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

Optimal. Leaf size=89 \[ -\frac{a+b \sin ^{-1}\left (c x^2\right )}{10 x^{10}}-\frac{3 b c^3 \sqrt{1-c^2 x^4}}{80 x^4}-\frac{b c \sqrt{1-c^2 x^4}}{40 x^8}-\frac{3}{80} b c^5 \tanh ^{-1}\left (\sqrt{1-c^2 x^4}\right ) \]

[Out]

-(b*c*Sqrt[1 - c^2*x^4])/(40*x^8) - (3*b*c^3*Sqrt[1 - c^2*x^4])/(80*x^4) - (a + b*ArcSin[c*x^2])/(10*x^10) - (
3*b*c^5*ArcTanh[Sqrt[1 - c^2*x^4]])/80

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Rubi [A]  time = 0.0632384, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {4842, 12, 266, 51, 63, 208} \[ -\frac{a+b \sin ^{-1}\left (c x^2\right )}{10 x^{10}}-\frac{3 b c^3 \sqrt{1-c^2 x^4}}{80 x^4}-\frac{b c \sqrt{1-c^2 x^4}}{40 x^8}-\frac{3}{80} b c^5 \tanh ^{-1}\left (\sqrt{1-c^2 x^4}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x^2])/x^11,x]

[Out]

-(b*c*Sqrt[1 - c^2*x^4])/(40*x^8) - (3*b*c^3*Sqrt[1 - c^2*x^4])/(80*x^4) - (a + b*ArcSin[c*x^2])/(10*x^10) - (
3*b*c^5*ArcTanh[Sqrt[1 - c^2*x^4]])/80

Rule 4842

Int[((a_.) + ArcSin[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcSin[
u]))/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/Sqrt[1 - u^2], x]
, x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)^(
m + 1), u, x] &&  !FunctionOfExponentialQ[u, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{a+b \sin ^{-1}\left (c x^2\right )}{x^{11}} \, dx &=-\frac{a+b \sin ^{-1}\left (c x^2\right )}{10 x^{10}}+\frac{1}{10} b \int \frac{2 c}{x^9 \sqrt{1-c^2 x^4}} \, dx\\ &=-\frac{a+b \sin ^{-1}\left (c x^2\right )}{10 x^{10}}+\frac{1}{5} (b c) \int \frac{1}{x^9 \sqrt{1-c^2 x^4}} \, dx\\ &=-\frac{a+b \sin ^{-1}\left (c x^2\right )}{10 x^{10}}+\frac{1}{20} (b c) \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{1-c^2 x}} \, dx,x,x^4\right )\\ &=-\frac{b c \sqrt{1-c^2 x^4}}{40 x^8}-\frac{a+b \sin ^{-1}\left (c x^2\right )}{10 x^{10}}+\frac{1}{80} \left (3 b c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1-c^2 x}} \, dx,x,x^4\right )\\ &=-\frac{b c \sqrt{1-c^2 x^4}}{40 x^8}-\frac{3 b c^3 \sqrt{1-c^2 x^4}}{80 x^4}-\frac{a+b \sin ^{-1}\left (c x^2\right )}{10 x^{10}}+\frac{1}{160} \left (3 b c^5\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-c^2 x}} \, dx,x,x^4\right )\\ &=-\frac{b c \sqrt{1-c^2 x^4}}{40 x^8}-\frac{3 b c^3 \sqrt{1-c^2 x^4}}{80 x^4}-\frac{a+b \sin ^{-1}\left (c x^2\right )}{10 x^{10}}-\frac{1}{80} \left (3 b c^3\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{c^2}-\frac{x^2}{c^2}} \, dx,x,\sqrt{1-c^2 x^4}\right )\\ &=-\frac{b c \sqrt{1-c^2 x^4}}{40 x^8}-\frac{3 b c^3 \sqrt{1-c^2 x^4}}{80 x^4}-\frac{a+b \sin ^{-1}\left (c x^2\right )}{10 x^{10}}-\frac{3}{80} b c^5 \tanh ^{-1}\left (\sqrt{1-c^2 x^4}\right )\\ \end{align*}

