3.6.6 \(\int \sqrt {d+c d x} \sqrt {f-c f x} (a+b \text {ArcSin}(c x)) \, dx\) [506]

Optimal. Leaf size=134 \[ -\frac {b c x^2 \sqrt {d+c d x} \sqrt {f-c f x}}{4 \sqrt {1-c^2 x^2}}+\frac {1}{2} x \sqrt {d+c d x} \sqrt {f-c f x} (a+b \text {ArcSin}(c x))+\frac {\sqrt {d+c d x} \sqrt {f-c f x} (a+b \text {ArcSin}(c x))^2}{4 b c \sqrt {1-c^2 x^2}} \]

[Out]

1/2*x*(a+b*arcsin(c*x))*(c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)-1/4*b*c*x^2*(c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)/(-c^2*x^
2+1)^(1/2)+1/4*(a+b*arcsin(c*x))^2*(c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)/b/c/(-c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.12, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {4763, 4741, 4737, 30} \begin {gather*} \frac {\sqrt {c d x+d} \sqrt {f-c f x} (a+b \text {ArcSin}(c x))^2}{4 b c \sqrt {1-c^2 x^2}}+\frac {1}{2} x \sqrt {c d x+d} \sqrt {f-c f x} (a+b \text {ArcSin}(c x))-\frac {b c x^2 \sqrt {c d x+d} \sqrt {f-c f x}}{4 \sqrt {1-c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(a + b*ArcSin[c*x]),x]

[Out]

-1/4*(b*c*x^2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/Sqrt[1 - c^2*x^2] + (x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(a + b*A
rcSin[c*x]))/2 + (Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(a + b*ArcSin[c*x])^2)/(4*b*c*Sqrt[1 - c^2*x^2])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 4737

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

Rule 4741

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

Rule 4763

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(q_), x_Symbol] :> D
ist[(d + e*x)^q*((f + g*x)^q/(1 - c^2*x^2)^q), Int[(d + e*x)^(p - q)*(1 - c^2*x^2)^q*(a + b*ArcSin[c*x])^n, x]
, x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^2 - e^2, 0] && HalfIntegerQ[p, q]
&& GeQ[p - q, 0]

Rubi steps

\begin {align*} \int \sqrt {d+c d x} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac {\left (\sqrt {d+c d x} \sqrt {f-c f x}\right ) \int \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\sqrt {1-c^2 x^2}}\\ &=\frac {1}{2} x \sqrt {d+c d x} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right )+\frac {\left (\sqrt {d+c d x} \sqrt {f-c f x}\right ) \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{2 \sqrt {1-c^2 x^2}}-\frac {\left (b c \sqrt {d+c d x} \sqrt {f-c f x}\right ) \int x \, dx}{2 \sqrt {1-c^2 x^2}}\\ &=-\frac {b c x^2 \sqrt {d+c d x} \sqrt {f-c f x}}{4 \sqrt {1-c^2 x^2}}+\frac {1}{2} x \sqrt {d+c d x} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right )+\frac {\sqrt {d+c d x} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c \sqrt {1-c^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.31, size = 207, normalized size = 1.54 \begin {gather*} \frac {1}{2} a x \sqrt {-f (-1+c x)} \sqrt {d (1+c x)}-\frac {a \sqrt {d} \sqrt {f} \text {ArcTan}\left (\frac {c x \sqrt {-f (-1+c x)} \sqrt {d (1+c x)}}{\sqrt {d} \sqrt {f} (-1+c x) (1+c x)}\right )}{2 c}+\frac {b \sqrt {d+c d x} \sqrt {f-c f x} \sqrt {-d f \left (1-c^2 x^2\right )} (\cos (2 \text {ArcSin}(c x))+2 \text {ArcSin}(c x) (\text {ArcSin}(c x)+\sin (2 \text {ArcSin}(c x))))}{8 c \sqrt {(-d-c d x) (f-c f x)} \sqrt {1-c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(a + b*ArcSin[c*x]),x]

[Out]

(a*x*Sqrt[-(f*(-1 + c*x))]*Sqrt[d*(1 + c*x)])/2 - (a*Sqrt[d]*Sqrt[f]*ArcTan[(c*x*Sqrt[-(f*(-1 + c*x))]*Sqrt[d*
(1 + c*x)])/(Sqrt[d]*Sqrt[f]*(-1 + c*x)*(1 + c*x))])/(2*c) + (b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*Sqrt[-(d*f*(1
- c^2*x^2))]*(Cos[2*ArcSin[c*x]] + 2*ArcSin[c*x]*(ArcSin[c*x] + Sin[2*ArcSin[c*x]])))/(8*c*Sqrt[(-d - c*d*x)*(
f - c*f*x)]*Sqrt[1 - c^2*x^2])

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Maple [F]
time = 0.12, size = 0, normalized size = 0.00 \[\int \sqrt {c d x +d}\, \left (a +b \arcsin \left (c x \right )\right ) \sqrt {-c f x +f}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d*x+d)^(1/2)*(a+b*arcsin(c*x))*(-c*f*x+f)^(1/2),x)

[Out]

int((c*d*x+d)^(1/2)*(a+b*arcsin(c*x))*(-c*f*x+f)^(1/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^(1/2)*(a+b*arcsin(c*x))*(-c*f*x+f)^(1/2),x, algorithm="maxima")

[Out]

b*sqrt(d)*sqrt(f)*integrate(sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)), x) + 1/2*
(sqrt(-c^2*d*f*x^2 + d*f)*x + d*f*arcsin(c*x)/(sqrt(d*f)*c))*a

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^(1/2)*(a+b*arcsin(c*x))*(-c*f*x+f)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*d*x + d)*sqrt(-c*f*x + f)*(b*arcsin(c*x) + a), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {d \left (c x + 1\right )} \sqrt {- f \left (c x - 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)**(1/2)*(a+b*asin(c*x))*(-c*f*x+f)**(1/2),x)

[Out]

Integral(sqrt(d*(c*x + 1))*sqrt(-f*(c*x - 1))*(a + b*asin(c*x)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^(1/2)*(a+b*arcsin(c*x))*(-c*f*x+f)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*d*x + d)*sqrt(-c*f*x + f)*(b*arcsin(c*x) + a), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\sqrt {d+c\,d\,x}\,\sqrt {f-c\,f\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))*(d + c*d*x)^(1/2)*(f - c*f*x)^(1/2),x)

[Out]

int((a + b*asin(c*x))*(d + c*d*x)^(1/2)*(f - c*f*x)^(1/2), x)

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