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

Optimal. Leaf size=273 \[ \frac{f \sqrt{c d x+d} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c \sqrt{1-c^2 x^2}}+\frac{f \left (1-c^2 x^2\right ) \sqrt{c d x+d} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right )}{3 c}+\frac{1}{2} f x \sqrt{c d x+d} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right )+\frac{b c^2 f x^3 \sqrt{c d x+d} \sqrt{f-c f x}}{9 \sqrt{1-c^2 x^2}}-\frac{b c f x^2 \sqrt{c d x+d} \sqrt{f-c f x}}{4 \sqrt{1-c^2 x^2}}-\frac{b f x \sqrt{c d x+d} \sqrt{f-c f x}}{3 \sqrt{1-c^2 x^2}} \]

[Out]

-(b*f*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(3*Sqrt[1 - c^2*x^2]) - (b*c*f*x^2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(
4*Sqrt[1 - c^2*x^2]) + (b*c^2*f*x^3*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(9*Sqrt[1 - c^2*x^2]) + (f*x*Sqrt[d + c*d
*x]*Sqrt[f - c*f*x]*(a + b*ArcSin[c*x]))/2 + (f*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(1 - c^2*x^2)*(a + b*ArcSin[c*
x]))/(3*c) + (f*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])

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Rubi [A]  time = 0.311758, antiderivative size = 273, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4673, 4763, 4647, 4641, 30, 4677} \[ \frac{f \sqrt{c d x+d} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c \sqrt{1-c^2 x^2}}+\frac{f \left (1-c^2 x^2\right ) \sqrt{c d x+d} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right )}{3 c}+\frac{1}{2} f x \sqrt{c d x+d} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right )+\frac{b c^2 f x^3 \sqrt{c d x+d} \sqrt{f-c f x}}{9 \sqrt{1-c^2 x^2}}-\frac{b c f x^2 \sqrt{c d x+d} \sqrt{f-c f x}}{4 \sqrt{1-c^2 x^2}}-\frac{b f x \sqrt{c d x+d} \sqrt{f-c f x}}{3 \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*f*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(3*Sqrt[1 - c^2*x^2]) - (b*c*f*x^2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(
4*Sqrt[1 - c^2*x^2]) + (b*c^2*f*x^3*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(9*Sqrt[1 - c^2*x^2]) + (f*x*Sqrt[d + c*d
*x]*Sqrt[f - c*f*x]*(a + b*ArcSin[c*x]))/2 + (f*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(1 - c^2*x^2)*(a + b*ArcSin[c*
x]))/(3*c) + (f*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 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 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 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 30

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

Rubi steps

\begin{align*} \int \sqrt{d+c d x} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac{\left (\sqrt{d+c d x} \sqrt{f-c f x}\right ) \int (f-c f x) \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\sqrt{1-c^2 x^2}}\\ &=\frac{\left (\sqrt{d+c d x} \sqrt{f-c f x}\right ) \int \left (f \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )-c f x \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )\right ) \, dx}{\sqrt{1-c^2 x^2}}\\ &=\frac{\left (f \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{\left (c f \sqrt{d+c d x} \sqrt{f-c f x}\right ) \int x \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\sqrt{1-c^2 x^2}}\\ &=\frac{1}{2} f x \sqrt{d+c d x} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right )+\frac{f \sqrt{d+c d x} \sqrt{f-c f x} \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c}+\frac{\left (f \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 f \sqrt{d+c d x} \sqrt{f-c f x}\right ) \int \left (1-c^2 x^2\right ) \, dx}{3 \sqrt{1-c^2 x^2}}-\frac{\left (b c f \sqrt{d+c d x} \sqrt{f-c f x}\right ) \int x \, dx}{2 \sqrt{1-c^2 x^2}}\\ &=-\frac{b f x \sqrt{d+c d x} \sqrt{f-c f x}}{3 \sqrt{1-c^2 x^2}}-\frac{b c f x^2 \sqrt{d+c d x} \sqrt{f-c f x}}{4 \sqrt{1-c^2 x^2}}+\frac{b c^2 f x^3 \sqrt{d+c d x} \sqrt{f-c f x}}{9 \sqrt{1-c^2 x^2}}+\frac{1}{2} f x \sqrt{d+c d x} \sqrt{f-c f x} \left (a+b \sin ^{-1}(c x)\right )+\frac{f \sqrt{d+c d x} \sqrt{f-c f x} \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c}+\frac{f \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*}

Mathematica [A]  time = 1.0032, size = 260, normalized size = 0.95 \[ \frac{f \sqrt{c d x+d} \sqrt{f-c f x} \left (12 a \sqrt{1-c^2 x^2} \left (-2 c^2 x^2+3 c x+2\right )+8 b c x \left (c^2 x^2-3\right )+9 b \cos \left (2 \sin ^{-1}(c x)\right )\right )-36 a \sqrt{d} f^{3/2} \sqrt{1-c^2 x^2} \tan ^{-1}\left (\frac{c x \sqrt{c d x+d} \sqrt{f-c f x}}{\sqrt{d} \sqrt{f} \left (c^2 x^2-1\right )}\right )+6 b f \sqrt{c d x+d} \sqrt{f-c f x} \left (4 \left (1-c^2 x^2\right )^{3/2}+3 \sin \left (2 \sin ^{-1}(c x)\right )\right ) \sin ^{-1}(c x)+18 b f \sqrt{c d x+d} \sqrt{f-c f x} \sin ^{-1}(c x)^2}{72 c \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(18*b*f*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]^2 - 36*a*Sqrt[d]*f^(3/2)*Sqrt[1 - c^2*x^2]*ArcTan[(c*x*Sqr
t[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]*(12*a*(2 +
3*c*x - 2*c^2*x^2)*Sqrt[1 - c^2*x^2] + 8*b*c*x*(-3 + c^2*x^2) + 9*b*Cos[2*ArcSin[c*x]]) + 6*b*f*Sqrt[d + c*d*
x]*Sqrt[f - c*f*x]*ArcSin[c*x]*(4*(1 - c^2*x^2)^(3/2) + 3*Sin[2*ArcSin[c*x]]))/(72*c*Sqrt[1 - c^2*x^2])

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Maple [F]  time = 0.236, size = 0, normalized size = 0. \begin{align*} \int \sqrt{cdx+d} \left ( -cfx+f \right ) ^{{\frac{3}{2}}} \left ( a+b\arcsin \left ( cx \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((c*d*x+d)^(1/2)*(-c*f*x+f)^(3/2)*(a+b*arcsin(c*x)),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)*(-c*f*x+f)^(3/2)*(a+b*arcsin(c*x)),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 (-{\left (a c f x - a f +{\left (b c f x - b f\right )} \arcsin \left (c x\right )\right )} \sqrt{c d x + d} \sqrt{-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)*(-c*f*x+f)^(3/2)*(a+b*arcsin(c*x)),x, algorithm="fricas")

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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