3.7.10 \(\int \frac {(d+e x^2)^2 (a+b \text {ArcSin}(c x))}{x} \, dx\) [610]

Optimal. Leaf size=229 \[ \frac {b d e x \sqrt {1-c^2 x^2}}{2 c}+\frac {3 b e^2 x \sqrt {1-c^2 x^2}}{32 c^3}+\frac {b e^2 x^3 \sqrt {1-c^2 x^2}}{16 c}-\frac {b d e \text {ArcSin}(c x)}{2 c^2}-\frac {3 b e^2 \text {ArcSin}(c x)}{32 c^4}-\frac {1}{2} i b d^2 \text {ArcSin}(c x)^2+d e x^2 (a+b \text {ArcSin}(c x))+\frac {1}{4} e^2 x^4 (a+b \text {ArcSin}(c x))+b d^2 \text {ArcSin}(c x) \log \left (1-e^{2 i \text {ArcSin}(c x)}\right )-b d^2 \text {ArcSin}(c x) \log (x)+d^2 (a+b \text {ArcSin}(c x)) \log (x)-\frac {1}{2} i b d^2 \text {PolyLog}\left (2,e^{2 i \text {ArcSin}(c x)}\right ) \]

[Out]

-1/2*b*d*e*arcsin(c*x)/c^2-3/32*b*e^2*arcsin(c*x)/c^4-1/2*I*b*d^2*arcsin(c*x)^2+d*e*x^2*(a+b*arcsin(c*x))+1/4*
e^2*x^4*(a+b*arcsin(c*x))+b*d^2*arcsin(c*x)*ln(1-(I*c*x+(-c^2*x^2+1)^(1/2))^2)-b*d^2*arcsin(c*x)*ln(x)+d^2*(a+
b*arcsin(c*x))*ln(x)-1/2*I*b*d^2*polylog(2,(I*c*x+(-c^2*x^2+1)^(1/2))^2)+1/2*b*d*e*x*(-c^2*x^2+1)^(1/2)/c+3/32
*b*e^2*x*(-c^2*x^2+1)^(1/2)/c^3+1/16*b*e^2*x^3*(-c^2*x^2+1)^(1/2)/c

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Rubi [A]
time = 0.24, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 12, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {272, 45, 4815, 6874, 327, 222, 2363, 4721, 3798, 2221, 2317, 2438} \begin {gather*} d^2 \log (x) (a+b \text {ArcSin}(c x))+d e x^2 (a+b \text {ArcSin}(c x))+\frac {1}{4} e^2 x^4 (a+b \text {ArcSin}(c x))-\frac {3 b e^2 \text {ArcSin}(c x)}{32 c^4}-\frac {b d e \text {ArcSin}(c x)}{2 c^2}-\frac {1}{2} i b d^2 \text {Li}_2\left (e^{2 i \text {ArcSin}(c x)}\right )-\frac {1}{2} i b d^2 \text {ArcSin}(c x)^2+b d^2 \text {ArcSin}(c x) \log \left (1-e^{2 i \text {ArcSin}(c x)}\right )-b d^2 \log (x) \text {ArcSin}(c x)+\frac {b d e x \sqrt {1-c^2 x^2}}{2 c}+\frac {b e^2 x^3 \sqrt {1-c^2 x^2}}{16 c}+\frac {3 b e^2 x \sqrt {1-c^2 x^2}}{32 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)^2*(a + b*ArcSin[c*x]))/x,x]

[Out]

(b*d*e*x*Sqrt[1 - c^2*x^2])/(2*c) + (3*b*e^2*x*Sqrt[1 - c^2*x^2])/(32*c^3) + (b*e^2*x^3*Sqrt[1 - c^2*x^2])/(16
*c) - (b*d*e*ArcSin[c*x])/(2*c^2) - (3*b*e^2*ArcSin[c*x])/(32*c^4) - (I/2)*b*d^2*ArcSin[c*x]^2 + d*e*x^2*(a +
b*ArcSin[c*x]) + (e^2*x^4*(a + b*ArcSin[c*x]))/4 + b*d^2*ArcSin[c*x]*Log[1 - E^((2*I)*ArcSin[c*x])] - b*d^2*Ar
cSin[c*x]*Log[x] + d^2*(a + b*ArcSin[c*x])*Log[x] - (I/2)*b*d^2*PolyLog[2, E^((2*I)*ArcSin[c*x])]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2363

