\(\int (d x)^{5/2} (a+b \arcsin (c x))^2 \, dx\) [209]

Optimal result

Integrand size = 18, antiderivative size = 109 \[ \int (d x)^{5/2} (a+b \arcsin (c x))^2 \, dx=\frac {2 (d x)^{7/2} (a+b \arcsin (c x))^2}{7 d}-\frac {8 b c (d x)^{9/2} (a+b \arcsin (c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {9}{4},\frac {13}{4},c^2 x^2\right )}{63 d^2}+\frac {16 b^2 c^2 (d x)^{11/2} \, _3F_2\left (1,\frac {11}{4},\frac {11}{4};\frac {13}{4},\frac {15}{4};c^2 x^2\right )}{693 d^3} \]

[Out]

2/7*(d*x)^(7/2)*(a+b*arcsin(c*x))^2/d-8/63*b*c*(d*x)^(9/2)*(a+b*arcsin(c*x))*hypergeom([1/2, 9/4],[13/4],c^2*x
^2)/d^2+16/693*b^2*c^2*(d*x)^(11/2)*hypergeom([1, 11/4, 11/4],[13/4, 15/4],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} \[ \int (d x)^{5/2} (a+b \arcsin (c x))^2 \, dx=\frac {16 b^2 c^2 (d x)^{11/2} \, _3F_2\left (1,\frac {11}{4},\frac {11}{4};\frac {13}{4},\frac {15}{4};c^2 x^2\right )}{693 d^3}-\frac {8 b c (d x)^{9/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {9}{4},\frac {13}{4},c^2 x^2\right ) (a+b \arcsin (c x))}{63 d^2}+\frac {2 (d x)^{7/2} (a+b \arcsin (c x))^2}{7 d} \]

[In]

Int[(d*x)^(5/2)*(a + b*ArcSin[c*x])^2,x]

[Out]

(2*(d*x)^(7/2)*(a + b*ArcSin[c*x])^2)/(7*d) - (8*b*c*(d*x)^(9/2)*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, 9/
4, 13/4, c^2*x^2])/(63*d^2) + (16*b^2*c^2*(d*x)^(11/2)*HypergeometricPFQ[{1, 11/4, 11/4}, {13/4, 15/4}, c^2*x^
2])/(693*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)^{7/2} (a+b \arcsin (c x))^2}{7 d}-\frac {(4 b c) \int \frac {(d x)^{7/2} (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}} \, dx}{7 d} \\ & = \frac {2 (d x)^{7/2} (a+b \arcsin (c x))^2}{7 d}-\frac {8 b c (d x)^{9/2} (a+b \arcsin (c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {9}{4},\frac {13}{4},c^2 x^2\right )}{63 d^2}+\frac {16 b^2 c^2 (d x)^{11/2} \, _3F_2\left (1,\frac {11}{4},\frac {11}{4};\frac {13}{4},\frac {15}{4};c^2 x^2\right )}{693 d^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.83 \[ \int (d x)^{5/2} (a+b \arcsin (c x))^2 \, dx=\frac {2}{693} x (d x)^{5/2} \left (11 (a+b \arcsin (c x)) \left (9 (a+b \arcsin (c x))-4 b c x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {9}{4},\frac {13}{4},c^2 x^2\right )\right )+8 b^2 c^2 x^2 \, _3F_2\left (1,\frac {11}{4},\frac {11}{4};\frac {13}{4},\frac {15}{4};c^2 x^2\right )\right ) \]

[In]

Integrate[(d*x)^(5/2)*(a + b*ArcSin[c*x])^2,x]

[Out]

(2*x*(d*x)^(5/2)*(11*(a + b*ArcSin[c*x])*(9*(a + b*ArcSin[c*x]) - 4*b*c*x*Hypergeometric2F1[1/2, 9/4, 13/4, c^
2*x^2]) + 8*b^2*c^2*x^2*HypergeometricPFQ[{1, 11/4, 11/4}, {13/4, 15/4}, c^2*x^2]))/693

Maple [F]

\[\int \left (d x \right )^{\frac {5}{2}} \left (a +b \arcsin \left (c x \right )\right )^{2}d x\]

[In]

int((d*x)^(5/2)*(a+b*arcsin(c*x))^2,x)

[Out]

int((d*x)^(5/2)*(a+b*arcsin(c*x))^2,x)

Fricas [F]

\[ \int (d x)^{5/2} (a+b \arcsin (c x))^2 \, dx=\int { \left (d x\right )^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2} \,d x } \]

[In]

integrate((d*x)^(5/2)*(a+b*arcsin(c*x))^2,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int (d x)^{5/2} (a+b \arcsin (c x))^2 \, dx=\text {Timed out} \]

[In]

integrate((d*x)**(5/2)*(a+b*asin(c*x))**2,x)

[Out]

Timed out

Maxima [F]

\[ \int (d x)^{5/2} (a+b \arcsin (c x))^2 \, dx=\int { \left (d x\right )^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2} \,d x } \]

[In]

integrate((d*x)^(5/2)*(a+b*arcsin(c*x))^2,x, algorithm="maxima")

[Out]

2/7*b^2*d^(5/2)*x^(7/2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2 + 1/42*a^2*c^2*d^(5/2)*(4*(3*c^2*x^(7/2)
+ 7*x^(3/2))/c^4 + 42*arctan(sqrt(c)*sqrt(x))/c^(11/2) + 21*log((c*sqrt(x) - sqrt(c))/(c*sqrt(x) + sqrt(c)))/c
^(11/2)) + 14*a*b*c^2*d^(5/2)*integrate(1/7*x^(9/2)*arctan(c*x/(sqrt(c*x + 1)*sqrt(-c*x + 1)))/(c^2*x^2 - 1),
x) + 4*b^2*c*d^(5/2)*integrate(1/7*sqrt(c*x + 1)*sqrt(-c*x + 1)*x^(7/2)*arctan(c*x/(sqrt(c*x + 1)*sqrt(-c*x +
1)))/(c^2*x^2 - 1), x) - 1/6*a^2*d^(5/2)*(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)) - 14*a*b*d^(5/2)*integrate(1/7*x^(5/2)*arctan(c*x/(sqrt(c*x + 1)*s
qrt(-c*x + 1)))/(c^2*x^2 - 1), x)

Giac [F(-2)]

Exception generated. \[ \int (d x)^{5/2} (a+b \arcsin (c x))^2 \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((d*x)^(5/2)*(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. \[ \int (d x)^{5/2} (a+b \arcsin (c x))^2 \, dx=\int {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,{\left (d\,x\right )}^{5/2} \,d x \]

[In]

int((a + b*asin(c*x))^2*(d*x)^(5/2),x)

[Out]

int((a + b*asin(c*x))^2*(d*x)^(5/2), x)