3.208 \(\int \frac{a+b \sin ^{-1}(c x)}{(d x)^{5/2}} \, dx\)

Optimal. Leaf size=125 \[ \frac{4 b c^{3/2} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right ),-1\right )}{3 d^{5/2}}-\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{3 d (d x)^{3/2}}-\frac{4 b c \sqrt{1-c^2 x^2}}{3 d^2 \sqrt{d x}}-\frac{4 b c^{3/2} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )\right |-1\right )}{3 d^{5/2}} \]

[Out]

(-4*b*c*Sqrt[1 - c^2*x^2])/(3*d^2*Sqrt[d*x]) - (2*(a + b*ArcSin[c*x]))/(3*d*(d*x)^(3/2)) - (4*b*c^(3/2)*Ellipt
icE[ArcSin[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]], -1])/(3*d^(5/2)) + (4*b*c^(3/2)*EllipticF[ArcSin[(Sqrt[c]*Sqrt[d*x])/
Sqrt[d]], -1])/(3*d^(5/2))

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Rubi [A]  time = 0.0968909, antiderivative size = 125, 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, 325, 329, 307, 221, 1199, 424} \[ -\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{3 d (d x)^{3/2}}-\frac{4 b c \sqrt{1-c^2 x^2}}{3 d^2 \sqrt{d x}}+\frac{4 b c^{3/2} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )\right |-1\right )}{3 d^{5/2}}-\frac{4 b c^{3/2} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )\right |-1\right )}{3 d^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*b*c*Sqrt[1 - c^2*x^2])/(3*d^2*Sqrt[d*x]) - (2*(a + b*ArcSin[c*x]))/(3*d*(d*x)^(3/2)) - (4*b*c^(3/2)*Ellipt
icE[ArcSin[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]], -1])/(3*d^(5/2)) + (4*b*c^(3/2)*EllipticF[ArcSin[(Sqrt[c]*Sqrt[d*x])/
Sqrt[d]], -1])/(3*d^(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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 \frac{a+b \sin ^{-1}(c x)}{(d x)^{5/2}} \, dx &=-\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{3 d (d x)^{3/2}}+\frac{(2 b c) \int \frac{1}{(d x)^{3/2} \sqrt{1-c^2 x^2}} \, dx}{3 d}\\ &=-\frac{4 b c \sqrt{1-c^2 x^2}}{3 d^2 \sqrt{d x}}-\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{3 d (d x)^{3/2}}-\frac{\left (2 b c^3\right ) \int \frac{\sqrt{d x}}{\sqrt{1-c^2 x^2}} \, dx}{3 d^3}\\ &=-\frac{4 b c \sqrt{1-c^2 x^2}}{3 d^2 \sqrt{d x}}-\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{3 d (d x)^{3/2}}-\frac{\left (4 b c^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-\frac{c^2 x^4}{d^2}}} \, dx,x,\sqrt{d x}\right )}{3 d^4}\\ &=-\frac{4 b c \sqrt{1-c^2 x^2}}{3 d^2 \sqrt{d x}}-\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{3 d (d x)^{3/2}}+\frac{\left (4 b c^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{c^2 x^4}{d^2}}} \, dx,x,\sqrt{d x}\right )}{3 d^3}-\frac{\left (4 b c^2\right ) \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 )}{3 d^3}\\ &=-\frac{4 b c \sqrt{1-c^2 x^2}}{3 d^2 \sqrt{d x}}-\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{3 d (d x)^{3/2}}+\frac{4 b c^{3/2} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )\right |-1\right )}{3 d^{5/2}}-\frac{\left (4 b c^2\right ) \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 )}{3 d^3}\\ &=-\frac{4 b c \sqrt{1-c^2 x^2}}{3 d^2 \sqrt{d x}}-\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{3 d (d x)^{3/2}}-\frac{4 b c^{3/2} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )\right |-1\right )}{3 d^{5/2}}+\frac{4 b c^{3/2} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )\right |-1\right )}{3 d^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.015386, size = 42, normalized size = 0.34 \[ -\frac{2 x \left (2 b c x \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},c^2 x^2\right )+a+b \sin ^{-1}(c x)\right )}{3 (d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.013, size = 129, normalized size = 1. \begin{align*} 2\,{\frac{1}{d} \left ( -1/3\,{\frac{a}{ \left ( dx \right ) ^{3/2}}}+b \left ( -1/3\,{\frac{\arcsin \left ( cx \right ) }{ \left ( dx \right ) ^{3/2}}}+2/3\,{\frac{c}{d} \left ( -{\frac{\sqrt{-{c}^{2}{x}^{2}+1}}{\sqrt{dx}}}+{\frac{c\sqrt{-cx+1}\sqrt{cx+1}}{d\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((a+b*arcsin(c*x))/(d*x)^(5/2),x)

[Out]

2/d*(-1/3*a/(d*x)^(3/2)+b*(-1/3/(d*x)^(3/2)*arcsin(c*x)+2/3*c/d*(-(-c^2*x^2+1)^(1/2)/(d*x)^(1/2)+c/d/(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)-EllipticE((d*x)^(1/2)
*(c/d)^(1/2),I)))))

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

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(d*x)*(b*arcsin(c*x) + a)/(d^3*x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \arcsin \left (c x\right ) + a}{\left (d x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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