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

Optimal. Leaf size=115 \[ -\frac{e \sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 a}{b}+2 \sin ^{-1}(c x)\right )}{2 b c^2}+\frac{e \cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \sin ^{-1}(c x)\right )}{2 b c^2}+\frac{d \cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )}{b c}+\frac{d \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )}{b c} \]

[Out]

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

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Rubi [A]  time = 0.306474, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {4747, 6742, 3303, 3299, 3302, 4406, 12} \[ -\frac{e \sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 a}{b}+2 \sin ^{-1}(c x)\right )}{2 b c^2}+\frac{e \cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \sin ^{-1}(c x)\right )}{2 b c^2}+\frac{d \cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )}{b c}+\frac{d \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )}{b c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)/(a + b*ArcSin[c*x]),x]

[Out]

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

Rule 4747

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

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

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]

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 12

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

Rubi steps

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

Mathematica [A]  time = 0.186046, size = 98, normalized size = 0.85 \[ \frac{2 c d \cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )-e \sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+2 c d \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )+e \cos \left (\frac{2 a}{b}\right ) \text{Si}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )}{2 b c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.036, size = 103, normalized size = 0.9 \begin{align*}{\frac{1}{c} \left ( -{\frac{e}{2\,bc} \left ({\it Ci} \left ( 2\,\arcsin \left ( cx \right ) +2\,{\frac{a}{b}} \right ) \sin \left ( 2\,{\frac{a}{b}} \right ) -{\it Si} \left ( 2\,\arcsin \left ( cx \right ) +2\,{\frac{a}{b}} \right ) \cos \left ( 2\,{\frac{a}{b}} \right ) \right ) }+{\frac{d}{b} \left ({\it Si} \left ( \arcsin \left ( cx \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ) +{\it Ci} \left ( \arcsin \left ( cx \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ) \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.26102, size = 192, normalized size = 1.67 \begin{align*} \frac{d \cos \left (\frac{a}{b}\right ) \operatorname{Ci}\left (\frac{a}{b} + \arcsin \left (c x\right )\right )}{b c} - \frac{\cos \left (\frac{a}{b}\right ) \operatorname{Ci}\left (\frac{2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right ) e \sin \left (\frac{a}{b}\right )}{b c^{2}} + \frac{\cos \left (\frac{a}{b}\right )^{2} e \operatorname{Si}\left (\frac{2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{b c^{2}} + \frac{d \sin \left (\frac{a}{b}\right ) \operatorname{Si}\left (\frac{a}{b} + \arcsin \left (c x\right )\right )}{b c} - \frac{e \operatorname{Si}\left (\frac{2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{2 \, b c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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