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

Optimal. Leaf size=141 \[ \frac {e^2 \cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a+b \sin ^{-1}(c+d x)}{b}\right )}{4 b d}-\frac {e^2 \cos \left (\frac {3 a}{b}\right ) \text {Ci}\left (\frac {3 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{4 b d}+\frac {e^2 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \sin ^{-1}(c+d x)}{b}\right )}{4 b d}-\frac {e^2 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{4 b d} \]

[Out]

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

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Rubi [A]  time = 0.28, antiderivative size = 137, normalized size of antiderivative = 0.97, number of steps used = 11, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4805, 12, 4635, 4406, 3303, 3299, 3302} \[ \frac {e^2 \cos \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )}{4 b d}-\frac {e^2 \cos \left (\frac {3 a}{b}\right ) \text {CosIntegral}\left (\frac {3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{4 b d}+\frac {e^2 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )}{4 b d}-\frac {e^2 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.21, size = 102, normalized size = 0.72 \[ \frac {e^2 \left (\cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )-\cos \left (\frac {3 a}{b}\right ) \text {Ci}\left (3 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )+\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )-\sin \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )\right )}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}{b \arcsin \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.25, size = 197, normalized size = 1.40 \[ -\frac {\cos \left (\frac {a}{b}\right )^{3} \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (d x + c\right )\right ) e^{2}}{b d} - \frac {\cos \left (\frac {a}{b}\right )^{2} e^{2} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (d x + c\right )\right )}{b d} + \frac {3 \, \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (d x + c\right )\right ) e^{2}}{4 \, b d} + \frac {\cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b} + \arcsin \left (d x + c\right )\right ) e^{2}}{4 \, b d} + \frac {e^{2} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (d x + c\right )\right )}{4 \, b d} + \frac {e^{2} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arcsin \left (d x + c\right )\right )}{4 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-cos(a/b)^3*cos_integral(3*a/b + 3*arcsin(d*x + c))*e^2/(b*d) - cos(a/b)^2*e^2*sin(a/b)*sin_integral(3*a/b + 3
*arcsin(d*x + c))/(b*d) + 3/4*cos(a/b)*cos_integral(3*a/b + 3*arcsin(d*x + c))*e^2/(b*d) + 1/4*cos(a/b)*cos_in
tegral(a/b + arcsin(d*x + c))*e^2/(b*d) + 1/4*e^2*sin(a/b)*sin_integral(3*a/b + 3*arcsin(d*x + c))/(b*d) + 1/4
*e^2*sin(a/b)*sin_integral(a/b + arcsin(d*x + c))/(b*d)

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maple [A]  time = 0.04, size = 103, normalized size = 0.73 \[ \frac {e^{2} \left (\Si \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )+\Ci \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )-\Si \left (3 \arcsin \left (d x +c \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )-\Ci \left (3 \arcsin \left (d x +c \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )\right )}{4 d b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,e+d\,e\,x\right )}^2}{a+b\,\mathrm {asin}\left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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