3.290 \(\int \frac{a+b \sin ^{-1}(c+d x)}{(c e+d e x)^{11/2}} \, dx\)

Optimal. Leaf size=139 \[ \frac{20 b \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{e (c+d x)}}{\sqrt{e}}\right ),-1\right )}{189 d e^{11/2}}-\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{9 d e (e (c+d x))^{9/2}}-\frac{20 b \sqrt{1-(c+d x)^2}}{189 d e^4 (e (c+d x))^{3/2}}-\frac{4 b \sqrt{1-(c+d x)^2}}{63 d e^2 (e (c+d x))^{7/2}} \]

[Out]

(-4*b*Sqrt[1 - (c + d*x)^2])/(63*d*e^2*(e*(c + d*x))^(7/2)) - (20*b*Sqrt[1 - (c + d*x)^2])/(189*d*e^4*(e*(c +
d*x))^(3/2)) - (2*(a + b*ArcSin[c + d*x]))/(9*d*e*(e*(c + d*x))^(9/2)) + (20*b*EllipticF[ArcSin[Sqrt[e*(c + d*
x)]/Sqrt[e]], -1])/(189*d*e^(11/2))

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Rubi [A]  time = 0.113851, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {4805, 4627, 325, 329, 221} \[ -\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{9 d e (e (c+d x))^{9/2}}-\frac{20 b \sqrt{1-(c+d x)^2}}{189 d e^4 (e (c+d x))^{3/2}}-\frac{4 b \sqrt{1-(c+d x)^2}}{63 d e^2 (e (c+d x))^{7/2}}+\frac{20 b F\left (\left .\sin ^{-1}\left (\frac{\sqrt{e (c+d x)}}{\sqrt{e}}\right )\right |-1\right )}{189 d e^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x])/(c*e + d*e*x)^(11/2),x]

[Out]

(-4*b*Sqrt[1 - (c + d*x)^2])/(63*d*e^2*(e*(c + d*x))^(7/2)) - (20*b*Sqrt[1 - (c + d*x)^2])/(189*d*e^4*(e*(c +
d*x))^(3/2)) - (2*(a + b*ArcSin[c + d*x]))/(9*d*e*(e*(c + d*x))^(9/2)) + (20*b*EllipticF[ArcSin[Sqrt[e*(c + d*
x)]/Sqrt[e]], -1])/(189*d*e^(11/2))

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 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 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]

Rubi steps

\begin{align*} \int \frac{a+b \sin ^{-1}(c+d x)}{(c e+d e x)^{11/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a+b \sin ^{-1}(x)}{(e x)^{11/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{9 d e (e (c+d x))^{9/2}}+\frac{(2 b) \operatorname{Subst}\left (\int \frac{1}{(e x)^{9/2} \sqrt{1-x^2}} \, dx,x,c+d x\right )}{9 d e}\\ &=-\frac{4 b \sqrt{1-(c+d x)^2}}{63 d e^2 (e (c+d x))^{7/2}}-\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{9 d e (e (c+d x))^{9/2}}+\frac{(10 b) \operatorname{Subst}\left (\int \frac{1}{(e x)^{5/2} \sqrt{1-x^2}} \, dx,x,c+d x\right )}{63 d e^3}\\ &=-\frac{4 b \sqrt{1-(c+d x)^2}}{63 d e^2 (e (c+d x))^{7/2}}-\frac{20 b \sqrt{1-(c+d x)^2}}{189 d e^4 (e (c+d x))^{3/2}}-\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{9 d e (e (c+d x))^{9/2}}+\frac{(10 b) \operatorname{Subst}\left (\int \frac{1}{\sqrt{e x} \sqrt{1-x^2}} \, dx,x,c+d x\right )}{189 d e^5}\\ &=-\frac{4 b \sqrt{1-(c+d x)^2}}{63 d e^2 (e (c+d x))^{7/2}}-\frac{20 b \sqrt{1-(c+d x)^2}}{189 d e^4 (e (c+d x))^{3/2}}-\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{9 d e (e (c+d x))^{9/2}}+\frac{(20 b) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^4}{e^2}}} \, dx,x,\sqrt{e (c+d x)}\right )}{189 d e^6}\\ &=-\frac{4 b \sqrt{1-(c+d x)^2}}{63 d e^2 (e (c+d x))^{7/2}}-\frac{20 b \sqrt{1-(c+d x)^2}}{189 d e^4 (e (c+d x))^{3/2}}-\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{9 d e (e (c+d x))^{9/2}}+\frac{20 b F\left (\left .\sin ^{-1}\left (\frac{\sqrt{e (c+d x)}}{\sqrt{e}}\right )\right |-1\right )}{189 d e^{11/2}}\\ \end{align*}

