3.506 \(\int \sqrt {d+c d x} \sqrt {f-c f x} (a+b \sin ^{-1}(c x)) \, dx\)

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

[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 = 0.17, 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 = {4673, 4647, 4641, 30} \[ \frac {\sqrt {c d x+d} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c \sqrt {1-c^2 x^2}}+\frac {1}{2} x \sqrt {c d x+d} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right )-\frac {b c x^2 \sqrt {c d x+d} \sqrt {f-c f x}}{4 \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*x^2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(4*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 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]

Rule 4647

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[Sqrt[d + e*x^2]/(2*Sqrt[1 - c^2*x^2]), Int[(a + b*ArcSin[c*x])^n/Sqrt[1 -
c^2*x^2], x], x] - Dist[(b*c*n*Sqrt[d + e*x^2])/(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 4673

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 = 0.65, size = 207, normalized size = 1.54 \[ \frac {1}{2} a x \sqrt {d (c x+1)} \sqrt {-f (c x-1)}-\frac {a \sqrt {d} \sqrt {f} \tan ^{-1}\left (\frac {c x \sqrt {d (c x+1)} \sqrt {-f (c x-1)}}{\sqrt {d} \sqrt {f} (c x-1) (c x+1)}\right )}{2 c}+\frac {b \sqrt {c d x+d} \sqrt {f-c f x} \sqrt {-d f \left (1-c^2 x^2\right )} \left (2 \sin ^{-1}(c x) \left (\sin ^{-1}(c x)+\sin \left (2 \sin ^{-1}(c x)\right )\right )+\cos \left (2 \sin ^{-1}(c x)\right )\right )}{8 c \sqrt {1-c^2 x^2} \sqrt {(-c d x-d) (f-c f x)}} \]

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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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {c d x + d} \sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )}, x\right ) \]

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

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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maple [F]  time = 0.28, 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 \[ b \sqrt {d} \sqrt {f} \int \sqrt {c x + 1} \sqrt {-c x + 1} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )\,{d x} + \frac {1}{2} \, {\left (\sqrt {-c^{2} d f x^{2} + d f} x + \frac {d f \arcsin \left (c x\right )}{\sqrt {d f} c}\right )} a \]

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

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

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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