3.513 \(\int \frac{(f-c f x)^{3/2} (a+b \sin ^{-1}(c x))}{\sqrt{d+c d x}} \, dx\)

Optimal. Leaf size=242 \[ \frac{3 f^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c \sqrt{c d x+d} \sqrt{f-c f x}}-\frac{f^2 x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{2 \sqrt{c d x+d} \sqrt{f-c f x}}+\frac{2 f^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{c \sqrt{c d x+d} \sqrt{f-c f x}}+\frac{b c f^2 x^2 \sqrt{1-c^2 x^2}}{4 \sqrt{c d x+d} \sqrt{f-c f x}}-\frac{2 b f^2 x \sqrt{1-c^2 x^2}}{\sqrt{c d x+d} \sqrt{f-c f x}} \]

[Out]

(-2*b*f^2*x*Sqrt[1 - c^2*x^2])/(Sqrt[d + c*d*x]*Sqrt[f - c*f*x]) + (b*c*f^2*x^2*Sqrt[1 - c^2*x^2])/(4*Sqrt[d +
c*d*x]*Sqrt[f - c*f*x]) + (2*f^2*(1 - c^2*x^2)*(a + b*ArcSin[c*x]))/(c*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]) - (f^
2*x*(1 - c^2*x^2)*(a + b*ArcSin[c*x]))/(2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]) + (3*f^2*Sqrt[1 - c^2*x^2]*(a + b*A
rcSin[c*x])^2)/(4*b*c*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])

________________________________________________________________________________________

Rubi [A]  time = 0.425293, antiderivative size = 242, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.233, Rules used = {4673, 4763, 4641, 4677, 8, 4707, 30} \[ \frac{3 f^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c \sqrt{c d x+d} \sqrt{f-c f x}}-\frac{f^2 x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{2 \sqrt{c d x+d} \sqrt{f-c f x}}+\frac{2 f^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{c \sqrt{c d x+d} \sqrt{f-c f x}}+\frac{b c f^2 x^2 \sqrt{1-c^2 x^2}}{4 \sqrt{c d x+d} \sqrt{f-c f x}}-\frac{2 b f^2 x \sqrt{1-c^2 x^2}}{\sqrt{c d x+d} \sqrt{f-c f x}} \]

Antiderivative was successfully verified.

[In]

Int[((f - c*f*x)^(3/2)*(a + b*ArcSin[c*x]))/Sqrt[d + c*d*x],x]

[Out]

(-2*b*f^2*x*Sqrt[1 - c^2*x^2])/(Sqrt[d + c*d*x]*Sqrt[f - c*f*x]) + (b*c*f^2*x^2*Sqrt[1 - c^2*x^2])/(4*Sqrt[d +
c*d*x]*Sqrt[f - c*f*x]) + (2*f^2*(1 - c^2*x^2)*(a + b*ArcSin[c*x]))/(c*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]) - (f^
2*x*(1 - c^2*x^2)*(a + b*ArcSin[c*x]))/(2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]) + (3*f^2*Sqrt[1 - c^2*x^2]*(a + b*A
rcSin[c*x])^2)/(4*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]

