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

Optimal. Leaf size=186 \[ \frac{e^2 \sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}(c+d x)}{b}\right )}{4 b^2 d}-\frac{3 e^2 \sin \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (\frac{3 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{4 b^2 d}-\frac{e^2 \cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \sin ^{-1}(c+d x)}{b}\right )}{4 b^2 d}+\frac{3 e^2 \cos \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{4 b^2 d}-\frac{e^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )} \]

[Out]

-((e^2*(c + d*x)^2*Sqrt[1 - (c + d*x)^2])/(b*d*(a + b*ArcSin[c + d*x]))) + (e^2*CosIntegral[(a + b*ArcSin[c +
d*x])/b]*Sin[a/b])/(4*b^2*d) - (3*e^2*CosIntegral[(3*(a + b*ArcSin[c + d*x]))/b]*Sin[(3*a)/b])/(4*b^2*d) - (e^
2*Cos[a/b]*SinIntegral[(a + b*ArcSin[c + d*x])/b])/(4*b^2*d) + (3*e^2*Cos[(3*a)/b]*SinIntegral[(3*(a + b*ArcSi
n[c + d*x]))/b])/(4*b^2*d)

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Rubi [A]  time = 0.264746, antiderivative size = 182, normalized size of antiderivative = 0.98, number of steps used = 10, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {4805, 12, 4631, 3303, 3299, 3302} \[ \frac{e^2 \sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )}{4 b^2 d}-\frac{3 e^2 \sin \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (\frac{3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{4 b^2 d}-\frac{e^2 \cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )}{4 b^2 d}+\frac{3 e^2 \cos \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{4 b^2 d}-\frac{e^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((e^2*(c + d*x)^2*Sqrt[1 - (c + d*x)^2])/(b*d*(a + b*ArcSin[c + d*x]))) + (e^2*CosIntegral[a/b + ArcSin[c + d
*x]]*Sin[a/b])/(4*b^2*d) - (3*e^2*CosIntegral[(3*a)/b + 3*ArcSin[c + d*x]]*Sin[(3*a)/b])/(4*b^2*d) - (e^2*Cos[
a/b]*SinIntegral[a/b + ArcSin[c + d*x]])/(4*b^2*d) + (3*e^2*Cos[(3*a)/b]*SinIntegral[(3*a)/b + 3*ArcSin[c + d*
x]])/(4*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)^2}{\left (a+b \sin ^{-1}(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^2 x^2}{\left (a+b \sin ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^2 \operatorname{Subst}\left (\int \frac{x^2}{\left (a+b \sin ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{d}\\ &=-\frac{e^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{e^2 \operatorname{Subst}\left (\int \left (-\frac{\sin (x)}{4 (a+b x)}+\frac{3 \sin (3 x)}{4 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{b d}\\ &=-\frac{e^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac{e^2 \operatorname{Subst}\left (\int \frac{\sin (x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{4 b d}+\frac{\left (3 e^2\right ) \operatorname{Subst}\left (\int \frac{\sin (3 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{4 b d}\\ &=-\frac{e^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac{\left (e^2 \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+d x)\right )}{4 b d}+\frac{\left (3 e^2 \cos \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 )}{4 b d}+\frac{\left (e^2 \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+d x)\right )}{4 b d}-\frac{\left (3 e^2 \sin \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 )}{4 b d}\\ &=-\frac{e^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{e^2 \text{Ci}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right ) \sin \left (\frac{a}{b}\right )}{4 b^2 d}-\frac{3 e^2 \text{Ci}\left (\frac{3 a}{b}+3 \sin ^{-1}(c+d x)\right ) \sin \left (\frac{3 a}{b}\right )}{4 b^2 d}-\frac{e^2 \cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )}{4 b^2 d}+\frac{3 e^2 \cos \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{4 b^2 d}\\ \end{align*}

Mathematica [A]  time = 0.783611, size = 140, normalized size = 0.75 \[ \frac{e^2 \left (\sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )-3 \sin \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (3 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )-\cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )+3 \cos \left (\frac{3 a}{b}\right ) \text{Si}\left (3 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )-\frac{4 b \sqrt{1-(c+d x)^2} (c+d x)^2}{a+b \sin ^{-1}(c+d x)}\right )}{4 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^2*((-4*b*(c + d*x)^2*Sqrt[1 - (c + d*x)^2])/(a + b*ArcSin[c + d*x]) + CosIntegral[a/b + ArcSin[c + d*x]]*Si
n[a/b] - 3*CosIntegral[3*(a/b + ArcSin[c + d*x])]*Sin[(3*a)/b] - Cos[a/b]*SinIntegral[a/b + ArcSin[c + d*x]] +
3*Cos[(3*a)/b]*SinIntegral[3*(a/b + ArcSin[c + d*x])]))/(4*b^2*d)

