3.3.22 \(\int \frac {(c e+d e x)^3}{(a+b \text {ArcSin}(c+d x))^2} \, dx\) [222]

Optimal. Leaf size=190 \[ -\frac {e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{b d (a+b \text {ArcSin}(c+d x))}+\frac {e^3 \cos \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 (a+b \text {ArcSin}(c+d x))}{b}\right )}{2 b^2 d}-\frac {e^3 \cos \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (\frac {4 (a+b \text {ArcSin}(c+d x))}{b}\right )}{2 b^2 d}+\frac {e^3 \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \text {ArcSin}(c+d x))}{b}\right )}{2 b^2 d}-\frac {e^3 \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \text {ArcSin}(c+d x))}{b}\right )}{2 b^2 d} \]

[Out]

1/2*e^3*Ci(2*(a+b*arcsin(d*x+c))/b)*cos(2*a/b)/b^2/d-1/2*e^3*Ci(4*(a+b*arcsin(d*x+c))/b)*cos(4*a/b)/b^2/d+1/2*
e^3*Si(2*(a+b*arcsin(d*x+c))/b)*sin(2*a/b)/b^2/d-1/2*e^3*Si(4*(a+b*arcsin(d*x+c))/b)*sin(4*a/b)/b^2/d-e^3*(d*x
+c)^3*(1-(d*x+c)^2)^(1/2)/b/d/(a+b*arcsin(d*x+c))

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Rubi [A]
time = 0.19, antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4889, 12, 4727, 3384, 3380, 3383} \begin {gather*} \frac {e^3 \cos \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 (a+b \text {ArcSin}(c+d x))}{b}\right )}{2 b^2 d}-\frac {e^3 \cos \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (\frac {4 (a+b \text {ArcSin}(c+d x))}{b}\right )}{2 b^2 d}+\frac {e^3 \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \text {ArcSin}(c+d x))}{b}\right )}{2 b^2 d}-\frac {e^3 \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \text {ArcSin}(c+d x))}{b}\right )}{2 b^2 d}-\frac {e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{b d (a+b \text {ArcSin}(c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^3/(a + b*ArcSin[c + d*x])^2,x]

[Out]

-((e^3*(c + d*x)^3*Sqrt[1 - (c + d*x)^2])/(b*d*(a + b*ArcSin[c + d*x]))) + (e^3*Cos[(2*a)/b]*CosIntegral[(2*(a
+ b*ArcSin[c + d*x]))/b])/(2*b^2*d) - (e^3*Cos[(4*a)/b]*CosIntegral[(4*(a + b*ArcSin[c + d*x]))/b])/(2*b^2*d)
+ (e^3*Sin[(2*a)/b]*SinIntegral[(2*(a + b*ArcSin[c + d*x]))/b])/(2*b^2*d) - (e^3*Sin[(4*a)/b]*SinIntegral[(4*
(a + b*ArcSin[c + d*x]))/b])/(2*b^2*d)

Rule 12

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

Rule 3380

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3383

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3384

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 4727

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

Rule 4889

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]

Rubi steps

\begin {align*} \int \frac {(c e+d e x)^3}{\left (a+b \sin ^{-1}(c+d x)\right )^2} \, dx &=\frac {\text {Subst}\left (\int \frac {e^3 x^3}{\left (a+b \sin ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 \text {Subst}\left (\int \frac {x^3}{\left (a+b \sin ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{d}\\ &=-\frac {e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {e^3 \text {Subst}\left (\int \left (\frac {\cos (2 x)}{2 (a+b x)}-\frac {\cos (4 x)}{2 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{b d}\\ &=-\frac {e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {e^3 \text {Subst}\left (\int \frac {\cos (2 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b d}-\frac {e^3 \text {Subst}\left (\int \frac {\cos (4 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b d}\\ &=-\frac {e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {\left (e^3 \cos \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b d}-\frac {\left (e^3 \cos \left (\frac {4 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b d}+\frac {\left (e^3 \sin \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b d}-\frac {\left (e^3 \sin \left (\frac {4 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b d}\\ &=-\frac {e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {e^3 \cos \left (\frac {2 a}{b}\right ) \text {Ci}\left (\frac {2 a}{b}+2 \sin ^{-1}(c+d x)\right )}{2 b^2 d}-\frac {e^3 \cos \left (\frac {4 a}{b}\right ) \text {Ci}\left (\frac {4 a}{b}+4 \sin ^{-1}(c+d x)\right )}{2 b^2 d}+\frac {e^3 \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sin ^{-1}(c+d x)\right )}{2 b^2 d}-\frac {e^3 \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 a}{b}+4 \sin ^{-1}(c+d x)\right )}{2 b^2 d}\\ \end {align*}