Mathematica [C]  time = 0.0196502, size = 63, normalized size = 0.71 \[ -\frac{1}{10} b c^5 \sqrt{1-c^2 x^4} \text{Hypergeometric2F1}\left (\frac{1}{2},3,\frac{3}{2},1-c^2 x^4\right )-\frac{a}{10 x^{10}}-\frac{b \sin ^{-1}\left (c x^2\right )}{10 x^{10}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(10*x^10) - (b*ArcSin[c*x^2])/(10*x^10) - (b*c^5*Sqrt[1 - c^2*x^4]*Hypergeometric2F1[1/2, 3, 3/2, 1 - c^2*x
^4])/10

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Maple [A]  time = 0.014, size = 84, normalized size = 0.9 \begin{align*} -{\frac{a}{10\,{x}^{10}}}+b \left ( -{\frac{\arcsin \left ( c{x}^{2} \right ) }{10\,{x}^{10}}}+{\frac{c}{5} \left ( -{\frac{1}{8\,{x}^{8}}\sqrt{-{c}^{2}{x}^{4}+1}}+{\frac{3\,{c}^{2}}{8} \left ( -{\frac{1}{2\,{x}^{4}}\sqrt{-{c}^{2}{x}^{4}+1}}-{\frac{{c}^{2}}{2}{\it Artanh} \left ({\frac{1}{\sqrt{-{c}^{2}{x}^{4}+1}}} \right ) } \right ) } \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(c*x^2))/x^11,x)

[Out]

-1/10*a/x^10+b*(-1/10/x^10*arcsin(c*x^2)+1/5*c*(-1/8/x^8*(-c^2*x^4+1)^(1/2)+3/8*c^2*(-1/2/x^4*(-c^2*x^4+1)^(1/
2)-1/2*c^2*arctanh(1/(-c^2*x^4+1)^(1/2)))))

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Maxima [A]  time = 1.43481, size = 169, normalized size = 1.9 \begin{align*} -\frac{1}{160} \,{\left ({\left (3 \, c^{4} \log \left (\sqrt{-c^{2} x^{4} + 1} + 1\right ) - 3 \, c^{4} \log \left (\sqrt{-c^{2} x^{4} + 1} - 1\right ) - \frac{2 \,{\left (3 \,{\left (-c^{2} x^{4} + 1\right )}^{\frac{3}{2}} c^{4} - 5 \, \sqrt{-c^{2} x^{4} + 1} c^{4}\right )}}{2 \, c^{2} x^{4} +{\left (c^{2} x^{4} - 1\right )}^{2} - 1}\right )} c + \frac{16 \, \arcsin \left (c x^{2}\right )}{x^{10}}\right )} b - \frac{a}{10 \, x^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x^2))/x^11,x, algorithm="maxima")

[Out]

-1/160*((3*c^4*log(sqrt(-c^2*x^4 + 1) + 1) - 3*c^4*log(sqrt(-c^2*x^4 + 1) - 1) - 2*(3*(-c^2*x^4 + 1)^(3/2)*c^4
- 5*sqrt(-c^2*x^4 + 1)*c^4)/(2*c^2*x^4 + (c^2*x^4 - 1)^2 - 1))*c + 16*arcsin(c*x^2)/x^10)*b - 1/10*a/x^10

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Fricas [A]  time = 3.49535, size = 238, normalized size = 2.67 \begin{align*} -\frac{3 \, b c^{5} x^{10} \log \left (\sqrt{-c^{2} x^{4} + 1} + 1\right ) - 3 \, b c^{5} x^{10} \log \left (\sqrt{-c^{2} x^{4} + 1} - 1\right ) + 16 \, b \arcsin \left (c x^{2}\right ) + 2 \,{\left (3 \, b c^{3} x^{6} + 2 \, b c x^{2}\right )} \sqrt{-c^{2} x^{4} + 1} + 16 \, a}{160 \, x^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x^2))/x^11,x, algorithm="fricas")

