\(\int \frac {(c e+d e x)^3}{a+b \arcsin (c+d x)} \, dx\) [216]

Optimal result

Integrand size = 23, antiderivative size = 145 \[ \int \frac {(c e+d e x)^3}{a+b \arcsin (c+d x)} \, dx=-\frac {e^3 \operatorname {CosIntegral}\left (\frac {2 (a+b \arcsin (c+d x))}{b}\right ) \sin \left (\frac {2 a}{b}\right )}{4 b d}+\frac {e^3 \operatorname {CosIntegral}\left (\frac {4 (a+b \arcsin (c+d x))}{b}\right ) \sin \left (\frac {4 a}{b}\right )}{8 b d}+\frac {e^3 \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c+d x))}{b}\right )}{4 b d}-\frac {e^3 \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arcsin (c+d x))}{b}\right )}{8 b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4889, 12, 4731, 4491, 3384, 3380, 3383} \[ \int \frac {(c e+d e x)^3}{a+b \arcsin (c+d x)} \, dx=-\frac {e^3 \sin \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arcsin (c+d x))}{b}\right )}{4 b d}+\frac {e^3 \sin \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {4 (a+b \arcsin (c+d x))}{b}\right )}{8 b d}+\frac {e^3 \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c+d x))}{b}\right )}{4 b d}-\frac {e^3 \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arcsin (c+d x))}{b}\right )}{8 b d} \]

[In]

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

[Out]

-1/4*(e^3*CosIntegral[(2*(a + b*ArcSin[c + d*x]))/b]*Sin[(2*a)/b])/(b*d) + (e^3*CosIntegral[(4*(a + b*ArcSin[c
+ d*x]))/b]*Sin[(4*a)/b])/(8*b*d) + (e^3*Cos[(2*a)/b]*SinIntegral[(2*(a + b*ArcSin[c + d*x]))/b])/(4*b*d) - (
e^3*Cos[(4*a)/b]*SinIntegral[(4*(a + b*ArcSin[c + d*x]))/b])/(8*b*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 4491

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 4731

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

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*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {e^3 x^3}{a+b \arcsin (x)} \, dx,x,c+d x\right )}{d} \\ & = \frac {e^3 \text {Subst}\left (\int \frac {x^3}{a+b \arcsin (x)} \, dx,x,c+d x\right )}{d} \\ & = -\frac {e^3 \text {Subst}\left (\int \frac {\cos \left (\frac {a}{b}-\frac {x}{b}\right ) \sin ^3\left (\frac {a}{b}-\frac {x}{b}\right )}{x} \, dx,x,a+b \arcsin (c+d x)\right )}{b d} \\ & = -\frac {e^3 \text {Subst}\left (\int \left (-\frac {\sin \left (\frac {4 a}{b}-\frac {4 x}{b}\right )}{8 x}+\frac {\sin \left (\frac {2 a}{b}-\frac {2 x}{b}\right )}{4 x}\right ) \, dx,x,a+b \arcsin (c+d x)\right )}{b d} \\ & = \frac {e^3 \text {Subst}\left (\int \frac {\sin \left (\frac {4 a}{b}-\frac {4 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c+d x)\right )}{8 b d}-\frac {e^3 \text {Subst}\left (\int \frac {\sin \left (\frac {2 a}{b}-\frac {2 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c+d x)\right )}{4 b d} \\ & = \frac {\left (e^3 \cos \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {2 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c+d x)\right )}{4 b d}-\frac {\left (e^3 \cos \left (\frac {4 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {4 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c+d x)\right )}{8 b d}-\frac {\left (e^3 \sin \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {2 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c+d x)\right )}{4 b d}+\frac {\left (e^3 \sin \left (\frac {4 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {4 x}{b}\right )}{x} \, dx,x,a+b \arcsin (c+d x)\right )}{8 b d} \\ & = -\frac {e^3 \operatorname {CosIntegral}\left (\frac {2 (a+b \arcsin (c+d x))}{b}\right ) \sin \left (\frac {2 a}{b}\right )}{4 b d}+\frac {e^3 \operatorname {CosIntegral}\left (\frac {4 (a+b \arcsin (c+d x))}{b}\right ) \sin \left (\frac {4 a}{b}\right )}{8 b d}+\frac {e^3 \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c+d x))}{b}\right )}{4 b d}-\frac {e^3 \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arcsin (c+d x))}{b}\right )}{8 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.75 \[ \int \frac {(c e+d e x)^3}{a+b \arcsin (c+d x)} \, dx=\frac {e^3 \left (-2 \operatorname {CosIntegral}\left (2 \left (\frac {a}{b}+\arcsin (c+d x)\right )\right ) \sin \left (\frac {2 a}{b}\right )+\operatorname {CosIntegral}\left (4 \left (\frac {a}{b}+\arcsin (c+d x)\right )\right ) \sin \left (\frac {4 a}{b}\right )+2 \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arcsin (c+d x)\right )\right )-\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\arcsin (c+d x)\right )\right )\right )}{8 b d} \]

