3.196 \(\int \frac{(a+b \sin ^{-1}(c+d x))^2}{(c e+d e x)^4} \, dx\)

Optimal. Leaf size=187 \[ \frac{i b^2 \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c+d x)}\right )}{3 d e^4}-\frac{i b^2 \text{PolyLog}\left (2,e^{i \sin ^{-1}(c+d x)}\right )}{3 d e^4}-\frac{b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}-\frac{2 b \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e^4}-\frac{b^2}{3 d e^4 (c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.245836, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {4805, 12, 4627, 4701, 4709, 4183, 2279, 2391, 30} \[ \frac{i b^2 \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c+d x)}\right )}{3 d e^4}-\frac{i b^2 \text{PolyLog}\left (2,e^{i \sin ^{-1}(c+d x)}\right )}{3 d e^4}-\frac{b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}-\frac{2 b \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e^4}-\frac{b^2}{3 d e^4 (c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x])^2/(c*e + d*e*x)^4,x]

[Out]

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

Rule 4805

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 4627

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSi
n[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 4701

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(
(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n)/(d*f*(m + 1)), x] + (Dist[(c^2*(m + 2*p + 3))/(f^2*(m
+ 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^F
racPart[p])/(f*(m + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x
])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[m, -1] && Inte
gerQ[m]

Rule 4709

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[1/(c^(m
+ 1)*Sqrt[d]), Subst[Int[(a + b*x)^n*Sin[x]^m, x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2
*d + e, 0] && GtQ[d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 4183

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

Rule 2279

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 2391

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 2.04051, size = 246, normalized size = 1.32 \[ -\frac{-4 i b^2 (c+d x)^3 \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c+d x)}\right )+b^2 \left (4 i (c+d x)^3 \text{PolyLog}\left (2,e^{i \sin ^{-1}(c+d x)}\right )+4 (c+d x)^2+4 \sin ^{-1}(c+d x)^2+\sin ^{-1}(c+d x) \left (2 \sin \left (2 \sin ^{-1}(c+d x)\right )+\left (\sin \left (3 \sin ^{-1}(c+d x)\right )-3 (c+d x)\right ) \left (\log \left (1-e^{i \sin ^{-1}(c+d x)}\right )-\log \left (1+e^{i \sin ^{-1}(c+d x)}\right )\right )\right )\right )+4 a^2+8 a b \sin ^{-1}(c+d x)+2 a b \sin \left (2 \sin ^{-1}(c+d x)\right )+a b \left (3 (c+d x)-\sin \left (3 \sin ^{-1}(c+d x)\right )\right ) \left (\log \left (\cos \left (\frac{1}{2} \sin ^{-1}(c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} \sin ^{-1}(c+d x)\right )\right )\right )}{12 d e^4 (c+d x)^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSin[c + d*x])^2/(c*e + d*e*x)^4,x]

[Out]

-(4*a^2 + 8*a*b*ArcSin[c + d*x] - (4*I)*b^2*(c + d*x)^3*PolyLog[2, -E^(I*ArcSin[c + d*x])] + 2*a*b*Sin[2*ArcSi
n[c + d*x]] + a*b*(Log[Cos[ArcSin[c + d*x]/2]] - Log[Sin[ArcSin[c + d*x]/2]])*(3*(c + d*x) - Sin[3*ArcSin[c +
d*x]]) + b^2*(4*(c + d*x)^2 + 4*ArcSin[c + d*x]^2 + (4*I)*(c + d*x)^3*PolyLog[2, E^(I*ArcSin[c + d*x])] + ArcS
in[c + d*x]*(2*Sin[2*ArcSin[c + d*x]] + (Log[1 - E^(I*ArcSin[c + d*x])] - Log[1 + E^(I*ArcSin[c + d*x])])*(-3*
(c + d*x) + Sin[3*ArcSin[c + d*x]]))))/(12*d*e^4*(c + d*x)^3)