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-e, 2]*(x/Sqr
t[d])]*((a + b*Log[c*x^n])/Rt[-e, 2]), x] - Dist[b*(n/Rt[-e, 2]), Int[ArcSin[Rt[-e, 2]*(x/Sqrt[d])]/x, x], x]
/; FreeQ[{a, b, c, d, e, n}, x] && GtQ[d, 0] && NegQ[e]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3798

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[I*((c + d*x)^(m + 1)/(d*(
m + 1))), x] - Dist[2*I, Int[(c + d*x)^m*E^(2*I*k*Pi)*(E^(2*I*(e + f*x))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x))))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 4721

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n*Cot[x], x], x, ArcSin[c*
x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 4815

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSin[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1 -
c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[c^2*d + e, 0] && IntegerQ[p] && (GtQ[p, 0] ||
(IGtQ[(m - 1)/2, 0] && LeQ[m + p, 0]))

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {\left (d+e x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{x} \, dx &=d e x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \sin ^{-1}(c x)\right )+d^2 \left (a+b \sin ^{-1}(c x)\right ) \log (x)-(b c) \int \frac {d e x^2+\frac {e^2 x^4}{4}+d^2 \log (x)}{\sqrt {1-c^2 x^2}} \, dx\\ &=d e x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \sin ^{-1}(c x)\right )+d^2 \left (a+b \sin ^{-1}(c x)\right ) \log (x)-(b c) \int \left (\frac {d e x^2}{\sqrt {1-c^2 x^2}}+\frac {e^2 x^4}{4 \sqrt {1-c^2 x^2}}+\frac {d^2 \log (x)}{\sqrt {1-c^2 x^2}}\right ) \, dx\\ &=d e x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \sin ^{-1}(c x)\right )+d^2 \left (a+b \sin ^{-1}(c x)\right ) \log (x)-\left (b c d^2\right ) \int \frac {\log (x)}{\sqrt {1-c^2 x^2}} \, dx-(b c d e) \int \frac {x^2}{\sqrt {1-c^2 x^2}} \, dx-\frac {1}{4} \left (b c e^2\right ) \int \frac {x^4}{\sqrt {1-c^2 x^2}} \, dx\\ &=\frac {b d e x \sqrt {1-c^2 x^2}}{2 c}+\frac {b e^2 x^3 \sqrt {1-c^2 x^2}}{16 c}+d e x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \sin ^{-1}(c x)\right )-b d^2 \sin ^{-1}(c x) \log (x)+d^2 \left (a+b \sin ^{-1}(c x)\right ) \log (x)+\left (b d^2\right ) \int \frac {\sin ^{-1}(c x)}{x} \, dx-\frac {(b d e) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{2 c}-\frac {\left (3 b e^2\right ) \int \frac {x^2}{\sqrt {1-c^2 x^2}} \, dx}{16 c}\\ &=\frac {b d e x \sqrt {1-c^2 x^2}}{2 c}+\frac {3 b e^2 x \sqrt {1-c^2 x^2}}{32 c^3}+\frac {b e^2 x^3 \sqrt {1-c^2 x^2}}{16 c}-\frac {b d e \sin ^{-1}(c x)}{2 c^2}+d e x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \sin ^{-1}(c x)\right )-b d^2 \sin ^{-1}(c x) \log (x)+d^2 \left (a+b \sin ^{-1}(c x)\right ) \log (x)+\left (b d^2\right ) \text {Subst}\left (\int x \cot (x) \, dx,x,\sin ^{-1}(c x)\right )-\frac {\left (3 b e^2\right ) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{32 