Mathematica [C]  time = 0.0446952, size = 66, normalized size = 0.47 \[ -\frac{2 \sqrt{e (c+d x)} \left (2 b (c+d x) \text{Hypergeometric2F1}\left (-\frac{7}{4},\frac{1}{2},-\frac{3}{4},(c+d x)^2\right )+7 \left (a+b \sin ^{-1}(c+d x)\right )\right )}{63 d e^6 (c+d x)^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c + d*x])/(c*e + d*e*x)^(11/2),x]

[Out]

(-2*Sqrt[e*(c + d*x)]*(7*(a + b*ArcSin[c + d*x]) + 2*b*(c + d*x)*Hypergeometric2F1[-7/4, 1/2, -3/4, (c + d*x)^
2]))/(63*d*e^6*(c + d*x)^5)

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Maple [A]  time = 0.017, size = 203, normalized size = 1.5 \begin{align*} 2\,{\frac{1}{de} \left ( -1/9\,{\frac{a}{ \left ( dex+ce \right ) ^{9/2}}}+b \left ( -1/9\,{\frac{1}{ \left ( dex+ce \right ) ^{9/2}}\arcsin \left ({\frac{dex+ce}{e}} \right ) }+2/9\,{\frac{1}{e} \left ( -1/7\,{\frac{1}{ \left ( dex+ce \right ) ^{7/2}}\sqrt{-{\frac{ \left ( dex+ce \right ) ^{2}}{{e}^{2}}}+1}}-{\frac{5}{21\,{e}^{2} \left ( dex+ce \right ) ^{3/2}}\sqrt{-{\frac{ \left ( dex+ce \right ) ^{2}}{{e}^{2}}}+1}}+{\frac{5\,{\it EllipticF} \left ( \sqrt{dex+ce}\sqrt{{e}^{-1}},i \right ) }{21\,{e}^{4}\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)^(11/2),x)

[Out]

2/d/e*(-1/9*a/(d*e*x+c*e)^(9/2)+b*(-1/9/(d*e*x+c*e)^(9/2)*arcsin((d*e*x+c*e)/e)+2/9/e*(-1/7*(-(d*e*x+c*e)^2/e^
2+1)^(1/2)/(d*e*x+c*e)^(7/2)-5/21/e^2*(-(d*e*x+c*e)^2/e^2+1)^(1/2)/(d*e*x+c*e)^(3/2)+5/21/e^4/(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))))

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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)^(11/2),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^{6} e^{6} x^{6} + 6 \, c d^{5} e^{6} x^{5} + 15 \, c^{2} d^{4} e^{6} x^{4} + 20 \, c^{3} d^{3} e^{6} x^{3} + 15 \, c^{4} d^{2} e^{6} x^{2} + 6 \, c^{5} d e^{6} x + c^{6} e^{6}}, 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)^(11/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*e*x + c*e)*(b*arcsin(d*x + c) + a)/(d^6*e^6*x^6 + 6*c*d^5*e^6*x^5 + 15*c^2*d^4*e^6*x^4 + 20*c^
3*d^3*e^6*x^3 + 15*c^4*d^2*e^6*x^2 + 6*c^5*d*e^6*x + c^6*e^6), 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)**(11/2),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)^(11/2),x, algorithm="giac")

[Out]

Timed out