3.204 \(\int (d x)^{3/2} (a+b \sin ^{-1}(c x)) \, dx\)

Optimal. Leaf size=124 \[ \frac{12 b d^{3/2} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right ),-1\right )}{25 c^{5/2}}+\frac{2 (d x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 d}-\frac{12 b d^{3/2} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )\right |-1\right )}{25 c^{5/2}}+\frac{4 b \sqrt{1-c^2 x^2} (d x)^{3/2}}{25 c} \]

[Out]

(4*b*(d*x)^(3/2)*Sqrt[1 - c^2*x^2])/(25*c) + (2*(d*x)^(5/2)*(a + b*ArcSin[c*x]))/(5*d) - (12*b*d^(3/2)*Ellipti
cE[ArcSin[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]], -1])/(25*c^(5/2)) + (12*b*d^(3/2)*EllipticF[ArcSin[(Sqrt[c]*Sqrt[d*x])
/Sqrt[d]], -1])/(25*c^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0984692, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {4627, 321, 329, 307, 221, 1199, 424} \[ \frac{2 (d x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 d}+\frac{12 b d^{3/2} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )\right |-1\right )}{25 c^{5/2}}-\frac{12 b d^{3/2} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )\right |-1\right )}{25 c^{5/2}}+\frac{4 b \sqrt{1-c^2 x^2} (d x)^{3/2}}{25 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b*(d*x)^(3/2)*Sqrt[1 - c^2*x^2])/(25*c) + (2*(d*x)^(5/2)*(a + b*ArcSin[c*x]))/(5*d) - (12*b*d^(3/2)*Ellipti
cE[ArcSin[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]], -1])/(25*c^(5/2)) + (12*b*d^(3/2)*EllipticF[ArcSin[(Sqrt[c]*Sqrt[d*x])
/Sqrt[d]], -1])/(25*c^(5/2))

Rule 4627

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 321

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

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 307

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-(b/a), 2]}, -Dist[q^(-1), Int[1/Sqrt[a + b*x^
4], x], x] + Dist[1/q, Int[(1 + q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/a]

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 1199

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[d/Sqrt[a], Int[Sqrt[1 + (e*x^2)/d]/Sqrt
[1 - (e*x^2)/d], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] && GtQ[a, 0]

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rubi steps

\begin{align*} \int (d x)^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac{2 (d x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 d}-\frac{(2 b c) \int \frac{(d x)^{5/2}}{\sqrt{1-c^2 x^2}} \, dx}{5 d}\\ &=\frac{4 b (d x)^{3/2} \sqrt{1-c^2 x^2}}{25 c}+\frac{2 (d x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 d}-\frac{(6 b d) \int \frac{\sqrt{d x}}{\sqrt{1-c^2 x^2}} \, dx}{25 c}\\ &=\frac{4 b (d x)^{3/2} \sqrt{1-c^2 x^2}}{25 c}+\frac{2 (d x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 d}-\frac{(12 b) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-\frac{c^2 x^4}{d^2}}} \, dx,x,\sqrt{d x}\right )}{25 c}\\ &=\frac{4 b (d x)^{3/2} \sqrt{1-c^2 x^2}}{25 c}+\frac{2 (d x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 d}+\frac{(12 b d) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{c^2 x^4}{d^2}}} \, dx,x,\sqrt{d x}\right )}{25 c^2}-\frac{(12 b d) \operatorname{Subst}\left (\int \frac{1+\frac{c x^2}{d}}{\sqrt{1-\frac{c^2 x^4}{d^2}}} \, dx,x,\sqrt{d x}\right )}{25 c^2}\\ &=\frac{4 b (d x)^{3/2} \sqrt{1-c^2 x^2}}{25 c}+\frac{2 (d x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 d}+\frac{12 b d^{3/2} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )\right |-1\right )}{25 c^{5/2}}-\frac{(12 b d) \operatorname{Subst}\left (\int \frac{\sqrt{1+\frac{c x^2}{d}}}{\sqrt{1-\frac{c x^2}{d}}} \, dx,x,\sqrt{d x}\right )}{25 c^2}\\ &=\frac{4 b (d x)^{3/2} \sqrt{1-c^2 x^2}}{25 c}+\frac{2 (d x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 d}-\frac{12 b d^{3/2} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )\right |-1\right )}{25 c^{5/2}}+\frac{12 b d^{3/2} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )\right |-1\right )}{25 c^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0248279, size = 66, normalized size = 0.53 \[ \frac{2 (d x)^{3/2} \left (-2 b \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},c^2 x^2\right )+5 a c x+2 b \sqrt{1-c^2 x^2}+5 b c x \sin ^{-1}(c x)\right )}{25 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d*x)^(3/2)*(5*a*c*x + 2*b*Sqrt[1 - c^2*x^2] + 5*b*c*x*ArcSin[c*x] - 2*b*Hypergeometric2F1[1/2, 3/4, 7/4, c
^2*x^2]))/(25*c)

