Integrand size = 21, antiderivative size = 707 \[ \int \frac {x^5 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx=-\frac {b c d \sqrt {1-\frac {1}{c^2 x^2}}}{8 e^2 \left (c^2 d+e\right ) \left (e+\frac {d}{x^2}\right ) x}-\frac {a+b \sec ^{-1}(c x)}{4 e \left (e+\frac {d}{x^2}\right )^2}-\frac {a+b \sec ^{-1}(c x)}{2 e^2 \left (e+\frac {d}{x^2}\right )}-\frac {b \arctan \left (\frac {\sqrt {c^2 d+e}}{c \sqrt {e} \sqrt {1-\frac {1}{c^2 x^2}} x}\right )}{2 e^{5/2} \sqrt {c^2 d+e}}-\frac {b \left (c^2 d+2 e\right ) \arctan \left (\frac {\sqrt {c^2 d+e}}{c \sqrt {e} \sqrt {1-\frac {1}{c^2 x^2}} x}\right )}{8 e^{5/2} \left (c^2 d+e\right )^{3/2}}+\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 e^3}+\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 e^3}+\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 e^3}+\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 e^3}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )}{e^3}-\frac {i b \operatorname {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 e^3}-\frac {i b \operatorname {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 e^3}-\frac {i b \operatorname {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 e^3}-\frac {i b \operatorname {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 e^3}+\frac {i b \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )}{2 e^3} \]
1/4*(-a-b*arcsec(c*x))/e/(e+d/x^2)^2+1/2*(-a-b*arcsec(c*x))/e^2/(e+d/x^2)- 1/8*b*(c^2*d+2*e)*arctan((c^2*d+e)^(1/2)/c/x/e^(1/2)/(1-1/c^2/x^2)^(1/2))/ e^(5/2)/(c^2*d+e)^(3/2)-(a+b*arcsec(c*x))*ln(1+(1/c/x+I*(1-1/c^2/x^2)^(1/2 ))^2)/e^3+1/2*(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)))/e^3+1/2*(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)))/e^3+1/2*(a+b*a rcsec(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)))/e^3+1/2*(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)))/e^3+1/2*I*b*polylog(2,-(1/c/x+I*(1 -1/c^2/x^2)^(1/2))^2)/e^3-1/2*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)))/e^3-1/2*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)))/e^3-1/2*I*b*pol ylog(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 )))/e^3-1/2*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)))/e^3-1/2*b*arctan((c^2*d+e)^(1/2)/c/x/e^(1/2)/(1-1/c^ 2/x^2)^(1/2))/e^(5/2)/(c^2*d+e)^(1/2)-1/8*b*c*d*(1-1/c^2/x^2)^(1/2)/e^2/(c ^2*d+e)/(e+d/x^2)/x
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1805\) vs. \(2(707)=1414\).
Time = 7.51 (sec) , antiderivative size = 1805, normalized size of antiderivative = 2.55 \[ \int \frac {x^5 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx =\text {Too large to display} \]
-1/4*(a*d^2)/(e^3*(d + e*x^2)^2) + (a*d)/(e^3*(d + e*x^2)) + (a*Log[d + e* x^2])/(2*e^3) + b*((((-7*I)/16)*Sqrt[d]*(-(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*Sqrt[e]*x))]/Sqrt[-(c^2*d) - e]))/Sqrt[d]))/e^(5/2) + (((7*I)/16)*Sqrt[d]*(-(ArcSec[c*x]/((-I)*Sqrt[d]*Sqrt[e] + e*x)) - (I*(A rcSin[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*Sqrt[e]*x))]/Sqrt[-(c^2*d) - e]))/Sqrt[d]))/e^(5/2) - (d*(-(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))/(16*e^(5/2)) - ( d*((I*c*Sqrt[e]*Sqrt[1 - 1/(c^2*x^2)]*x)/(Sqrt[d]*(c^2*d + e)*(I*Sqrt[d] + Sqrt[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))))/(16*e^(5/2) ) + ((I/4)*(8*ArcSin[Sqrt[1 + (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*ArcTan[...
