3.289 \(\int \frac{a+b \sin ^{-1}(c+d x)}{(c e+d e x)^{9/2}} \, dx\)

Optimal. Leaf size=159 \[ -\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{7 d e (e (c+d x))^{7/2}}-\frac{12 b \sqrt{1-(c+d x)^2}}{35 d e^4 \sqrt{e (c+d x)}}-\frac{4 b \sqrt{1-(c+d x)^2}}{35 d e^2 (e (c+d x))^{5/2}}+\frac{12 b \sqrt{e (c+d x)} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{-c-d x+1}}{\sqrt{2}}\right )\right |2\right )}{35 d e^5 \sqrt{c+d x}} \]

[Out]

(-4*b*Sqrt[1 - (c + d*x)^2])/(35*d*e^2*(e*(c + d*x))^(5/2)) - (12*b*Sqrt[1 - (c + d*x)^2])/(35*d*e^4*Sqrt[e*(c
+ d*x)]) - (2*(a + b*ArcSin[c + d*x]))/(7*d*e*(e*(c + d*x))^(7/2)) + (12*b*Sqrt[e*(c + d*x)]*EllipticE[ArcSin
[Sqrt[1 - c - d*x]/Sqrt[2]], 2])/(35*d*e^5*Sqrt[c + d*x])

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Rubi [A]  time = 0.127513, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {4805, 4627, 325, 320, 318, 424} \[ -\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{7 d e (e (c+d x))^{7/2}}-\frac{12 b \sqrt{1-(c+d x)^2}}{35 d e^4 \sqrt{e (c+d x)}}-\frac{4 b \sqrt{1-(c+d x)^2}}{35 d e^2 (e (c+d x))^{5/2}}+\frac{12 b \sqrt{e (c+d x)} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{-c-d x+1}}{\sqrt{2}}\right )\right |2\right )}{35 d e^5 \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x])/(c*e + d*e*x)^(9/2),x]

[Out]

(-4*b*Sqrt[1 - (c + d*x)^2])/(35*d*e^2*(e*(c + d*x))^(5/2)) - (12*b*Sqrt[1 - (c + d*x)^2])/(35*d*e^4*Sqrt[e*(c
+ d*x)]) - (2*(a + b*ArcSin[c + d*x]))/(7*d*e*(e*(c + d*x))^(7/2)) + (12*b*Sqrt[e*(c + d*x)]*EllipticE[ArcSin
[Sqrt[1 - c - d*x]/Sqrt[2]], 2])/(35*d*e^5*Sqrt[c + d*x])

Rule 4805

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

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 320

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

Rule 318

Int[Sqrt[x_]/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Dist[-2/(Sqrt[a]*(-(b/a))^(3/4)), Subst[Int[Sqrt[1 - 2*x^
2]/Sqrt[1 - x^2], x], x, Sqrt[1 - Sqrt[-(b/a)]*x]/Sqrt[2]], x] /; FreeQ[{a, b}, x] && GtQ[-(b/a), 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+d x)}{(c e+d e x)^{9/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a+b \sin ^{-1}(x)}{(e x)^{9/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{7 d e (e (c+d x))^{7/2}}+\frac{(2 b) \operatorname{Subst}\left (\int \frac{1}{(e x)^{7/2} \sqrt{1-x^2}} \, dx,x,c+d x\right )}{7 d e}\\ &=-\frac{4 b \sqrt{1-(c+d x)^2}}{35 d e^2 (e (c+d x))^{5/2}}-\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{7 d e (e (c+d x))^{7/2}}+\frac{(6 b) \operatorname{Subst}\left (\int \frac{1}{(e x)^{3/2} \sqrt{1-x^2}} \, dx,x,c+d x\right )}{35 d e^3}\\ &=-\frac{4 b \sqrt{1-(c+d x)^2}}{35 d e^2 (e (c+d x))^{5/2}}-\frac{12 b \sqrt{1-(c+d x)^2}}{35 d e^4 \sqrt{e (c+d x)}}-\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{7 d e (e (c+d x))^{7/2}}-\frac{(6 b) \operatorname{Subst}\left (\int \frac{\sqrt{e x}}{\sqrt{1-x^2}} \, dx,x,c+d x\right )}{35 d e^5}\\ &=-\frac{4 b \sqrt{1-(c+d x)^2}}{35 d e^2 (e (c+d x))^{5/2}}-\frac{12 b \sqrt{1-(c+d x)^2}}{35 d e^4 \sqrt{e (c+d x)}}-\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{7 d e (e (c+d x))^{7/2}}-\frac{\left (6 b \sqrt{e (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{x}}{\sqrt{1-x^2}} \, dx,x,c+d x\right )}{35 d e^5 \sqrt{c+d x}}\\ &=-\frac{4 b \sqrt{1-(c+d x)^2}}{35 d e^2 (e (c+d x))^{5/2}}-\frac{12 b \sqrt{1-(c+d x)^2}}{35 d e^4 \sqrt{e (c+d x)}}-\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{7 d e (e (c+d x))^{7/2}}+\frac{\left (12 b \sqrt{e (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1-2 x^2}}{\sqrt{1-x^2}} \, dx,x,\frac{\sqrt{1-c-d x}}{\sqrt{2}}\right )}{35 d e^5 \sqrt{c+d x}}\\ &=-\frac{4 b \sqrt{1-(c+d x)^2}}{35 d e^2 (e (c+d x))^{5/2}}-\frac{12 b \sqrt{1-(c+d x)^2}}{35 d e^4 \sqrt{e (c+d x)}}-\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{7 d e (e (c+d x))^{7/2}}+\frac{12 b \sqrt{e (c+d x)} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{1-c-d x}}{\sqrt{2}}\right )\right |2\right )}{35 d e^5 \sqrt{c+d x}}\\ \end{align*}