c^3}\\ &=\frac {b d e x \sqrt {1-c^2 x^2}}{2 c}+\frac {3 b e^2 x \sqrt {1-c^2 x^2}}{32 c^3}+\frac {b e^2 x^3 \sqrt {1-c^2 x^2}}{16 c}-\frac {b d e \sin ^{-1}(c x)}{2 c^2}-\frac {3 b e^2 \sin ^{-1}(c x)}{32 c^4}-\frac {1}{2} i b d^2 \sin ^{-1}(c x)^2+d e x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \sin ^{-1}(c x)\right )-b d^2 \sin ^{-1}(c x) \log (x)+d^2 \left (a+b \sin ^{-1}(c x)\right ) \log (x)-\left (2 i b d^2\right ) \text {Subst}\left (\int \frac {e^{2 i x} x}{1-e^{2 i x}} \, dx,x,\sin ^{-1}(c x)\right )\\ &=\frac {b d e x \sqrt {1-c^2 x^2}}{2 c}+\frac {3 b e^2 x \sqrt {1-c^2 x^2}}{32 c^3}+\frac {b e^2 x^3 \sqrt {1-c^2 x^2}}{16 c}-\frac {b d e \sin ^{-1}(c x)}{2 c^2}-\frac {3 b e^2 \sin ^{-1}(c x)}{32 c^4}-\frac {1}{2} i b d^2 \sin ^{-1}(c x)^2+d e x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \sin ^{-1}(c x)\right )+b d^2 \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-b d^2 \sin ^{-1}(c x) \log (x)+d^2 \left (a+b \sin ^{-1}(c x)\right ) \log (x)-\left (b d^2\right ) \text {Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )\\ &=\frac {b d e x \sqrt {1-c^2 x^2}}{2 c}+\frac {3 b e^2 x \sqrt {1-c^2 x^2}}{32 c^3}+\frac {b e^2 x^3 \sqrt {1-c^2 x^2}}{16 c}-\frac {b d e \sin ^{-1}(c x)}{2 c^2}-\frac {3 b e^2 \sin ^{-1}(c x)}{32 c^4}-\frac {1}{2} i b d^2 \sin ^{-1}(c x)^2+d e x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \sin ^{-1}(c x)\right )+b d^2 \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-b d^2 \sin ^{-1}(c x) \log (x)+d^2 \left (a+b \sin ^{-1}(c x)\right ) \log (x)+\frac {1}{2} \left (i b d^2\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )\\ &=\frac {b d e x \sqrt {1-c^2 x^2}}{2 c}+\frac {3 b e^2 x \sqrt {1-c^2 x^2}}{32 c^3}+\frac {b e^2 x^3 \sqrt {1-c^2 x^2}}{16 c}-\frac {b d e \sin ^{-1}(c x)}{2 c^2}-\frac {3 b e^2 \sin ^{-1}(c x)}{32 c^4}-\frac {1}{2} i b d^2 \sin ^{-1}(c x)^2+d e x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \sin ^{-1}(c x)\right )+b d^2 \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-b d^2 \sin ^{-1}(c x) \log (x)+d^2 \left (a+b \sin ^{-1}(c x)\right ) \log (x)-\frac {1}{2} i b d^2 \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 218, normalized size = 0.95 \begin {gather*} \frac {1}{4} \left (4 a d e x^2+a e^2 x^4+4 b d e x^2 \text {ArcSin}(c x)+b e^2 x^4 \text {ArcSin}(c x)+\frac {b e^2 \left (c x \sqrt {1-c^2 x^2} \left (3+2 c^2 x^2\right )-6 \text {ArcTan}\left (\frac {c x}{-1+\sqrt {1-c^2 x^2}}\right )\right )}{8 c^4}+\frac {2 b d e \left (c x \sqrt {1-c^2 x^2}-2 \text {ArcTan}\left (\frac {c x}{-1+\sqrt {1-c^2 x^2}}\right )\right )}{c^2}+4 b d^2 \text {ArcSin}(c x) \log \left (1-e^{2 i \text {ArcSin}(c x)}\right )+4 a d^2 \log (x)-2 i b d^2 \left (\text {ArcSin}(c x)^2+\text {PolyLog}\left (2,e^{2 i \text {ArcSin}(c x)}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)^2*(a + b*ArcSin[c*x]))/x,x]