[In]

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

[Out]

(e^3*(-2*CosIntegral[2*(a/b + ArcSin[c + d*x])]*Sin[(2*a)/b] + CosIntegral[4*(a/b + ArcSin[c + d*x])]*Sin[(4*a
)/b] + 2*Cos[(2*a)/b]*SinIntegral[2*(a/b + ArcSin[c + d*x])] - Cos[(4*a)/b]*SinIntegral[4*(a/b + ArcSin[c + d*
x])]))/(8*b*d)

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(-\frac {e^{3} \left (\operatorname {Si}\left (4 \arcsin \left (d x +c \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right )-\operatorname {Ci}\left (4 \arcsin \left (d x +c \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right )-2 \,\operatorname {Si}\left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right )+2 \,\operatorname {Ci}\left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right )\right )}{8 d b}\)	\(112\)
default	\(-\frac {e^{3} \left (\operatorname {Si}\left (4 \arcsin \left (d x +c \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right )-\operatorname {Ci}\left (4 \arcsin \left (d x +c \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right )-2 \,\operatorname {Si}\left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right )+2 \,\operatorname {Ci}\left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right )\right )}{8 d b}\)	\(112\)

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {(c e+d e x)^3}{a+b \arcsin (c+d x)} \, dx=\int { \frac {{\left (d e x + c e\right )}^{3}}{b \arcsin \left (d x + c\right ) + a} \,d x } \]

[In]

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

Sympy [F]

\[ \int \frac {(c e+d e x)^3}{a+b \arcsin (c+d x)} \, dx=e^{3} \left (\int \frac {c^{3}}{a + b \operatorname {asin}{\left (c + d x \right )}}\, dx + \int \frac {d^{3} x^{3}}{a + b \operatorname {asin}{\left (c + d x \right )}}\, dx + \int \frac {3 c d^{2} x^{2}}{a + b \operatorname {asin}{\left (c + d x \right )}}\, dx + \int \frac {3 c^{2} d x}{a + b \operatorname {asin}{\left (c + d x \right )}}\, dx\right ) \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {(c e+d e x)^3}{a+b \arcsin (c+d x)} \, dx=\int { \frac {{\left (d e x + c e\right )}^{3}}{b \arcsin \left (d x + c\right ) + a} \,d x } \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (137) = 274\).

Time = 0.33 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.91 \[ \int \frac {(c e+d e x)^3}{a+b \arcsin (c+d x)} \, dx=\frac {e^{3} \cos \left (\frac {a}{b}\right )^{3} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (d x + c\right )\right ) \sin \left (\frac {a}{b}\right )}{b d} - \frac {e^{3} \cos \left (\frac {a}{b}\right )^{4} \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (d x + c\right )\right )}{b d} - \frac {e^{3} \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (d x + c\right )\right ) \sin \left (\frac {a}{b}\right )}{2 \, b d} - \frac {e^{3} \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (d x + c\right )\right ) \sin \left (\frac {a}{b}\right )}{2 \, b d} + \frac {e^{3} \cos \left (\frac {a}{b}\right )^{2} \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (d x + c\right )\right )}{b d} + \frac {e^{3} \cos \left (\frac {a}{b}\right )^{2} \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (d x + c\right )\right )}{2 \, b d} - \frac {e^{3} \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (d x + c\right )\right )}{8 \, b d} - \frac {e^{3} \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (d x + c\right )\right )}{4 \, b d} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(c e+d e x)^3}{a+b \arcsin (c+d x)} \, dx=\int \frac {{\left (c\,e+d\,e\,x\right )}^3}{a+b\,\mathrm {asin}\left (c+d\,x\right )} \,d x \]

[In]

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

[Out]

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