Integrand size = 21, antiderivative size = 727 \[ \int \frac {a+b \arcsin (c x)}{x \left (d+e x^2\right )^3} \, dx=-\frac {b c e x \sqrt {1-c^2 x^2}}{8 d^2 \left (c^2 d+e\right ) \left (d+e x^2\right )}+\frac {a+b \arcsin (c x)}{4 d \left (d+e x^2\right )^2}+\frac {a+b \arcsin (c x)}{2 d^2 \left (d+e x^2\right )}-\frac {b c \arctan \left (\frac {\sqrt {c^2 d+e} x}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{2 d^{5/2} \sqrt {c^2 d+e}}-\frac {b c \left (2 c^2 d+e\right ) \arctan \left (\frac {\sqrt {c^2 d+e} x}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{8 d^{5/2} \left (c^2 d+e\right )^{3/2}}-\frac {(a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 d^3}-\frac {(a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 d^3}-\frac {(a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 d^3}-\frac {(a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 d^3}+\frac {(a+b \arcsin (c x)) \log \left (1-e^{2 i \arcsin (c x)}\right )}{d^3}+\frac {i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 d^3}+\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 d^3}+\frac {i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 d^3}+\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 d^3}-\frac {i b \operatorname {PolyLog}\left (2,e^{2 i \arcsin (c x)}\right )}{2 d^3} \]
1/4*(a+b*arcsin(c*x))/d/(e*x^2+d)^2+1/2*(a+b*arcsin(c*x))/d^2/(e*x^2+d)-1/ 8*b*c*(2*c^2*d+e)*arctan(x*(c^2*d+e)^(1/2)/d^(1/2)/(-c^2*x^2+1)^(1/2))/d^( 5/2)/(c^2*d+e)^(3/2)+(a+b*arcsin(c*x))*ln(1-(I*c*x+(-c^2*x^2+1)^(1/2))^2)/ d^3-1/2*(a+b*arcsin(c*x))*ln(1-(I*c*x+(-c^2*x^2+1)^(1/2))*e^(1/2)/(I*c*(-d )^(1/2)-(c^2*d+e)^(1/2)))/d^3-1/2*(a+b*arcsin(c*x))*ln(1+(I*c*x+(-c^2*x^2+ 1)^(1/2))*e^(1/2)/(I*c*(-d)^(1/2)-(c^2*d+e)^(1/2)))/d^3-1/2*(a+b*arcsin(c* x))*ln(1-(I*c*x+(-c^2*x^2+1)^(1/2))*e^(1/2)/(I*c*(-d)^(1/2)+(c^2*d+e)^(1/2 )))/d^3-1/2*(a+b*arcsin(c*x))*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))*e^(1/2)/(I*c *(-d)^(1/2)+(c^2*d+e)^(1/2)))/d^3-1/2*I*b*polylog(2,(I*c*x+(-c^2*x^2+1)^(1 /2))^2)/d^3+1/2*I*b*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))*e^(1/2)/(I*c*(-d )^(1/2)-(c^2*d+e)^(1/2)))/d^3+1/2*I*b*polylog(2,(I*c*x+(-c^2*x^2+1)^(1/2)) *e^(1/2)/(I*c*(-d)^(1/2)-(c^2*d+e)^(1/2)))/d^3+1/2*I*b*polylog(2,-(I*c*x+( -c^2*x^2+1)^(1/2))*e^(1/2)/(I*c*(-d)^(1/2)+(c^2*d+e)^(1/2)))/d^3+1/2*I*b*p olylog(2,(I*c*x+(-c^2*x^2+1)^(1/2))*e^(1/2)/(I*c*(-d)^(1/2)+(c^2*d+e)^(1/2 )))/d^3-1/2*b*c*arctan(x*(c^2*d+e)^(1/2)/d^(1/2)/(-c^2*x^2+1)^(1/2))/d^(5/ 2)/(c^2*d+e)^(1/2)-1/8*b*c*e*x*(-c^2*x^2+1)^(1/2)/d^2/(c^2*d+e)/(e*x^2+d)
Time = 3.89 (sec) , antiderivative size = 1022, normalized size of antiderivative = 1.41 \[ \int \frac {a+b \arcsin (c x)}{x \left (d+e x^2\right )^3} \, dx=\frac {-\frac {b c d \sqrt {e} \sqrt {1-c^2 x^2}}{\left (c^2 d+e\right ) \left (-i \sqrt {d}+\sqrt {e} x\right )}-\frac {b c d \sqrt {e} \sqrt {1-c^2 x^2}}{\left (c^2 d+e\right ) \left (i \sqrt {d}+\sqrt {e} x\right )}+\frac {4 a d^2}{\left (d+e x^2\right )^2}+\frac {8 a d}{d+e x^2}+\frac {b d \arcsin (c x)}{\left (\sqrt {d}-i \sqrt {e} x\right )^2}+\frac {5 b \sqrt {d} \arcsin (c x)}{\sqrt {d}-i \sqrt {e} x}+\frac {b d \arcsin (c x)}{\left (\sqrt {d}+i \sqrt {e} x\right )^2}+\frac {5 b \sqrt {d} \arcsin (c x)}{\sqrt {d}+i \sqrt {e} x}-\frac {5 b c \sqrt {d} \arctan \left (\frac {i \sqrt {e}+c^2 \sqrt {d} x}{\sqrt {c^2 d+e} \sqrt {1-c^2 x^2}}\right )}{\sqrt {c^2 d+e}}+\frac {5 i b c \sqrt {d} \text {arctanh}\left (\frac {\sqrt {e}+i c^2 \sqrt {d} x}{\sqrt {c^2 d+e} \sqrt {1-c^2 x^2}}\right )}{\sqrt {c^2 d+e}}-8 b \arcsin (c x) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}-\sqrt {c^2 d+e}}\right )-8 b \arcsin (c x) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{-c \sqrt {d}+\sqrt {c^2 d+e}}\right )-8 b \arcsin (c x) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )-8 b \arcsin (c x) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )+16 b \arcsin (c x) \log \left (1-e^{2 i \arcsin (c x)}\right )+16 a \log (x)-8 a \log \left (d+e x^2\right )-\frac {i b c^3 d^{3/2} \log \left (\frac {e \sqrt {c^2 d+e} \left (\sqrt {e}-i c^2 \sqrt {d} x+\sqrt {c^2 d+e} \sqrt {1-c^2 x^2}\right )}{c^3 \left (d+i \sqrt {d} \sqrt {e} x\right )}\right )}{\left (c^2 d+e\right )^{3/2}}+\frac {i b c^3 d^{3/2} \log \left (\frac {e \sqrt {c^2 d+e} \left (\sqrt {e}+i c^2 \sqrt {d} x+\sqrt {c^2 d+e} \sqrt {1-c^2 x^2}\right )}{c^3 \left (d-i \sqrt {d} \sqrt {e} x\right )}\right )}{\left (c^2 d+e\right )^{3/2}}+8 i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}-\sqrt {c^2 d+e}}\right )+8 i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{-c \sqrt {d}+\sqrt {c^2 d+e}}\right )+8 i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )+8 i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )-8 i b \operatorname {PolyLog}\left (2,e^{2 i \arcsin (c x)}\right )}{16 d^3} \]
(-((b*c*d*Sqrt[e]*Sqrt[1 - c^2*x^2])/((c^2*d + e)*((-I)*Sqrt[d] + Sqrt[e]* x))) - (b*c*d*Sqrt[e]*Sqrt[1 - c^2*x^2])/((c^2*d + e)*(I*Sqrt[d] + Sqrt[e] *x)) + (4*a*d^2)/(d + e*x^2)^2 + (8*a*d)/(d + e*x^2) + (b*d*ArcSin[c*x])/( Sqrt[d] - I*Sqrt[e]*x)^2 + (5*b*Sqrt[d]*ArcSin[c*x])/(Sqrt[d] - I*Sqrt[e]* x) + (b*d*ArcSin[c*x])/(Sqrt[d] + I*Sqrt[e]*x)^2 + (5*b*Sqrt[d]*ArcSin[c*x ])/(Sqrt[d] + I*Sqrt[e]*x) - (5*b*c*Sqrt[d]*ArcTan[(I*Sqrt[e] + c^2*Sqrt[d ]*x)/(Sqrt[c^2*d + e]*Sqrt[1 - c^2*x^2])])/Sqrt[c^2*d + e] + ((5*I)*b*c*Sq rt[d]*ArcTanh[(Sqrt[e] + I*c^2*Sqrt[d]*x)/(Sqrt[c^2*d + e]*Sqrt[1 - c^2*x^ 2])])/Sqrt[c^2*d + e] - 8*b*ArcSin[c*x]*Log[1 + (Sqrt[e]*E^(I*ArcSin[c*x]) )/(c*Sqrt[d] - Sqrt[c^2*d + e])] - 8*b*ArcSin[c*x]*Log[1 + (Sqrt[e]*E^(I*A rcSin[c*x]))/(-(c*Sqrt[d]) + Sqrt[c^2*d + e])] - 8*b*ArcSin[c*x]*Log[1 - ( Sqrt[e]*E^(I*ArcSin[c*x]))/(c*Sqrt[d] + Sqrt[c^2*d + e])] - 8*b*ArcSin[c*x ]*Log[1 + (Sqrt[e]*E^(I*ArcSin[c*x]))/(c*Sqrt[d] + Sqrt[c^2*d + e])] + 16* b*ArcSin[c*x]*Log[1 - E^((2*I)*ArcSin[c*x])] + 16*a*Log[x] - 8*a*Log[d + e *x^2] - (I*b*c^3*d^(3/2)*Log[(e*Sqrt[c^2*d + e]*(Sqrt[e] - I*c^2*Sqrt[d]*x + Sqrt[c^2*d + e]*Sqrt[1 - c^2*x^2]))/(c^3*(d + I*Sqrt[d]*Sqrt[e]*x))])/( c^2*d + e)^(3/2) + (I*b*c^3*d^(3/2)*Log[(e*Sqrt[c^2*d + e]*(Sqrt[e] + I*c^ 2*Sqrt[d]*x + Sqrt[c^2*d + e]*Sqrt[1 - c^2*x^2]))/(c^3*(d - I*Sqrt[d]*Sqrt [e]*x))])/(c^2*d + e)^(3/2) + (8*I)*b*PolyLog[2, (Sqrt[e]*E^(I*ArcSin[c*x] ))/(c*Sqrt[d] - Sqrt[c^2*d + e])] + (8*I)*b*PolyLog[2, (Sqrt[e]*E^(I*Ar...
Time = 1.58 (sec) , antiderivative size = 727, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {5232, 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.
