3.76 \(\int \frac{(d-c^2 d x^2)^{3/2} (a+b \sin ^{-1}(c x))}{x^{12}} \, dx\)

Optimal. Leaf size=385 \[ -\frac{16 c^6 \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{1155 d x^5}-\frac{8 c^4 \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{231 d x^7}-\frac{2 c^2 \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{33 d x^9}-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{11 d x^{11}}-\frac{4 b c^9 d \sqrt{d-c^2 d x^2}}{1155 x^2 \sqrt{1-c^2 x^2}}-\frac{b c^7 d \sqrt{d-c^2 d x^2}}{770 x^4 \sqrt{1-c^2 x^2}}-\frac{b c^5 d \sqrt{d-c^2 d x^2}}{1386 x^6 \sqrt{1-c^2 x^2}}+\frac{b c^3 d \sqrt{d-c^2 d x^2}}{66 x^8 \sqrt{1-c^2 x^2}}-\frac{b c d \sqrt{d-c^2 d x^2}}{110 x^{10} \sqrt{1-c^2 x^2}}+\frac{16 b c^{11} d \log (x) \sqrt{d-c^2 d x^2}}{1155 \sqrt{1-c^2 x^2}} \]

[Out]

-(b*c*d*Sqrt[d - c^2*d*x^2])/(110*x^10*Sqrt[1 - c^2*x^2]) + (b*c^3*d*Sqrt[d - c^2*d*x^2])/(66*x^8*Sqrt[1 - c^2
*x^2]) - (b*c^5*d*Sqrt[d - c^2*d*x^2])/(1386*x^6*Sqrt[1 - c^2*x^2]) - (b*c^7*d*Sqrt[d - c^2*d*x^2])/(770*x^4*S
qrt[1 - c^2*x^2]) - (4*b*c^9*d*Sqrt[d - c^2*d*x^2])/(1155*x^2*Sqrt[1 - c^2*x^2]) - ((d - c^2*d*x^2)^(5/2)*(a +
b*ArcSin[c*x]))/(11*d*x^11) - (2*c^2*(d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/(33*d*x^9) - (8*c^4*(d - c^2*
d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/(231*d*x^7) - (16*c^6*(d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/(1155*d*x^5
) + (16*b*c^11*d*Sqrt[d - c^2*d*x^2]*Log[x])/(1155*Sqrt[1 - c^2*x^2])

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Rubi [A]  time = 0.301133, antiderivative size = 385, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {271, 264, 4691, 12, 1799, 1620} \[ -\frac{16 c^6 \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{1155 d x^5}-\frac{8 c^4 \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{231 d x^7}-\frac{2 c^2 \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{33 d x^9}-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{11 d x^{11}}-\frac{4 b c^9 d \sqrt{d-c^2 d x^2}}{1155 x^2 \sqrt{1-c^2 x^2}}-\frac{b c^7 d \sqrt{d-c^2 d x^2}}{770 x^4 \sqrt{1-c^2 x^2}}-\frac{b c^5 d \sqrt{d-c^2 d x^2}}{1386 x^6 \sqrt{1-c^2 x^2}}+\frac{b c^3 d \sqrt{d-c^2 d x^2}}{66 x^8 \sqrt{1-c^2 x^2}}-\frac{b c d \sqrt{d-c^2 d x^2}}{110 x^{10} \sqrt{1-c^2 x^2}}+\frac{16 b c^{11} d \log (x) \sqrt{d-c^2 d x^2}}{1155 \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]))/x^12,x]

[Out]

-(b*c*d*Sqrt[d - c^2*d*x^2])/(110*x^10*Sqrt[1 - c^2*x^2]) + (b*c^3*d*Sqrt[d - c^2*d*x^2])/(66*x^8*Sqrt[1 - c^2
*x^2]) - (b*c^5*d*Sqrt[d - c^2*d*x^2])/(1386*x^6*Sqrt[1 - c^2*x^2]) - (b*c^7*d*Sqrt[d - c^2*d*x^2])/(770*x^4*S
qrt[1 - c^2*x^2]) - (4*b*c^9*d*Sqrt[d - c^2*d*x^2])/(1155*x^2*Sqrt[1 - c^2*x^2]) - ((d - c^2*d*x^2)^(5/2)*(a +
b*ArcSin[c*x]))/(11*d*x^11) - (2*c^2*(d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/(33*d*x^9) - (8*c^4*(d - c^2*
d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/(231*d*x^7) - (16*c^6*(d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/(1155*d*x^5
) + (16*b*c^11*d*Sqrt[d - c^2*d*x^2]*Log[x])/(1155*Sqrt[1 - c^2*x^2])

