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

Optimal. Leaf size=179 \[ \frac{e \cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}(c x)}{b}\right )}{4 b c^3}-\frac{e \cos \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (\frac{3 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{4 b c^3}+\frac{e \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \sin ^{-1}(c x)}{b}\right )}{4 b c^3}-\frac{e \sin \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{4 b c^3}+\frac{d \cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}(c x)}{b}\right )}{b c}+\frac{d \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \sin ^{-1}(c x)}{b}\right )}{b c} \]

[Out]

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

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Rubi [A]  time = 0.335467, antiderivative size = 175, normalized size of antiderivative = 0.98, number of steps used = 15, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {4667, 4623, 3303, 3299, 3302, 4635, 4406} \[ \frac{e \cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )}{4 b c^3}-\frac{e \cos \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (\frac{3 a}{b}+3 \sin ^{-1}(c x)\right )}{4 b c^3}+\frac{e \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )}{4 b c^3}-\frac{e \sin \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 a}{b}+3 \sin ^{-1}(c x)\right )}{4 b c^3}+\frac{d \cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}(c x)}{b}\right )}{b c}+\frac{d \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \sin ^{-1}(c x)}{b}\right )}{b c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^2)/(a + b*ArcSin[c*x]),x]

[Out]

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

Rule 4667

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a
+ b*ArcSin[c*x])^n, (d + e*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && NeQ[c^2*d + e, 0] && IntegerQ[p]
&& (GtQ[p, 0] || IGtQ[n, 0])

Rule 4623

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

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 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]

Rubi steps

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

Mathematica [A]  time = 0.267668, size = 125, normalized size = 0.7 \[ \frac{\cos \left (\frac{a}{b}\right ) \left (4 c^2 d+e\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )+4 c^2 d \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )-e \cos \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (3 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+e \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )-e \sin \left (\frac{3 a}{b}\right ) \text{Si}\left (3 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )}{4 b c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x^2)/(a + b*ArcSin[c*x]),x]

[Out]

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

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Maple [A]  time = 0.038, size = 142, normalized size = 0.8 \begin{align*} -{\frac{1}{4\,{c}^{3}b} \left ( -4\,{\it Si} \left ( \arcsin \left ( cx \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ){c}^{2}d-4\,{\it Ci} \left ( \arcsin \left ( cx \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ){c}^{2}d+{\it Si} \left ( 3\,\arcsin \left ( cx \right ) +3\,{\frac{a}{b}} \right ) \sin \left ( 3\,{\frac{a}{b}} \right ) e+{\it Ci} \left ( 3\,\arcsin \left ( cx \right ) +3\,{\frac{a}{b}} \right ) \cos \left ( 3\,{\frac{a}{b}} \right ) e-{\it Si} \left ( \arcsin \left ( cx \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ) e-{\it Ci} \left ( \arcsin \left ( cx \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ) e \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)/(a+b*arcsin(c*x)),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e x^{2} + 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^2+d)/(a+b*arcsin(c*x)),x, algorithm="maxima")

[Out]

integrate((e*x^2 + 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^{2} + 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^2+d)/(a+b*arcsin(c*x)),x, algorithm="fricas")

[Out]

integral((e*x^2 + 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^{2}}{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**2+d)/(a+b*asin(c*x)),x)

[Out]

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

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Giac [A]  time = 1.24264, size = 317, normalized size = 1.77 \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 )^{3} \operatorname{Ci}\left (\frac{3 \, a}{b} + 3 \, \arcsin \left (c x\right )\right ) e}{b c^{3}} - \frac{\cos \left (\frac{a}{b}\right )^{2} e \sin \left (\frac{a}{b}\right ) \operatorname{Si}\left (\frac{3 \, a}{b} + 3 \, \arcsin \left (c x\right )\right )}{b c^{3}} + \frac{d \sin \left (\frac{a}{b}\right ) \operatorname{Si}\left (\frac{a}{b} + \arcsin \left (c x\right )\right )}{b c} + \frac{3 \, \cos \left (\frac{a}{b}\right ) \operatorname{Ci}\left (\frac{3 \, a}{b} + 3 \, \arcsin \left (c x\right )\right ) e}{4 \, b c^{3}} + \frac{\cos \left (\frac{a}{b}\right ) \operatorname{Ci}\left (\frac{a}{b} + \arcsin \left (c x\right )\right ) e}{4 \, b c^{3}} + \frac{e \sin \left (\frac{a}{b}\right ) \operatorname{Si}\left (\frac{3 \, a}{b} + 3 \, \arcsin \left (c x\right )\right )}{4 \, b c^{3}} + \frac{e \sin \left (\frac{a}{b}\right ) \operatorname{Si}\left (\frac{a}{b} + \arcsin \left (c x\right )\right )}{4 \, b c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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