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

Optimal. Leaf size=104 \[ \frac{e \cos \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{b^2 d}+\frac{e \sin \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{b^2 d}-\frac{e \sqrt{1-(c+d x)^2} (c+d x)}{b d \left (a+b \sin ^{-1}(c+d x)\right )} \]

[Out]

-((e*(c + d*x)*Sqrt[1 - (c + d*x)^2])/(b*d*(a + b*ArcSin[c + d*x]))) + (e*Cos[(2*a)/b]*CosIntegral[(2*(a + b*A
rcSin[c + d*x]))/b])/(b^2*d) + (e*Sin[(2*a)/b]*SinIntegral[(2*(a + b*ArcSin[c + d*x]))/b])/(b^2*d)

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Rubi [A]  time = 0.14108, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {4805, 12, 4631, 3303, 3299, 3302} \[ \frac{e \cos \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 a}{b}+2 \sin ^{-1}(c+d x)\right )}{b^2 d}+\frac{e \sin \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \sin ^{-1}(c+d x)\right )}{b^2 d}-\frac{e \sqrt{1-(c+d x)^2} (c+d x)}{b d \left (a+b \sin ^{-1}(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((e*(c + d*x)*Sqrt[1 - (c + d*x)^2])/(b*d*(a + b*ArcSin[c + d*x]))) + (e*Cos[(2*a)/b]*CosIntegral[(2*a)/b + 2
*ArcSin[c + d*x]])/(b^2*d) + (e*Sin[(2*a)/b]*SinIntegral[(2*a)/b + 2*ArcSin[c + d*x]])/(b^2*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 4631

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^m*Sqrt[1 - c^2*x^2]*(a + b*ArcSin
[c*x])^(n + 1))/(b*c*(n + 1)), x] - Dist[1/(b*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n + 1)
, Sin[x]^(m - 1)*(m - (m + 1)*Sin[x]^2), x], x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && G
eQ[n, -2] && LtQ[n, -1]

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

Mathematica [A]  time = 0.294227, size = 99, normalized size = 0.95 \[ \frac{e \left (-\frac{b \sqrt{-c^2-2 c d x-d^2 x^2+1} (c+d x)}{a+b \sin ^{-1}(c+d x)}+\cos \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )+\sin \left (\frac{2 a}{b}\right ) \text{Si}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )\right )}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.033, size = 151, normalized size = 1.5 \begin{align*}{\frac{e}{2\,d \left ( a+b\arcsin \left ( dx+c \right ) \right ){b}^{2}} \left ( 2\,\arcsin \left ( dx+c \right ){\it Si} \left ( 2\,\arcsin \left ( dx+c \right ) +2\,{\frac{a}{b}} \right ) \sin \left ( 2\,{\frac{a}{b}} \right ) b+2\,\arcsin \left ( dx+c \right ){\it Ci} \left ( 2\,\arcsin \left ( dx+c \right ) +2\,{\frac{a}{b}} \right ) \cos \left ( 2\,{\frac{a}{b}} \right ) b+2\,{\it Si} \left ( 2\,\arcsin \left ( dx+c \right ) +2\,{\frac{a}{b}} \right ) \sin \left ( 2\,{\frac{a}{b}} \right ) a+2\,{\it Ci} \left ( 2\,\arcsin \left ( dx+c \right ) +2\,{\frac{a}{b}} \right ) \cos \left ( 2\,{\frac{a}{b}} \right ) a-\sin \left ( 2\,\arcsin \left ( dx+c \right ) \right ) b \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{d e x + c e}{b^{2} \arcsin \left (d x + c\right )^{2} + 2 \, a b \arcsin \left (d x + c\right ) + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [B]  time = 1.44294, size = 470, normalized size = 4.52 \begin{align*} \frac{2 \, b \arcsin \left (d x + c\right ) \cos \left (\frac{a}{b}\right )^{2} \operatorname{Ci}\left (\frac{2 \, a}{b} + 2 \, \arcsin \left (d x + c\right )\right ) e}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} + \frac{2 \, b \arcsin \left (d x + c\right ) \cos \left (\frac{a}{b}\right ) e \sin \left (\frac{a}{b}\right ) \operatorname{Si}\left (\frac{2 \, a}{b} + 2 \, \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} + \frac{2 \, a \cos \left (\frac{a}{b}\right )^{2} \operatorname{Ci}\left (\frac{2 \, a}{b} + 2 \, \arcsin \left (d x + c\right )\right ) e}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} + \frac{2 \, a \cos \left (\frac{a}{b}\right ) e \sin \left (\frac{a}{b}\right ) \operatorname{Si}\left (\frac{2 \, a}{b} + 2 \, \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} - \frac{b \arcsin \left (d x + c\right ) \operatorname{Ci}\left (\frac{2 \, a}{b} + 2 \, \arcsin \left (d x + c\right )\right ) e}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} - \frac{\sqrt{-{\left (d x + c\right )}^{2} + 1}{\left (d x + c\right )} b e}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} - \frac{a \operatorname{Ci}\left (\frac{2 \, a}{b} + 2 \, \arcsin \left (d x + c\right )\right ) e}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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