3.2.31 \(\int \frac {x^4 (a+b \text {ArcSin}(c x))}{(d-c^2 d x^2)^{5/2}} \, dx\) [131]

Optimal. Leaf size=212 \[ -\frac {b}{6 c^5 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {x^3 (a+b \text {ArcSin}(c x))}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {x (a+b \text {ArcSin}(c x))}{c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))^2}{2 b c^5 d^2 \sqrt {d-c^2 d x^2}}-\frac {2 b \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{3 c^5 d^2 \sqrt {d-c^2 d x^2}} \]

[Out]

1/3*x^3*(a+b*arcsin(c*x))/c^2/d/(-c^2*d*x^2+d)^(3/2)-x*(a+b*arcsin(c*x))/c^4/d^2/(-c^2*d*x^2+d)^(1/2)-1/6*b/c^
5/d^2/(-c^2*x^2+1)^(1/2)/(-c^2*d*x^2+d)^(1/2)+1/2*(a+b*arcsin(c*x))^2*(-c^2*x^2+1)^(1/2)/b/c^5/d^2/(-c^2*d*x^2
+d)^(1/2)-2/3*b*ln(-c^2*x^2+1)*(-c^2*x^2+1)^(1/2)/c^5/d^2/(-c^2*d*x^2+d)^(1/2)

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Rubi [A]
time = 0.20, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {4791, 4737, 266, 272, 45} \begin {gather*} \frac {x^3 (a+b \text {ArcSin}(c x))}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {\sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))^2}{2 b c^5 d^2 \sqrt {d-c^2 d x^2}}-\frac {x (a+b \text {ArcSin}(c x))}{c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {b}{6 c^5 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}-\frac {2 b \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{3 c^5 d^2 \sqrt {d-c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

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

Rule 45

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_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 272

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 4737

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

Rule 4791

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/(2*c*(p + 1)))*Simp[(d
+ e*x^2)^p/(1 - c^2*x^2)^p], Int[(f*x)^(m - 1)*(1 - c^2*x^2)^(p + 1/2)*(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] && LtQ[p, -1] && IGtQ[m, 1]

Rubi steps

\begin {align*} \int \frac {x^4 \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^{5/2}} \, dx &=\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {\int \frac {x^2 \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^{3/2}} \, dx}{c^2 d}-\frac {\left (b \sqrt {1-c^2 x^2}\right ) \int \frac {x^3}{\left (1-c^2 x^2\right )^2} \, dx}{3 c d^2 \sqrt {d-c^2 d x^2}}\\ &=\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {x \left (a+b \sin ^{-1}(c x)\right )}{c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\int \frac {a+b \sin ^{-1}(c x)}{\sqrt {d-c^2 d x^2}} \, dx}{c^4 d^2}+\frac {\left (b \sqrt {1-c^2 x^2}\right ) \int \frac {x}{1-c^2 x^2} \, dx}{c^3 d^2 \sqrt {d-c^2 d x^2}}-\frac {\left (b \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int \frac {x}{\left (1-c^2 x\right )^2} \, dx,x,x^2\right )}{6 c d^2 \sqrt {d-c^2 d x^2}}\\ &=\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {x \left (a+b \sin ^{-1}(c x)\right )}{c^4 d^2 \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^2 \sqrt {d-c^2 d x^2}}+\frac {\sqrt {1-c^2 x^2} \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {\left (b \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int \left (\frac {1}{c^2 \left (-1+c^2 x\right )^2}+\frac {1}{c^2 \left (-1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{6 c d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {b}{6 c^5 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {x \left (a+b \sin ^{-1}(c x)\right )}{c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c^5 d^2 \sqrt {d-c^2 d x^2}}-\frac {2 b \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.29, size = 213, normalized size = 1.00 \begin {gather*} \frac {-3 b \sqrt {d} \left (1-c^2 x^2\right )^{3/2} \text {ArcSin}(c x)^2-6 a \left (-1+c^2 x^2\right ) \sqrt {d-c^2 d x^2} \text {ArcTan}\left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (-1+c^2 x^2\right )}\right )+\sqrt {d} \left (6 a c x-8 a c^3 x^3+b \sqrt {1-c^2 x^2}+4 b \left (1-c^2 x^2\right )^{3/2} \log \left (1-c^2 x^2\right )\right )+2 b \sqrt {d} \text {ArcSin}(c x) \sin (3 \text {ArcSin}(c x))}{6 c^5 d^{5/2} \left (-1+c^2 x^2\right ) \sqrt {d-c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.38, size = 531, normalized size = 2.50

method	result	size
default	\(\frac {a \,x^{3}}{3 c^{2} d \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {a x}{c^{4} d^{2} \sqrt {-c^{2} d \,x^{2}+d}}+\frac {a \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{c^{4} d^{2} \sqrt {c^{2} d}}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{2}}{2 c^{5} d^{3} \left (c^{2} x^{2}-1\right )}-\frac {8 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )}{3 c^{5} d^{3} \left (c^{2} x^{2}-1\right )}-\frac {4 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{3 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right ) c^{5}}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x}{d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right ) c^{4}}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}}{6 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right ) c^{5}}+\frac {4 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{3 c^{5} d^{3} \left (c^{2} x^{2}-1\right )}+\frac {4 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, x^{2}}{3 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right ) c^{3}}+\frac {4 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x^{3}}{3 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right ) c^{2}}\)	\(531\)

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)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/3*a*x^3/c^2/d/(-c^2*d*x^2+d)^(3/2)-a/c^4/d^2*x/(-c^2*d*x^2+d)^(1/2)+a/c^4/d^2/(c^2*d)^(1/2)*arctan((c^2*d)^(
1/2)*x/(-c^2*d*x^2+d)^(1/2))-1/2*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^5/d^3/(c^2*x^2-1)*arcsin(c*x)^2
-8/3*I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^5/d^3/(c^2*x^2-1)*arcsin(c*x)-4/3*I*b*(-d*(c^2*x^2-1))^(1
/2)/d^3/(c^4*x^4-2*c^2*x^2+1)/c^5*arcsin(c*x)*(-c^2*x^2+1)^(1/2)-b*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^4*x^4-2*c^2*x
^2+1)/c^4*arcsin(c*x)*x-1/6*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^4*x^4-2*c^2*x^2+1)/c^5*(-c^2*x^2+1)^(1/2)+4/3*b*(-
d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^5/d^3/(c^2*x^2-1)*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)+4/3*I*b*(-d*(c^
2*x^2-1))^(1/2)/d^3/(c^4*x^4-2*c^2*x^2+1)/c^3*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x^2+4/3*b*(-d*(c^2*x^2-1))^(1/2)/
d^3/(c^4*x^4-2*c^2*x^2+1)/c^2*arcsin(c*x)*x^3

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^(5/2),x, algorithm="fricas")

[Out]

integral(-(b*x^4*arcsin(c*x) + a*x^4)*sqrt(-c^2*d*x^2 + d)/(c^6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 - d^3)
, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {5}{2}}}\, dx \end {gather*}

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)**(5/2),x)

[Out]

Integral(x**4*(a + b*asin(c*x))/(-d*(c*x - 1)*(c*x + 1))**(5/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x) + a)*x^4/(-c^2*d*x^2 + d)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^4\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^{5/2}} \,d x \end {gather*}

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)^(5/2),x)

[Out]

int((x^4*(a + b*asin(c*x)))/(d - c^2*d*x^2)^(5/2), x)

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