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

Optimal. Leaf size=213 \[ \frac{e^4 \cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}(c+d x)}{b}\right )}{8 b d}-\frac{3 e^4 \cos \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (\frac{3 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{16 b d}+\frac{e^4 \cos \left (\frac{5 a}{b}\right ) \text{CosIntegral}\left (\frac{5 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{16 b d}+\frac{e^4 \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \sin ^{-1}(c+d x)}{b}\right )}{8 b d}-\frac{3 e^4 \sin \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{16 b d}+\frac{e^4 \sin \left (\frac{5 a}{b}\right ) \text{Si}\left (\frac{5 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{16 b d} \]

[Out]

(e^4*Cos[a/b]*CosIntegral[(a + b*ArcSin[c + d*x])/b])/(8*b*d) - (3*e^4*Cos[(3*a)/b]*CosIntegral[(3*(a + b*ArcS
in[c + d*x]))/b])/(16*b*d) + (e^4*Cos[(5*a)/b]*CosIntegral[(5*(a + b*ArcSin[c + d*x]))/b])/(16*b*d) + (e^4*Sin
[a/b]*SinIntegral[(a + b*ArcSin[c + d*x])/b])/(8*b*d) - (3*e^4*Sin[(3*a)/b]*SinIntegral[(3*(a + b*ArcSin[c + d
*x]))/b])/(16*b*d) + (e^4*Sin[(5*a)/b]*SinIntegral[(5*(a + b*ArcSin[c + d*x]))/b])/(16*b*d)

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Rubi [A]  time = 0.408438, antiderivative size = 209, normalized size of antiderivative = 0.98, number of steps used = 14, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {4805, 12, 4635, 4406, 3303, 3299, 3302} \[ \frac{e^4 \cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )}{8 b d}-\frac{3 e^4 \cos \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (\frac{3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{16 b d}+\frac{e^4 \cos \left (\frac{5 a}{b}\right ) \text{CosIntegral}\left (\frac{5 a}{b}+5 \sin ^{-1}(c+d x)\right )}{16 b d}+\frac{e^4 \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )}{8 b d}-\frac{3 e^4 \sin \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{16 b d}+\frac{e^4 \sin \left (\frac{5 a}{b}\right ) \text{Si}\left (\frac{5 a}{b}+5 \sin ^{-1}(c+d x)\right )}{16 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^4*Cos[a/b]*CosIntegral[a/b + ArcSin[c + d*x]])/(8*b*d) - (3*e^4*Cos[(3*a)/b]*CosIntegral[(3*a)/b + 3*ArcSin
[c + d*x]])/(16*b*d) + (e^4*Cos[(5*a)/b]*CosIntegral[(5*a)/b + 5*ArcSin[c + d*x]])/(16*b*d) + (e^4*Sin[a/b]*Si
nIntegral[a/b + ArcSin[c + d*x]])/(8*b*d) - (3*e^4*Sin[(3*a)/b]*SinIntegral[(3*a)/b + 3*ArcSin[c + d*x]])/(16*
b*d) + (e^4*Sin[(5*a)/b]*SinIntegral[(5*a)/b + 5*ArcSin[c + d*x]])/(16*b*d)

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 4635

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

Rule 4406

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 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 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]

