Integrand size = 18, antiderivative size = 1114 \[ \int \frac {a+b \sec ^{-1}(c x)}{\left (d+e x^2\right )^3} \, dx=\frac {b c \sqrt {e} \sqrt {1-\frac {1}{c^2 x^2}}}{16 (-d)^{3/2} \left (c^2 d+e\right ) \left (\sqrt {-d} \sqrt {e}-\frac {d}{x}\right )}+\frac {b c \sqrt {e} \sqrt {1-\frac {1}{c^2 x^2}}}{16 (-d)^{3/2} \left (c^2 d+e\right ) \left (\sqrt {-d} \sqrt {e}+\frac {d}{x}\right )}+\frac {\sqrt {e} \left (a+b \sec ^{-1}(c x)\right )}{16 (-d)^{3/2} \left (\sqrt {-d} \sqrt {e}-\frac {d}{x}\right )^2}-\frac {5 \left (a+b \sec ^{-1}(c x)\right )}{16 d^2 \left (\sqrt {-d} \sqrt {e}-\frac {d}{x}\right )}-\frac {\sqrt {e} \left (a+b \sec ^{-1}(c x)\right )}{16 (-d)^{3/2} \left (\sqrt {-d} \sqrt {e}+\frac {d}{x}\right )^2}+\frac {5 \left (a+b \sec ^{-1}(c x)\right )}{16 d^2 \left (\sqrt {-d} \sqrt {e}+\frac {d}{x}\right )}+\frac {b e \text {arctanh}\left (\frac {c^2 d-\frac {\sqrt {-d} \sqrt {e}}{x}}{c \sqrt {d} \sqrt {c^2 d+e} \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{16 d^{5/2} \left (c^2 d+e\right )^{3/2}}-\frac {5 b \text {arctanh}\left (\frac {c^2 d-\frac {\sqrt {-d} \sqrt {e}}{x}}{c \sqrt {d} \sqrt {c^2 d+e} \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{16 d^{5/2} \sqrt {c^2 d+e}}+\frac {b e \text {arctanh}\left (\frac {c^2 d+\frac {\sqrt {-d} \sqrt {e}}{x}}{c \sqrt {d} \sqrt {c^2 d+e} \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{16 d^{5/2} \left (c^2 d+e\right )^{3/2}}-\frac {5 b \text {arctanh}\left (\frac {c^2 d+\frac {\sqrt {-d} \sqrt {e}}{x}}{c \sqrt {d} \sqrt {c^2 d+e} \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{16 d^{5/2} \sqrt {c^2 d+e}}+\frac {3 \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 )}{16 (-d)^{5/2} \sqrt {e}}-\frac {3 \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 )}{16 (-d)^{5/2} \sqrt {e}}+\frac {3 \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 )}{16 (-d)^{5/2} \sqrt {e}}-\frac {3 \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 )}{16 (-d)^{5/2} \sqrt {e}}+\frac {3 i b \operatorname {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{16 (-d)^{5/2} \sqrt {e}}-\frac {3 i b \operatorname {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{16 (-d)^{5/2} \sqrt {e}}+\frac {3 i b \operatorname {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{16 (-d)^{5/2} \sqrt {e}}-\frac {3 i b \operatorname {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{16 (-d)^{5/2} \sqrt {e}} \]
