\(\int (b x)^m \arcsin (a x) \, dx\) [122]

Optimal result

Integrand size = 10, antiderivative size = 69 \[ \int (b x)^m \arcsin (a x) \, dx=\frac {(b x)^{1+m} \arcsin (a x)}{b (1+m)}-\frac {a (b x)^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},a^2 x^2\right )}{b^2 (1+m) (2+m)} \]

[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)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4723, 371} \[ \int (b x)^m \arcsin (a x) \, dx=\frac {\arcsin (a x) (b x)^{m+1}}{b (m+1)}-\frac {a (b x)^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},a^2 x^2\right )}{b^2 (m+1) (m+2)} \]

[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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 4723

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*} \text {integral}& = \frac {(b x)^{1+m} \arcsin (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} \arcsin (a x)}{b (1+m)}-\frac {a (b x)^{2+m} \operatorname {Hypergeometric2F1}\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] (verified)

Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.81 \[ \int (b x)^m \arcsin (a x) \, dx=-\frac {x (b x)^m \left (-((2+m) \arcsin (a x))+a x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},a^2 x^2\right )\right )}{(1+m) (2+m)} \]

[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)))

Maple [F]

\[\int \left (b x \right )^{m} \arcsin \left (a x \right )d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int (b x)^m \arcsin (a x) \, dx=\int { \left (b x\right )^{m} \arcsin \left (a x\right ) \,d x } \]

[In]

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

[Out]

integral((b*x)^m*arcsin(a*x), x)

Sympy [F]

\[ \int (b x)^m \arcsin (a x) \, dx=\int \left (b x\right )^{m} \operatorname {asin}{\left (a x \right )}\, dx \]

[In]

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

[Out]

Integral((b*x)**m*asin(a*x), x)

Maxima [F]

\[ \int (b x)^m \arcsin (a x) \, dx=\int { \left (b x\right )^{m} \arcsin \left (a x\right ) \,d x } \]

[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)

Giac [F]

\[ \int (b x)^m \arcsin (a x) \, dx=\int { \left (b x\right )^{m} \arcsin \left (a x\right ) \,d x } \]

[In]

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

[Out]

integrate((b*x)^m*arcsin(a*x), x)

Mupad [F(-1)]

Timed out. \[ \int (b x)^m \arcsin (a x) \, dx=\int \mathrm {asin}\left (a\,x\right )\,{\left (b\,x\right )}^m \,d x \]

[In]

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

[Out]

int(asin(a*x)*(b*x)^m, x)