3.120 \(\int \frac {x^4 (a+b \sin ^{-1}(c x))}{(d-c^2 d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=214 \[ \frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {3 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c^5 d \sqrt {d-c^2 d x^2}}+\frac {3 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c^4 d^2}+\frac {b \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{2 c^5 d \sqrt {d-c^2 d x^2}}-\frac {b x^2 \sqrt {1-c^2 x^2}}{4 c^3 d \sqrt {d-c^2 d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {4703, 4707, 4643, 4641, 30, 266, 43} \[ \frac {3 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c^4 d^2}+\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {3 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c^5 d \sqrt {d-c^2 d x^2}}-\frac {b x^2 \sqrt {1-c^2 x^2}}{4 c^3 d \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{2 c^5 d \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(3/2),x]

[Out]

-(b*x^2*Sqrt[1 - c^2*x^2])/(4*c^3*d*Sqrt[d - c^2*d*x^2]) + (x^3*(a + b*ArcSin[c*x]))/(c^2*d*Sqrt[d - c^2*d*x^2
]) + (3*x*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(2*c^4*d^2) - (3*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/(
4*b*c^5*d*Sqrt[d - c^2*d*x^2]) + (b*Sqrt[1 - c^2*x^2]*Log[1 - c^2*x^2])/(2*c^5*d*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 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 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 4703

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(
f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n)/(2*e*(p + 1)), x] + (-Dist[(f^2*(m - 1))/(2*e*(p +
1)), Int[(f*x)^(m - 2)*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n, x], x] + Dist[(b*f*n*d^IntPart[p]*(d + e*x^2
)^FracPart[p])/(2*c*(p + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSi
n[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] && Gt
Q[m, 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]

Rubi steps

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

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Mathematica [A]  time = 0.52, size = 173, normalized size = 0.81 \[ \frac {-4 a c \sqrt {d} x \left (c^2 x^2-3\right )+12 a \sqrt {d-c^2 d x^2} \tan ^{-1}\left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (c^2 x^2-1\right )}\right )+b \sqrt {d} \left (\sqrt {1-c^2 x^2} \left (4 \log \left (1-c^2 x^2\right )-6 \sin ^{-1}(c x)^2+2 \sin \left (2 \sin ^{-1}(c x)\right ) \sin ^{-1}(c x)+\cos \left (2 \sin ^{-1}(c x)\right )\right )+8 c x \sin ^{-1}(c x)\right )}{8 c^5 d^{3/2} \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(3/2),x]

[Out]

(-4*a*c*Sqrt[d]*x*(-3 + c^2*x^2) + 12*a*Sqrt[d - c^2*d*x^2]*ArcTan[(c*x*Sqrt[d - c^2*d*x^2])/(Sqrt[d]*(-1 + c^
2*x^2))] + b*Sqrt[d]*(8*c*x*ArcSin[c*x] + Sqrt[1 - c^2*x^2]*(-6*ArcSin[c*x]^2 + Cos[2*ArcSin[c*x]] + 4*Log[1 -
c^2*x^2] + 2*ArcSin[c*x]*Sin[2*ArcSin[c*x]])))/(8*c^5*d^(3/2)*Sqrt[d - c^2*d*x^2])

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{4} \arcsin \left (c x\right ) + a x^{4}\right )} \sqrt {-c^{2} d x^{2} + d}}{c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}}, x\right ) \]

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)^(3/2),x, algorithm="fricas")

[Out]

integral((b*x^4*arcsin(c*x) + a*x^4)*sqrt(-c^2*d*x^2 + d)/(c^4*d^2*x^4 - 2*c^2*d^2*x^2 + d^2), 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(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep^4-1)]index.cc index_m i_lex_is_greater Error: Bad Argument Value

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maple [C]  time = 0.60, size = 432, normalized size = 2.02 \[ -\frac {a \,x^{3}}{2 c^{2} d \sqrt {-c^{2} d \,x^{2}+d}}+\frac {3 a x}{2 c^{4} d \sqrt {-c^{2} d \,x^{2}+d}}-\frac {3 a \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{2 c^{4} d \sqrt {c^{2} d}}+\frac {3 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{2}}{4 c^{5} d^{2} \left (c^{2} x^{2}-1\right )}+\frac {i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )}{c^{5} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{c^{5} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}}{16 c^{5} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {9 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x}{8 d^{2} \left (c^{2} x^{2}-1\right ) c^{4}}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \cos \left (3 \arcsin \left (c x \right )\right )}{16 c^{5} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \sin \left (3 \arcsin \left (c x \right )\right )}{8 c^{5} d^{2} \left (c^{2} x^{2}-1\right )} \]

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)^(3/2),x)

[Out]

-1/2*a*x^3/c^2/d/(-c^2*d*x^2+d)^(1/2)+3/2*a/c^4*x/d/(-c^2*d*x^2+d)^(1/2)-3/2*a/c^4/d/(c^2*d)^(1/2)*arctan((c^2
*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))+3/4*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^5/d^2/(c^2*x^2-1)*arcsin(c
*x)^2+I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^5/d^2/(c^2*x^2-1)*arcsin(c*x)-b*(-d*(c^2*x^2-1))^(1/2)*(
-c^2*x^2+1)^(1/2)/c^5/d^2/(c^2*x^2-1)*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)-1/16*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x
^2+1)^(1/2)/c^5/d^2/(c^2*x^2-1)-9/8*b*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)/c^4*arcsin(c*x)*x-1/16*b*(-d*(c^2
*x^2-1))^(1/2)/c^5/d^2/(c^2*x^2-1)*cos(3*arcsin(c*x))-1/8*b*(-d*(c^2*x^2-1))^(1/2)/c^5/d^2/(c^2*x^2-1)*arcsin(
c*x)*sin(3*arcsin(c*x))

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maxima [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(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(3/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^{3/2}} \,d x \]

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)^(3/2),x)

[Out]

int((x^4*(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 \[ \int \frac {x^{4} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

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)**(3/2),x)

[Out]

Integral(x**4*(a + b*asin(c*x))/(-d*(c*x - 1)*(c*x + 1))**(3/2), x)

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