Integrand size = 21, antiderivative size = 1124 \[ \int \frac {x^4 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx =\text {Too large to display} \] Output:
1/16*b*c*(-d)^(1/2)*(1-1/c^2/x^2)^(1/2)/e^(3/2)/(c^2*d+e)/((-d)^(1/2)*e^(1 /2)-d/x)+1/16*b*c*(-d)^(1/2)*(1-1/c^2/x^2)^(1/2)/e^(3/2)/(c^2*d+e)/((-d)^( 1/2)*e^(1/2)+d/x)+1/16*(-d)^(1/2)*(a+b*arcsec(c*x))/e^(3/2)/((-d)^(1/2)*e^ (1/2)-d/x)^2+3/16*(a+b*arcsec(c*x))/e^2/((-d)^(1/2)*e^(1/2)-d/x)-1/16*(-d) ^(1/2)*(a+b*arcsec(c*x))/e^(3/2)/((-d)^(1/2)*e^(1/2)+d/x)^2-3/16*(a+b*arcs ec(c*x))/e^2/((-d)^(1/2)*e^(1/2)+d/x)+1/16*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^(1/2)/e/(c^2*d+e) ^(3/2)+3/16*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^(1/2)/e^2/(c^2*d+e)^(1/2)+1/16*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^(1/ 2)/e/(c^2*d+e)^(3/2)+3/16*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^(1/2)/e^2/(c^2*d+e)^(1/2)+3/16*(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)^(1/2)/e^(5/2)-3/16*(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)^(1/2)/e^(5 /2)+3/16*(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)^(1/2)/e^(5/2)-3/16*(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 )^(1/2)/e^(5/2)-3/16*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)^(1/2)/e^(5/2)-3/16*I*b*polylog(2,c*...
Time = 6.07 (sec) , antiderivative size = 1819, normalized size of antiderivative = 1.62 \[ \int \frac {x^4 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx =\text {Too large to display} \] Input:
Integrate[(x^4*(a + b*ArcSec[c*x]))/(d + e*x^2)^3,x]
Output:
(a*d*x)/(4*e^2*(d + e*x^2)^2) - (5*a*x)/(8*e^2*(d + e*x^2)) + (3*a*ArcTan[ (Sqrt[e]*x)/Sqrt[d]])/(8*Sqrt[d]*e^(5/2)) + b*((5*(-(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]) )/(16*e^2) + (5*(-(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] + Sqr t[-(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]))/(16*e^2) + ((I/16)*Sqrt[d]*( -(ArcSec[c*x]/(Sqrt[e]*((-I)*Sqrt[d] + Sqrt[e]*x)^2)) + (ArcSin[1/(c*x)]/S qrt[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))/e^2 - (( I/16)*Sqrt[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)))) /e^2 + (3*(8*ArcSin[Sqrt[1 + (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*ArcTan[(...
Time = 2.25 (sec) , antiderivative size = 1188, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5763, 5173, 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.
\(\displaystyle \int \frac {x^4 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 5763 |
\(\displaystyle -\int \frac {a+b \arccos \left (\frac {1}{c x}\right )}{\left (\frac {d}{x^2}+e\right )^3}d\frac {1}{x}\) |
\(\Big \downarrow \) 5173 |
\(\displaystyle -\int \left (-\frac {\left (a+b \arccos \left (\frac {1}{c x}\right )\right ) d^3}{8 (-d)^{3/2} e^{3/2} \left (\sqrt {-d} \sqrt {e}-\frac {d}{x}\right )^3}-\frac {\left (a+b \arccos \left (\frac {1}{c x}\right )\right ) d^3}{8 (-d)^{3/2} e^{3/2} \left (\frac {d}{x}+\sqrt {-d} \sqrt {e}\right )^3}-\frac {3 \left (a+b \arccos \left (\frac {1}{c x}\right )\right ) d}{8 e^2 \left (-\frac {d^2}{x^2}-e d\right )}-\frac {3 \left (a+b \arccos \left (\frac {1}{c x}\right )\right ) d}{16 e^2 \left (\sqrt {-d} \sqrt {e}-\frac {d}{x}\right )^2}-\frac {3 \left (a+b \arccos \left (\frac {1}{c x}\right )\right ) d}{16 e^2 \left (\frac {d}{x}+\sqrt {-d} \sqrt {e}\right )^2}\right )d\frac {1}{x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b \sqrt {-d} \sqrt {1-\frac {1}{c^2 x^2}} c}{16 e^{3/2} \left (d c^2+e\right ) \left (\sqrt {-d} \sqrt {e}-\frac {d}{x}\right )}+\frac {b \sqrt {-d} \sqrt {1-\frac {1}{c^2 x^2}} c}{16 e^{3/2} \left (d c^2+e\right ) \left (\frac {d}{x}+\sqrt {-d} \sqrt {e}\right )}+\frac {3 \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{16 e^2 \left (\sqrt {-d} \sqrt {e}-\frac {d}{x}\right )}-\frac {3 \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{16 e^2 \left (\frac {d}{x}+\sqrt {-d} \sqrt {e}\right )}+\frac {\sqrt {-d} \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{16 