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

Optimal. Leaf size=183 \[ \frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 c^4 d^2}-\frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d}-\frac {b c x^5 \sqrt {d-c^2 d x^2}}{25 \sqrt {1-c^2 x^2}}+\frac {b x^3 \sqrt {d-c^2 d x^2}}{45 c \sqrt {1-c^2 x^2}}+\frac {2 b x \sqrt {d-c^2 d x^2}}{15 c^3 \sqrt {1-c^2 x^2}} \]

[Out]

-1/3*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))/c^4/d+1/5*(-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(c*x))/c^4/d^2+2/15*b*x*
(-c^2*d*x^2+d)^(1/2)/c^3/(-c^2*x^2+1)^(1/2)+1/45*b*x^3*(-c^2*d*x^2+d)^(1/2)/c/(-c^2*x^2+1)^(1/2)-1/25*b*c*x^5*
(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {266, 43, 4691, 12} \[ \frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 c^4 d^2}-\frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d}-\frac {b c x^5 \sqrt {d-c^2 d x^2}}{25 \sqrt {1-c^2 x^2}}+\frac {b x^3 \sqrt {d-c^2 d x^2}}{45 c \sqrt {1-c^2 x^2}}+\frac {2 b x \sqrt {d-c^2 d x^2}}{15 c^3 \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*x*Sqrt[d - c^2*d*x^2])/(15*c^3*Sqrt[1 - c^2*x^2]) + (b*x^3*Sqrt[d - c^2*d*x^2])/(45*c*Sqrt[1 - c^2*x^2])
- (b*c*x^5*Sqrt[d - c^2*d*x^2])/(25*Sqrt[1 - c^2*x^2]) - ((d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]))/(3*c^4*d)
+ ((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/(5*c^4*d^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 4691

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[x^
m*(1 - c^2*x^2)^p, x]}, Dist[a + b*ArcSin[c*x], Int[x^m*(d + e*x^2)^p, x], x] - Dist[(b*c*d^(p - 1/2)*Sqrt[d +
e*x^2])/Sqrt[1 - c^2*x^2], Int[SimplifyIntegrand[u/Sqrt[1 - c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x
] && EqQ[c^2*d + e, 0] && IGtQ[p + 1/2, 0] && (IGtQ[(m + 1)/2, 0] || ILtQ[(m + 2*p + 3)/2, 0])

Rubi steps

\begin {align*} \int x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx &=-\frac {\left (b c \sqrt {d-c^2 d x^2}\right ) \int \frac {-2-c^2 x^2+3 c^4 x^4}{15 c^4} \, dx}{\sqrt {1-c^2 x^2}}+\left (a+b \sin ^{-1}(c x)\right ) \int x^3 \sqrt {d-c^2 d x^2} \, dx\\ &=-\frac {\left (b \sqrt {d-c^2 d x^2}\right ) \int \left (-2-c^2 x^2+3 c^4 x^4\right ) \, dx}{15 c^3 \sqrt {1-c^2 x^2}}+\frac {1}{2} \left (a+b \sin ^{-1}(c x)\right ) \operatorname {Subst}\left (\int x \sqrt {d-c^2 d x} \, dx,x,x^2\right )\\ &=\frac {2 b x \sqrt {d-c^2 d x^2}}{15 c^3 \sqrt {1-c^2 x^2}}+\frac {b x^3 \sqrt {d-c^2 d x^2}}{45 c \sqrt {1-c^2 x^2}}-\frac {b c x^5 \sqrt {d-c^2 d x^2}}{25 \sqrt {1-c^2 x^2}}+\frac {1}{2} \left (a+b \sin ^{-1}(c x)\right ) \operatorname {Subst}\left (\int \left (\frac {\sqrt {d-c^2 d x}}{c^2}-\frac {\left (d-c^2 d x\right )^{3/2}}{c^2 d}\right ) \, dx,x,x^2\right )\\ &=\frac {2 b x \sqrt {d-c^2 d x^2}}{15 c^3 \sqrt {1-c^2 x^2}}+\frac {b x^3 \sqrt {d-c^2 d x^2}}{45 c \sqrt {1-c^2 x^2}}-\frac {b c x^5 \sqrt {d-c^2 d x^2}}{25 \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d}+\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 c^4 d^2}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 134, normalized size = 0.73 \[ \frac {\sqrt {d-c^2 d x^2} \left (15 a \sqrt {1-c^2 x^2} \left (3 c^4 x^4-c^2 x^2-2\right )+b \left (-9 c^5 x^5+5 c^3 x^3+30 c x\right )+15 b \sqrt {1-c^2 x^2} \left (3 c^4 x^4-c^2 x^2-2\right ) \sin ^{-1}(c x)\right )}{225 c^4 \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d - c^2*d*x^2]*(15*a*Sqrt[1 - c^2*x^2]*(-2 - c^2*x^2 + 3*c^4*x^4) + b*(30*c*x + 5*c^3*x^3 - 9*c^5*x^5) +
15*b*Sqrt[1 - c^2*x^2]*(-2 - c^2*x^2 + 3*c^4*x^4)*ArcSin[c*x]))/(225*c^4*Sqrt[1 - c^2*x^2])

