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

Optimal. Leaf size=69 \[ \frac {\sin ^{-1}(a x) (b x)^{m+1}}{b (m+1)}-\frac {a (b x)^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};a^2 x^2\right )}{b^2 (m+1) (m+2)} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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])

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]

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*}

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Mathematica [A]  time = 0.02, size = 56, normalized size = 0.81 \[ -\frac {x (b x)^m \left (a x \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};a^2 x^2\right )-(m+2) \sin ^{-1}(a x)\right )}{(m+1) (m+2)} \]

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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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (b x\right )^{m} \arcsin \left (a x\right ), x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b x\right )^{m} \arcsin \left (a x\right )\,{d x} \]

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)

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maple [F]  time = 0.80, size = 0, normalized size = 0.00 \[ \int \left (b x \right )^{m} \arcsin \left (a x \right )\, dx \]

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]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {b^{m} x x^{m} \arctan \left (a x, \sqrt {a x + 1} \sqrt {-a x + 1}\right ) + \frac {{\left (a b^{m} m + a b^{m}\right )} \int \frac {\sqrt {-a x + 1} x x^{m}}{\sqrt {a x + 1} {\left (a x - 1\right )}}\,{d x}}{m + 1}}{m + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^m*x*x^m*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1)) + (a*b^m*m + a*b^m)*integrate(sqrt(a*x + 1)*sqrt(-a*x +
1)*x*x^m/((a^2*m + a^2)*x^2 - m - 1), x))/(m + 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {asin}\left (a\,x\right )\,{\left (b\,x\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(asin(a*x)*(b*x)^m, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b x\right )^{m} \operatorname {asin}{\left (a x \right )}\, dx \]

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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