3.222 \(\int \frac {(c e+d e x)^3}{(a+b \sin ^{-1}(c+d x))^2} \, dx\)

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

[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 = 0.29, 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 = {4805, 12, 4631, 3303, 3299, 3302} \[ \frac {e^3 \cos \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 a}{b}+2 \sin ^{-1}(c+d x)\right )}{2 b^2 d}-\frac {e^3 \cos \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (\frac {4 a}{b}+4 \sin ^{-1}(c+d x)\right )}{2 b^2 d}+\frac {e^3 \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sin ^{-1}(c+d x)\right )}{2 b^2 d}-\frac {e^3 \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 a}{b}+4 \sin ^{-1}(c+d x)\right )}{2 b^2 d}-\frac {e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )} \]

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 + 2*ArcSin[c + d*x]])/(2*b^2*d) - (e^3*Cos[(4*a)/b]*CosIntegral[(4*a)/b + 4*ArcSin[c + d*x]])/(2*b^2*d) + (
e^3*Sin[(2*a)/b]*SinIntegral[(2*a)/b + 2*ArcSin[c + d*x]])/(2*b^2*d) - (e^3*Sin[(4*a)/b]*SinIntegral[(4*a)/b +
4*ArcSin[c + d*x]])/(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 3299

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 3302

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 3303

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 4631

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*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n + 1)
, Sin[x]^(m - 1)*(m - (m + 1)*Sin[x]^2), x], x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && G
eQ[n, -2] && LtQ[n, -1]

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]

Rubi steps

\begin {align*} \int \frac {(c e+d e x)^3}{\left (a+b \sin ^{-1}(c+d x)\right )^2} \, dx &=\frac {\operatorname {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 \operatorname {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 \operatorname {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 \operatorname {Subst}\left (\int \frac {\cos (2 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b d}-\frac {e^3 \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 = 0.95, size = 220, normalized size = 1.16 \[ -\frac {e^3 \left (3 \left (\cos \left (\frac {2 a}{b}\right ) \text {Ci}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )+\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )-\log \left (a+b \sin ^{-1}(c+d x)\right )\right )-4 \cos \left (\frac {2 a}{b}\right ) \text {Ci}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )+\cos \left (\frac {4 a}{b}\right ) \text {Ci}\left (4 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )-4 \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )+\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )+\frac {2 b \sqrt {1-(c+d x)^2} (c+d x)^3}{a+b \sin ^{-1}(c+d x)}+3 \log \left (a+b \sin ^{-1}(c+d x)\right )\right )}{2 b^2 d} \]

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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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}}{b^{2} \arcsin \left (d x + c\right )^{2} + 2 \, a b \arcsin \left (d x + c\right ) + a^{2}}, x\right ) \]

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*e^3*x^3 + 3*c*d^2*e^3*x^2 + 3*c^2*d*e^3*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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giac [B]  time = 0.42, size = 910, normalized size = 4.79 \[ \text {result too large to display} \]

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*arcsin(d*x + c)*cos(a/b)^4*cos_integral(4*a/b + 4*arcsin(d*x + c))*e^3/(b^3*d*arcsin(d*x + c) + a*b^2*d)
- 4*b*arcsin(d*x + c)*cos(a/b)^3*e^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*cos(a/b)^4*cos_integral(4*a/b + 4*arcsin(d*x + c))*e^3/(b^3*d*arcsin(d*x + c) + a*b^2*d) - 4*a
*cos(a/b)^3*e^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*arcsi
n(d*x + c)*cos(a/b)^2*cos_integral(4*a/b + 4*arcsin(d*x + c))*e^3/(b^3*d*arcsin(d*x + c) + a*b^2*d) + b*arcsin
(d*x + c)*cos(a/b)^2*cos_integral(2*a/b + 2*arcsin(d*x + c))*e^3/(b^3*d*arcsin(d*x + c) + a*b^2*d) + 2*b*arcsi
n(d*x + c)*cos(a/b)*e^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) + b
*arcsin(d*x + c)*cos(a/b)*e^3*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*cos(a/b)^2*cos_integral(4*a/b + 4*arcsin(d*x + c))*e^3/(b^3*d*arcsin(d*x + c) + a*b^2*d) + a*cos(a/b)
^2*cos_integral(2*a/b + 2*arcsin(d*x + c))*e^3/(b^3*d*arcsin(d*x + c) + a*b^2*d) + 2*a*cos(a/b)*e^3*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*cos(a/b)*e^3*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*arcsin(d*x + c)*cos_integral(4*a/b + 4*arcsin(d*x + c))*e^3/(b^3*d*arcsi
n(d*x + c) + a*b^2*d) - 1/2*b*arcsin(d*x + c)*cos_integral(2*a/b + 2*arcsin(d*x + c))*e^3/(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*cos_integral(
4*a/b + 4*arcsin(d*x + c))*e^3/(b^3*d*arcsin(d*x + c) + a*b^2*d) - 1/2*a*cos_integral(2*a/b + 2*arcsin(d*x + c
))*e^3/(b^3*d*arcsin(d*x + c) + a*b^2*d)

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maple [A]  time = 0.05, size = 281, normalized size = 1.48 \[ -\frac {e^{3} \left (4 \arcsin \left (d x +c \right ) \Si \left (4 \arcsin \left (d x +c \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) b +4 \arcsin \left (d x +c \right ) \Ci \left (4 \arcsin \left (d x +c \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) b -4 \arcsin \left (d x +c \right ) \Si \left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) b -4 \arcsin \left (d x +c \right ) \Ci \left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) b +4 \Si \left (4 \arcsin \left (d x +c \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) a +4 \Ci \left (4 \arcsin \left (d x +c \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) a -4 \Si \left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) a -4 \Ci \left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) a -\sin \left (4 \arcsin \left (d x +c \right )\right ) b +2 \sin \left (2 \arcsin \left (d x +c \right )\right ) b \right )}{8 d \left (a +b \arcsin \left (d x +c \right )\right ) b^{2}} \]

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)

[Out]

-1/8/d*e^3*(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)*c
os(4*a/b)*b-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)*c
os(2*a/b)*b+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-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-sin(4*arcsin(d*x+c))*b+2*sin(2*arcsin(d*x+c)
)*b)/(a+b*arcsin(d*x+c))/b^2

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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

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

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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