3.122 \(\int (b x)^m \sin ^{-1}(a x) \, dx\)

Optimal. Leaf size=69 \[ \frac{\sin ^{-1}(a x) (b x)^{m+1}}{b (m+1)}-\frac{a (b x)^{m+2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},a^2 x^2\right )}{b^2 (m+1) (m+2)} \]

[Out]

((b*x)^(1 + m)*ArcSin[a*x])/(b*(1 + m)) - (a*(b*x)^(2 + m)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, a^2*x^
2])/(b^2*(1 + m)*(2 + m))

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Rubi [A]  time = 0.0252512, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4627, 364} \[ \frac{\sin ^{-1}(a x) (b x)^{m+1}}{b (m+1)}-\frac{a (b x)^{m+2} \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};a^2 x^2\right )}{b^2 (m+1) (m+2)} \]

Antiderivative was successfully verified.

[In]

Int[(b*x)^m*ArcSin[a*x],x]

[Out]

((b*x)^(1 + m)*ArcSin[a*x])/(b*(1 + m)) - (a*(b*x)^(2 + m)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, a^2*x^
2])/(b^2*(1 + m)*(2 + m))

Rule 4627

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSi
n[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int (b x)^m \sin ^{-1}(a x) \, dx &=\frac{(b x)^{1+m} \sin ^{-1}(a x)}{b (1+m)}-\frac{a \int \frac{(b x)^{1+m}}{\sqrt{1-a^2 x^2}} \, dx}{b (1+m)}\\ &=\frac{(b x)^{1+m} \sin ^{-1}(a x)}{b (1+m)}-\frac{a (b x)^{2+m} \, _2F_1\left (\frac{1}{2},\frac{2+m}{2};\frac{4+m}{2};a^2 x^2\right )}{b^2 (1+m) (2+m)}\\ \end{align*}

Mathematica [A]  time = 0.0203494, size = 56, normalized size = 0.81 \[ -\frac{x (b x)^m \left (a x \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},a^2 x^2\right )-(m+2) \sin ^{-1}(a x)\right )}{(m+1) (m+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x)^m*ArcSin[a*x],x]

[Out]

-((x*(b*x)^m*(-((2 + m)*ArcSin[a*x]) + a*x*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, a^2*x^2]))/((1 + m)*(2
+ m)))

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Maple [F]  time = 0.531, size = 0, normalized size = 0. \begin{align*} \int \left ( bx \right ) ^{m}\arcsin \left ( ax \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x)^m*arcsin(a*x),x)

[Out]

int((b*x)^m*arcsin(a*x),x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x)^m*arcsin(a*x),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (b x\right )^{m} \arcsin \left (a x\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x)^m*arcsin(a*x),x, algorithm="fricas")

[Out]

integral((b*x)^m*arcsin(a*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b x\right )^{m} \operatorname{asin}{\left (a x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x)**m*asin(a*x),x)

[Out]

Integral((b*x)**m*asin(a*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b x\right )^{m} \arcsin \left (a x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x)^m*arcsin(a*x),x, algorithm="giac")

[Out]

integrate((b*x)^m*arcsin(a*x), x)