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

Optimal. Leaf size=244 \[ \frac {e^2 \cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{4 b c^3}-\frac {e^2 \cos \left (\frac {3 a}{b}\right ) \text {Ci}\left (\frac {3 a}{b}+3 \sin ^{-1}(c x)\right )}{4 b c^3}+\frac {e^2 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{4 b c^3}-\frac {e^2 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \sin ^{-1}(c x)\right )}{4 b c^3}-\frac {d e \sin \left (\frac {2 a}{b}\right ) \text {Ci}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{b c^2}+\frac {d e \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{b c^2}+\frac {d^2 \cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{b c}+\frac {d^2 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{b c} \]

[Out]

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

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Rubi [A]  time = 0.67, antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4747, 6742, 3303, 3299, 3302, 4406} \[ -\frac {d e \sin \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{b c^2}+\frac {e^2 \cos \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{4 b c^3}-\frac {e^2 \cos \left (\frac {3 a}{b}\right ) \text {CosIntegral}\left (\frac {3 a}{b}+3 \sin ^{-1}(c x)\right )}{4 b c^3}+\frac {d e \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{b c^2}+\frac {e^2 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{4 b c^3}-\frac {e^2 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \sin ^{-1}(c x)\right )}{4 b c^3}+\frac {d^2 \cos \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{b c}+\frac {d^2 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 4747

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

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.48, size = 187, normalized size = 0.77 \[ \frac {\cos \left (\frac {a}{b}\right ) \left (4 c^2 d^2+e^2\right ) \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )+4 c^2 d^2 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )-4 c d e \sin \left (\frac {2 a}{b}\right ) \text {Ci}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )-e^2 \cos \left (\frac {3 a}{b}\right ) \text {Ci}\left (3 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+4 c d e \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+e^2 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )-e^2 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )}{4 b c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {e^{2} x^{2} + 2 \, d e x + d^{2}}{b \arcsin \left (c x\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.37, size = 334, normalized size = 1.37 \[ \frac {d^{2} \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b} + \arcsin \left (c x\right )\right )}{b c} - \frac {2 \, d \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right ) e \sin \left (\frac {a}{b}\right )}{b c^{2}} + \frac {2 \, d \cos \left (\frac {a}{b}\right )^{2} e \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{b c^{2}} + \frac {d^{2} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arcsin \left (c x\right )\right )}{b c} - \frac {\cos \left (\frac {a}{b}\right )^{3} \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (c x\right )\right ) e^{2}}{b c^{3}} - \frac {\cos \left (\frac {a}{b}\right )^{2} e^{2} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (c x\right )\right )}{b c^{3}} - \frac {d e \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{b c^{2}} + \frac {3 \, \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (c x\right )\right ) e^{2}}{4 \, b c^{3}} + \frac {\cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b} + \arcsin \left (c x\right )\right ) e^{2}}{4 \, b c^{3}} + \frac {e^{2} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (c x\right )\right )}{4 \, b c^{3}} + \frac {e^{2} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arcsin \left (c x\right )\right )}{4 \, b c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^2*cos(a/b)*cos_integral(a/b + arcsin(c*x))/(b*c) - 2*d*cos(a/b)*cos_integral(2*a/b + 2*arcsin(c*x))*e*sin(a/
b)/(b*c^2) + 2*d*cos(a/b)^2*e*sin_integral(2*a/b + 2*arcsin(c*x))/(b*c^2) + d^2*sin(a/b)*sin_integral(a/b + ar
csin(c*x))/(b*c) - cos(a/b)^3*cos_integral(3*a/b + 3*arcsin(c*x))*e^2/(b*c^3) - cos(a/b)^2*e^2*sin(a/b)*sin_in
tegral(3*a/b + 3*arcsin(c*x))/(b*c^3) - d*e*sin_integral(2*a/b + 2*arcsin(c*x))/(b*c^2) + 3/4*cos(a/b)*cos_int
egral(3*a/b + 3*arcsin(c*x))*e^2/(b*c^3) + 1/4*cos(a/b)*cos_integral(a/b + arcsin(c*x))*e^2/(b*c^3) + 1/4*e^2*
sin(a/b)*sin_integral(3*a/b + 3*arcsin(c*x))/(b*c^3) + 1/4*e^2*sin(a/b)*sin_integral(a/b + arcsin(c*x))/(b*c^3
)

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maple [A]  time = 0.08, size = 206, normalized size = 0.84 \[ \frac {4 \Si \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) c^{2} d^{2}+4 \Ci \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) c^{2} d^{2}+4 \Si \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) c d e -4 \Ci \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) c d e +\Si \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) e^{2}+\Ci \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) e^{2}-\Si \left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) e^{2}-\Ci \left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) e^{2}}{4 c^{3} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{2}}{b \arcsin \left (c x\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (d+e\,x\right )}^2}{a+b\,\mathrm {asin}\left (c\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^2/(a + b*asin(c*x)),x)

[Out]

int((d + e*x)^2/(a + b*asin(c*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d + e x\right )^{2}}{a + b \operatorname {asin}{\left (c x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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