Mathematica [C]  time = 0.0436403, size = 66, normalized size = 0.42 \[ -\frac{2 \sqrt{e (c+d x)} \left (2 b (c+d x) \text{Hypergeometric2F1}\left (-\frac{5}{4},\frac{1}{2},-\frac{1}{4},(c+d x)^2\right )+5 \left (a+b \sin ^{-1}(c+d x)\right )\right )}{35 d e^5 (c+d x)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c + d*x])/(c*e + d*e*x)^(9/2),x]

[Out]

(-2*Sqrt[e*(c + d*x)]*(5*(a + b*ArcSin[c + d*x]) + 2*b*(c + d*x)*Hypergeometric2F1[-5/4, 1/2, -1/4, (c + d*x)^
2]))/(35*d*e^5*(c + d*x)^4)

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Maple [C]  time = 0.015, size = 225, normalized size = 1.4 \begin{align*} 2\,{\frac{1}{de} \left ( -1/7\,{\frac{a}{ \left ( dex+ce \right ) ^{7/2}}}+b \left ( -1/7\,{\frac{1}{ \left ( dex+ce \right ) ^{7/2}}\arcsin \left ({\frac{dex+ce}{e}} \right ) }+2/7\,{\frac{1}{e} \left ( -1/5\,{\frac{1}{ \left ( dex+ce \right ) ^{5/2}}\sqrt{-{\frac{ \left ( dex+ce \right ) ^{2}}{{e}^{2}}}+1}}-3/5\,{\frac{1}{{e}^{2}\sqrt{dex+ce}}\sqrt{-{\frac{ \left ( dex+ce \right ) ^{2}}{{e}^{2}}}+1}}+3/5\,{\frac{{\it EllipticF} \left ( \sqrt{dex+ce}\sqrt{{e}^{-1}},i \right ) -{\it EllipticE} \left ( \sqrt{dex+ce}\sqrt{{e}^{-1}},i \right ) }{{e}^{3}\sqrt{{e}^{-1}}}\sqrt{1-{\frac{dex+ce}{e}}}\sqrt{{\frac{dex+ce}{e}}+1}{\frac{1}{\sqrt{-{\frac{ \left ( dex+ce \right ) ^{2}}{{e}^{2}}}+1}}}} \right ) } \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(d*x+c))/(d*e*x+c*e)^(9/2),x)

[Out]

2/d/e*(-1/7*a/(d*e*x+c*e)^(7/2)+b*(-1/7/(d*e*x+c*e)^(7/2)*arcsin((d*e*x+c*e)/e)+2/7/e*(-1/5*(-(d*e*x+c*e)^2/e^
2+1)^(1/2)/(d*e*x+c*e)^(5/2)-3/5/e^2*(-(d*e*x+c*e)^2/e^2+1)^(1/2)/(d*e*x+c*e)^(1/2)+3/5/e^3/(1/e)^(1/2)*(1-(d*
e*x+c*e)/e)^(1/2)*((d*e*x+c*e)/e+1)^(1/2)/(-(d*e*x+c*e)^2/e^2+1)^(1/2)*(EllipticF((d*e*x+c*e)^(1/2)*(1/e)^(1/2
),I)-EllipticE((d*e*x+c*e)^(1/2)*(1/e)^(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(d*x+c))/(d*e*x+c*e)^(9/2),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 e x + c e}{\left (b \arcsin \left (d x + c\right ) + a\right )}}{d^{5} e^{5} x^{5} + 5 \, c d^{4} e^{5} x^{4} + 10 \, c^{2} d^{3} e^{5} x^{3} + 10 \, c^{3} d^{2} e^{5} x^{2} + 5 \, c^{4} d e^{5} x + c^{5} e^{5}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))/(d*e*x+c*e)^(9/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*e*x + c*e)*(b*arcsin(d*x + c) + a)/(d^5*e^5*x^5 + 5*c*d^4*e^5*x^4 + 10*c^2*d^3*e^5*x^3 + 10*c^
3*d^2*e^5*x^2 + 5*c^4*d*e^5*x + c^5*e^5), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(d*x+c))/(d*e*x+c*e)**(9/2),x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))/(d*e*x+c*e)^(9/2),x, algorithm="giac")

[Out]

Timed out