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Mathematica [A]
time = 0.62, size = 220, normalized size = 1.16 \begin {gather*} -\frac {e^3 \left (\frac {2 b (c+d x)^3 \sqrt {1-(c+d x)^2}}{a+b \text {ArcSin}(c+d x)}-4 \cos \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (2 \left (\frac {a}{b}+\text {ArcSin}(c+d x)\right )\right )+\cos \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (4 \left (\frac {a}{b}+\text {ArcSin}(c+d x)\right )\right )+3 \log (a+b \text {ArcSin}(c+d x))-4 \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\text {ArcSin}(c+d x)\right )\right )+3 \left (\cos \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (2 \left (\frac {a}{b}+\text {ArcSin}(c+d x)\right )\right )-\log (a+b \text {ArcSin}(c+d x))+\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\text {ArcSin}(c+d x)\right )\right )\right )+\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\text {ArcSin}(c+d x)\right )\right )\right )}{2 b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^3/(a + b*ArcSin[c + d*x])^2,x]

[Out]

-1/2*(e^3*((2*b*(c + d*x)^3*Sqrt[1 - (c + d*x)^2])/(a + b*ArcSin[c + d*x]) - 4*Cos[(2*a)/b]*CosIntegral[2*(a/b
+ ArcSin[c + d*x])] + Cos[(4*a)/b]*CosIntegral[4*(a/b + ArcSin[c + d*x])] + 3*Log[a + b*ArcSin[c + d*x]] - 4*
Sin[(2*a)/b]*SinIntegral[2*(a/b + ArcSin[c + d*x])] + 3*(Cos[(2*a)/b]*CosIntegral[2*(a/b + ArcSin[c + d*x])] -
Log[a + b*ArcSin[c + d*x]] + Sin[(2*a)/b]*SinIntegral[2*(a/b + ArcSin[c + d*x])]) + Sin[(4*a)/b]*SinIntegral[
4*(a/b + ArcSin[c + d*x])]))/(b^2*d)

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Maple [A]
time = 0.13, size = 280, normalized size = 1.47

