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

Optimal. Leaf size=226 \[ \frac {3 x (c d x+d)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{8 \left (1-c^2 x^2\right )}+\frac {3 (c d x+d)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c \left (1-c^2 x^2\right )^{3/2}}+\frac {1}{4} x (c d x+d)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )-\frac {5 b c x^2 (c d x+d)^{3/2} (f-c f x)^{3/2}}{16 \left (1-c^2 x^2\right )^{3/2}}+\frac {b c^3 x^4 (c d x+d)^{3/2} (f-c f x)^{3/2}}{16 \left (1-c^2 x^2\right )^{3/2}} \]

[Out]

-5/16*b*c*x^2*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)/(-c^2*x^2+1)^(3/2)+1/16*b*c^3*x^4*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3
/2)/(-c^2*x^2+1)^(3/2)+1/4*x*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)*(a+b*arcsin(c*x))+3/8*x*(c*d*x+d)^(3/2)*(-c*f*x+
f)^(3/2)*(a+b*arcsin(c*x))/(-c^2*x^2+1)+3/16*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)*(a+b*arcsin(c*x))^2/b/c/(-c^2*x^
2+1)^(3/2)

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Rubi [A]  time = 0.22, antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4673, 4649, 4647, 4641, 30, 14} \[ \frac {3 x (c d x+d)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{8 \left (1-c^2 x^2\right )}+\frac {3 (c d x+d)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c \left (1-c^2 x^2\right )^{3/2}}+\frac {1}{4} x (c d x+d)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )+\frac {b c^3 x^4 (c d x+d)^{3/2} (f-c f x)^{3/2}}{16 \left (1-c^2 x^2\right )^{3/2}}-\frac {5 b c x^2 (c d x+d)^{3/2} (f-c f x)^{3/2}}{16 \left (1-c^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*b*c*x^2*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2))/(16*(1 - c^2*x^2)^(3/2)) + (b*c^3*x^4*(d + c*d*x)^(3/2)*(f -
c*f*x)^(3/2))/(16*(1 - c^2*x^2)^(3/2)) + (x*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)*(a + b*ArcSin[c*x]))/4 + (3*x*
(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)*(a + b*ArcSin[c*x]))/(8*(1 - c^2*x^2)) + (3*(d + c*d*x)^(3/2)*(f - c*f*x)^
(3/2)*(a + b*ArcSin[c*x])^2)/(16*b*c*(1 - c^2*x^2)^(3/2))