Rubi steps

\begin{align*} \int \frac{(c e+d e x)^4}{a+b \sin ^{-1}(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^4 x^4}{a+b \sin ^{-1}(x)} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^4 \operatorname{Subst}\left (\int \frac{x^4}{a+b \sin ^{-1}(x)} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^4 \operatorname{Subst}\left (\int \frac{\cos (x) \sin ^4(x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{d}\\ &=\frac{e^4 \operatorname{Subst}\left (\int \left (\frac{\cos (x)}{8 (a+b x)}-\frac{3 \cos (3 x)}{16 (a+b x)}+\frac{\cos (5 x)}{16 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d}\\ &=\frac{e^4 \operatorname{Subst}\left (\int \frac{\cos (5 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{16 d}+\frac{e^4 \operatorname{Subst}\left (\int \frac{\cos (x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 d}-\frac{\left (3 e^4\right ) \operatorname{Subst}\left (\int \frac{\cos (3 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{16 d}\\ &=\frac{\left (e^4 \cos \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 d}-\frac{\left (3 e^4 \cos \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{16 d}+\frac{\left (e^4 \cos \left (\frac{5 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{16 d}+\frac{\left (e^4 \sin \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 d}-\frac{\left (3 e^4 \sin \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{16 d}+\frac{\left (e^4 \sin \left (\frac{5 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{16 d}\\ &=\frac{e^4 \cos \left (\frac{a}{b}\right ) \text{Ci}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )}{8 b d}-\frac{3 e^4 \cos \left (\frac{3 a}{b}\right ) \text{Ci}\left (\frac{3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{16 b d}+\frac{e^4 \cos \left (\frac{5 a}{b}\right ) \text{Ci}\left (\frac{5 a}{b}+5 \sin ^{-1}(c+d x)\right )}{16 b d}+\frac{e^4 \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )}{8 b d}-\frac{3 e^4 \sin \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{16 b d}+\frac{e^4 \sin \left (\frac{5 a}{b}\right ) \text{Si}\left (\frac{5 a}{b}+5 \sin ^{-1}(c+d x)\right )}{16 b d}\\ \end{align*}

Mathematica [A]  time = 0.323096, size = 150, normalized size = 0.7 \[ \frac{e^4 \left (2 \cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )-3 \cos \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (3 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )+\cos \left (\frac{5 a}{b}\right ) \text{CosIntegral}\left (5 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )+2 \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )-3 \sin \left (\frac{3 a}{b}\right ) \text{Si}\left (3 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )+\sin \left (\frac{5 a}{b}\right ) \text{Si}\left (5 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )\right )}{16 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^4*(2*Cos[a/b]*CosIntegral[a/b + ArcSin[c + d*x]] - 3*Cos[(3*a)/b]*CosIntegral[3*(a/b + ArcSin[c + d*x])] +
Cos[(5*a)/b]*CosIntegral[5*(a/b + ArcSin[c + d*x])] + 2*Sin[a/b]*SinIntegral[a/b + ArcSin[c + d*x]] - 3*Sin[(3
*a)/b]*SinIntegral[3*(a/b + ArcSin[c + d*x])] + Sin[(5*a)/b]*SinIntegral[5*(a/b + ArcSin[c + d*x])]))/(16*b*d)