________________________________________________________________________________________

Maple [A]  time = 0.01, size = 138, normalized size = 1.1 \begin{align*} 2\,{\frac{1}{d} \left ( 1/5\, \left ( dx \right ) ^{5/2}a+b \left ( 1/5\, \left ( dx \right ) ^{5/2}\arcsin \left ( cx \right ) -2/5\,{\frac{c}{d} \left ( -1/5\,{\frac{{d}^{2} \left ( dx \right ) ^{3/2}\sqrt{-{c}^{2}{x}^{2}+1}}{{c}^{2}}}-3/5\,{\frac{{d}^{3}\sqrt{-cx+1}\sqrt{cx+1}}{{c}^{3}\sqrt{-{c}^{2}{x}^{2}+1}} \left ({\it EllipticF} \left ( \sqrt{dx}\sqrt{{\frac{c}{d}}},i \right ) -{\it EllipticE} \left ( \sqrt{dx}\sqrt{{\frac{c}{d}}},i \right ) \right ){\frac{1}{\sqrt{{\frac{c}{d}}}}}} \right ) } \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*(1/5*(d*x)^(5/2)*a+b*(1/5*(d*x)^(5/2)*arcsin(c*x)-2/5*c/d*(-1/5/c^2*d^2*(d*x)^(3/2)*(-c^2*x^2+1)^(1/2)-3/5
/c^3*d^3/(c/d)^(1/2)*(-c*x+1)^(1/2)*(c*x+1)^(1/2)/(-c^2*x^2+1)^(1/2)*(EllipticF((d*x)^(1/2)*(c/d)^(1/2),I)-Ell
ipticE((d*x)^(1/2)*(c/d)^(1/2),I)))))

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b d x \arcsin \left (c x\right ) + a d x\right )} \sqrt{d x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 163.098, size = 82, normalized size = 0.66 \begin{align*} a \left (\begin{cases} 0 & \text{for}\: d = 0 \\\frac{2 \left (d x\right )^{\frac{5}{2}}}{5 d} & \text{otherwise} \end{cases}\right ) - b c \left (\begin{cases} 0 & \text{for}\: d = 0 \\\frac{d^{\frac{3}{2}} x^{\frac{7}{2}} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{c^{2} x^{2} e^{2 i \pi }} \right )}}{5 \Gamma \left (\frac{11}{4}\right )} & \text{otherwise} \end{cases}\right ) + b \left (\begin{cases} 0 & \text{for}\: d = 0 \\\frac{2 \left (d x\right )^{\frac{5}{2}}}{5 d} & \text{otherwise} \end{cases}\right ) \operatorname{asin}{\left (c x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*Piecewise((0, Eq(d, 0)), (2*(d*x)**(5/2)/(5*d), True)) - b*c*Piecewise((0, Eq(d, 0)), (d**(3/2)*x**(7/2)*gam
ma(7/4)*hyper((1/2, 7/4), (11/4,), c**2*x**2*exp_polar(2*I*pi))/(5*gamma(11/4)), True)) + b*Piecewise((0, Eq(d
, 0)), (2*(d*x)**(5/2)/(5*d), True))*asin(c*x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d x\right )^{\frac{3}{2}}{\left (b \arcsin \left (c x\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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