\(\int (b x)^m \arcsin (a x)^2 \, dx\) [121]

Optimal result

Integrand size = 12, antiderivative size = 150 \[ \int (b x)^m \arcsin (a x)^2 \, dx=\frac {(b x)^{1+m} \arcsin (a x)^2}{b (1+m)}-\frac {2 a (b x)^{2+m} \arcsin (a x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},a^2 x^2\right )}{b^2 (1+m) (2+m)}+\frac {2 a^2 (b x)^{3+m} \, _3F_2\left (1,\frac {3}{2}+\frac {m}{2},\frac {3}{2}+\frac {m}{2};2+\frac {m}{2},\frac {5}{2}+\frac {m}{2};a^2 x^2\right )}{b^3 (1+m) (2+m) (3+m)} \]

[Out]

(b*x)^(1+m)*arcsin(a*x)^2/b/(1+m)-2*a*(b*x)^(2+m)*arcsin(a*x)*hypergeom([1/2, 1+1/2*m],[2+1/2*m],a^2*x^2)/b^2/
(1+m)/(2+m)+2*a^2*(b*x)^(3+m)*hypergeom([1, 3/2+1/2*m, 3/2+1/2*m],[2+1/2*m, 5/2+1/2*m],a^2*x^2)/b^3/(3+m)/(m^2
+3*m+2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4723, 4805} \[ \int (b x)^m \arcsin (a x)^2 \, dx=\frac {2 a^2 (b x)^{m+3} \, _3F_2\left (1,\frac {m}{2}+\frac {3}{2},\frac {m}{2}+\frac {3}{2};\frac {m}{2}+2,\frac {m}{2}+\frac {5}{2};a^2 x^2\right )}{b^3 (m+1) (m+2) (m+3)}-\frac {2 a \arcsin (a x) (b x)^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},a^2 x^2\right )}{b^2 (m+1) (m+2)}+\frac {\arcsin (a x)^2 (b x)^{m+1}}{b (m+1)} \]

[In]

Int[(b*x)^m*ArcSin[a*x]^2,x]

[Out]

((b*x)^(1 + m)*ArcSin[a*x]^2)/(b*(1 + m)) - (2*a*(b*x)^(2 + m)*ArcSin[a*x]*Hypergeometric2F1[1/2, (2 + m)/2, (
4 + m)/2, a^2*x^2])/(b^2*(1 + m)*(2 + m)) + (2*a^2*(b*x)^(3 + m)*HypergeometricPFQ[{1, 3/2 + m/2, 3/2 + m/2},
{2 + m/2, 5/2 + m/2}, a^2*x^2])/(b^3*(1 + m)*(2 + m)*(3 + m))

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]

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*} \text {integral}& = \frac {(b x)^{1+m} \arcsin (a x)^2}{b (1+m)}-\frac {(2 a) \int \frac {(b x)^{1+m} \arcsin (a x)}{\sqrt {1-a^2 x^2}} \, dx}{b (1+m)} \\ & = \frac {(b x)^{1+m} \arcsin (a x)^2}{b (1+m)}-\frac {2 a (b x)^{2+m} \arcsin (a x) \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)}+\frac {2 a^2 (b x)^{3+m} \, _3F_2\left (1,\frac {3}{2}+\frac {m}{2},\frac {3}{2}+\frac {m}{2};2+\frac {m}{2},\frac {5}{2}+\frac {m}{2};a^2 x^2\right )}{b^3 (1+m) (2+m) (3+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.81 \[ \int (b x)^m \arcsin (a x)^2 \, dx=\frac {x (b x)^m \left ((3+m) \arcsin (a x) \left ((2+m) \arcsin (a x)-2 a x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},a^2 x^2\right )\right )+2 a^2 x^2 \, _3F_2\left (1,\frac {3}{2}+\frac {m}{2},\frac {3}{2}+\frac {m}{2};2+\frac {m}{2},\frac {5}{2}+\frac {m}{2};a^2 x^2\right )\right )}{(1+m) (2+m) (3+m)} \]

[In]

Integrate[(b*x)^m*ArcSin[a*x]^2,x]

[Out]

(x*(b*x)^m*((3 + m)*ArcSin[a*x]*((2 + m)*ArcSin[a*x] - 2*a*x*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, a^2*
x^2]) + 2*a^2*x^2*HypergeometricPFQ[{1, 3/2 + m/2, 3/2 + m/2}, {2 + m/2, 5/2 + m/2}, a^2*x^2]))/((1 + m)*(2 +
m)*(3 + m))

Maple [F]

\[\int \left (b x \right )^{m} \arcsin \left (a x \right )^{2}d x\]

[In]

int((b*x)^m*arcsin(a*x)^2,x)

[Out]

int((b*x)^m*arcsin(a*x)^2,x)

Fricas [F]

\[ \int (b x)^m \arcsin (a x)^2 \, dx=\int { \left (b x\right )^{m} \arcsin \left (a x\right )^{2} \,d x } \]

[In]

integrate((b*x)^m*arcsin(a*x)^2,x, algorithm="fricas")

[Out]

integral((b*x)^m*arcsin(a*x)^2, x)

Sympy [F]

\[ \int (b x)^m \arcsin (a x)^2 \, dx=\int \left (b x\right )^{m} \operatorname {asin}^{2}{\left (a x \right )}\, dx \]

[In]

integrate((b*x)**m*asin(a*x)**2,x)

[Out]

Integral((b*x)**m*asin(a*x)**2, x)

Maxima [F]

\[ \int (b x)^m \arcsin (a x)^2 \, dx=\int { \left (b x\right )^{m} \arcsin \left (a x\right )^{2} \,d x } \]

[In]

integrate((b*x)^m*arcsin(a*x)^2,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int (b x)^m \arcsin (a x)^2 \, dx=\int { \left (b x\right )^{m} \arcsin \left (a x\right )^{2} \,d x } \]

[In]

integrate((b*x)^m*arcsin(a*x)^2,x, algorithm="giac")

[Out]

integrate((b*x)^m*arcsin(a*x)^2, x)

Mupad [F(-1)]

Timed out. \[ \int (b x)^m \arcsin (a x)^2 \, dx=\int {\mathrm {asin}\left (a\,x\right )}^2\,{\left (b\,x\right )}^m \,d x \]

[In]

int(asin(a*x)^2*(b*x)^m,x)

[Out]

int(asin(a*x)^2*(b*x)^m, x)