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Maple [A]  time = 0.038, size = 155, normalized size = 0.7 \begin{align*} -{\frac{{e}^{4}}{16\,bd} \left ( 3\,{\it Si} \left ( 3\,\arcsin \left ( dx+c \right ) +3\,{\frac{a}{b}} \right ) \sin \left ( 3\,{\frac{a}{b}} \right ) +3\,{\it Ci} \left ( 3\,\arcsin \left ( dx+c \right ) +3\,{\frac{a}{b}} \right ) \cos \left ( 3\,{\frac{a}{b}} \right ) -{\it Si} \left ( 5\,\arcsin \left ( dx+c \right ) +5\,{\frac{a}{b}} \right ) \sin \left ( 5\,{\frac{a}{b}} \right ) -{\it Ci} \left ( 5\,\arcsin \left ( dx+c \right ) +5\,{\frac{a}{b}} \right ) \cos \left ( 5\,{\frac{a}{b}} \right ) -2\,{\it Si} \left ( \arcsin \left ( dx+c \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ) -2\,{\it Ci} \left ( \arcsin \left ( dx+c \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16/d*e^4*(3*Si(3*arcsin(d*x+c)+3*a/b)*sin(3*a/b)+3*Ci(3*arcsin(d*x+c)+3*a/b)*cos(3*a/b)-Si(5*arcsin(d*x+c)+
5*a/b)*sin(5*a/b)-Ci(5*arcsin(d*x+c)+5*a/b)*cos(5*a/b)-2*Si(arcsin(d*x+c)+a/b)*sin(a/b)-2*Ci(arcsin(d*x+c)+a/b
)*cos(a/b))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((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)/(b*arcsin(d*x + c) + a)
, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.26859, size = 549, normalized size = 2.58 \begin{align*} \frac{\cos \left (\frac{a}{b}\right )^{5} \operatorname{Ci}\left (\frac{5 \, a}{b} + 5 \, \arcsin \left (d x + c\right )\right ) e^{4}}{b d} + \frac{\cos \left (\frac{a}{b}\right )^{4} e^{4} \sin \left (\frac{a}{b}\right ) \operatorname{Si}\left (\frac{5 \, a}{b} + 5 \, \arcsin \left (d x + c\right )\right )}{b d} - \frac{5 \, \cos \left (\frac{a}{b}\right )^{3} \operatorname{Ci}\left (\frac{5 \, a}{b} + 5 \, \arcsin \left (d x + c\right )\right ) e^{4}}{4 \, b d} - \frac{3 \, \cos \left (\frac{a}{b}\right )^{3} \operatorname{Ci}\left (\frac{3 \, a}{b} + 3 \, \arcsin \left (d x + c\right )\right ) e^{4}}{4 \, b d} - \frac{3 \, \cos \left (\frac{a}{b}\right )^{2} e^{4} \sin \left (\frac{a}{b}\right ) \operatorname{Si}\left (\frac{5 \, a}{b} + 5 \, \arcsin \left (d x + c\right )\right )}{4 \, b d} - \frac{3 \, \cos \left (\frac{a}{b}\right )^{2} e^{4} \sin \left (\frac{a}{b}\right ) \operatorname{Si}\left (\frac{3 \, a}{b} + 3 \, \arcsin \left (d x + c\right )\right )}{4 \, b d} + \frac{5 \, \cos \left (\frac{a}{b}\right ) \operatorname{Ci}\left (\frac{5 \, a}{b} + 5 \, \arcsin \left (d x + c\right )\right ) e^{4}}{16 \, b d} + \frac{9 \, \cos \left (\frac{a}{b}\right ) \operatorname{Ci}\left (\frac{3 \, a}{b} + 3 \, \arcsin \left (d x + c\right )\right ) e^{4}}{16 \, b d} + \frac{\cos \left (\frac{a}{b}\right ) \operatorname{Ci}\left (\frac{a}{b} + \arcsin \left (d x + c\right )\right ) e^{4}}{8 \, b d} + \frac{e^{4} \sin \left (\frac{a}{b}\right ) \operatorname{Si}\left (\frac{5 \, a}{b} + 5 \, \arcsin \left (d x + c\right )\right )}{16 \, b d} + \frac{3 \, e^{4} \sin \left (\frac{a}{b}\right ) \operatorname{Si}\left (\frac{3 \, a}{b} + 3 \, \arcsin \left (d x + c\right )\right )}{16 \, b d} + \frac{e^{4} \sin \left (\frac{a}{b}\right ) \operatorname{Si}\left (\frac{a}{b} + \arcsin \left (d x + c\right )\right )}{8 \, b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

cos(a/b)^5*cos_integral(5*a/b + 5*arcsin(d*x + c))*e^4/(b*d) + cos(a/b)^4*e^4*sin(a/b)*sin_integral(5*a/b + 5*
arcsin(d*x + c))/(b*d) - 5/4*cos(a/b)^3*cos_integral(5*a/b + 5*arcsin(d*x + c))*e^4/(b*d) - 3/4*cos(a/b)^3*cos
_integral(3*a/b + 3*arcsin(d*x + c))*e^4/(b*d) - 3/4*cos(a/b)^2*e^4*sin(a/b)*sin_integral(5*a/b + 5*arcsin(d*x
+ c))/(b*d) - 3/4*cos(a/b)^2*e^4*sin(a/b)*sin_integral(3*a/b + 3*arcsin(d*x + c))/(b*d) + 5/16*cos(a/b)*cos_i
ntegral(5*a/b + 5*arcsin(d*x + c))*e^4/(b*d) + 9/16*cos(a/b)*cos_integral(3*a/b + 3*arcsin(d*x + c))*e^4/(b*d)
+ 1/8*cos(a/b)*cos_integral(a/b + arcsin(d*x + c))*e^4/(b*d) + 1/16*e^4*sin(a/b)*sin_integral(5*a/b + 5*arcsi
n(d*x + c))/(b*d) + 3/16*e^4*sin(a/b)*sin_integral(3*a/b + 3*arcsin(d*x + c))/(b*d) + 1/8*e^4*sin(a/b)*sin_int
egral(a/b + arcsin(d*x + c))/(b*d)