Rule 271

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

Rule 264

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] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 4691

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[x^
m*(1 - c^2*x^2)^p, x]}, Dist[a + b*ArcSin[c*x], Int[x^m*(d + e*x^2)^p, x], x] - Dist[(b*c*d^(p - 1/2)*Sqrt[d +
e*x^2])/Sqrt[1 - c^2*x^2], Int[SimplifyIntegrand[u/Sqrt[1 - c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x
] && EqQ[c^2*d + e, 0] && IGtQ[p + 1/2, 0] && (IGtQ[(m + 1)/2, 0] || ILtQ[(m + 2*p + 3)/2, 0])

Rule 12

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

Rule 1799

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*SubstFor[x^2,
Pq, x]*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && IntegerQ[(m - 1)/2]

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

\begin{align*} \int \frac{\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{x^{12}} \, dx &=-\frac{\left (b c d \sqrt{d-c^2 d x^2}\right ) \int \frac{\left (1-c^2 x^2\right )^2 \left (-105-70 c^2 x^2-40 c^4 x^4-16 c^6 x^6\right )}{1155 x^{11}} \, dx}{\sqrt{1-c^2 x^2}}+\left (a+b \sin ^{-1}(c x)\right ) \int \frac{\left (d-c^2 d x^2\right )^{3/2}}{x^{12}} \, dx\\ &=-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{11 d x^{11}}-\frac{\left (b c d \sqrt{d-c^2 d x^2}\right ) \int \frac{\left (1-c^2 x^2\right )^2 \left (-105-70 c^2 x^2-40 c^4 x^4-16 c^6 x^6\right )}{x^{11}} \, dx}{1155 \sqrt{1-c^2 x^2}}+\frac{1}{11} \left (6 c^2 \left (a+b \sin ^{-1}(c x)\right )\right ) \int \frac{\left (d-c^2 d x^2\right )^{3/2}}{x^{10}} \, dx\\ &=-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{11 d x^{11}}-\frac{2 c^2 \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{33 d x^9}-\frac{\left (b c d \sqrt{d-c^2 d x^2}\right ) \operatorname{Subst}\left (\int \frac{\left (1-c^2 x\right )^2 \left (-105-70 c^2 x-40 c^4 x^2-16 c^6 x^3\right )}{x^6} \, dx,x,x^2\right )}{2310 \sqrt{1-c^2 x^2}}+\frac{1}{33} \left (8 c^4 \left (a+b \sin ^{-1}(c x)\right )\right ) \int \frac{\left (d-c^2 d x^2\right )^{3/2}}{x^8} \, dx\\ &=-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{11 d x^{11}}-\frac{2 c^2 \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{33 d x^9}-\frac{8 c^4 \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{231 d x^7}-\frac{\left (b c d \sqrt{d-c^2 d x^2}\right ) \operatorname{Subst}\left (\int \left (-\frac{105}{x^6}+\frac{140 c^2}{x^5}-\frac{5 c^4}{x^4}-\frac{6 c^6}{x^3}-\frac{8 c^8}{x^2}-\frac{16 c^{10}}{x}\right ) \, dx,x,x^2\right )}{2310 \sqrt{1-c^2 x^2}}+\frac{1}{231} \left (16 c^6 \left (a+b \sin ^{-1}(c x)\right )\right ) \int \frac{\left (d-c^2 d x^2\right )^{3/2}}{x^6} \, dx\\ &=-\frac{b c d \sqrt{d-c^2 d x^2}}{110 x^{10} \sqrt{1-c^2 x^2}}+\frac{b c^3 d \sqrt{d-c^2 d x^2}}{66 x^8 \sqrt{1-c^2 x^2}}-\frac{b c^5 d \sqrt{d-c^2 d x^2}}{1386 x^6 \sqrt{1-c^2 x^2}}-\frac{b c^7 d \sqrt{d-c^2 d x^2}}{770 x^4 \sqrt{1-c^2 x^2}}-\frac{4 b c^9 d \sqrt{d-c^2 d x^2}}{1155 x^2 \sqrt{1-c^2 x^2}}-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{11 d x^{11}}-\frac{2 c^2 \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{33 d x^9}-\frac{8 c^4 \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{231 d x^7}-\frac{16 c^6 \left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{1155 d x^5}+\frac{16 b c^{11} d \sqrt{d-c^2 d x^2} \log (x)}{1155 \sqrt{1-c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.240391, size = 221, normalized size = 0.57 \[ \frac{16 b c^{11} d \log (x) \sqrt{d-c^2 d x^2}}{1155 \sqrt{1-c^2 x^2}}-\frac{d \sqrt{d-c^2 d x^2} \left (630 a \left (16 c^6 x^6+40 c^4 x^4+70 c^2 x^2+105\right ) \left (c^2 x^2-1\right )^3-b c x \sqrt{1-c^2 x^2} \left (29524 c^{10} x^{10}+2520 c^8 x^8+945 c^6 x^6+525 c^4 x^4-11025 c^2 x^2+6615\right )+630 b \left (16 c^6 x^6+40 c^4 x^4+70 c^2 x^2+105\right ) \left (c^2 x^2-1\right )^3 \sin ^{-1}(c x)\right )}{727650 x^{11} \left (c^2 x^2-1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[((d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]))/x^12,x]

