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*f^2*(-c^2*x^2+1)*(a+b*arcsin(c*x))/c/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)-1/2*f^2*x*(-c^2*x^2+1)*(a+b*arcsin(c*x
))/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)-2*b*f^2*x*(-c^2*x^2+1)^(1/2)/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)+1/4*b*c*f^2*
x^2*(-c^2*x^2+1)^(1/2)/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)+3/4*f^2*(a+b*arcsin(c*x))^2*(-c^2*x^2+1)^(1/2)/b/c/(c*
d*x+d)^(1/2)/(-c*f*x+f)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.43, antiderivative size = 242, normalized size of antiderivative = 1.00, 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 8

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

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 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 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 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 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))

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 = 1.32, 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])

________________________________________________________________________________________

fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a c f x - a f + {\left (b c f x - b f\right )} \arcsin \left (c x\right )\right )} \sqrt {-c f x + f}}{\sqrt {c d x + d}}, x\right ) \]

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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-c f x + f\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt {c d x + d}}\,{d x} \]

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)

________________________________________________________________________________________

maple [F]  time = 0.32, size = 0, normalized size = 0.00 \[ \int \frac {\left (-c f x +f \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )}{\sqrt {c d x +d}}\, dx \]

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]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, {\left (\frac {\sqrt {-c^{2} d f x^{2} + d f} f x}{d} - \frac {3 \, f^{2} \arcsin \left (c x\right )}{\sqrt {d f} c} - \frac {4 \, \sqrt {-c^{2} d f x^{2} + d f} f}{c d}\right )} a - \frac {-\frac {1}{4} \, {\left (\frac {2 \, \sqrt {c x + 1} {\left (-c x + 1\right )}^{\frac {3}{2}} f \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{c} + \frac {{\left (c x - 1\right )}^{2} f}{c} - 6 \, f x - \frac {12 \, f \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {-c x + 1}\right )^{2}}{c} + \frac {6 \, \sqrt {c x + 1} \sqrt {-c x + 1} f \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{c} - \frac {12 \, f \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {-c x + 1}\right ) \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{c} + \frac {6 \, f}{c}\right )} b \sqrt {f}}{\sqrt {d}} \]

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]

-1/2*(sqrt(-c^2*d*f*x^2 + d*f)*f*x/d - 3*f^2*arcsin(c*x)/(sqrt(d*f)*c) - 4*sqrt(-c^2*d*f*x^2 + d*f)*f/(c*d))*a
- b*sqrt(f)*integrate((c*f*x - f)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/sqrt(c*x + 1), x)
/sqrt(d)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (f-c\,f\,x\right )}^{3/2}}{\sqrt {d+c\,d\,x}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- f \left (c x - 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\sqrt {d \left (c x + 1\right )}}\, dx \]

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]

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

________________________________________________________________________________________