Rule 4707

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

Rule 30

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

Rubi steps

\begin{align*} \int \frac{(f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{d+c d x}} \, dx &=\frac{\sqrt{1-c^2 x^2} \int \frac{(f-c f x)^2 \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{f^2 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{1-c^2 x^2}}-\frac{2 c f^2 x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{1-c^2 x^2}}+\frac{c^2 f^2 x^2 \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 (f^2 \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 (2 c f^2 \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{\left (c^2 f^2 \sqrt{1-c^2 x^2}\right ) \int \frac{x^2 \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{2 f^2 \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{f^2 x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{2 \sqrt{d+c d x} \sqrt{f-c f x}}+\frac{f^2 \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 (f^2 \sqrt{1-c^2 x^2}\right ) \int \frac{a+b \sin ^{-1}(c x)}{\sqrt{1-c^2 x^2}} \, dx}{2 \sqrt{d+c d x} \sqrt{f-c f x}}-\frac{\left (2 b f^2 \sqrt{1-c^2 x^2}\right ) \int 1 \, dx}{\sqrt{d+c d x} \sqrt{f-c f x}}+\frac{\left (b c f^2 \sqrt{1-c^2 x^2}\right ) \int x \, dx}{2 \sqrt{d+c d x} \sqrt{f-c f x}}\\ &=-\frac{2 b f^2 x \sqrt{1-c^2 x^2}}{\sqrt{d+c d x} \sqrt{f-c f x}}+\frac{b c f^2 x^2 \sqrt{1-c^2 x^2}}{4 \sqrt{d+c d x} \sqrt{f-c f x}}+\frac{2 f^2 \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{f^2 x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{2 \sqrt{d+c d x} \sqrt{f-c f x}}+\frac{3 f^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c \sqrt{d+c d x} \sqrt{f-c f x}}\\ \end{align*}

Mathematica [A]  time = 1.22425, size = 238, normalized size = 0.98 \[ \frac{-f \sqrt{c d x+d} \sqrt{f-c f x} \left (4 a (c x-4) \sqrt{1-c^2 x^2}+16 b c x+b \cos \left (2 \sin ^{-1}(c x)\right )\right )-12 a \sqrt{d} f^{3/2} \sqrt{1-c^2 x^2} \tan ^{-1}\left (\frac{c x \sqrt{c d x+d} \sqrt{f-c f x}}{\sqrt{d} \sqrt{f} \left (c^2 x^2-1\right )}\right )-4 b f (c x-4) \sqrt{1-c^2 x^2} \sqrt{c d x+d} \sqrt{f-c f x} \sin ^{-1}(c x)+6 b f \sqrt{c d x+d} \sqrt{f-c f x} \sin ^{-1}(c x)^2}{8 c d \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((f - c*f*x)^(3/2)*(a + b*ArcSin[c*x]))/Sqrt[d + c*d*x],x]

[Out]

(-4*b*f*(-4 + c*x)*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*Sqrt[1 - c^2*x^2]*ArcSin[c*x] + 6*b*f*Sqrt[d + c*d*x]*Sqrt[
f - c*f*x]*ArcSin[c*x]^2 - 12*a*Sqrt[d]*f^(3/2)*Sqrt[1 - c^2*x^2]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])
/(Sqrt[d]*Sqrt[f]*(-1 + c^2*x^2))] - f*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(16*b*c*x + 4*a*(-4 + c*x)*Sqrt[1 - c^2
*x^2] + b*Cos[2*ArcSin[c*x]]))/(8*c*d*Sqrt[1 - c^2*x^2])

________________________________________________________________________________________

Maple [F]  time = 0.23, size = 0, normalized size = 0. \begin{align*} \int{(a+b\arcsin \left ( cx \right ) ) \left ( -cfx+f \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{cdx+d}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-c*f*x+f)^(3/2)*(a+b*arcsin(c*x))/(c*d*x+d)^(1/2),x)

[Out]

int((-c*f*x+f)^(3/2)*(a+b*arcsin(c*x))/(c*d*x+d)^(1/2),x)

________________________________________________________________________________________

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*f*x+f)^(3/2)*(a+b*arcsin(c*x))/(c*d*x+d)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c*f*x+f)^(3/2)*(a+b*arcsin(c*x))/(c*d*x+d)^(1/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c*f*x+f)**(3/2)*(a+b*asin(c*x))/(c*d*x+d)**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-c f x + f\right )}^{\frac{3}{2}}{\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt{c d x + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c*f*x+f)^(3/2)*(a+b*arcsin(c*x))/(c*d*x+d)^(1/2),x, algorithm="giac")

[Out]

integrate((-c*f*x + f)^(3/2)*(b*arcsin(c*x) + a)/sqrt(c*d*x + d), x)