1/16*b*e*arctanh((c^2*d-(-d)^(1/2)*e^(1/2)/x)/c/d^(1/2)/(c^2*d+e)^(1/2)/(1 -1/c^2/x^2)^(1/2))/d^(5/2)/(c^2*d+e)^(3/2)+1/16*b*e*arctanh((c^2*d+(-d)^(1 /2)*e^(1/2)/x)/c/d^(1/2)/(c^2*d+e)^(1/2)/(1-1/c^2/x^2)^(1/2))/d^(5/2)/(c^2 *d+e)^(3/2)+3/16*(a+b*arcsec(c*x))*ln(1-c*(1/c/x+I*(1-1/c^2/x^2)^(1/2))*(- d)^(1/2)/(e^(1/2)-(c^2*d+e)^(1/2)))/(-d)^(5/2)/e^(1/2)-3/16*(a+b*arcsec(c* x))*ln(1+c*(1/c/x+I*(1-1/c^2/x^2)^(1/2))*(-d)^(1/2)/(e^(1/2)-(c^2*d+e)^(1/ 2)))/(-d)^(5/2)/e^(1/2)+3/16*(a+b*arcsec(c*x))*ln(1-c*(1/c/x+I*(1-1/c^2/x^ 2)^(1/2))*(-d)^(1/2)/(e^(1/2)+(c^2*d+e)^(1/2)))/(-d)^(5/2)/e^(1/2)-3/16*(a +b*arcsec(c*x))*ln(1+c*(1/c/x+I*(1-1/c^2/x^2)^(1/2))*(-d)^(1/2)/(e^(1/2)+( c^2*d+e)^(1/2)))/(-d)^(5/2)/e^(1/2)+3/16*I*b*polylog(2,-c*(1/c/x+I*(1-1/c^ 2/x^2)^(1/2))*(-d)^(1/2)/(e^(1/2)-(c^2*d+e)^(1/2)))/(-d)^(5/2)/e^(1/2)-3/1 6*I*b*polylog(2,c*(1/c/x+I*(1-1/c^2/x^2)^(1/2))*(-d)^(1/2)/(e^(1/2)-(c^2*d +e)^(1/2)))/(-d)^(5/2)/e^(1/2)+3/16*I*b*polylog(2,-c*(1/c/x+I*(1-1/c^2/x^2 )^(1/2))*(-d)^(1/2)/(e^(1/2)+(c^2*d+e)^(1/2)))/(-d)^(5/2)/e^(1/2)-3/16*I*b *polylog(2,c*(1/c/x+I*(1-1/c^2/x^2)^(1/2))*(-d)^(1/2)/(e^(1/2)+(c^2*d+e)^( 1/2)))/(-d)^(5/2)/e^(1/2)+1/16*(a+b*arcsec(c*x))*e^(1/2)/(-d)^(3/2)/(-d/x+ (-d)^(1/2)*e^(1/2))^2-5/16*(a+b*arcsec(c*x))/d^2/(-d/x+(-d)^(1/2)*e^(1/2)) -1/16*(a+b*arcsec(c*x))*e^(1/2)/(-d)^(3/2)/(d/x+(-d)^(1/2)*e^(1/2))^2+5/16 *(a+b*arcsec(c*x))/d^2/(d/x+(-d)^(1/2)*e^(1/2))-5/16*b*arctanh((c^2*d-(-d) ^(1/2)*e^(1/2)/x)/c/d^(1/2)/(c^2*d+e)^(1/2)/(1-1/c^2/x^2)^(1/2))/d^(5/2...
Time = 6.04 (sec) , antiderivative size = 1812, normalized size of antiderivative = 1.63 \[ \int \frac {a+b \sec ^{-1}(c x)}{\left (d+e x^2\right )^3} \, dx =\text {Too large to display} \]
(a*x)/(4*d*(d + e*x^2)^2) + (3*a*x)/(8*d^2*(d + e*x^2)) + (3*a*ArcTan[(Sqr t[e]*x)/Sqrt[d]])/(8*d^(5/2)*Sqrt[e]) + b*((-3*(-(ArcSec[c*x]/(I*Sqrt[d]*S qrt[e] + e*x)) + (I*(ArcSin[1/(c*x)]/Sqrt[e] - Log[(2*Sqrt[d]*Sqrt[e]*(Sqr t[e] + c*(I*c*Sqrt[d] - Sqrt[-(c^2*d) - e]*Sqrt[1 - 1/(c^2*x^2)])*x))/(Sqr t[-(c^2*d) - e]*(Sqrt[d] - I*Sqrt[e]*x))]/Sqrt[-(c^2*d) - e]))/Sqrt[d]))/( 16*d^2) - (3*(-(ArcSec[c*x]/((-I)*Sqrt[d]*Sqrt[e] + e*x)) - (I*(ArcSin[1/( c*x)]/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*S qrt[e]*x))]/Sqrt[-(c^2*d) - e]))/Sqrt[d]))/(16*d^2) + ((I/16)*(-(ArcSec[c* x]/(Sqrt[e]*((-I)*Sqrt[d] + Sqrt[e]*x)^2)) + (ArcSin[1/(c*x)]/Sqrt[e] - I* ((c*Sqrt[d]*Sqrt[e]*Sqrt[1 - 1/(c^2*x^2)]*x)/((c^2*d + e)*((-I)*Sqrt[d] + Sqrt[e]*x)) + ((2*c^2*d + e)*Log[(-4*d*Sqrt[e]*Sqrt[c^2*d + e]*(I*Sqrt[e] + c*(c*Sqrt[d] - Sqrt[c^2*d + e]*Sqrt[1 - 1/(c^2*x^2)])*x))/((2*c^2*d + e) *((-I)*Sqrt[d] + Sqrt[e]*x))])/(c^2*d + e)^(3/2)))/d))/d^(3/2) - ((I/16)*( (I*c*Sqrt[e]*Sqrt[1 - 1/(c^2*x^2)]*x)/(Sqrt[d]*(c^2*d + e)*(I*Sqrt[d] + Sq rt[e]*x)) - ArcSec[c*x]/(Sqrt[e]*(I*Sqrt[d] + Sqrt[e]*x)^2) + ArcSin[1/(c* x)]/(d*Sqrt[e]) - (I*(2*c^2*d + e)*Log[(4*d*Sqrt[e]*Sqrt[c^2*d + e]*((-I)* Sqrt[e] + c*(c*Sqrt[d] + Sqrt[c^2*d + e]*Sqrt[1 - 1/(c^2*x^2)])*x))/((2*c^ 2*d + e)*(I*Sqrt[d] + Sqrt[e]*x))])/(d*(c^2*d + e)^(3/2))))/d^(3/2) + (3*( 8*ArcSin[Sqrt[1 + (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*ArcTan[((I*c*Sqrt[d...