[Out]

(4*a*d*e*x^2 + a*e^2*x^4 + 4*b*d*e*x^2*ArcSin[c*x] + b*e^2*x^4*ArcSin[c*x] + (b*e^2*(c*x*Sqrt[1 - c^2*x^2]*(3
+ 2*c^2*x^2) - 6*ArcTan[(c*x)/(-1 + Sqrt[1 - c^2*x^2])]))/(8*c^4) + (2*b*d*e*(c*x*Sqrt[1 - c^2*x^2] - 2*ArcTan
[(c*x)/(-1 + Sqrt[1 - c^2*x^2])]))/c^2 + 4*b*d^2*ArcSin[c*x]*Log[1 - E^((2*I)*ArcSin[c*x])] + 4*a*d^2*Log[x] -
(2*I)*b*d^2*(ArcSin[c*x]^2 + PolyLog[2, E^((2*I)*ArcSin[c*x])]))/4

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Maple [A]
time = 0.31, size = 262, normalized size = 1.14

method	result	size
derivativedivides	\(a d e \,x^{2}+\frac {a \,e^{2} x^{4}}{4}+a \,d^{2} \ln \left (c x \right )-\frac {i b \,d^{2} \arcsin \left (c x \right )^{2}}{2}+b \,d^{2} \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )+b \,d^{2} \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-i b \,d^{2} \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )-i b \,d^{2} \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )+\frac {b \arcsin \left (c x \right ) e^{2} \cos \left (4 \arcsin \left (c x \right )\right )}{32 c^{4}}-\frac {b \,e^{2} \sin \left (4 \arcsin \left (c x \right )\right )}{128 c^{4}}-\frac {b \arcsin \left (c x \right ) \cos \left (2 \arcsin \left (c x \right )\right ) d e}{2 c^{2}}-\frac {b \arcsin \left (c x \right ) \cos \left (2 \arcsin \left (c x \right )\right ) e^{2}}{8 c^{4}}+\frac {b \sin \left (2 \arcsin \left (c x \right )\right ) d e}{4 c^{2}}+\frac {b \sin \left (2 \arcsin \left (c x \right )\right ) e^{2}}{16 c^{4}}\)	\(262\)
default	\(a d e \,x^{2}+\frac {a \,e^{2} x^{4}}{4}+a \,d^{2} \ln \left (c x \right )-\frac {i b \,d^{2} \arcsin \left (c x \right )^{2}}{2}+b \,d^{2} \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )+b \,d^{2} \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-i b \,d^{2} \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )-i b \,d^{2} \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )+\frac {b \arcsin \left (c x \right ) e^{2} \cos \left (4 \arcsin \left (c x \right )\right )}{32 c^{4}}-\frac {b \,e^{2} \sin \left (4 \arcsin \left (c x \right )\right )}{128 c^{4}}-\frac {b \arcsin \left (c x \right ) \cos \left (2 \arcsin \left (c x \right )\right ) d e}{2 c^{2}}-\frac {b \arcsin \left (c x \right ) \cos \left (2 \arcsin \left (c x \right )\right ) e^{2}}{8 c^{4}}+\frac {b \sin \left (2 \arcsin \left (c x \right )\right ) d e}{4 c^{2}}+\frac {b \sin \left (2 \arcsin \left (c x \right )\right ) e^{2}}{16 c^{4}}\)	\(262\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^2*(a+b*arcsin(c*x))/x,x,method=_RETURNVERBOSE)

[Out]

a*d*e*x^2+1/4*a*e^2*x^4+a*d^2*ln(c*x)-1/2*I*b*d^2*arcsin(c*x)^2+b*d^2*arcsin(c*x)*ln(1-I*c*x-(-c^2*x^2+1)^(1/2
))+b*d^2*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-I*b*d^2*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))-I*b*d^2*polyl
og(2,I*c*x+(-c^2*x^2+1)^(1/2))+1/32*b/c^4*arcsin(c*x)*e^2*cos(4*arcsin(c*x))-1/128*b/c^4*e^2*sin(4*arcsin(c*x)
)-1/2*b/c^2*arcsin(c*x)*cos(2*arcsin(c*x))*d*e-1/8*b/c^4*arcsin(c*x)*cos(2*arcsin(c*x))*e^2+1/4*b/c^2*sin(2*ar
csin(c*x))*d*e+1/16*b/c^4*sin(2*arcsin(c*x))*e^2

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^2*(a+b*arcsin(c*x))/x,x, algorithm="maxima")

[Out]

1/4*a*x^4*e^2 + a*d*x^2*e + a*d^2*log(x) + integrate((b*x^4*e^2 + 2*b*d*x^2*e + b*d^2)*arctan2(c*x, sqrt(c*x +
1)*sqrt(-c*x + 1))/x, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^2*(a+b*arcsin(c*x))/x,x, algorithm="fricas")

[Out]

integral((a*x^4*e^2 + 2*a*d*x^2*e + a*d^2 + (b*x^4*e^2 + 2*b*d*x^2*e + b*d^2)*arcsin(c*x))/x, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{2}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**2*(a+b*asin(c*x))/x,x)

[Out]

Integral((a + b*asin(c*x))*(d + e*x**2)**2/x, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^2*(a+b*arcsin(c*x))/x,x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^2*(b*arcsin(c*x) + a)/x, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^2}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*asin(c*x))*(d + e*x^2)^2)/x,x)

[Out]

int(((a + b*asin(c*x))*(d + e*x^2)^2)/x, x)

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