(e*(c^2*d+e))^(1/2)/d/e/(c^2* d+e)*c^2*arcsec(c*x)*ln(1-d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d+2* (e*(c^2*d+e))^(1/2)-2*e))-1/4*I*(e*(c^2*d+e))^(1/2)/d/e/(c^2*d+e)*arcsec(c *x)^2*c^2-1/8*I*(e*(c^2*d+e))^(1/2)/d/e/(c^2*d+e)*polylog(2,d*c^2*(1/c/x+I *(1-1/c^2/x^2)^(1/2))^2/(-c^2*d+2*(e*(c^2*d+e))^(1/2)-2*e))*c^2-I*(-(e*(c^ 2*d+e))^(1/2)*c^2*d+2*c^2*d*e-2*(e*(c^2*d+e))^(1/2)*e+2*e^2)*e*arcsec(c*x) ^2/d^4/(c^2*d+e)/c^4+(-(e*(c^2*d+e))^(1/2)*c^2*d+2*c^2*d*e-2*(e*(c^2*d+e)) ^(1/2)*e+2*e^2)/d^4*e/(c^2*d+e)/c^4*ln(1-d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2 ))^2/(-c^2*d-2*(e*(c^2*d+e))^(1/2)-2*e))*arcsec(c*x)-1/2*I*(-(e*(c^2*d+e)) ^(1/2)*c^2*d+2*c^2*d*e-2*(e*(c^2*d+e))^(1/2)*e+2*e^2)*e*polylog(2,d*c^2*(1 /c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+e))^(1/2)-2*e))/d^4/(c^2 *d+e)/c^4+I*(c^2*d-2*(e*(c^2*d+e))^(1/2)+2*e)*arcsec(c*x)^2*e/d^4/c^4+(-(e *(c^2*d+e))^(1/2)*c^2*d+2*c^2*d*e-2*(e*(c^2*d+e))^(1/2)*e+2*e^2)/d^3/(c^2* d+e)/c^2*ln(1-d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+e) )^(1/2)-2*e))*arcsec(c*x)-I*(-(e*(c^2*d+e))^(1/2)*c^2*d+2*c^2*d*e-2*(e*...
\[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^2} \, dx=\int { \frac {b \operatorname {arcsec}\left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{2} x} \,d x } \] Input:
integrate((a+b*arcsec(c*x))/x/(e*x^2+d)^2,x, algorithm="fricas")
Output:
integral((b*arcsec(c*x) + a)/(e^2*x^5 + 2*d*e*x^3 + d^2*x), x)
Timed out. \[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^2} \, dx=\text {Timed out} \] Input:
integrate((a+b*asec(c*x))/x/(e*x**2+d)**2,x)
Output:
Timed out
\[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^2} \, dx=\int { \frac {b \operatorname {arcsec}\left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{2} x} \,d x } \] Input:
integrate((a+b*arcsec(c*x))/x/(e*x^2+d)^2,x, algorithm="maxima")
Output:
1/2*a*(1/(d*e*x^2 + d^2) - log(e*x^2 + d)/d^2 + 2*log(x)/d^2) + b*integrat e(arctan(sqrt(c*x + 1)*sqrt(c*x - 1))/(e^2*x^5 + 2*d*e*x^3 + d^2*x), x)
Exception generated. \[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^2} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate((a+b*arcsec(c*x))/x/(e*x^2+d)^2,x, algorithm="giac")
Output:
Exception raised: RuntimeError >> an error occurred running a Giac command :INPUT:sage2OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const ve cteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^2} \, dx=\int \frac {a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )}{x\,{\left (e\,x^2+d\right )}^2} \,d x \] Input:
int((a + b*acos(1/(c*x)))/(x*(d + e*x^2)^2),x)
Output:
int((a + b*acos(1/(c*x)))/(x*(d + e*x^2)^2), x)
\[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^2} \, dx=\frac {2 \left (\int \frac {\mathit {asec} \left (c x \right )}{e^{2} x^{5}+2 d e \,x^{3}+d^{2} x}d x \right ) b \,d^{3}+2 \left (\int \frac {\mathit {asec} \left (c x \right )}{e^{2} x^{5}+2 d e \,x^{3}+d^{2} x}d x \right ) b \,d^{2} e \,x^{2}-\mathrm {log}\left (e \,x^{2}+d \right ) a d -\mathrm {log}\left (e \,x^{2}+d \right ) a e \,x^{2}+2 \,\mathrm {log}\left (x \right ) a d +2 \,\mathrm {log}\left (x \right ) a e \,x^{2}-a e \,x^{2}}{2 d^{2} \left (e \,x^{2}+d \right )} \] Input:
int((a+b*asec(c*x))/x/(e*x^2+d)^2,x)
Output:
(2*int(asec(c*x)/(d**2*x + 2*d*e*x**3 + e**2*x**5),x)*b*d**3 + 2*int(asec( c*x)/(d**2*x + 2*d*e*x**3 + e**2*x**5),x)*b*d**2*e*x**2 - log(d + e*x**2)* a*d - log(d + e*x**2)*a*e*x**2 + 2*log(x)*a*d + 2*log(x)*a*e*x**2 - a*e*x* *2)/(2*d**2*(d + e*x**2))