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Maple [A]  time = 0.129, size = 336, normalized size = 1.8 \begin{align*} -{\frac{{a}^{2}}{3\,d{e}^{4} \left ( dx+c \right ) ^{3}}}-{\frac{{b}^{2}\arcsin \left ( dx+c \right ) }{3\,d{e}^{4} \left ( dx+c \right ) ^{2}}\sqrt{1- \left ( dx+c \right ) ^{2}}}-{\frac{{b}^{2} \left ( \arcsin \left ( dx+c \right ) \right ) ^{2}}{3\,d{e}^{4} \left ( dx+c \right ) ^{3}}}-{\frac{{b}^{2}}{3\,d{e}^{4} \left ( dx+c \right ) }}-{\frac{{b}^{2}\arcsin \left ( dx+c \right ) }{3\,d{e}^{4}}\ln \left ( 1+i \left ( dx+c \right ) +\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }+{\frac{{\frac{i}{3}}{b}^{2}}{d{e}^{4}}{\it polylog} \left ( 2,-i \left ( dx+c \right ) -\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }+{\frac{{b}^{2}\arcsin \left ( dx+c \right ) }{3\,d{e}^{4}}\ln \left ( 1-i \left ( dx+c \right ) -\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }-{\frac{{\frac{i}{3}}{b}^{2}}{d{e}^{4}}{\it polylog} \left ( 2,i \left ( dx+c \right ) +\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }-{\frac{2\,ab\arcsin \left ( dx+c \right ) }{3\,d{e}^{4} \left ( dx+c \right ) ^{3}}}-{\frac{ab}{3\,d{e}^{4} \left ( dx+c \right ) ^{2}}\sqrt{1- \left ( dx+c \right ) ^{2}}}-{\frac{ab}{3\,d{e}^{4}}{\it Artanh} \left ({\frac{1}{\sqrt{1- \left ( dx+c \right ) ^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/d*a^2/e^4/(d*x+c)^3-1/3/d*b^2/e^4/(d*x+c)^2*arcsin(d*x+c)*(1-(d*x+c)^2)^(1/2)-1/3/d*b^2/e^4/(d*x+c)^3*arc
sin(d*x+c)^2-1/3*b^2/d/e^4/(d*x+c)-1/3/d*b^2/e^4*arcsin(d*x+c)*ln(1+I*(d*x+c)+(1-(d*x+c)^2)^(1/2))+1/3*I*b^2*p
olylog(2,-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))/d/e^4+1/3/d*b^2/e^4*arcsin(d*x+c)*ln(1-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))
-1/3*I*b^2*polylog(2,I*(d*x+c)+(1-(d*x+c)^2)^(1/2))/d/e^4-2/3/d*a*b/e^4/(d*x+c)^3*arcsin(d*x+c)-1/3/d*a*b/e^4/
(d*x+c)^2*(1-(d*x+c)^2)^(1/2)-1/3/d*a*b/e^4*arctanh(1/(1-(d*x+c)^2)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{a^{2}}{3 \,{\left (d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}\right )}} - \frac{b^{2} \arctan \left (d x + c, \sqrt{d x + c + 1} \sqrt{-d x - c + 1}\right )^{2} + 2 \,{\left (d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}\right )} \int \frac{{\left (b^{2} d x + b^{2} c\right )} \sqrt{d x + c + 1} \sqrt{-d x - c + 1} \arctan \left (d x + c, \sqrt{d x + c + 1} \sqrt{-d x - c + 1}\right ) - 3 \,{\left (a b d^{2} x^{2} + 2 \, a b c d x + a b c^{2} - a b\right )} \arctan \left (d x + c, \sqrt{d x + c + 1} \sqrt{-d x - c + 1}\right )}{d^{6} e^{4} x^{6} + 6 \, c d^{5} e^{4} x^{5} +{\left (15 \, c^{2} - 1\right )} d^{4} e^{4} x^{4} + 4 \,{\left (5 \, c^{3} - c\right )} d^{3} e^{4} x^{3} + 3 \,{\left (5 \, c^{4} - 2 \, c^{2}\right )} d^{2} e^{4} x^{2} + 2 \,{\left (3 \, c^{5} - 2 \, c^{3}\right )} d e^{4} x +{\left (c^{6} - c^{4}\right )} e^{4}}\,{d x}}{3 \,{\left (d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^4,x, algorithm="maxima")

[Out]

-1/3*a^2/(d^4*e^4*x^3 + 3*c*d^3*e^4*x^2 + 3*c^2*d^2*e^4*x + c^3*d*e^4) - 1/3*(b^2*arctan2(d*x + c, sqrt(d*x +
c + 1)*sqrt(-d*x - c + 1))^2 + 3*(d^4*e^4*x^3 + 3*c*d^3*e^4*x^2 + 3*c^2*d^2*e^4*x + c^3*d*e^4)*integrate(2/3*(
(b^2*d*x + b^2*c)*sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1))
- 3*(a*b*d^2*x^2 + 2*a*b*c*d*x + a*b*c^2 - a*b)*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)))/(d^6*e
^4*x^6 + 6*c*d^5*e^4*x^5 + (15*c^2 - 1)*d^4*e^4*x^4 + 4*(5*c^3 - c)*d^3*e^4*x^3 + 3*(5*c^4 - 2*c^2)*d^2*e^4*x^
2 + 2*(3*c^5 - 2*c^3)*d*e^4*x + (c^6 - c^4)*e^4), x))/(d^4*e^4*x^3 + 3*c*d^3*e^4*x^2 + 3*c^2*d^2*e^4*x + c^3*d
*e^4)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \arcsin \left (d x + c\right )^{2} + 2 \, a b \arcsin \left (d x + c\right ) + a^{2}}{d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^4,x, algorithm="fricas")

[Out]

integral((b^2*arcsin(d*x + c)^2 + 2*a*b*arcsin(d*x + c) + a^2)/(d^4*e^4*x^4 + 4*c*d^3*e^4*x^3 + 6*c^2*d^2*e^4*
x^2 + 4*c^3*d*e^4*x + c^4*e^4), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{2}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac{b^{2} \operatorname{asin}^{2}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac{2 a b \operatorname{asin}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx}{e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(d*x+c))**2/(d*e*x+c*e)**4,x)

[Out]

(Integral(a**2/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(b**2*asin(c +
d*x)**2/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(2*a*b*asin(c + d*x)
/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x))/e**4

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^4,x, algorithm="giac")

[Out]

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