\(\int (d x)^{3/2} (a+b \arcsin (c x))^2 \, dx\) [210]

Optimal result

Integrand size = 18, antiderivative size = 109 \[ \int (d x)^{3/2} (a+b \arcsin (c x))^2 \, dx=\frac {2 (d x)^{5/2} (a+b \arcsin (c x))^2}{5 d}-\frac {8 b c (d x)^{7/2} (a+b \arcsin (c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{4},\frac {11}{4},c^2 x^2\right )}{35 d^2}+\frac {16 b^2 c^2 (d x)^{9/2} \, _3F_2\left (1,\frac {9}{4},\frac {9}{4};\frac {11}{4},\frac {13}{4};c^2 x^2\right )}{315 d^3} \]

[Out]

2/5*(d*x)^(5/2)*(a+b*arcsin(c*x))^2/d-8/35*b*c*(d*x)^(7/2)*(a+b*arcsin(c*x))*hypergeom([1/2, 7/4],[11/4],c^2*x
^2)/d^2+16/315*b^2*c^2*(d*x)^(9/2)*hypergeom([1, 9/4, 9/4],[11/4, 13/4],c^2*x^2)/d^3

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4723, 4805} \[ \int (d x)^{3/2} (a+b \arcsin (c x))^2 \, dx=\frac {16 b^2 c^2 (d x)^{9/2} \, _3F_2\left (1,\frac {9}{4},\frac {9}{4};\frac {11}{4},\frac {13}{4};c^2 x^2\right )}{315 d^3}-\frac {8 b c (d x)^{7/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{4},\frac {11}{4},c^2 x^2\right ) (a+b \arcsin (c x))}{35 d^2}+\frac {2 (d x)^{5/2} (a+b \arcsin (c x))^2}{5 d} \]

[In]

Int[(d*x)^(3/2)*(a + b*ArcSin[c*x])^2,x]

[Out]

(2*(d*x)^(5/2)*(a + b*ArcSin[c*x])^2)/(5*d) - (8*b*c*(d*x)^(7/2)*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, 7/
4, 11/4, c^2*x^2])/(35*d^2) + (16*b^2*c^2*(d*x)^(9/2)*HypergeometricPFQ[{1, 9/4, 9/4}, {11/4, 13/4}, c^2*x^2])
/(315*d^3)

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 {2 (d x)^{5/2} (a+b \arcsin (c x))^2}{5 d}-\frac {(4 b c) \int \frac {(d x)^{5/2} (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}} \, dx}{5 d} \\ & = \frac {2 (d x)^{5/2} (a+b \arcsin (c x))^2}{5 d}-\frac {8 b c (d x)^{7/2} (a+b \arcsin (c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{4},\frac {11}{4},c^2 x^2\right )}{35 d^2}+\frac {16 b^2 c^2 (d x)^{9/2} \, _3F_2\left (1,\frac {9}{4},\frac {9}{4};\frac {11}{4},\frac {13}{4};c^2 x^2\right )}{315 d^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.83 \[ \int (d x)^{3/2} (a+b \arcsin (c x))^2 \, dx=\frac {2}{315} x (d x)^{3/2} \left (9 (a+b \arcsin (c x)) \left (7 (a+b \arcsin (c x))-4 b c x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{4},\frac {11}{4},c^2 x^2\right )\right )+8 b^2 c^2 x^2 \, _3F_2\left (1,\frac {9}{4},\frac {9}{4};\frac {11}{4},\frac {13}{4};c^2 x^2\right )\right ) \]

[In]

Integrate[(d*x)^(3/2)*(a + b*ArcSin[c*x])^2,x]

[Out]

(2*x*(d*x)^(3/2)*(9*(a + b*ArcSin[c*x])*(7*(a + b*ArcSin[c*x]) - 4*b*c*x*Hypergeometric2F1[1/2, 7/4, 11/4, c^2
*x^2]) + 8*b^2*c^2*x^2*HypergeometricPFQ[{1, 9/4, 9/4}, {11/4, 13/4}, c^2*x^2]))/315

Maple [F]

\[\int \left (d x \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )^{2}d x\]

[In]

int((d*x)^(3/2)*(a+b*arcsin(c*x))^2,x)

[Out]

int((d*x)^(3/2)*(a+b*arcsin(c*x))^2,x)

Fricas [F]

\[ \int (d x)^{3/2} (a+b \arcsin (c x))^2 \, dx=\int { \left (d x\right )^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2} \,d x } \]

[In]

integrate((d*x)^(3/2)*(a+b*arcsin(c*x))^2,x, algorithm="fricas")

[Out]

integral((b^2*d*x*arcsin(c*x)^2 + 2*a*b*d*x*arcsin(c*x) + a^2*d*x)*sqrt(d*x), x)

Sympy [F]

\[ \int (d x)^{3/2} (a+b \arcsin (c x))^2 \, dx=\int \left (d x\right )^{\frac {3}{2}} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}\, dx \]

[In]

integrate((d*x)**(3/2)*(a+b*asin(c*x))**2,x)

[Out]

Integral((d*x)**(3/2)*(a + b*asin(c*x))**2, x)

Maxima [F]

\[ \int (d x)^{3/2} (a+b \arcsin (c x))^2 \, dx=\int { \left (d x\right )^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2} \,d x } \]

[In]

integrate((d*x)^(3/2)*(a+b*arcsin(c*x))^2,x, algorithm="maxima")

[Out]

2/5*b^2*d^(3/2)*x^(5/2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2 + 1/10*a^2*c^2*d^(3/2)*(4*(c^2*x^(5/2) +
5*sqrt(x))/c^4 - 10*arctan(sqrt(c)*sqrt(x))/c^(9/2) + 5*log((c*sqrt(x) - sqrt(c))/(c*sqrt(x) + sqrt(c)))/c^(9/
2)) + 10*a*b*c^2*d^(3/2)*integrate(1/5*x^(7/2)*arctan(c*x/(sqrt(c*x + 1)*sqrt(-c*x + 1)))/(c^2*x^2 - 1), x) +
4*b^2*c*d^(3/2)*integrate(1/5*sqrt(c*x + 1)*sqrt(-c*x + 1)*x^(5/2)*arctan(c*x/(sqrt(c*x + 1)*sqrt(-c*x + 1)))/
(c^2*x^2 - 1), x) - 1/2*a^2*d^(3/2)*(4*sqrt(x)/c^2 - 2*arctan(sqrt(c)*sqrt(x))/c^(5/2) + log((c*sqrt(x) - sqrt
(c))/(c*sqrt(x) + sqrt(c)))/c^(5/2)) - 10*a*b*d^(3/2)*integrate(1/5*x^(3/2)*arctan(c*x/(sqrt(c*x + 1)*sqrt(-c*
x + 1)))/(c^2*x^2 - 1), x)

Giac [F(-2)]

Exception generated. \[ \int (d x)^{3/2} (a+b \arcsin (c x))^2 \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((d*x)^(3/2)*(a+b*arcsin(c*x))^2,x, algorithm="giac")

[Out]

Exception raised: RuntimeError >> an error occurred running a Giac command:INPUT:sage2OUTPUT:sym2poly/r2sym(co
nst gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int (d x)^{3/2} (a+b \arcsin (c x))^2 \, dx=\int {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,{\left (d\,x\right )}^{3/2} \,d x \]

[In]

int((a + b*asin(c*x))^2*(d*x)^(3/2),x)

[Out]

int((a + b*asin(c*x))^2*(d*x)^(3/2), x)