[Out]

-1/160*(3*b*c^5*x^10*log(sqrt(-c^2*x^4 + 1) + 1) - 3*b*c^5*x^10*log(sqrt(-c^2*x^4 + 1) - 1) + 16*b*arcsin(c*x^
2) + 2*(3*b*c^3*x^6 + 2*b*c*x^2)*sqrt(-c^2*x^4 + 1) + 16*a)/x^10

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Sympy [A]  time = 29.2096, size = 201, normalized size = 2.26 \begin{align*} - \frac{a}{10 x^{10}} + \frac{b c \left (\begin{cases} - \frac{3 c^{4} \operatorname{acosh}{\left (\frac{1}{c x^{2}} \right )}}{16} + \frac{3 c^{3}}{16 x^{2} \sqrt{-1 + \frac{1}{c^{2} x^{4}}}} - \frac{c}{16 x^{6} \sqrt{-1 + \frac{1}{c^{2} x^{4}}}} - \frac{1}{8 c x^{10} \sqrt{-1 + \frac{1}{c^{2} x^{4}}}} & \text{for}\: \frac{1}{\left |{c^{2} x^{4}}\right |} > 1 \\\frac{3 i c^{4} \operatorname{asin}{\left (\frac{1}{c x^{2}} \right )}}{16} - \frac{3 i c^{3}}{16 x^{2} \sqrt{1 - \frac{1}{c^{2} x^{4}}}} + \frac{i c}{16 x^{6} \sqrt{1 - \frac{1}{c^{2} x^{4}}}} + \frac{i}{8 c x^{10} \sqrt{1 - \frac{1}{c^{2} x^{4}}}} & \text{otherwise} \end{cases}\right )}{5} - \frac{b \operatorname{asin}{\left (c x^{2} \right )}}{10 x^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(c*x**2))/x**11,x)

[Out]

-a/(10*x**10) + b*c*Piecewise((-3*c**4*acosh(1/(c*x**2))/16 + 3*c**3/(16*x**2*sqrt(-1 + 1/(c**2*x**4))) - c/(1
6*x**6*sqrt(-1 + 1/(c**2*x**4))) - 1/(8*c*x**10*sqrt(-1 + 1/(c**2*x**4))), 1/Abs(c**2*x**4) > 1), (3*I*c**4*as
in(1/(c*x**2))/16 - 3*I*c**3/(16*x**2*sqrt(1 - 1/(c**2*x**4))) + I*c/(16*x**6*sqrt(1 - 1/(c**2*x**4))) + I/(8*
c*x**10*sqrt(1 - 1/(c**2*x**4))), True))/5 - b*asin(c*x**2)/(10*x**10)