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fricas [A]  time = 0.57, size = 150, normalized size = 0.82 \[ \frac {{\left (9 \, b c^{5} x^{5} - 5 \, b c^{3} x^{3} - 30 \, b c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} + 15 \, {\left (3 \, a c^{6} x^{6} - 4 \, a c^{4} x^{4} - a c^{2} x^{2} + {\left (3 \, b c^{6} x^{6} - 4 \, b c^{4} x^{4} - b c^{2} x^{2} + 2 \, b\right )} \arcsin \left (c x\right ) + 2 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{225 \, {\left (c^{6} x^{2} - c^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/225*((9*b*c^5*x^5 - 5*b*c^3*x^3 - 30*b*c*x)*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1) + 15*(3*a*c^6*x^6 - 4*a*
c^4*x^4 - a*c^2*x^2 + (3*b*c^6*x^6 - 4*b*c^4*x^4 - b*c^2*x^2 + 2*b)*arcsin(c*x) + 2*a)*sqrt(-c^2*d*x^2 + d))/(
c^6*x^2 - c^4)

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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(x^3*(-c^2*d*x^2+d)^(1/2)*(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 = 0.29, size = 544, normalized size = 2.97 \[ a \left (-\frac {x^{2} \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{5 c^{2} d}-\frac {2 \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{15 d \,c^{4}}\right )+b \left (\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (16 c^{6} x^{6}-28 c^{4} x^{4}-16 i \sqrt {-c^{2} x^{2}+1}\, x^{5} c^{5}+13 c^{2} x^{2}+20 i \sqrt {-c^{2} x^{2}+1}\, x^{3} c^{3}-5 i \sqrt {-c^{2} x^{2}+1}\, x c -1\right ) \left (i+5 \arcsin \left (c x \right )\right )}{800 c^{4} \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (c^{2} x^{2}-i \sqrt {-c^{2} x^{2}+1}\, x c -1\right ) \left (i+\arcsin \left (c x \right )\right )}{16 c^{4} \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, x c +c^{2} x^{2}-1\right ) \left (\arcsin \left (c x \right )-i\right )}{16 c^{4} \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (4 i \sqrt {-c^{2} x^{2}+1}\, x^{3} c^{3}+4 c^{4} x^{4}-3 i \sqrt {-c^{2} x^{2}+1}\, x c -5 c^{2} x^{2}+1\right ) \left (-i+3 \arcsin \left (c x \right )\right )}{288 c^{4} \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, x c +c^{2} x^{2}-1\right ) \left (17 i+15 \arcsin \left (c x \right )\right ) \cos \left (4 \arcsin \left (c x \right )\right )}{3600 c^{4} \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i c^{2} x^{2}-c x \sqrt {-c^{2} x^{2}+1}-i\right ) \left (2 i+15 \arcsin \left (c x \right )\right ) \sin \left (4 \arcsin \left (c x \right )\right )}{900 c^{4} \left (c^{2} x^{2}-1\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(-1/5*x^2*(-c^2*d*x^2+d)^(3/2)/c^2/d-2/15/d/c^4*(-c^2*d*x^2+d)^(3/2))+b*(1/800*(-d*(c^2*x^2-1))^(1/2)*(16*c^
6*x^6-28*c^4*x^4-16*I*(-c^2*x^2+1)^(1/2)*x^5*c^5+13*c^2*x^2+20*I*(-c^2*x^2+1)^(1/2)*x^3*c^3-5*I*(-c^2*x^2+1)^(
1/2)*x*c-1)*(I+5*arcsin(c*x))/c^4/(c^2*x^2-1)-1/16*(-d*(c^2*x^2-1))^(1/2)*(c^2*x^2-I*(-c^2*x^2+1)^(1/2)*x*c-1)
*(I+arcsin(c*x))/c^4/(c^2*x^2-1)-1/16*(-d*(c^2*x^2-1))^(1/2)*(I*(-c^2*x^2+1)^(1/2)*x*c+c^2*x^2-1)*(arcsin(c*x)
-I)/c^4/(c^2*x^2-1)+1/288*(-d*(c^2*x^2-1))^(1/2)*(4*I*(-c^2*x^2+1)^(1/2)*x^3*c^3+4*c^4*x^4-3*I*(-c^2*x^2+1)^(1
/2)*x*c-5*c^2*x^2+1)*(-I+3*arcsin(c*x))/c^4/(c^2*x^2-1)-1/3600*(-d*(c^2*x^2-1))^(1/2)*(I*(-c^2*x^2+1)^(1/2)*x*
c+c^2*x^2-1)*(17*I+15*arcsin(c*x))*cos(4*arcsin(c*x))/c^4/(c^2*x^2-1)-1/900*(-d*(c^2*x^2-1))^(1/2)*(I*c^2*x^2-
c*x*(-c^2*x^2+1)^(1/2)-I)*(2*I+15*arcsin(c*x))*sin(4*arcsin(c*x))/c^4/(c^2*x^2-1))

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maxima [A]  time = 0.51, size = 138, normalized size = 0.75 \[ -\frac {1}{15} \, b {\left (\frac {3 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x^{2}}{c^{2} d} + \frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}}}{c^{4} d}\right )} \arcsin \left (c x\right ) - \frac {1}{15} \, a {\left (\frac {3 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x^{2}}{c^{2} d} + \frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}}}{c^{4} d}\right )} - \frac {{\left (9 \, c^{4} \sqrt {d} x^{5} - 5 \, c^{2} \sqrt {d} x^{3} - 30 \, \sqrt {d} x\right )} b}{225 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*b*(3*(-c^2*d*x^2 + d)^(3/2)*x^2/(c^2*d) + 2*(-c^2*d*x^2 + d)^(3/2)/(c^4*d))*arcsin(c*x) - 1/15*a*(3*(-c^
2*d*x^2 + d)^(3/2)*x^2/(c^2*d) + 2*(-c^2*d*x^2 + d)^(3/2)/(c^4*d)) - 1/225*(9*c^4*sqrt(d)*x^5 - 5*c^2*sqrt(d)*
x^3 - 30*sqrt(d)*x)*b/c^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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