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

Optimal. Leaf size=181 \[ \frac{e \cos \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{b^2 c^2}+\frac{e \sin \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{b^2 c^2}+\frac{d \sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}(c x)}{b}\right )}{b^2 c}-\frac{d \cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \sin ^{-1}(c x)}{b}\right )}{b^2 c}-\frac{d \sqrt{1-c^2 x^2}}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac{e x \sqrt{1-c^2 x^2}}{b c \left (a+b \sin ^{-1}(c x)\right )} \]

[Out]

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

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Rubi [A]  time = 0.312024, antiderivative size = 177, normalized size of antiderivative = 0.98, 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 = {4745, 4621, 4723, 3303, 3299, 3302, 4631} \[ \frac{e \cos \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 a}{b}+2 \sin ^{-1}(c x)\right )}{b^2 c^2}+\frac{e \sin \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \sin ^{-1}(c x)\right )}{b^2 c^2}+\frac{d \sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )}{b^2 c}-\frac{d \cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )}{b^2 c}-\frac{d \sqrt{1-c^2 x^2}}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac{e x \sqrt{1-c^2 x^2}}{b c \left (a+b \sin ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4745

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

Rule 4621

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^(n + 1))
/(b*c*(n + 1)), x] + Dist[c/(b*(n + 1)), Int[(x*(a + b*ArcSin[c*x])^(n + 1))/Sqrt[1 - c^2*x^2], x], x] /; Free
Q[{a, b, c}, x] && LtQ[n, -1]

Rule 4723

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^(
m + 1), Subst[Int[(a + b*x)^n*Sin[x]^m*Cos[x]^(2*p + 1), x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e, n},
x] && EqQ[c^2*d + e, 0] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 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]

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]

Rubi steps

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

Mathematica [A]  time = 0.685827, size = 149, normalized size = 0.82 \[ \frac{-\frac{b c \sqrt{1-c^2 x^2} (d+e x)}{a+b \sin ^{-1}(c x)}+c d \left (\sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )-\cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+e \left (\cos \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+\sin \left (\frac{2 a}{b}\right ) \text{Si}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )-\log \left (a+b \sin ^{-1}(c x)\right )\right )+e \log \left (a+b \sin ^{-1}(c x)\right )}{b^2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.051, size = 257, normalized size = 1.4 \begin{align*}{\frac{1}{c} \left ({\frac{e}{2\,c \left ( a+b\arcsin \left ( cx \right ) \right ){b}^{2}} \left ( 2\,\arcsin \left ( cx \right ){\it Si} \left ( 2\,\arcsin \left ( cx \right ) +2\,{\frac{a}{b}} \right ) \sin \left ( 2\,{\frac{a}{b}} \right ) b+2\,\arcsin \left ( cx \right ){\it Ci} \left ( 2\,\arcsin \left ( cx \right ) +2\,{\frac{a}{b}} \right ) \cos \left ( 2\,{\frac{a}{b}} \right ) b+2\,{\it Si} \left ( 2\,\arcsin \left ( cx \right ) +2\,{\frac{a}{b}} \right ) \sin \left ( 2\,{\frac{a}{b}} \right ) a+2\,{\it Ci} \left ( 2\,\arcsin \left ( cx \right ) +2\,{\frac{a}{b}} \right ) \cos \left ( 2\,{\frac{a}{b}} \right ) a-\sin \left ( 2\,\arcsin \left ( cx \right ) \right ) b \right ) }-{\frac{d}{ \left ( a+b\arcsin \left ( cx \right ) \right ){b}^{2}} \left ( \arcsin \left ( cx \right ){\it Si} \left ( \arcsin \left ( cx \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ) b-\arcsin \left ( cx \right ){\it Ci} \left ( \arcsin \left ( cx \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ) b+{\it Si} \left ( \arcsin \left ( cx \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ) a-{\it Ci} \left ( \arcsin \left ( cx \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ) a+\sqrt{-{c}^{2}{x}^{2}+1}b \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(1/2/c*e*(2*arcsin(c*x)*Si(2*arcsin(c*x)+2*a/b)*sin(2*a/b)*b+2*arcsin(c*x)*Ci(2*arcsin(c*x)+2*a/b)*cos(2*a
/b)*b+2*Si(2*arcsin(c*x)+2*a/b)*sin(2*a/b)*a+2*Ci(2*arcsin(c*x)+2*a/b)*cos(2*a/b)*a-sin(2*arcsin(c*x))*b)/(a+b
*arcsin(c*x))/b^2-d*(arcsin(c*x)*Si(arcsin(c*x)+a/b)*cos(a/b)*b-arcsin(c*x)*Ci(arcsin(c*x)+a/b)*sin(a/b)*b+Si(
arcsin(c*x)+a/b)*cos(a/b)*a-Ci(arcsin(c*x)+a/b)*sin(a/b)*a+(-c^2*x^2+1)^(1/2)*b)/(a+b*arcsin(c*x))/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((e*x+d)/(a+b*arcsin(c*x))^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{e x + d}{b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.39965, size = 757, normalized size = 4.18 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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