3.230 \(\int \frac{c e+d e x}{(a+b \sin ^{-1}(c+d x))^3} \, dx\)

Optimal. Leaf size=157 \[ \frac{e \sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{b^3 d}-\frac{e \cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{b^3 d}+\frac{e (c+d x)^2}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac{e}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac{e \sqrt{1-(c+d x)^2} (c+d x)}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.329673, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.476, Rules used = {4805, 12, 4633, 4719, 4635, 4406, 3303, 3299, 3302, 4641} \[ \frac{e \sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 a}{b}+2 \sin ^{-1}(c+d x)\right )}{b^3 d}-\frac{e \cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \sin ^{-1}(c+d x)\right )}{b^3 d}+\frac{e (c+d x)^2}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac{e}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac{e \sqrt{1-(c+d x)^2} (c+d x)}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)/(a + b*ArcSin[c + d*x])^3,x]

[Out]

-(e*(c + d*x)*Sqrt[1 - (c + d*x)^2])/(2*b*d*(a + b*ArcSin[c + d*x])^2) - e/(2*b^2*d*(a + b*ArcSin[c + d*x])) +
(e*(c + d*x)^2)/(b^2*d*(a + b*ArcSin[c + d*x])) + (e*CosIntegral[(2*a)/b + 2*ArcSin[c + d*x]]*Sin[(2*a)/b])/(
b^3*d) - (e*Cos[(2*a)/b]*SinIntegral[(2*a)/b + 2*ArcSin[c + d*x]])/(b^3*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 4633

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[(c*(m + 1))/(b*(n + 1)), Int[(x^(m + 1)*(a + b*ArcSin[c*x])^(n + 1))
/Sqrt[1 - c^2*x^2], x], x] - Dist[m/(b*c*(n + 1)), Int[(x^(m - 1)*(a + b*ArcSin[c*x])^(n + 1))/Sqrt[1 - c^2*x^
2], x], x]) /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && LtQ[n, -2]

Rule 4719

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[
((f*x)^m*(a + b*ArcSin[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] - Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), Int[(f*x)^
(m - 1)*(a + b*ArcSin[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && LtQ[n,
-1] && GtQ[d, 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]

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 4641

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSin[c*x])^
(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0] && NeQ[n,
-1]

Rubi steps

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

Mathematica [A]  time = 0.57071, size = 107, normalized size = 0.68 \[ -\frac{e \left (-4 \sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )+4 \cos \left (\frac{2 a}{b}\right ) \text{Si}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )+\frac{b \left (2 \cos \left (2 \sin ^{-1}(c+d x)\right ) \left (a+b \sin ^{-1}(c+d x)\right )+b \sin \left (2 \sin ^{-1}(c+d x)\right )\right )}{\left (a+b \sin ^{-1}(c+d x)\right )^2}\right )}{4 b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)/(a + b*ArcSin[c + d*x])^3,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.035, size = 263, normalized size = 1.7 \begin{align*} -{\frac{e}{4\,d \left ( a+b\arcsin \left ( dx+c \right ) \right ) ^{2}{b}^{3}} \left ( 4\, \left ( \arcsin \left ( dx+c \right ) \right ) ^{2}{\it Si} \left ( 2\,\arcsin \left ( dx+c \right ) +2\,{\frac{a}{b}} \right ) \cos \left ( 2\,{\frac{a}{b}} \right ){b}^{2}-4\, \left ( \arcsin \left ( dx+c \right ) \right ) ^{2}{\it Ci} \left ( 2\,\arcsin \left ( dx+c \right ) +2\,{\frac{a}{b}} \right ) \sin \left ( 2\,{\frac{a}{b}} \right ){b}^{2}+8\,\arcsin \left ( dx+c \right ){\it Si} \left ( 2\,\arcsin \left ( dx+c \right ) +2\,{\frac{a}{b}} \right ) \cos \left ( 2\,{\frac{a}{b}} \right ) ab-8\,\arcsin \left ( dx+c \right ){\it Ci} \left ( 2\,\arcsin \left ( dx+c \right ) +2\,{\frac{a}{b}} \right ) \sin \left ( 2\,{\frac{a}{b}} \right ) ab+2\,\arcsin \left ( dx+c \right ) \cos \left ( 2\,\arcsin \left ( dx+c \right ) \right ){b}^{2}+4\,{\it Si} \left ( 2\,\arcsin \left ( dx+c \right ) +2\,{\frac{a}{b}} \right ) \cos \left ( 2\,{\frac{a}{b}} \right ){a}^{2}-4\,{\it Ci} \left ( 2\,\arcsin \left ( dx+c \right ) +2\,{\frac{a}{b}} \right ) \sin \left ( 2\,{\frac{a}{b}} \right ){a}^{2}+\sin \left ( 2\,\arcsin \left ( dx+c \right ) \right ){b}^{2}+2\,\cos \left ( 2\,\arcsin \left ( dx+c \right ) \right ) ab \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))^3,x)

