3.110 \(\int \frac {x^4 (a+b \sin ^{-1}(c x))}{\sqrt {d-c^2 d x^2}} \, dx\)

Optimal. Leaf size=200 \[ -\frac {x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d}+\frac {3 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c^5 \sqrt {d-c^2 d x^2}}-\frac {3 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 c^4 d}+\frac {b x^4 \sqrt {1-c^2 x^2}}{16 c \sqrt {d-c^2 d x^2}}+\frac {3 b x^2 \sqrt {1-c^2 x^2}}{16 c^3 \sqrt {d-c^2 d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 200, 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 = {4707, 4643, 4641, 30} \[ -\frac {x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d}-\frac {3 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 c^4 d}+\frac {3 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c^5 \sqrt {d-c^2 d x^2}}+\frac {b x^4 \sqrt {1-c^2 x^2}}{16 c \sqrt {d-c^2 d x^2}}+\frac {3 b x^2 \sqrt {1-c^2 x^2}}{16 c^3 \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcSin[c*x]))/Sqrt[d - c^2*d*x^2],x]

[Out]

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

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 4643

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 - c^2*x^2]/Sq
rt[d + e*x^2], Int[(a + b*ArcSin[c*x])^n/Sqrt[1 - c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*
d + e, 0] &&  !GtQ[d, 0]

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]

Rubi steps

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

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Mathematica [A]  time = 0.97, size = 161, normalized size = 0.80 \[ \frac {-\frac {16 a c x \left (2 c^2 x^2+3\right ) \sqrt {d-c^2 d x^2}}{d}-\frac {48 a \tan ^{-1}\left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (c^2 x^2-1\right )}\right )}{\sqrt {d}}+\frac {b \sqrt {1-c^2 x^2} \left (4 \sin ^{-1}(c x) \left (6 \sin ^{-1}(c x)-8 \sin \left (2 \sin ^{-1}(c x)\right )+\sin \left (4 \sin ^{-1}(c x)\right )\right )-16 \cos \left (2 \sin ^{-1}(c x)\right )+\cos \left (4 \sin ^{-1}(c x)\right )\right )}{\sqrt {d-c^2 d x^2}}}{128 c^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(a + b*ArcSin[c*x]))/Sqrt[d - c^2*d*x^2],x]

[Out]

((-16*a*c*x*(3 + 2*c^2*x^2)*Sqrt[d - c^2*d*x^2])/d - (48*a*ArcTan[(c*x*Sqrt[d - c^2*d*x^2])/(Sqrt[d]*(-1 + c^2
*x^2))])/Sqrt[d] + (b*Sqrt[1 - c^2*x^2]*(-16*Cos[2*ArcSin[c*x]] + Cos[4*ArcSin[c*x]] + 4*ArcSin[c*x]*(6*ArcSin
[c*x] - 8*Sin[2*ArcSin[c*x]] + Sin[4*ArcSin[c*x]])))/Sqrt[d - c^2*d*x^2])/(128*c^5)

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fricas [F]  time = 1.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (b x^{4} \arcsin \left (c x\right ) + a x^{4}\right )} \sqrt {-c^{2} d x^{2} + d}}{c^{2} d x^{2} - d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

integral(-(b*x^4*arcsin(c*x) + a*x^4)*sqrt(-c^2*d*x^2 + d)/(c^2*d*x^2 - d), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{4}}{\sqrt {-c^{2} d x^{2} + d}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.60, size = 381, normalized size = 1.90 \[ -\frac {a \,x^{3} \sqrt {-c^{2} d \,x^{2}+d}}{4 c^{2} d}-\frac {3 a x \sqrt {-c^{2} d \,x^{2}+d}}{8 c^{4} d}+\frac {3 a \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{8 c^{4} \sqrt {c^{2} d}}-\frac {3 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{2}}{16 c^{5} d \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-c^{2} x^{2}+1}}{16 c^{5} \sqrt {-d \left (c^{2} x^{2}-1\right )}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x}{8 c^{4} d \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \cos \left (5 \arcsin \left (c x \right )\right )}{256 c^{5} d \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \sin \left (5 \arcsin \left (c x \right )\right )}{64 c^{5} d \left (c^{2} x^{2}-1\right )}+\frac {15 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \cos \left (3 \arcsin \left (c x \right )\right )}{256 c^{5} d \left (c^{2} x^{2}-1\right )}+\frac {7 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \sin \left (3 \arcsin \left (c x \right )\right )}{64 c^{5} d \left (c^{2} x^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{8} \, a {\left (\frac {2 \, \sqrt {-c^{2} d x^{2} + d} x^{3}}{c^{2} d} + \frac {3 \, \sqrt {-c^{2} d x^{2} + d} x}{c^{4} d} - \frac {3 \, \arcsin \left (c x\right )}{c^{5} \sqrt {d}}\right )} + \frac {b \int \frac {x^{4} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{\sqrt {c x + 1} \sqrt {-c x + 1}}\,{d x}}{\sqrt {d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

-1/8*a*(2*sqrt(-c^2*d*x^2 + d)*x^3/(c^2*d) + 3*sqrt(-c^2*d*x^2 + d)*x/(c^4*d) - 3*arcsin(c*x)/(c^5*sqrt(d))) +
b*integrate(x^4*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/(sqrt(c*x + 1)*sqrt(-c*x + 1)), x)/sqrt(d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{\sqrt {d-c^2\,d\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a + b*asin(c*x)))/(d - c^2*d*x^2)^(1/2),x)

[Out]

int((x^4*(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 \[ \int \frac {x^{4} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\sqrt {- d \left (c x - 1\right ) \left (c x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+b*asin(c*x))/(-c**2*d*x**2+d)**(1/2),x)

[Out]

Integral(x**4*(a + b*asin(c*x))/sqrt(-d*(c*x - 1)*(c*x + 1)), x)

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