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

Optimal. Leaf size=145 \[ -\frac {e^3 \text {CosIntegral}\left (\frac {2 (a+b \text {ArcSin}(c+d x))}{b}\right ) \sin \left (\frac {2 a}{b}\right )}{4 b d}+\frac {e^3 \text {CosIntegral}\left (\frac {4 (a+b \text {ArcSin}(c+d x))}{b}\right ) \sin \left (\frac {4 a}{b}\right )}{8 b d}+\frac {e^3 \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \text {ArcSin}(c+d x))}{b}\right )}{4 b d}-\frac {e^3 \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \text {ArcSin}(c+d x))}{b}\right )}{8 b d} \]

[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

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Rubi [A]
time = 0.22, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4889, 12, 4731, 4491, 3384, 3380, 3383} \begin {gather*} -\frac {e^3 \sin \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 (a+b \text {ArcSin}(c+d x))}{b}\right )}{4 b d}+\frac {e^3 \sin \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (\frac {4 (a+b \text {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 \text {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 \text {ArcSin}(c+d x))}{b}\right )}{8 b d} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {(c e+d e x)^3}{a+b \sin ^{-1}(c+d x)} \, dx &=\frac {\text {Subst}\left (\int \frac {e^3 x^3}{a+b \sin ^{-1}(x)} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 \text {Subst}\left (\int \frac {x^3}{a+b \sin ^{-1}(x)} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 \text {Subst}\left (\int \frac {\cos (x) \sin ^3(x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{d}\\ &=\frac {e^3 \text {Subst}\left (\int \left (\frac {\sin (2 x)}{4 (a+b x)}-\frac {\sin (4 x)}{8 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d}\\ &=-\frac {e^3 \text {Subst}\left (\int \frac {\sin (4 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 d}+\frac {e^3 \text {Subst}\left (\int \frac {\sin (2 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{4 d}\\ &=\frac {\left (e^3 \cos \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 )}{4 d}-\frac {\left (e^3 \cos \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 )}{8 d}-\frac {\left (e^3 \sin \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 )}{4 d}+\frac {\left (e^3 \sin \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 )}{8 d}\\ &=-\frac {e^3 \text {Ci}\left (\frac {2 a}{b}+2 \sin ^{-1}(c+d x)\right ) \sin \left (\frac {2 a}{b}\right )}{4 b d}+\frac {e^3 \text {Ci}\left (\frac {4 a}{b}+4 \sin ^{-1}(c+d x)\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}+2 \sin ^{-1}(c+d x)\right )}{4 b d}-\frac {e^3 \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 a}{b}+4 \sin ^{-1}(c+d x)\right )}{8 b d}\\ \end {align*}

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Mathematica [A]
time = 0.18, size = 109, normalized size = 0.75 \begin {gather*} \frac {e^3 \left (-2 \text {CosIntegral}\left (2 \left (\frac {a}{b}+\text {ArcSin}(c+d x)\right )\right ) \sin \left (\frac {2 a}{b}\right )+\text {CosIntegral}\left (4 \left (\frac {a}{b}+\text {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}+\text {ArcSin}(c+d x)\right )\right )-\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\text {ArcSin}(c+d x)\right )\right )\right )}{8 b d} \end {gather*}

Antiderivative was successfully verified.

[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)

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Maple [A]
time = 0.12, size = 112, normalized size = 0.77

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)),x, algorithm="maxima")

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (137) = 274\).
time = 0.44, size = 277, normalized size = 1.91 \begin {gather*} \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} \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)),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)

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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}{a+b\,\mathrm {asin}\left (c+d\,x\right )} \,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)),x)

[Out]

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

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