Time = 1.90 (sec) , antiderivative size = 775, 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.
| \(\displaystyle \int \frac {x^5 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 5763 |
| \(\displaystyle -\int \frac {x \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{\left (\frac {d}{x^2}+e\right )^3}d\frac {1}{x}\) |
| \(\Big \downarrow \) 5233 |
| \(\displaystyle -\int \left (\frac {x \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{e^3}-\frac {d \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{e^3 \left (\frac {d}{x^2}+e\right ) x}-\frac {d \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{e^2 \left (\frac {d}{x^2}+e\right )^2 x}-\frac {d \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{e \left (\frac {d}{x^2}+e\right )^3 x}\right )d\frac {1}{x}\) |
| \(\Big \downarrow \) 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 e^3}+\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 e^3}+\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 e^3}+\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 e^3}-\frac {a+b \arccos \left (\frac {1}{c x}\right )}{2 e^2 \left (\frac {d}{x^2}+e\right )}-\frac {a+b \arccos \left (\frac {1}{c x}\right )}{4 e \left (\frac {d}{x^2}+e\right )^2}-\frac {\log \left (1+e^{2 i \arccos \left (\frac {1}{c x}\right )}\right ) \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{e^3}-\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 e^3}-\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 e^3}-\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 e^3}-\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 e^3}+\frac {i b \operatorname {PolyLog}\left (2,-e^{2 i \arccos \left (\frac {1}{c x}\right )}\right )}{2 e^3}-\frac {b \left (c^2 d+2 e\right ) \arctan \left (\frac {\sqrt {c^2 d+e}}{c \sqrt {e} x \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{8 e^{5/2} \left (c^2 d+e\right )^{3/2}}-\frac {b \arctan \left (\frac {\sqrt {c^2 d+e}}{c \sqrt {e} x \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{2 e^{5/2} \sqrt {c^2 d+e}}-\frac {b c d \sqrt {1-\frac {1}{c^2 x^2}}}{8 e^2 x \left (c^2 d+e\right ) \left (\frac {d}{x^2}+e\right )}\) |
-1/8*(b*c*d*Sqrt[1 - 1/(c^2*x^2)])/(e^2*(c^2*d + e)*(e + d/x^2)*x) - (a + b*ArcCos[1/(c*x)])/(4*e*(e + d/x^2)^2) - (a + b*ArcCos[1/(c*x)])/(2*e^2*(e + d/x^2)) - (b*ArcTan[Sqrt[c^2*d + e]/(c*Sqrt[e]*Sqrt[1 - 1/(c^2*x^2)]*x) ])/(2*e^(5/2)*Sqrt[c^2*d + e]) - (b*(c^2*d + 2*e)*ArcTan[Sqrt[c^2*d + e]/( c*Sqrt[e]*Sqrt[1 - 1/(c^2*x^2)]*x)])/(8*e^(5/2)*(c^2*d + e)^(3/2)) + ((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*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])])/(2*e^3) + ((a + b*Arc Cos[1/(c*x)])*Log[1 - (c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] + Sqrt[c ^2*d + e])])/(2*e^3) + ((a + b*ArcCos[1/(c*x)])*Log[1 + (c*Sqrt[-d]*E^(I*A rcCos[1/(c*x)]))/(Sqrt[e] + Sqrt[c^2*d + e])])/(2*e^3) - ((a + b*ArcCos[1/ (c*x)])*Log[1 + E^((2*I)*ArcCos[1/(c*x)])])/e^3 - ((I/2)*b*PolyLog[2, -((c *Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] - Sqrt[c^2*d + e]))])/e^3 - ((I/ 2)*b*PolyLog[2, (c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] - Sqrt[c^2*d + e])])/e^3 - ((I/2)*b*PolyLog[2, -((c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqr t[e] + Sqrt[c^2*d + e]))])/e^3 - ((I/2)*b*PolyLog[2, (c*Sqrt[-d]*E^(I*ArcC os[1/(c*x)]))/(Sqrt[e] + Sqrt[c^2*d + e])])/e^3 + ((I/2)*b*PolyLog[2, -E^( (2*I)*ArcCos[1/(c*x)])])/e^3
3.2.4.3.1 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_.)*(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 = 8.69 (sec) , antiderivative size = 1396, normalized size of antiderivative = 1.97
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1396\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(1409\) |
| default | \(\text {Expression too large to display}\) | \(1409\) |
a*(-1/4*d^2/e^3/(e*x^2+d)^2+1/2/e^3*ln(e*x^2+d)+d/e^3/(e*x^2+d))+b/c^6*(-1 /8*c^6*(4*c^6*d^2*arcsec(c*x)*x^2+6*c^6*d*e*arcsec(c*x)*x^4+((c^2*x^2-1)/c ^2/x^2)^(1/2)*c^5*d^2*x+((c^2*x^2-1)/c^2/x^2)^(1/2)*c^5*d*e*x^3+4*c^4*d*e* arcsec(c*x)*x^2+6*arcsec(c*x)*e^2*c^4*x^4+I*c^4*d^2+2*I*c^4*d*e*x^2+I*e^2* c^4*x^4)/e^2/(c^2*d+e)/(c^2*e*x^2+c^2*d)^2+I/(c^2*d+e)/e^2*c^6*dilog(1+I*( 1/c/x+I*(1-1/c^2/x^2)^(1/2)))+I/(c^2*d+e)/e^2*c^6*dilog(1-I*(1/c/x+I*(1-1/ c^2/x^2)^(1/2)))-1/(c^2*d+e)/e^2*c^6*arcsec(c*x)*ln(1-I*(1/c/x+I*(1-1/c^2/ x^2)^(1/2)))-1/(c^2*d+e)/e^2*c^6*arcsec(c*x)*ln(1+I*(1/c/x+I*(1-1/c^2/x^2) ^(1/2)))-1/4*I/(c^2*d+e)/e^2*c^6*sum((_R1^2*c^2*d+c^2*d+4*e)/(_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))-1/4*I/(c^2*d+e)/e^2*c^8*d*sum((_R1^2+1)/(_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/(c^2*d+e)/e^3*c^8*d*arcsec(c*x)*ln(1+I*(1/c/x+I*(1-1/c^2/x^2)^( 1/2)))-1/(c^2*d+e)/e^3*c^8*d*arcsec(c*x)*ln(1-I*(1/c/x+I*(1-1/c^2/x^2)^(1/ 2)))-1/4*I/(c^2*d+e)/e^3*c^8*d*sum((_R1^2*c^2*d+c^2*d+4*e)/(_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))-1/4*I/(c^2*d+e)/e^3*c^10*d^2*sum((_R1^2+1)/(_R1^2*c^2*d+c^...
\[ \int \frac {x^5 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx=\int { \frac {{\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} x^{5}}{{\left (e x^{2} + d\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {x^5 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx=\text {Timed out} \]
\[ \int \frac {x^5 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx=\int { \frac {{\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} x^{5}}{{\left (e x^{2} + d\right )}^{3}} \,d x } \]
1/4*a*((4*d*e*x^2 + 3*d^2)/(e^5*x^4 + 2*d*e^4*x^2 + d^2*e^3) + 2*log(e*x^2 + d)/e^3) + b*integrate(x^5*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))/(e^3*x^6 + 3*d*e^2*x^4 + 3*d^2*e*x^2 + d^3), x)
Timed out. \[ \int \frac {x^5 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {x^5 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx=\int \frac {x^5\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}{{\left (e\,x^2+d\right )}^3} \,d x \]