e^{3/2} \left (\sqrt {-d} \sqrt {e}-\frac {d}{x}\right )^2}-\frac {\sqrt {-d} \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{16 e^{3/2} \left (\frac {d}{x}+\sqrt {-d} \sqrt {e}\right )^2}+\frac {3 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 \sqrt {d} e^2 \sqrt {d c^2+e}}+\frac {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 \sqrt {d} e \left (d c^2+e\right )^{3/2}}+\frac {3 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 \sqrt {d} e^2 \sqrt {d c^2+e}}+\frac {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 \sqrt {d} e \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 \sqrt {-d} e^{5/2}}-\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 \sqrt {-d} e^{5/2}}+\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 \sqrt {-d} e^{5/2}}-\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 \sqrt {-d} e^{5/2}}+\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 \sqrt {-d} e^{5/2}}-\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 \sqrt {-d} e^{5/2}}+\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 \sqrt {-d} e^{5/2}}-\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 \sqrt {-d} e^{5/2}}\) |
Input:
Int[(x^4*(a + b*ArcSec[c*x]))/(d + e*x^2)^3,x]
Output:
(b*c*Sqrt[-d]*Sqrt[1 - 1/(c^2*x^2)])/(16*e^(3/2)*(c^2*d + e)*(Sqrt[-d]*Sqr t[e] - d/x)) + (b*c*Sqrt[-d]*Sqrt[1 - 1/(c^2*x^2)])/(16*e^(3/2)*(c^2*d + e )*(Sqrt[-d]*Sqrt[e] + d/x)) + (Sqrt[-d]*(a + b*ArcCos[1/(c*x)]))/(16*e^(3/ 2)*(Sqrt[-d]*Sqrt[e] - d/x)^2) + (3*(a + b*ArcCos[1/(c*x)]))/(16*e^2*(Sqrt [-d]*Sqrt[e] - d/x)) - (Sqrt[-d]*(a + b*ArcCos[1/(c*x)]))/(16*e^(3/2)*(Sqr t[-d]*Sqrt[e] + d/x)^2) - (3*(a + b*ArcCos[1/(c*x)]))/(16*e^2*(Sqrt[-d]*Sq rt[e] + d/x)) + (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*Sqrt[d]*e*(c^2*d + e)^(3/2)) + (3* 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*Sqrt[d]*e^2*Sqrt[c^2*d + e]) + (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*Sqrt[d]*e*(c^2*d + e)^(3/2)) + (3*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*Sqrt[d]*e^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*Sqrt[-d]*e^(5/2)) - (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*Sqrt[-d]*e^(5/2)) + (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* Sqrt[-d]*e^(5/2)) - (3*(a + b*ArcCos[1/(c*x)])*Log[1 + (c*Sqrt[-d]*E^(I*Ar cCos[1/(c*x)]))/(Sqrt[e] + Sqrt[c^2*d + e])])/(16*Sqrt[-d]*e^(5/2)) + (...
Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x _Symbol] :> Int[ExpandIntegrand[(a + b*ArcCos[c*x])^n, (d + e*x^2)^p, x], x ] /; FreeQ[{a, b, c, d, e, n}, x] && NeQ[c^2*d + e, 0] && IntegerQ[p] && (G tQ[p, 0] || IGtQ[n, 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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 72.63 (sec) , antiderivative size = 1822, normalized size of antiderivative = 1.62
method | result | size |
parts | \(\text {Expression too large to display}\) | \(1822\) |
derivativedivides | \(\text {Expression too large to display}\) | \(1845\) |
default | \(\text {Expression too large to display}\) | \(1845\) |
Input:
int(x^4*(a+b*arcsec(c*x))/(e*x^2+d)^3,x,method=_RETURNVERBOSE)
Output:
a*((-5/8/e*x^3-3/8*d/e^2*x)/(e*x^2+d)^2+3/8/e^2/(d*e)^(1/2)*arctan(e*x/(d* e)^(1/2)))+b/c^5*(-1/8*x*c^7*(3*d^2*c^4*arcsec(c*x)+5*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+3*c^2*d*e*arcsec(c*x)+5*e^2*arcsec(c*x)*c^2*x^2)/e^2/(c^2*d+e)/(c^ 2*e*x^2+c^2*d)^2-3/16*I/(c^2*d+e)/e^2*c^8*d*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)*c*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)/e/d^3+3/8*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)*c^3*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/e^2/d^2-1/2*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)*c*ar ctan(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)/e/d^3+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)*c*a rctanh(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/e/d^3-3/8*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)*c^3*arctanh(c*d*(1/c/x+I*(1-1...