\(\displaystyle \int \frac {a+b \arcsin (c x)}{x \left (d+e x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 5232 |
\(\displaystyle \int \left (-\frac {e x (a+b \arcsin (c x))}{d^3 \left (d+e x^2\right )}+\frac {a+b \arcsin (c x)}{d^3 x}-\frac {e x (a+b \arcsin (c x))}{d^2 \left (d+e x^2\right )^2}-\frac {e x (a+b \arcsin (c x))}{d \left (d+e x^2\right )^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {(a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d^3}-\frac {(a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d^3}-\frac {(a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d^3}-\frac {(a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d^3}+\frac {\log \left (1-e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{d^3}+\frac {a+b \arcsin (c x)}{2 d^2 \left (d+e x^2\right )}+\frac {a+b \arcsin (c x)}{4 d \left (d+e x^2\right )^2}+\frac {i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {d c^2+e}}\right )}{2 d^3}+\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {d c^2+e}}\right )}{2 d^3}+\frac {i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i \sqrt {-d} c+\sqrt {d c^2+e}}\right )}{2 d^3}+\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i \sqrt {-d} c+\sqrt {d c^2+e}}\right )}{2 d^3}-\frac {i b \operatorname {PolyLog}\left (2,e^{2 i \arcsin (c x)}\right )}{2 d^3}-\frac {b c \left (2 c^2 d+e\right ) \arctan \left (\frac {x \sqrt {c^2 d+e}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{8 d^{5/2} \left (c^2 d+e\right )^{3/2}}-\frac {b c \arctan \left (\frac {x \sqrt {c^2 d+e}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{2 d^{5/2} \sqrt {c^2 d+e}}-\frac {b c e x \sqrt {1-c^2 x^2}}{8 d^2 \left (c^2 d+e\right ) \left (d+e x^2\right )}\) |
-1/8*(b*c*e*x*Sqrt[1 - c^2*x^2])/(d^2*(c^2*d + e)*(d + e*x^2)) + (a + b*Ar cSin[c*x])/(4*d*(d + e*x^2)^2) + (a + b*ArcSin[c*x])/(2*d^2*(d + e*x^2)) - (b*c*ArcTan[(Sqrt[c^2*d + e]*x)/(Sqrt[d]*Sqrt[1 - c^2*x^2])])/(2*d^(5/2)* Sqrt[c^2*d + e]) - (b*c*(2*c^2*d + e)*ArcTan[(Sqrt[c^2*d + e]*x)/(Sqrt[d]* Sqrt[1 - c^2*x^2])])/(8*d^(5/2)*(c^2*d + e)^(3/2)) - ((a + b*ArcSin[c*x])* Log[1 - (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c^2*d + e])])/(2* d^3) - ((a + b*ArcSin[c*x])*Log[1 + (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[ -d] - Sqrt[c^2*d + e])])/(2*d^3) - ((a + b*ArcSin[c*x])*Log[1 - (Sqrt[e]*E ^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e])])/(2*d^3) - ((a + b*Arc Sin[c*x])*Log[1 + (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e])])/(2*d^3) + ((a + b*ArcSin[c*x])*Log[1 - E^((2*I)*ArcSin[c*x])])/d^3 + ((I/2)*b*PolyLog[2, -((Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c ^2*d + e]))])/d^3 + ((I/2)*b*PolyLog[2, (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*S qrt[-d] - Sqrt[c^2*d + e])])/d^3 + ((I/2)*b*PolyLog[2, -((Sqrt[e]*E^(I*Arc Sin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e]))])/d^3 + ((I/2)*b*PolyLog[2, ( Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e])])/d^3 - ((I/2) *b*PolyLog[2, E^((2*I)*ArcSin[c*x])])/d^3
