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

Optimal. Leaf size=188 \[ \frac {1}{4} x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )+\frac {3}{8} d x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {3 d \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c \sqrt {1-c^2 x^2}}-\frac {5 b c d x^2 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}}+\frac {b c^3 d x^4 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {4649, 4647, 4641, 30, 14} \[ \frac {1}{4} x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )+\frac {3}{8} d x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {3 d \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c \sqrt {1-c^2 x^2}}+\frac {b c^3 d x^4 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}}-\frac {5 b c d x^2 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac {1}{4} x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} (3 d) \int \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx-\frac {\left (b c d \sqrt {d-c^2 d x^2}\right ) \int x \left (1-c^2 x^2\right ) \, dx}{4 \sqrt {1-c^2 x^2}}\\ &=\frac {3}{8} d x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )+\frac {\left (3 d \sqrt {d-c^2 d x^2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{8 \sqrt {1-c^2 x^2}}-\frac {\left (b c d \sqrt {d-c^2 d x^2}\right ) \int \left (x-c^2 x^3\right ) \, dx}{4 \sqrt {1-c^2 x^2}}-\frac {\left (3 b c d \sqrt {d-c^2 d x^2}\right ) \int x \, dx}{8 \sqrt {1-c^2 x^2}}\\ &=-\frac {5 b c d x^2 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}}+\frac {b c^3 d x^4 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}}+\frac {3}{8} d x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )+\frac {3 d \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c \sqrt {1-c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.62, size = 210, normalized size = 1.12 \[ \frac {d \sqrt {d-c^2 d x^2} \left (16 a c x \sqrt {1-c^2 x^2} \left (5-2 c^2 x^2\right )+16 b \cos \left (2 \sin ^{-1}(c x)\right )+b \cos \left (4 \sin ^{-1}(c x)\right )\right )-48 a d^{3/2} \sqrt {1-c^2 x^2} \tan ^{-1}\left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (c^2 x^2-1\right )}\right )+24 b d \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)^2+4 b d \sqrt {d-c^2 d x^2} \left (8 \sin \left (2 \sin ^{-1}(c x)\right )+\sin \left (4 \sin ^{-1}(c x)\right )\right ) \sin ^{-1}(c x)}{128 c \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(24*b*d*Sqrt[d - c^2*d*x^2]*ArcSin[c*x]^2 - 48*a*d^(3/2)*Sqrt[1 - c^2*x^2]*ArcTan[(c*x*Sqrt[d - c^2*d*x^2])/(S
qrt[d]*(-1 + c^2*x^2))] + d*Sqrt[d - c^2*d*x^2]*(16*a*c*x*(5 - 2*c^2*x^2)*Sqrt[1 - c^2*x^2] + 16*b*Cos[2*ArcSi
n[c*x]] + b*Cos[4*ArcSin[c*x]]) + 4*b*d*Sqrt[d - c^2*d*x^2]*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 = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a c^{2} d x^{2} - a d + {\left (b c^{2} d x^{2} - b d\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [C]  time = 0.17, size = 1167, normalized size = 6.21 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-17/256*b*(-d*(c^2*x^2-1))^(1/2)*cos(3*arcsin(c*x))*d/c/(c^2*x^2-1)+7/64*I*b*(-d*(c^2*x^2-1))^(1/2)*d/(c^2*x^2
-1)*x-5/16*b*(-d*(c^2*x^2-1))^(1/2)*d/(c^2*x^2-1)*arcsin(c*x)*x-17/256*b*(-d*(c^2*x^2-1))^(1/2)*d/c/(c^2*x^2-1
)*(-c^2*x^2+1)^(1/2)+5/32*b*(-d*(c^2*x^2-1))^(1/2)*d*c/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^2-15/256*b*(-d*(c^2*x^
2-1))^(1/2)*sin(3*arcsin(c*x))*d/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x+17/256*b*(-d*(c^2*x^2-1))^(1/2)*cos(3*arcsin
(c*x))*d*c/(c^2*x^2-1)*x^2-1/32*I*b*(-d*(c^2*x^2-1))^(1/2)*d*c^4/(c^2*x^2-1)*x^5-5/64*I*b*(-d*(c^2*x^2-1))^(1/
2)*d*c^2/(c^2*x^2-1)*x^3-15/256*I*b*(-d*(c^2*x^2-1))^(1/2)*sin(3*arcsin(c*x))*d/c/(c^2*x^2-1)-1/8*b*(-d*(c^2*x
^2-1))^(1/2)*d*c^4/(c^2*x^2-1)*arcsin(c*x)*x^5+7/16*b*(-d*(c^2*x^2-1))^(1/2)*d*c^2/(c^2*x^2-1)*arcsin(c*x)*x^3
-9/64*b*(-d*(c^2*x^2-1))^(1/2)*sin(3*arcsin(c*x))*d/c/(c^2*x^2-1)*arcsin(c*x)-3/16*b*(-d*(c^2*x^2-1))^(1/2)*(-
c^2*x^2+1)^(1/2)/c/(c^2*x^2-1)*arcsin(c*x)^2*d-1/32*b*(-d*(c^2*x^2-1))^(1/2)*d*c^3/(c^2*x^2-1)*(-c^2*x^2+1)^(1
/2)*x^4+1/8*I*b*(-d*(c^2*x^2-1))^(1/2)*d*c/(c^2*x^2-1)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x^2-7/64*I*b*(-d*(c^2*x^
2-1))^(1/2)*cos(3*arcsin(c*x))*d*c/(c^2*x^2-1)*arcsin(c*x)*x^2+9/64*I*b*(-d*(c^2*x^2-1))^(1/2)*sin(3*arcsin(c*
x))*d/(c^2*x^2-1)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x+1/8*I*b*(-d*(c^2*x^2-1))^(1/2)*d*c^3/(c^2*x^2-1)*arcsin(c*x
)*(-c^2*x^2+1)^(1/2)*x^4+3/8*a*d*x*(-c^2*d*x^2+d)^(1/2)+3/8*a*d^2/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2)*x/(-c^2*d
*x^2+d)^(1/2))+1/4*a*x*(-c^2*d*x^2+d)^(3/2)+15/256*I*b*(-d*(c^2*x^2-1))^(1/2)*sin(3*arcsin(c*x))*d*c/(c^2*x^2-
1)*x^2-7/64*I*b*(-d*(c^2*x^2-1))^(1/2)*d/c/(c^2*x^2-1)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)+17/256*I*b*(-d*(c^2*x^2-
1))^(1/2)*cos(3*arcsin(c*x))*d/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x+7/64*I*b*(-d*(c^2*x^2-1))^(1/2)*cos(3*arcsin(c
*x))*d/c/(c^2*x^2-1)*arcsin(c*x)+7/64*b*(-d*(c^2*x^2-1))^(1/2)*cos(3*arcsin(c*x))*d/(c^2*x^2-1)*arcsin(c*x)*(-
c^2*x^2+1)^(1/2)*x+9/64*b*(-d*(c^2*x^2-1))^(1/2)*sin(3*arcsin(c*x))*d*c/(c^2*x^2-1)*arcsin(c*x)*x^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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