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Maple [A]  time = 0.048, size = 266, normalized size = 1.4 \begin{align*} -{\frac{{e}^{2}}{4\,d \left ( a+b\arcsin \left ( dx+c \right ) \right ){b}^{2}} \left ( \arcsin \left ( dx+c \right ){\it Si} \left ( \arcsin \left ( dx+c \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ) b-\arcsin \left ( dx+c \right ){\it Ci} \left ( \arcsin \left ( dx+c \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ) b-3\,\arcsin \left ( dx+c \right ){\it Si} \left ( 3\,\arcsin \left ( dx+c \right ) +3\,{\frac{a}{b}} \right ) \cos \left ( 3\,{\frac{a}{b}} \right ) b+3\,\arcsin \left ( dx+c \right ){\it Ci} \left ( 3\,\arcsin \left ( dx+c \right ) +3\,{\frac{a}{b}} \right ) \sin \left ( 3\,{\frac{a}{b}} \right ) b+{\it Si} \left ( \arcsin \left ( dx+c \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ) a-{\it Ci} \left ( \arcsin \left ( dx+c \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ) a-3\,{\it Si} \left ( 3\,\arcsin \left ( dx+c \right ) +3\,{\frac{a}{b}} \right ) \cos \left ( 3\,{\frac{a}{b}} \right ) a+3\,{\it Ci} \left ( 3\,\arcsin \left ( dx+c \right ) +3\,{\frac{a}{b}} \right ) \sin \left ( 3\,{\frac{a}{b}} \right ) a+\sqrt{1- \left ( dx+c \right ) ^{2}}b-\cos \left ( 3\,\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)^2/(a+b*arcsin(d*x+c))^2,x)

[Out]

-1/4/d*e^2*(arcsin(d*x+c)*Si(arcsin(d*x+c)+a/b)*cos(a/b)*b-arcsin(d*x+c)*Ci(arcsin(d*x+c)+a/b)*sin(a/b)*b-3*ar
csin(d*x+c)*Si(3*arcsin(d*x+c)+3*a/b)*cos(3*a/b)*b+3*arcsin(d*x+c)*Ci(3*arcsin(d*x+c)+3*a/b)*sin(3*a/b)*b+Si(a
rcsin(d*x+c)+a/b)*cos(a/b)*a-Ci(arcsin(d*x+c)+a/b)*sin(a/b)*a-3*Si(3*arcsin(d*x+c)+3*a/b)*cos(3*a/b)*a+3*Ci(3*
arcsin(d*x+c)+3*a/b)*sin(3*a/b)*a+(1-(d*x+c)^2)^(1/2)*b-cos(3*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)^2/(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^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}{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)^2/(a+b*arcsin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

[Out]

e**2*(Integral(c**2/(a**2 + 2*a*b*asin(c + d*x) + b**2*asin(c + d*x)**2), x) + Integral(d**2*x**2/(a**2 + 2*a*
b*asin(c + d*x) + b**2*asin(c + d*x)**2), x) + Integral(2*c*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 = 1.46423, size = 923, normalized size = 4.96 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*b*arcsin(d*x + c)*cos(a/b)^2*cos_integral(3*a/b + 3*arcsin(d*x + c))*e^2*sin(a/b)/(b^3*d*arcsin(d*x + c) +
a*b^2*d) + 3*b*arcsin(d*x + c)*cos(a/b)^3*e^2*sin_integral(3*a/b + 3*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) +
a*b^2*d) - 3*a*cos(a/b)^2*cos_integral(3*a/b + 3*arcsin(d*x + c))*e^2*sin(a/b)/(b^3*d*arcsin(d*x + c) + a*b^2
*d) + 3*a*cos(a/b)^3*e^2*sin_integral(3*a/b + 3*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) + 3/4*b*arc
sin(d*x + c)*cos_integral(3*a/b + 3*arcsin(d*x + c))*e^2*sin(a/b)/(b^3*d*arcsin(d*x + c) + a*b^2*d) + 1/4*b*ar
csin(d*x + c)*cos_integral(a/b + arcsin(d*x + c))*e^2*sin(a/b)/(b^3*d*arcsin(d*x + c) + a*b^2*d) - 9/4*b*arcsi
n(d*x + c)*cos(a/b)*e^2*sin_integral(3*a/b + 3*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) - 1/4*b*arcs
in(d*x + c)*cos(a/b)*e^2*sin_integral(a/b + arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) + 3/4*a*cos_int
egral(3*a/b + 3*arcsin(d*x + c))*e^2*sin(a/b)/(b^3*d*arcsin(d*x + c) + a*b^2*d) + 1/4*a*cos_integral(a/b + arc
sin(d*x + c))*e^2*sin(a/b)/(b^3*d*arcsin(d*x + c) + a*b^2*d) - 9/4*a*cos(a/b)*e^2*sin_integral(3*a/b + 3*arcsi
n(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) - 1/4*a*cos(a/b)*e^2*sin_integral(a/b + arcsin(d*x + c))/(b^3*d*
arcsin(d*x + c) + a*b^2*d) + (-(d*x + c)^2 + 1)^(3/2)*b*e^2/(b^3*d*arcsin(d*x + c) + a*b^2*d) - sqrt(-(d*x + c
)^2 + 1)*b*e^2/(b^3*d*arcsin(d*x + c) + a*b^2*d)