Time = 4.27 (sec) , antiderivative size = 1178, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5753, 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.
| \(\displaystyle \int \frac {a+b \sec ^{-1}(c x)}{\left (d+e x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 5753 |
| \(\displaystyle -\int \frac {a+b \arccos \left (\frac {1}{c x}\right )}{\left (\frac {d}{x^2}+e\right )^3 x^4}d\frac {1}{x}\) |
| \(\Big \downarrow \) 5233 |
| \(\displaystyle -\int \left (\frac {\left (a+b \arccos \left (\frac {1}{c x}\right )\right ) e^2}{d^2 \left (\frac {d}{x^2}+e\right )^3}-\frac {2 \left (a+b \arccos \left (\frac {1}{c x}\right )\right ) e}{d^2 \left (\frac {d}{x^2}+e\right )^2}+\frac {a+b \arccos \left (\frac {1}{c x}\right )}{d^2 \left (\frac {d}{x^2}+e\right )}\right )d\frac {1}{x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b \sqrt {e} \sqrt {1-\frac {1}{c^2 x^2}} c}{16 (-d)^{3/2} \left (d c^2+e\right ) \left (\sqrt {-d} \sqrt {e}-\frac {d}{x}\right )}+\frac {b \sqrt {e} \sqrt {1-\frac {1}{c^2 x^2}} c}{16 (-d)^{3/2} \left (d c^2+e\right ) \left (\frac {d}{x}+\sqrt {-d} \sqrt {e}\right )}-\frac {5 \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{16 d^2 \left (\sqrt {-d} \sqrt {e}-\frac {d}{x}\right )}+\frac {5 \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{16 d^2 \left (\frac {d}{x}+\sqrt {-d} \sqrt {e}\right )}+\frac {\sqrt {e} \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{16 (-d)^{3/2} \left (\sqrt {-d} \sqrt {e}-\frac {d}{x}\right )^2}-\frac {\sqrt {e} \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{16 (-d)^{3/2} \left (\frac {d}{x}+\sqrt {-d} \sqrt {e}\right )^2}-\frac {5 b \text {arctanh}\left (\frac {c^2 d-\frac {\sqrt {-d} \sqrt {e}}{x}}{c \sqrt {d} \sqrt {d c^2+e} \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{16 d^{5/2} \sqrt {d c^2+e}}+\frac {b e \text {arctanh}\left (\frac {c^2 d-\frac {\sqrt {-d} \sqrt {e}}{x}}{c \sqrt {d} \sqrt {d c^2+e} \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{16 d^{5/2} \left (d c^2+e\right )^{3/2}}-\frac {5 b \text {arctanh}\left (\frac {d c^2+\frac {\sqrt {-d} \sqrt {e}}{x}}{c \sqrt {d} \sqrt {d c^2+e} \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{16 d^{5/2} \sqrt {d c^2+e}}+\frac {b e \text {arctanh}\left (\frac {d c^2+\frac {\sqrt {-d} \sqrt {e}}{x}}{c \sqrt {d} \sqrt {d c^2+e} \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{16 d^{5/2} \left (d c^2+e\right )^{3/2}}+\frac {3 \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 {d c^2+e}}\right )}{16 (-d)^{5/2} \sqrt {e}}-\frac {3 \left (a+b \arccos \left (\frac {1}{c x}\right )\right ) \log \left (\frac {\sqrt {-d} e^{i \arccos \left (\frac {1}{c x}\right )} c}{\sqrt {e}-\sqrt {d c^2+e}}+1\right )}{16 (-d)^{5/2} \sqrt {e}}+\frac {3 \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 {d c^2+e}}\right )}{16 (-d)^{5/2} \sqrt {e}}-\frac {3 \left (a+b \arccos \left (\frac {1}{c x}\right )\right ) \log \left (\frac {\sqrt {-d} e^{i \arccos \left (\frac {1}{c x}\right )} c}{\sqrt {e}+\sqrt {d c^2+e}}+1\right )}{16 (-d)^{5/2} \sqrt {e}}+\frac {3 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 )}{16 (-d)^{5/2} \sqrt {e}}-\frac {3 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 )}{16 (-d)^{5/2} \sqrt {e}}+\frac {3 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 )}{16 (-d)^{5/2} \sqrt {e}}-\frac {3 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 )}{16 (-d)^{5/2} \sqrt {e}}\) |
(b*c*Sqrt[e]*Sqrt[1 - 1/(c^2*x^2)])/(16*(-d)^(3/2)*(c^2*d + e)*(Sqrt[-d]*S qrt[e] - d/x)) + (b*c*Sqrt[e]*Sqrt[1 - 1/(c^2*x^2)])/(16*(-d)^(3/2)*(c^2*d + e)*(Sqrt[-d]*Sqrt[e] + d/x)) + (Sqrt[e]*(a + b*ArcCos[1/(c*x)]))/(16*(- d)^(3/2)*(Sqrt[-d]*Sqrt[e] - d/x)^2) - (5*(a + b*ArcCos[1/(c*x)]))/(16*d^2 *(Sqrt[-d]*Sqrt[e] - d/x)) - (Sqrt[e]*(a + b*ArcCos[1/(c*x)]))/(16*(-d)^(3 /2)*(Sqrt[-d]*Sqrt[e] + d/x)^2) + (5*(a + b*ArcCos[1/(c*x)]))/(16*d^2*(Sqr t[-d]*Sqrt[e] + d/x)) + (b*e*ArcTanh[(c^2*d - (Sqrt[-d]*Sqrt[e])/x)/(c*Sqr t[d]*Sqrt[c^2*d + e]*Sqrt[1 - 1/(c^2*x^2)])])/(16*d^(5/2)*(c^2*d + e)^(3/2 )) - (5*b*ArcTanh[(c^2*d - (Sqrt[-d]*Sqrt[e])/x)/(c*Sqrt[d]*Sqrt[c^2*d + e ]*Sqrt[1 - 1/(c^2*x^2)])])/(16*d^(5/2)*Sqrt[c^2*d + e]) + (b*e*ArcTanh[(c^ 2*d + (Sqrt[-d]*Sqrt[e])/x)/(c*Sqrt[d]*Sqrt[c^2*d + e]*Sqrt[1 - 1/(c^2*x^2 )])])/(16*d^(5/2)*(c^2*d + e)^(3/2)) - (5*b*ArcTanh[(c^2*d + (Sqrt[-d]*Sqr t[e])/x)/(c*Sqrt[d]*Sqrt[c^2*d + e]*Sqrt[1 - 1/(c^2*x^2)])])/(16*d^(5/2)*S qrt[c^2*d + e]) + (3*(a + b*ArcCos[1/(c*x)])*Log[1 - (c*Sqrt[-d]*E^(I*ArcC os[1/(c*x)]))/(Sqrt[e] - Sqrt[c^2*d + e])])/(16*(-d)^(5/2)*Sqrt[e]) - (3*( a + b*ArcCos[1/(c*x)])*Log[1 + (c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] - Sqrt[c^2*d + e])])/(16*(-d)^(5/2)*Sqrt[e]) + (3*(a + b*ArcCos[1/(c*x)]) *Log[1 - (c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] + Sqrt[c^2*d + e])])/ (16*(-d)^(5/2)*Sqrt[e]) - (3*(a + b*ArcCos[1/(c*x)])*Log[1 + (c*Sqrt[-d]*E ^(I*ArcCos[1/(c*x)]))/(Sqrt[e] + Sqrt[c^2*d + e])])/(16*(-d)^(5/2)*Sqrt...