[Out]

-(d*Sqrt[d - c^2*d*x^2]*(630*a*(-1 + c^2*x^2)^3*(105 + 70*c^2*x^2 + 40*c^4*x^4 + 16*c^6*x^6) - b*c*x*Sqrt[1 -
c^2*x^2]*(6615 - 11025*c^2*x^2 + 525*c^4*x^4 + 945*c^6*x^6 + 2520*c^8*x^8 + 29524*c^10*x^10) + 630*b*(-1 + c^2
*x^2)^3*(105 + 70*c^2*x^2 + 40*c^4*x^4 + 16*c^6*x^6)*ArcSin[c*x]))/(727650*x^11*(-1 + c^2*x^2)) + (16*b*c^11*d
*Sqrt[d - c^2*d*x^2]*Log[x])/(1155*Sqrt[1 - c^2*x^2])

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Maple [C]  time = 0.598, size = 5881, normalized size = 15.3 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))/x^12,x)

[Out]

result too large to display

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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((-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))/x^12,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.96817, size = 1702, normalized size = 4.42 \begin{align*} \left [\frac{48 \,{\left (b c^{13} d x^{13} - b c^{11} d x^{11}\right )} \sqrt{d} \log \left (\frac{c^{2} d x^{6} + c^{2} d x^{2} - d x^{4} - \sqrt{-c^{2} d x^{2} + d} \sqrt{-c^{2} x^{2} + 1}{\left (x^{4} - 1\right )} \sqrt{d} - d}{c^{2} x^{4} - x^{2}}\right ) +{\left (24 \, b c^{9} d x^{9} + 9 \, b c^{7} d x^{7} -{\left (24 \, b c^{9} + 9 \, b c^{7} + 5 \, b c^{5} - 105 \, b c^{3} + 63 \, b c\right )} d x^{11} + 5 \, b c^{5} d x^{5} - 105 \, b c^{3} d x^{3} + 63 \, b c d x\right )} \sqrt{-c^{2} d x^{2} + d} \sqrt{-c^{2} x^{2} + 1} - 6 \,{\left (16 \, a c^{12} d x^{12} - 8 \, a c^{10} d x^{10} - 2 \, a c^{8} d x^{8} - a c^{6} d x^{6} - 145 \, a c^{4} d x^{4} + 245 \, a c^{2} d x^{2} - 105 \, a d +{\left (16 \, b c^{12} d x^{12} - 8 \, b c^{10} d x^{10} - 2 \, b c^{8} d x^{8} - b c^{6} d x^{6} - 145 \, b c^{4} d x^{4} + 245 \, b c^{2} d x^{2} - 105 \, b d\right )} \arcsin \left (c x\right )\right )} \sqrt{-c^{2} d x^{2} + d}}{6930 \,{\left (c^{2} x^{13} - x^{11}\right )}}, \frac{96 \,{\left (b c^{13} d x^{13} - b c^{11} d x^{11}\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{-c^{2} d x^{2} + d} \sqrt{-c^{2} x^{2} + 1}{\left (x^{2} + 1\right )} \sqrt{-d}}{c^{2} d x^{4} -{\left (c^{2} + 1\right )} d x^{2} + d}\right ) +{\left (24 \, b c^{9} d x^{9} + 9 \, b c^{7} d x^{7} -{\left (24 \, b c^{9} + 9 \, b c^{7} + 5 \, b c^{5} - 105 \, b c^{3} + 63 \, b c\right )} d x^{11} + 5 \, b c^{5} d x^{5} - 105 \, b c^{3} d x^{3} + 63 \, b c d x\right )} \sqrt{-c^{2} d x^{2} + d} \sqrt{-c^{2} x^{2} + 1} - 6 \,{\left (16 \, a c^{12} d x^{12} - 8 \, a c^{10} d x^{10} - 2 \, a c^{8} d x^{8} - a c^{6} d x^{6} - 145 \, a c^{4} d x^{4} + 245 \, a c^{2} d x^{2} - 105 \, a d +{\left (16 \, b c^{12} d x^{12} - 8 \, b c^{10} d x^{10} - 2 \, b c^{8} d x^{8} - b c^{6} d x^{6} - 145 \, b c^{4} d x^{4} + 245 \, b c^{2} d x^{2} - 105 \, b d\right )} \arcsin \left (c x\right )\right )} \sqrt{-c^{2} d x^{2} + d}}{6930 \,{\left (c^{2} x^{13} - x^{11}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))/x^12,x, algorithm="fricas")