[Out]

-1/4/d*e*(4*arcsin(d*x+c)^2*Si(2*arcsin(d*x+c)+2*a/b)*cos(2*a/b)*b^2-4*arcsin(d*x+c)^2*Ci(2*arcsin(d*x+c)+2*a/
b)*sin(2*a/b)*b^2+8*arcsin(d*x+c)*Si(2*arcsin(d*x+c)+2*a/b)*cos(2*a/b)*a*b-8*arcsin(d*x+c)*Ci(2*arcsin(d*x+c)+
2*a/b)*sin(2*a/b)*a*b+2*arcsin(d*x+c)*cos(2*arcsin(d*x+c))*b^2+4*Si(2*arcsin(d*x+c)+2*a/b)*cos(2*a/b)*a^2-4*Ci
(2*arcsin(d*x+c)+2*a/b)*sin(2*a/b)*a^2+sin(2*arcsin(d*x+c))*b^2+2*cos(2*arcsin(d*x+c))*a*b)/(a+b*arcsin(d*x+c)
)^2/b^3

________________________________________________________________________________________

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))^3,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{d e x + c e}{b^{3} \arcsin \left (d x + c\right )^{3} + 3 \, a b^{2} \arcsin \left (d x + c\right )^{2} + 3 \, a^{2} b \arcsin \left (d x + c\right ) + a^{3}}, 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))^3,x, algorithm="fricas")

[Out]

integral((d*e*x + c*e)/(b^3*arcsin(d*x + c)^3 + 3*a*b^2*arcsin(d*x + c)^2 + 3*a^2*b*arcsin(d*x + c) + a^3), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} e \left (\int \frac{c}{a^{3} + 3 a^{2} b \operatorname{asin}{\left (c + d x \right )} + 3 a b^{2} \operatorname{asin}^{2}{\left (c + d x \right )} + b^{3} \operatorname{asin}^{3}{\left (c + d x \right )}}\, dx + \int \frac{d x}{a^{3} + 3 a^{2} b \operatorname{asin}{\left (c + d x \right )} + 3 a b^{2} \operatorname{asin}^{2}{\left (c + d x \right )} + b^{3} \operatorname{asin}^{3}{\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))**3,x)

[Out]

e*(Integral(c/(a**3 + 3*a**2*b*asin(c + d*x) + 3*a*b**2*asin(c + d*x)**2 + b**3*asin(c + d*x)**3), x) + Integr
al(d*x/(a**3 + 3*a**2*b*asin(c + d*x) + 3*a*b**2*asin(c + d*x)**2 + b**3*asin(c + d*x)**3), x))

________________________________________________________________________________________

Giac [B]  time = 1.84671, size = 1218, normalized size = 7.76 \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)/(a+b*arcsin(d*x+c))^3,x, algorithm="giac")

[Out]

2*b^2*arcsin(d*x + c)^2*cos(a/b)*cos_integral(2*a/b + 2*arcsin(d*x + c))*e*sin(a/b)/(b^5*d*arcsin(d*x + c)^2 +
2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 2*b^2*arcsin(d*x + c)^2*cos(a/b)^2*e*sin_integral(2*a/b + 2*arcsin(d
*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 4*a*b*arcsin(d*x + c)*cos(a/b)*co
s_integral(2*a/b + 2*arcsin(d*x + c))*e*sin(a/b)/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^
3*d) - 4*a*b*arcsin(d*x + c)*cos(a/b)^2*e*sin_integral(2*a/b + 2*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2
*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 2*a^2*cos(a/b)*cos_integral(2*a/b + 2*arcsin(d*x + c))*e*sin(a/b)/(b^5
*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + b^2*arcsin(d*x + c)^2*e*sin_integral(2*a/b + 2
*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 2*a^2*cos(a/b)^2*e*sin_i
ntegral(2*a/b + 2*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + ((d*x +
c)^2 - 1)*b^2*arcsin(d*x + c)*e/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 2*a*b*arc
sin(d*x + c)*e*sin_integral(2*a/b + 2*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) +
a^2*b^3*d) - 1/2*sqrt(-(d*x + c)^2 + 1)*(d*x + c)*b^2*e/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) +
a^2*b^3*d) + ((d*x + c)^2 - 1)*a*b*e/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 1/2*
b^2*arcsin(d*x + c)*e/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + a^2*e*sin_integral(2
*a/b + 2*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 1/2*a*b*e/(b^5*d
*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d)