3.2.10.3.1 Defintions of rubi rules used
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_.)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> -Subst[Int[(e + d*x^2)^p*((a + b*ArcCos[x/c])^n/x^(2*(p + 1))) , x], x, 1/x] /; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[n, 0] && IntegerQ[p]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 94.04 (sec) , antiderivative size = 1812, normalized size of antiderivative = 1.63
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1812\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(1837\) |
| default | \(\text {Expression too large to display}\) | \(1837\) |
1/4*a*x/d/(e*x^2+d)^2+3/8*a/d^2*x/(e*x^2+d)+3/8*a/d^2/(d*e)^(1/2)*arctan(e *x/(d*e)^(1/2))+b/c*(1/8*x*c^3*(5*d^2*c^4*arcsec(c*x)+3*c^4*d*e*arcsec(c*x )*x^2+((c^2*x^2-1)/c^2/x^2)^(1/2)*c^3*d*e*x+((c^2*x^2-1)/c^2/x^2)^(1/2)*e^ 2*c^3*x^3+5*c^2*d*e*arcsec(c*x)+3*e^2*arcsec(c*x)*c^2*x^2)/d^2/(c^2*d+e)/( c^2*e*x^2+c^2*d)^2-5/8*I*(-(c^2*d-2*(e*(c^2*d+e))^(1/2)+2*e)*d)^(1/2)*((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)*arctanh(c* d*(1/c/x+I*(1-1/c^2/x^2)^(1/2))/((-c^2*d+2*(e*(c^2*d+e))^(1/2)-2*e)*d)^(1/ 2))/(c^2*d+e)^2/d^4/c-3/16*I/(c^2*d+e)/d^2*c^2*e*sum(1/_R1/(_R1^2*c^2*d+c^ 2*d+2*e)*(I*arcsec(c*x)*ln((_R1-1/c/x-I*(1-1/c^2/x^2)^(1/2))/_R1)+dilog((_ R1-1/c/x-I*(1-1/c^2/x^2)^(1/2))/_R1)),_R1=RootOf(c^2*d*_Z^4+(2*c^2*d+4*e)* _Z^2+c^2*d))+1/2*I*(-(c^2*d-2*(e*(c^2*d+e))^(1/2)+2*e)*d)^(1/2)*(c^2*d+2*( e*(c^2*d+e))^(1/2)+2*e)*arctanh(c*d*(1/c/x+I*(1-1/c^2/x^2)^(1/2))/((-c^2*d +2*(e*(c^2*d+e))^(1/2)-2*e)*d)^(1/2))*e/c^3/d^5/(c^2*d+e)-1/2*I*((c^2*d+2* (e*(c^2*d+e))^(1/2)+2*e)*d)^(1/2)*(-(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*arctan(c*d*(1/c/x+I*(1-1/c^2/x^2)^(1/2))/(( c^2*d+2*(e*(c^2*d+e))^(1/2)+2*e)*d)^(1/2))/(c^2*d+e)^2/d^5/c^3+5/8*I*((c^2 *d+2*(e*(c^2*d+e))^(1/2)+2*e)*d)^(1/2)*(c^2*d-2*(e*(c^2*d+e))^(1/2)+2*e)*a rctan(c*d*(1/c/x+I*(1-1/c^2/x^2)^(1/2))/((c^2*d+2*(e*(c^2*d+e))^(1/2)+2*e) *d)^(1/2))/c/d^4/(c^2*d+e)+5/8*I*arctanh(c*d*(1/c/x+I*(1-1/c^2/x^2)^(1/2)) /((-c^2*d+2*(e*(c^2*d+e))^(1/2)-2*e)*d)^(1/2))*(-(c^2*d-2*(e*(c^2*d+e))...
\[ \int \frac {a+b \sec ^{-1}(c x)}{\left (d+e x^2\right )^3} \, dx=\int { \frac {b \operatorname {arcsec}\left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {a+b \sec ^{-1}(c x)}{\left (d+e x^2\right )^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {a+b \sec ^{-1}(c x)}{\left (d+e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
Exception generated. \[ \int \frac {a+b \sec ^{-1}(c x)}{\left (d+e x^2\right )^3} \, dx=\text {Exception raised: RuntimeError} \]
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. \[ \int \frac {a+b \sec ^{-1}(c x)}{\left (d+e x^2\right )^3} \, dx=\int \frac {a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )}{{\left (e\,x^2+d\right )}^3} \,d x \]