[Out]

[1/6930*(48*(b*c^13*d*x^13 - b*c^11*d*x^11)*sqrt(d)*log((c^2*d*x^6 + c^2*d*x^2 - d*x^4 - sqrt(-c^2*d*x^2 + d)*
sqrt(-c^2*x^2 + 1)*(x^4 - 1)*sqrt(d) - d)/(c^2*x^4 - x^2)) + (24*b*c^9*d*x^9 + 9*b*c^7*d*x^7 - (24*b*c^9 + 9*b
*c^7 + 5*b*c^5 - 105*b*c^3 + 63*b*c)*d*x^11 + 5*b*c^5*d*x^5 - 105*b*c^3*d*x^3 + 63*b*c*d*x)*sqrt(-c^2*d*x^2 +
d)*sqrt(-c^2*x^2 + 1) - 6*(16*a*c^12*d*x^12 - 8*a*c^10*d*x^10 - 2*a*c^8*d*x^8 - a*c^6*d*x^6 - 145*a*c^4*d*x^4
+ 245*a*c^2*d*x^2 - 105*a*d + (16*b*c^12*d*x^12 - 8*b*c^10*d*x^10 - 2*b*c^8*d*x^8 - b*c^6*d*x^6 - 145*b*c^4*d*
x^4 + 245*b*c^2*d*x^2 - 105*b*d)*arcsin(c*x))*sqrt(-c^2*d*x^2 + d))/(c^2*x^13 - x^11), 1/6930*(96*(b*c^13*d*x^
13 - b*c^11*d*x^11)*sqrt(-d)*arctan(sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1)*(x^2 + 1)*sqrt(-d)/(c^2*d*x^4 - (c
^2 + 1)*d*x^2 + d)) + (24*b*c^9*d*x^9 + 9*b*c^7*d*x^7 - (24*b*c^9 + 9*b*c^7 + 5*b*c^5 - 105*b*c^3 + 63*b*c)*d*
x^11 + 5*b*c^5*d*x^5 - 105*b*c^3*d*x^3 + 63*b*c*d*x)*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1) - 6*(16*a*c^12*d*
x^12 - 8*a*c^10*d*x^10 - 2*a*c^8*d*x^8 - a*c^6*d*x^6 - 145*a*c^4*d*x^4 + 245*a*c^2*d*x^2 - 105*a*d + (16*b*c^1
2*d*x^12 - 8*b*c^10*d*x^10 - 2*b*c^8*d*x^8 - b*c^6*d*x^6 - 145*b*c^4*d*x^4 + 245*b*c^2*d*x^2 - 105*b*d)*arcsin
(c*x))*sqrt(-c^2*d*x^2 + d))/(c^2*x^13 - x^11)]

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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((-c**2*d*x**2+d)**(3/2)*(a+b*asin(c*x))/x**12,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}}{\left (b \arcsin \left (c x\right ) + a\right )}}{x^{12}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))/x^12,x, algorithm="giac")

[Out]

integrate((-c^2*d*x^2 + d)^(3/2)*(b*arcsin(c*x) + a)/x^12, x)