3.7.44.3.1 Defintions of rubi rules used
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_.)*((d_) + (e_ .)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcSin[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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 2.10 (sec) , antiderivative size = 1130, normalized size of antiderivative = 1.55
method | result | size |
parts | \(\text {Expression too large to display}\) | \(1130\) |
derivativedivides | \(\text {Expression too large to display}\) | \(1174\) |
default | \(\text {Expression too large to display}\) | \(1174\) |
a/d^3*ln(x)+1/2*a/d^2/(e*x^2+d)-1/2*a/d^3*ln(e*x^2+d)+1/4*a/d/(e*x^2+d)^2+ b*(1/8*c^2*(6*c^4*d^2*arcsin(c*x)+4*arcsin(c*x)*c^4*d*e*x^2+I*c^4*d^2+2*I* c^4*d*e*x^2+I*e^2*c^4*x^4-(-c^2*x^2+1)^(1/2)*c^3*d*e*x-(-c^2*x^2+1)^(1/2)* e^2*c^3*x^3+6*c^2*d*e*arcsin(c*x)+4*arcsin(c*x)*e^2*c^2*x^2)/d^2/(c^2*e*x^ 2+c^2*d)^2/(c^2*d+e)+1/4*I/(c^2*d+e)*c^2/d^2*e*sum((_R1^2-1)/(_R1^2*e-2*c^ 2*d-e)*(I*arcsin(c*x)*ln((_R1-I*c*x-(-c^2*x^2+1)^(1/2))/_R1)+dilog((_R1-I* c*x-(-c^2*x^2+1)^(1/2))/_R1)),_R1=RootOf(e*_Z^4+(-4*c^2*d-2*e)*_Z^2+e))+I/ (c^2*d+e)*c^2/d^2*dilog(I*c*x+(-c^2*x^2+1)^(1/2))-I/(c^2*d+e)*c^2/d^2*dilo g(1+I*c*x+(-c^2*x^2+1)^(1/2))+1/(c^2*d+e)*c^2/d^2*arcsin(c*x)*ln(1+I*c*x+( -c^2*x^2+1)^(1/2))+1/4*I/(c^2*d+e)/d^3*e^2*sum((_R1^2-1)/(_R1^2*e-2*c^2*d- e)*(I*arcsin(c*x)*ln((_R1-I*c*x-(-c^2*x^2+1)^(1/2))/_R1)+dilog((_R1-I*c*x- (-c^2*x^2+1)^(1/2))/_R1)),_R1=RootOf(e*_Z^4+(-4*c^2*d-2*e)*_Z^2+e))+3/4*I* (d*c^2*(c^2*d+e))^(1/2)/(c^2*d+e)^2/d^2*c^2*arctanh(1/4*(2*e*(I*c*x+(-c^2* x^2+1)^(1/2))^2-4*c^2*d-2*e)/(c^4*d^2+c^2*d*e)^(1/2))+I/(c^2*d+e)/d^3*e*di log(I*c*x+(-c^2*x^2+1)^(1/2))-I/(c^2*d+e)/d^3*e*dilog(1+I*c*x+(-c^2*x^2+1) ^(1/2))+5/8*I*(d*c^2*(c^2*d+e))^(1/2)/(c^2*d+e)^2/d^3*arctanh(1/4*(2*e*(I* c*x+(-c^2*x^2+1)^(1/2))^2-4*c^2*d-2*e)/(c^4*d^2+c^2*d*e)^(1/2))*e+1/4*I/(c ^2*d+e)*c^2/d^2*sum((_R1^2*e-4*c^2*d-e)/(_R1^2*e-2*c^2*d-e)*(I*arcsin(c*x) *ln((_R1-I*c*x-(-c^2*x^2+1)^(1/2))/_R1)+dilog((_R1-I*c*x-(-c^2*x^2+1)^(1/2 ))/_R1)),_R1=RootOf(e*_Z^4+(-4*c^2*d-2*e)*_Z^2+e))+1/(c^2*d+e)/d^3*e*ar...
\[ \int \frac {a+b \arcsin (c x)}{x \left (d+e x^2\right )^3} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{3} x} \,d x } \]
\[ \int \frac {a+b \arcsin (c x)}{x \left (d+e x^2\right )^3} \, dx=\int \frac {a + b \operatorname {asin}{\left (c x \right )}}{x \left (d + e x^{2}\right )^{3}}\, dx \]
\[ \int \frac {a+b \arcsin (c x)}{x \left (d+e x^2\right )^3} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{3} x} \,d x } \]
1/4*a*((2*e*x^2 + 3*d)/(d^2*e^2*x^4 + 2*d^3*e*x^2 + d^4) - 2*log(e*x^2 + d )/d^3 + 4*log(x)/d^3) + b*integrate(arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/(e^3*x^7 + 3*d*e^2*x^5 + 3*d^2*e*x^3 + d^3*x), x)
Timed out. \[ \int \frac {a+b \arcsin (c x)}{x \left (d+e x^2\right )^3} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {a+b \arcsin (c x)}{x \left (d+e x^2\right )^3} \, dx=\int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{x\,{\left (e\,x^2+d\right )}^3} \,d x \]