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

Optimal. Leaf size=141 \[ \frac{d \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c \sqrt{c d x+d} \sqrt{f-c f x}}-\frac{d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{c \sqrt{c d x+d} \sqrt{f-c f x}}+\frac{b d x \sqrt{1-c^2 x^2}}{\sqrt{c d x+d} \sqrt{f-c f x}} \]

[Out]

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

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Rubi [A]  time = 0.257583, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {4673, 4763, 4641, 4677, 8} \[ \frac{d \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c \sqrt{c d x+d} \sqrt{f-c f x}}-\frac{d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{c \sqrt{c d x+d} \sqrt{f-c f x}}+\frac{b d x \sqrt{1-c^2 x^2}}{\sqrt{c d x+d} \sqrt{f-c f x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 4763

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g},
x] && EqQ[c^2*d + e, 0] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ[n, 0] && (m == 1 || p > 0 ||
(n == 1 && p > -1) || (m == 2 && p < -2))

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 4677

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.705165, size = 200, normalized size = 1.42 \[ \frac{\frac{2 \sqrt{c d x+d} \sqrt{f-c f x} \left (b c x-a \sqrt{1-c^2 x^2}\right )}{\sqrt{1-c^2 x^2}}-2 a \sqrt{d} \sqrt{f} \tan ^{-1}\left (\frac{c x \sqrt{c d x+d} \sqrt{f-c f x}}{\sqrt{d} \sqrt{f} \left (c^2 x^2-1\right )}\right )+\frac{b \sqrt{c d x+d} \sqrt{f-c f x} \sin ^{-1}(c x)^2}{\sqrt{1-c^2 x^2}}-2 b \sqrt{c d x+d} \sqrt{f-c f x} \sin ^{-1}(c x)}{2 c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(b*c*x - a*Sqrt[1 - c^2*x^2]))/Sqrt[1 - c^2*x^2] - 2*b*Sqrt[d + c*d*x]*Sqr
t[f - c*f*x]*ArcSin[c*x] + (b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]^2)/Sqrt[1 - c^2*x^2] - 2*a*Sqrt[d]*S
qrt[f]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(Sqrt[d]*Sqrt[f]*(-1 + c^2*x^2))])/(2*c*f)

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Maple [F]  time = 0.243, size = 0, normalized size = 0. \begin{align*} \int{(a+b\arcsin \left ( cx \right ) )\sqrt{cdx+d}{\frac{1}{\sqrt{-cfx+f}}}}\, dx \end{align*}

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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{c d x + d} \sqrt{-c f x + f}{\left (b \arcsin \left (c x\right ) + a\right )}}{c f x - f}, x\right ) \end{align*}

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)/(c*f*x - f), x)

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

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c d x + d}{\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt{-c f x + f}}\,{d x} \end{align*}

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