\[ \int \frac {x^4 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx=\int { \frac {{\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} x^{4}}{{\left (e x^{2} + d\right )}^{3}} \,d x } \] Input:
integrate(x^4*(a+b*arcsec(c*x))/(e*x^2+d)^3,x, algorithm="fricas")
Output:
integral((b*x^4*arcsec(c*x) + a*x^4)/(e^3*x^6 + 3*d*e^2*x^4 + 3*d^2*e*x^2 + d^3), x)
Timed out. \[ \int \frac {x^4 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx=\text {Timed out} \] Input:
integrate(x**4*(a+b*asec(c*x))/(e*x**2+d)**3,x)
Output:
Timed out
Exception generated. \[ \int \frac {x^4 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^4*(a+b*arcsec(c*x))/(e*x^2+d)^3,x, algorithm="maxima")
Output:
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. \[ \int \frac {x^4 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(x^4*(a+b*arcsec(c*x))/(e*x^2+d)^3,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. \[ \int \frac {x^4 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx=\int \frac {x^4\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}{{\left (e\,x^2+d\right )}^3} \,d x \] Input:
int((x^4*(a + b*acos(1/(c*x))))/(d + e*x^2)^3,x)
Output:
int((x^4*(a + b*acos(1/(c*x))))/(d + e*x^2)^3, x)
\[ \int \frac {x^4 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx=\frac {3 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a \,d^{2}+6 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a d e \,x^{2}+3 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a \,e^{2} x^{4}+8 \left (\int \frac {\mathit {asec} \left (c x \right ) x^{4}}{e^{3} x^{6}+3 d \,e^{2} x^{4}+3 d^{2} e \,x^{2}+d^{3}}d x \right ) b \,d^{3} e^{3}+16 \left (\int \frac {\mathit {asec} \left (c x \right ) x^{4}}{e^{3} x^{6}+3 d \,e^{2} x^{4}+3 d^{2} e \,x^{2}+d^{3}}d x \right ) b \,d^{2} e^{4} x^{2}+8 \left (\int \frac {\mathit {asec} \left (c x \right ) x^{4}}{e^{3} x^{6}+3 d \,e^{2} x^{4}+3 d^{2} e \,x^{2}+d^{3}}d x \right ) b d \,e^{5} x^{4}-3 a \,d^{2} e x -5 a d \,e^{2} x^{3}}{8 d \,e^{3} \left (e^{2} x^{4}+2 d e \,x^{2}+d^{2}\right )} \] Input:
int(x^4*(a+b*asec(c*x))/(e*x^2+d)^3,x)
Output:
(3*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a*d**2 + 6*sqrt(e)*sqrt(d )*atan((e*x)/(sqrt(e)*sqrt(d)))*a*d*e*x**2 + 3*sqrt(e)*sqrt(d)*atan((e*x)/ (sqrt(e)*sqrt(d)))*a*e**2*x**4 + 8*int((asec(c*x)*x**4)/(d**3 + 3*d**2*e*x **2 + 3*d*e**2*x**4 + e**3*x**6),x)*b*d**3*e**3 + 16*int((asec(c*x)*x**4)/ (d**3 + 3*d**2*e*x**2 + 3*d*e**2*x**4 + e**3*x**6),x)*b*d**2*e**4*x**2 + 8 *int((asec(c*x)*x**4)/(d**3 + 3*d**2*e*x**2 + 3*d*e**2*x**4 + e**3*x**6),x )*b*d*e**5*x**4 - 3*a*d**2*e*x - 5*a*d*e**2*x**3)/(8*d*e**3*(d**2 + 2*d*e* x**2 + e**2*x**4))