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 4649

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(x*(d + e*x^2)^p*(
a + b*ArcSin[c*x])^n)/(2*p + 1), x] + (Dist[(2*d*p)/(2*p + 1), Int[(d + e*x^2)^(p - 1)*(a + b*ArcSin[c*x])^n,
x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/((2*p + 1)*(1 - c^2*x^2)^FracPart[p]), Int[x*(1 - c
^2*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && Gt
Q[n, 0] && GtQ[p, 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 (d+c d x)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac {\left ((d+c d x)^{3/2} (f-c f x)^{3/2}\right ) \int \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\left (1-c^2 x^2\right )^{3/2}}\\ &=\frac {1}{4} x (d+c d x)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )+\frac {\left (3 (d+c d x)^{3/2} (f-c f x)^{3/2}\right ) \int \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{4 \left (1-c^2 x^2\right )^{3/2}}-\frac {\left (b c (d+c d x)^{3/2} (f-c f x)^{3/2}\right ) \int x \left (1-c^2 x^2\right ) \, dx}{4 \left (1-c^2 x^2\right )^{3/2}}\\ &=\frac {1}{4} x (d+c d x)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )+\frac {3 x (d+c d x)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{8 \left (1-c^2 x^2\right )}+\frac {\left (3 (d+c d x)^{3/2} (f-c f x)^{3/2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{8 \left (1-c^2 x^2\right )^{3/2}}-\frac {\left (b c (d+c d x)^{3/2} (f-c f x)^{3/2}\right ) \int \left (x-c^2 x^3\right ) \, dx}{4 \left (1-c^2 x^2\right )^{3/2}}-\frac {\left (3 b c (d+c d x)^{3/2} (f-c f x)^{3/2}\right ) \int x \, dx}{8 \left (1-c^2 x^2\right )^{3/2}}\\ &=-\frac {5 b c x^2 (d+c d x)^{3/2} (f-c f x)^{3/2}}{16 \left (1-c^2 x^2\right )^{3/2}}+\frac {b c^3 x^4 (d+c d x)^{3/2} (f-c f x)^{3/2}}{16 \left (1-c^2 x^2\right )^{3/2}}+\frac {1}{4} x (d+c d x)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )+\frac {3 x (d+c d x)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{8 \left (1-c^2 x^2\right )}+\frac {3 (d+c d x)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c \left (1-c^2 x^2\right )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 1.13, size = 247, normalized size = 1.09 \[ \frac {d f \sqrt {c d x+d} \sqrt {f-c f x} \left (16 a c x \sqrt {1-c^2 x^2} \left (5-2 c^2 x^2\right )+16 b \cos \left (2 \sin ^{-1}(c x)\right )+b \cos \left (4 \sin ^{-1}(c x)\right )\right )-48 a d^{3/2} f^{3/2} \sqrt {1-c^2 x^2} \tan ^{-1}\left (\frac {c x \sqrt {c d x+d} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (c^2 x^2-1\right )}\right )+24 b d f \sqrt {c d x+d} \sqrt {f-c f x} \sin ^{-1}(c x)^2+4 b d f \sqrt {c d x+d} \sqrt {f-c f x} \left (8 \sin \left (2 \sin ^{-1}(c x)\right )+\sin \left (4 \sin ^{-1}(c x)\right )\right ) \sin ^{-1}(c x)}{128 c \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(24*b*d*f*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]^2 - 48*a*d^(3/2)*f^(3/2)*Sqrt[1 - c^2*x^2]*ArcTan[(c*x*S
qrt[d + c*d*x]*Sqrt[f - c*f*x])/(Sqrt[d]*Sqrt[f]*(-1 + c^2*x^2))] + d*f*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(16*a*
c*x*(5 - 2*c^2*x^2)*Sqrt[1 - c^2*x^2] + 16*b*Cos[2*ArcSin[c*x]] + b*Cos[4*ArcSin[c*x]]) + 4*b*d*f*Sqrt[d + c*d
*x]*Sqrt[f - c*f*x]*ArcSin[c*x]*(8*Sin[2*ArcSin[c*x]] + Sin[4*ArcSin[c*x]]))/(128*c*Sqrt[1 - c^2*x^2])

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a c^{2} d f x^{2} - a d f + {\left (b c^{2} d f x^{2} - b d f\right )} \arcsin \left (c x\right )\right )} \sqrt {c d x + d} \sqrt {-c f x + f}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.27, size = 0, normalized size = 0.00 \[ \int \left (c d x +d \right )^{\frac {3}{2}} \left (-c f x +f \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b \sqrt {d} \sqrt {f} \int -{\left (c^{2} d f x^{2} - d f\right )} \sqrt {c x + 1} \sqrt {-c x + 1} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )\,{d x} + \frac {1}{8} \, {\left (3 \, \sqrt {-c^{2} d f x^{2} + d f} d f x + \frac {3 \, d^{2} f^{2} \arcsin \left (c x\right )}{\sqrt {d f} c} + 2 \, {\left (-c^{2} d f x^{2} + d f\right )}^{\frac {3}{2}} x\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*sqrt(d)*sqrt(f)*integrate(-(c^2*d*f*x^2 - d*f)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(
-c*x + 1)), x) + 1/8*(3*sqrt(-c^2*d*f*x^2 + d*f)*d*f*x + 3*d^2*f^2*arcsin(c*x)/(sqrt(d*f)*c) + 2*(-c^2*d*f*x^2
+ d*f)^(3/2)*x)*a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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