method	result	size
derivativedivides	\(\frac {e^{3} \left (4 \arcsin \left (d x +c \right ) \sinIntegral \left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) b +4 \arcsin \left (d x +c \right ) \cosineIntegral \left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) b -4 \arcsin \left (d x +c \right ) \sinIntegral \left (4 \arcsin \left (d x +c \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) b -4 \arcsin \left (d x +c \right ) \cosineIntegral \left (4 \arcsin \left (d x +c \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) b +4 \sinIntegral \left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) a +4 \cosineIntegral \left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) a -4 \sinIntegral \left (4 \arcsin \left (d x +c \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) a -4 \cosineIntegral \left (4 \arcsin \left (d x +c \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) a -2 \sin \left (2 \arcsin \left (d x +c \right )\right ) b +\sin \left (4 \arcsin \left (d x +c \right )\right ) b \right )}{8 d \left (a +b \arcsin \left (d x +c \right )\right ) b^{2}}\)	\(280\)
default	\(\frac {e^{3} \left (4 \arcsin \left (d x +c \right ) \sinIntegral \left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) b +4 \arcsin \left (d x +c \right ) \cosineIntegral \left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) b -4 \arcsin \left (d x +c \right ) \sinIntegral \left (4 \arcsin \left (d x +c \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) b -4 \arcsin \left (d x +c \right ) \cosineIntegral \left (4 \arcsin \left (d x +c \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) b +4 \sinIntegral \left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) a +4 \cosineIntegral \left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) a -4 \sinIntegral \left (4 \arcsin \left (d x +c \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) a -4 \cosineIntegral \left (4 \arcsin \left (d x +c \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) a -2 \sin \left (2 \arcsin \left (d x +c \right )\right ) b +\sin \left (4 \arcsin \left (d x +c \right )\right ) b \right )}{8 d \left (a +b \arcsin \left (d x +c \right )\right ) b^{2}}\)	\(280\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^3/(a+b*arcsin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/8/d*e^3*(4*arcsin(d*x+c)*Si(2*arcsin(d*x+c)+2*a/b)*sin(2*a/b)*b+4*arcsin(d*x+c)*Ci(2*arcsin(d*x+c)+2*a/b)*co
s(2*a/b)*b-4*arcsin(d*x+c)*Si(4*arcsin(d*x+c)+4*a/b)*sin(4*a/b)*b-4*arcsin(d*x+c)*Ci(4*arcsin(d*x+c)+4*a/b)*co
s(4*a/b)*b+4*Si(2*arcsin(d*x+c)+2*a/b)*sin(2*a/b)*a+4*Ci(2*arcsin(d*x+c)+2*a/b)*cos(2*a/b)*a-4*Si(4*arcsin(d*x
+c)+4*a/b)*sin(4*a/b)*a-4*Ci(4*arcsin(d*x+c)+4*a/b)*cos(4*a/b)*a-2*sin(2*arcsin(d*x+c))*b+sin(4*arcsin(d*x+c))
*b)/(a+b*arcsin(d*x+c))/b^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((d*e*x+c*e)^3/(a+b*arcsin(d*x+c))^2,x, algorithm="maxima")

[Out]

-((d^3*x^3*e^3 + 3*c*d^2*x^2*e^3 + 3*c^2*d*x*e^3 + c^3*e^3)*sqrt(d*x + c + 1)*sqrt(-d*x - c + 1) - (b^2*d*arct
an2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + a*b*d)*integrate((4*d^4*x^4*e^3 + 16*c*d^3*x^3*e^3 + 3*(8
*c^2*e^3 - e^3)*d^2*x^2 + 4*c^4*e^3 + 2*(8*c^3*e^3 - 3*c*e^3)*d*x - 3*c^2*e^3)*sqrt(d*x + c + 1)*sqrt(-d*x - c
+ 1)/(a*b*d^2*x^2 + 2*a*b*c*d*x + a*b*c^2 - a*b + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 - b^2)*arctan2(d*x + c
, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1))), x))/(b^2*d*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) +
a*b*d)

