\(\int \sqrt {d x} (a+b \arcsin (c x))^2 \, dx\) [211]

Optimal result

Integrand size = 18, antiderivative size = 109 \[ \int \sqrt {d x} (a+b \arcsin (c x))^2 \, dx=\frac {2 (d x)^{3/2} (a+b \arcsin (c x))^2}{3 d}-\frac {8 b c (d x)^{5/2} (a+b \arcsin (c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{4},\frac {9}{4},c^2 x^2\right )}{15 d^2}+\frac {16 b^2 c^2 (d x)^{7/2} \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};c^2 x^2\right )}{105 d^3} \]

[Out]

2/3*(d*x)^(3/2)*(a+b*arcsin(c*x))^2/d-8/15*b*c*(d*x)^(5/2)*(a+b*arcsin(c*x))*hypergeom([1/2, 5/4],[9/4],c^2*x^
2)/d^2+16/105*b^2*c^2*(d*x)^(7/2)*hypergeom([1, 7/4, 7/4],[9/4, 11/4],c^2*x^2)/d^3

Rubi [A] (verified)

Time = 0.08 (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} \[ \int \sqrt {d x} (a+b \arcsin (c x))^2 \, dx=\frac {16 b^2 c^2 (d x)^{7/2} \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};c^2 x^2\right )}{105 d^3}-\frac {8 b c (d x)^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{4},\frac {9}{4},c^2 x^2\right ) (a+b \arcsin (c x))}{15 d^2}+\frac {2 (d x)^{3/2} (a+b \arcsin (c x))^2}{3 d} \]

[In]

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

[Out]

(2*(d*x)^(3/2)*(a + b*ArcSin[c*x])^2)/(3*d) - (8*b*c*(d*x)^(5/2)*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, 5/
4, 9/4, c^2*x^2])/(15*d^2) + (16*b^2*c^2*(d*x)^(7/2)*HypergeometricPFQ[{1, 7/4, 7/4}, {9/4, 11/4}, c^2*x^2])/(
105*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)^{3/2} (a+b \arcsin (c x))^2}{3 d}-\frac {(4 b c) \int \frac {(d x)^{3/2} (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}} \, dx}{3 d} \\ & = \frac {2 (d x)^{3/2} (a+b \arcsin (c x))^2}{3 d}-\frac {8 b c (d x)^{5/2} (a+b \arcsin (c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{4},\frac {9}{4},c^2 x^2\right )}{15 d^2}+\frac {16 b^2 c^2 (d x)^{7/2} \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};c^2 x^2\right )}{105 d^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.83 \[ \int \sqrt {d x} (a+b \arcsin (c x))^2 \, dx=\frac {2}{105} x \sqrt {d x} \left (7 (a+b \arcsin (c x)) \left (5 (a+b \arcsin (c x))-4 b c x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{4},\frac {9}{4},c^2 x^2\right )\right )+8 b^2 c^2 x^2 \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};c^2 x^2\right )\right ) \]

[In]

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

[Out]

(2*x*Sqrt[d*x]*(7*(a + b*ArcSin[c*x])*(5*(a + b*ArcSin[c*x]) - 4*b*c*x*Hypergeometric2F1[1/2, 5/4, 9/4, c^2*x^
2]) + 8*b^2*c^2*x^2*HypergeometricPFQ[{1, 7/4, 7/4}, {9/4, 11/4}, c^2*x^2]))/105

Maple [F]

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

[In]

int((d*x)^(1/2)*(a+b*arcsin(c*x))^2,x)

[Out]

int((d*x)^(1/2)*(a+b*arcsin(c*x))^2,x)

Fricas [F]

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

[In]

integrate((d*x)^(1/2)*(a+b*arcsin(c*x))^2,x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

integrate((d*x)**(1/2)*(a+b*asin(c*x))**2,x)

[Out]

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

Maxima [F]

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

[In]

integrate((d*x)^(1/2)*(a+b*arcsin(c*x))^2,x, algorithm="maxima")

[Out]

2/3*b^2*sqrt(d)*x^(3/2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2 + 1/6*a^2*c^2*sqrt(d)*(4*x^(3/2)/c^2 + 6*
arctan(sqrt(c)*sqrt(x))/c^(7/2) + 3*log((c*sqrt(x) - sqrt(c))/(c*sqrt(x) + sqrt(c)))/c^(7/2)) + 6*a*b*c^2*sqrt
(d)*integrate(1/3*x^(5/2)*arctan(c*x/(sqrt(c*x + 1)*sqrt(-c*x + 1)))/(c^2*x^2 - 1), x) + 4*b^2*c*sqrt(d)*integ
rate(1/3*sqrt(c*x + 1)*sqrt(-c*x + 1)*x^(3/2)*arctan(c*x/(sqrt(c*x + 1)*sqrt(-c*x + 1)))/(c^2*x^2 - 1), x) - 1
/2*a^2*sqrt(d)*(2*arctan(sqrt(c)*sqrt(x))/c^(3/2) + log((c*sqrt(x) - sqrt(c))/(c*sqrt(x) + sqrt(c)))/c^(3/2))
- 6*a*b*sqrt(d)*integrate(1/3*sqrt(x)*arctan(c*x/(sqrt(c*x + 1)*sqrt(-c*x + 1)))/(c^2*x^2 - 1), x)

Giac [F(-2)]

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

[In]

integrate((d*x)^(1/2)*(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. \[ \int \sqrt {d x} (a+b \arcsin (c x))^2 \, dx=\int {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,\sqrt {d\,x} \,d x \]

[In]

int((a + b*asin(c*x))^2*(d*x)^(1/2),x)

[Out]

int((a + b*asin(c*x))^2*(d*x)^(1/2), x)