3.2.55 \(\int \frac {x^m \text {ArcSin}(a x)}{\sqrt {1-a^2 x^2}} \, dx\) [155]

Optimal. Leaf size=100 \[ \frac {x^{1+m} \text {ArcSin}(a x) \text {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )}{1+m}-\frac {a x^{2+m} \text {HypergeometricPFQ}\left (\left \{1,1+\frac {m}{2},1+\frac {m}{2}\right \},\left \{\frac {3}{2}+\frac {m}{2},2+\frac {m}{2}\right \},a^2 x^2\right )}{2+3 m+m^2} \]

[Out]

x^(1+m)*arcsin(a*x)*hypergeom([1/2, 1/2+1/2*m],[3/2+1/2*m],a^2*x^2)/(1+m)-a*x^(2+m)*hypergeom([1, 1+1/2*m, 1+1
/2*m],[2+1/2*m, 3/2+1/2*m],a^2*x^2)/(m^2+3*m+2)

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Rubi [A]
time = 0.05, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {4805} \begin {gather*} \frac {x^{m+1} \text {ArcSin}(a x) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};a^2 x^2\right )}{m+1}-\frac {a x^{m+2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;a^2 x^2\right )}{m^2+3 m+2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^m*ArcSin[a*x])/Sqrt[1 - a^2*x^2],x]

[Out]

(x^(1 + m)*ArcSin[a*x]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, a^2*x^2])/(1 + m) - (a*x^(2 + m)*Hypergeom
etricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, a^2*x^2])/(2 + 3*m + m^2)

Rule 4805

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[((f*x)
^(m + 1)/(f*(m + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, (1 +
m)/2, (3 + m)/2, c^2*x^2], x] - Simp[b*c*((f*x)^(m + 2)/(f^2*(m + 1)*(m + 2)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d +
e*x^2]]*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2], x] /; FreeQ[{a, b, c, d, e,
f, m}, x] && EqQ[c^2*d + e, 0] &&  !IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {x^m \sin ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx &=\frac {x^{1+m} \sin ^{-1}(a x) \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};a^2 x^2\right )}{1+m}-\frac {a x^{2+m} \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};a^2 x^2\right )}{2+3 m+m^2}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 95, normalized size = 0.95 \begin {gather*} \frac {x^{1+m} \left ((2+m) \text {ArcSin}(a x) \text {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )-a x \text {HypergeometricPFQ}\left (\left \{1,1+\frac {m}{2},1+\frac {m}{2}\right \},\left \{\frac {3}{2}+\frac {m}{2},2+\frac {m}{2}\right \},a^2 x^2\right )\right )}{(1+m) (2+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^m*ArcSin[a*x])/Sqrt[1 - a^2*x^2],x]

[Out]

(x^(1 + m)*((2 + m)*ArcSin[a*x]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, a^2*x^2] - a*x*HypergeometricPFQ[
{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, a^2*x^2]))/((1 + m)*(2 + m))

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Maple [F]
time = 0.44, size = 0, normalized size = 0.00 \[\int \frac {x^{m} \arcsin \left (a x \right )}{\sqrt {-a^{2} x^{2}+1}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*arcsin(a*x)/(-a^2*x^2+1)^(1/2),x)

[Out]

int(x^m*arcsin(a*x)/(-a^2*x^2+1)^(1/2),x)

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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^m*arcsin(a*x)/(-a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^m*arcsin(a*x)/sqrt(-a^2*x^2 + 1), x)

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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^m*arcsin(a*x)/(-a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-a^2*x^2 + 1)*x^m*arcsin(a*x)/(a^2*x^2 - 1), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{m} \operatorname {asin}{\left (a x \right )}}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*asin(a*x)/(-a**2*x**2+1)**(1/2),x)

[Out]

Integral(x**m*asin(a*x)/sqrt(-(a*x - 1)*(a*x + 1)), 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^m*arcsin(a*x)/(-a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate(x^m*arcsin(a*x)/sqrt(-a^2*x^2 + 1), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^m\,\mathrm {asin}\left (a\,x\right )}{\sqrt {1-a^2\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^m*asin(a*x))/(1 - a^2*x^2)^(1/2),x)

[Out]

int((x^m*asin(a*x))/(1 - a^2*x^2)^(1/2), x)

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