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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((d*e*x+c*e)^3/(a+b*arcsin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} e^{3} \left (\int \frac {c^{3}}{a^{2} + 2 a b \operatorname {asin}{\left (c + d x \right )} + b^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}\, dx + \int \frac {d^{3} x^{3}}{a^{2} + 2 a b \operatorname {asin}{\left (c + d x \right )} + b^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}\, dx + \int \frac {3 c d^{2} x^{2}}{a^{2} + 2 a b \operatorname {asin}{\left (c + d x \right )} + b^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}\, dx + \int \frac {3 c^{2} d x}{a^{2} + 2 a b \operatorname {asin}{\left (c + d x \right )} + b^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**3/(a+b*asin(d*x+c))**2,x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 928 vs. \(2 (180) = 360\).
time = 0.49, size = 928, normalized size = 4.88 \begin {gather*} -\frac {4 \, b e^{3} \arcsin \left (d x + c\right ) \cos \left (\frac {a}{b}\right )^{4} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} - \frac {4 \, b e^{3} \arcsin \left (d x + c\right ) \cos \left (\frac {a}{b}\right )^{3} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} - \frac {4 \, a e^{3} \cos \left (\frac {a}{b}\right )^{4} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} - \frac {4 \, a e^{3} \cos \left (\frac {a}{b}\right )^{3} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} + \frac {4 \, b e^{3} \arcsin \left (d x + c\right ) \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} + \frac {b e^{3} \arcsin \left (d x + c\right ) \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} + \frac {2 \, b e^{3} \arcsin \left (d x + c\right ) \cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} + \frac {b e^{3} \arcsin \left (d x + c\right ) \cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} + \frac {4 \, a e^{3} \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} + \frac {a e^{3} \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} + \frac {2 \, a e^{3} \cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} + \frac {a e^{3} \cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} + \frac {{\left (-{\left (d x + c\right )}^{2} + 1\right )}^{\frac {3}{2}} {\left (d x + c\right )} b e^{3}}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} - \frac {b e^{3} \arcsin \left (d x + c\right ) \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (d x + c\right )\right )}{2 \, {\left (b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d\right )}} - \frac {b e^{3} \arcsin \left (d x + c\right ) \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (d x + c\right )\right )}{2 \, {\left (b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d\right )}} - \frac {\sqrt {-{\left (d x + c\right )}^{2} + 1} {\left (d x + c\right )} b e^{3}}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} - \frac {a e^{3} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (d x + c\right )\right )}{2 \, {\left (b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d\right )}} - \frac {a e^{3} \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (d x + c\right )\right )}{2 \, {\left (b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^3/(a+b*arcsin(d*x+c))^2,x, algorithm="giac")

[Out]

-4*b*e^3*arcsin(d*x + c)*cos(a/b)^4*cos_integral(4*a/b + 4*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d)
- 4*b*e^3*arcsin(d*x + c)*cos(a/b)^3*sin(a/b)*sin_integral(4*a/b + 4*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) +
a*b^2*d) - 4*a*e^3*cos(a/b)^4*cos_integral(4*a/b + 4*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) - 4*a
*e^3*cos(a/b)^3*sin(a/b)*sin_integral(4*a/b + 4*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) + 4*b*e^3*a
rcsin(d*x + c)*cos(a/b)^2*cos_integral(4*a/b + 4*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) + b*e^3*ar
csin(d*x + c)*cos(a/b)^2*cos_integral(2*a/b + 2*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) + 2*b*e^3*a
rcsin(d*x + c)*cos(a/b)*sin(a/b)*sin_integral(4*a/b + 4*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) + b
*e^3*arcsin(d*x + c)*cos(a/b)*sin(a/b)*sin_integral(2*a/b + 2*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*
d) + 4*a*e^3*cos(a/b)^2*cos_integral(4*a/b + 4*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) + a*e^3*cos(
a/b)^2*cos_integral(2*a/b + 2*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) + 2*a*e^3*cos(a/b)*sin(a/b)*s
in_integral(4*a/b + 4*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) + a*e^3*cos(a/b)*sin(a/b)*sin_integra
l(2*a/b + 2*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) + (-(d*x + c)^2 + 1)^(3/2)*(d*x + c)*b*e^3/(b^3
*d*arcsin(d*x + c) + a*b^2*d) - 1/2*b*e^3*arcsin(d*x + c)*cos_integral(4*a/b + 4*arcsin(d*x + c))/(b^3*d*arcsi
n(d*x + c) + a*b^2*d) - 1/2*b*e^3*arcsin(d*x + c)*cos_integral(2*a/b + 2*arcsin(d*x + c))/(b^3*d*arcsin(d*x +
c) + a*b^2*d) - sqrt(-(d*x + c)^2 + 1)*(d*x + c)*b*e^3/(b^3*d*arcsin(d*x + c) + a*b^2*d) - 1/2*a*e^3*cos_integ
ral(4*a/b + 4*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) - 1/2*a*e^3*cos_integral(2*a/b + 2*arcsin(d*x
+ c))/(b^3*d*arcsin(d*x + c) + a*b^2*d)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,e+d\,e\,x\right )}^3}{{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)^3/(a + b*asin(c + d*x))^2,x)

[Out]

int((c*e + d*e*x)^3/(a + b*asin(c + d*x))^2, x)

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