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Giac [B]  time = 4.11999, size = 630, normalized size = 7.08 \begin{align*} -\frac{\frac{2 \, b c^{11} x^{10} \arcsin \left (c x^{2}\right )}{{\left (\sqrt{-c^{2} x^{4} + 1} + 1\right )}^{5}} + \frac{2 \, a c^{11} x^{10}}{{\left (\sqrt{-c^{2} x^{4} + 1} + 1\right )}^{5}} - \frac{b c^{10} x^{8}}{{\left (\sqrt{-c^{2} x^{4} + 1} + 1\right )}^{4}} + \frac{10 \, b c^{9} x^{6} \arcsin \left (c x^{2}\right )}{{\left (\sqrt{-c^{2} x^{4} + 1} + 1\right )}^{3}} + \frac{10 \, a c^{9} x^{6}}{{\left (\sqrt{-c^{2} x^{4} + 1} + 1\right )}^{3}} - \frac{8 \, b c^{8} x^{4}}{{\left (\sqrt{-c^{2} x^{4} + 1} + 1\right )}^{2}} + \frac{20 \, b c^{7} x^{2} \arcsin \left (c x^{2}\right )}{\sqrt{-c^{2} x^{4} + 1} + 1} + \frac{20 \, a c^{7} x^{2}}{\sqrt{-c^{2} x^{4} + 1} + 1} - 24 \, b c^{6} \log \left (x^{2}{\left | c \right |}\right ) + 24 \, b c^{6} \log \left (\sqrt{-c^{2} x^{4} + 1} + 1\right ) + \frac{20 \, b c^{5}{\left (\sqrt{-c^{2} x^{4} + 1} + 1\right )} \arcsin \left (c x^{2}\right )}{x^{2}} + \frac{20 \, a c^{5}{\left (\sqrt{-c^{2} x^{4} + 1} + 1\right )}}{x^{2}} + \frac{8 \, b c^{4}{\left (\sqrt{-c^{2} x^{4} + 1} + 1\right )}^{2}}{x^{4}} + \frac{10 \, b c^{3}{\left (\sqrt{-c^{2} x^{4} + 1} + 1\right )}^{3} \arcsin \left (c x^{2}\right )}{x^{6}} + \frac{10 \, a c^{3}{\left (\sqrt{-c^{2} x^{4} + 1} + 1\right )}^{3}}{x^{6}} + \frac{b c^{2}{\left (\sqrt{-c^{2} x^{4} + 1} + 1\right )}^{4}}{x^{8}} + \frac{2 \, b c{\left (\sqrt{-c^{2} x^{4} + 1} + 1\right )}^{5} \arcsin \left (c x^{2}\right )}{x^{10}} + \frac{2 \, a c{\left (\sqrt{-c^{2} x^{4} + 1} + 1\right )}^{5}}{x^{10}}}{640 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x^2))/x^11,x, algorithm="giac")

[Out]

-1/640*(2*b*c^11*x^10*arcsin(c*x^2)/(sqrt(-c^2*x^4 + 1) + 1)^5 + 2*a*c^11*x^10/(sqrt(-c^2*x^4 + 1) + 1)^5 - b*
c^10*x^8/(sqrt(-c^2*x^4 + 1) + 1)^4 + 10*b*c^9*x^6*arcsin(c*x^2)/(sqrt(-c^2*x^4 + 1) + 1)^3 + 10*a*c^9*x^6/(sq
rt(-c^2*x^4 + 1) + 1)^3 - 8*b*c^8*x^4/(sqrt(-c^2*x^4 + 1) + 1)^2 + 20*b*c^7*x^2*arcsin(c*x^2)/(sqrt(-c^2*x^4 +
1) + 1) + 20*a*c^7*x^2/(sqrt(-c^2*x^4 + 1) + 1) - 24*b*c^6*log(x^2*abs(c)) + 24*b*c^6*log(sqrt(-c^2*x^4 + 1)
+ 1) + 20*b*c^5*(sqrt(-c^2*x^4 + 1) + 1)*arcsin(c*x^2)/x^2 + 20*a*c^5*(sqrt(-c^2*x^4 + 1) + 1)/x^2 + 8*b*c^4*(
sqrt(-c^2*x^4 + 1) + 1)^2/x^4 + 10*b*c^3*(sqrt(-c^2*x^4 + 1) + 1)^3*arcsin(c*x^2)/x^6 + 10*a*c^3*(sqrt(-c^2*x^
4 + 1) + 1)^3/x^6 + b*c^2*(sqrt(-c^2*x^4 + 1) + 1)^4/x^8 + 2*b*c*(sqrt(-c^2*x^4 + 1) + 1)^5*arcsin(c*x^2)/x^10
+ 2*a*c*(sqrt(-c^2*x^4 + 1) + 1)^5/x^10)/c