3.1.60 \(\int \frac {\sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))}{x^6} \, dx\) [60]

Optimal. Leaf size=187 \[ -\frac {b c \sqrt {d-c^2 d x^2}}{20 x^4 \sqrt {1-c^2 x^2}}+\frac {b c^3 \sqrt {d-c^2 d x^2}}{30 x^2 \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \text {ArcSin}(c x))}{5 d x^5}-\frac {2 c^2 \left (d-c^2 d x^2\right )^{3/2} (a+b \text {ArcSin}(c x))}{15 d x^3}-\frac {2 b c^5 \sqrt {d-c^2 d x^2} \log (x)}{15 \sqrt {1-c^2 x^2}} \]

[Out]

-1/5*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))/d/x^5-2/15*c^2*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))/d/x^3-1/20*b
*c*(-c^2*d*x^2+d)^(1/2)/x^4/(-c^2*x^2+1)^(1/2)+1/30*b*c^3*(-c^2*d*x^2+d)^(1/2)/x^2/(-c^2*x^2+1)^(1/2)-2/15*b*c
^5*ln(x)*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {277, 270, 4779, 12, 14} \begin {gather*} -\frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \text {ArcSin}(c x))}{5 d x^5}-\frac {2 c^2 \left (d-c^2 d x^2\right )^{3/2} (a+b \text {ArcSin}(c x))}{15 d x^3}-\frac {b c \sqrt {d-c^2 d x^2}}{20 x^4 \sqrt {1-c^2 x^2}}-\frac {2 b c^5 \log (x) \sqrt {d-c^2 d x^2}}{15 \sqrt {1-c^2 x^2}}+\frac {b c^3 \sqrt {d-c^2 d x^2}}{30 x^2 \sqrt {1-c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/x^6,x]

[Out]

-1/20*(b*c*Sqrt[d - c^2*d*x^2])/(x^4*Sqrt[1 - c^2*x^2]) + (b*c^3*Sqrt[d - c^2*d*x^2])/(30*x^2*Sqrt[1 - c^2*x^2
]) - ((d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]))/(5*d*x^5) - (2*c^2*(d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]))
/(15*d*x^3) - (2*b*c^5*Sqrt[d - c^2*d*x^2]*Log[x])/(15*Sqrt[1 - c^2*x^2])

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 270

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 277

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 4779

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

Rubi steps

\begin {align*} \int \frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{x^6} \, dx &=-\frac {\left (b c \sqrt {d-c^2 d x^2}\right ) \int \frac {-3+c^2 x^2+2 c^4 x^4}{15 x^5} \, dx}{\sqrt {1-c^2 x^2}}+\left (a+b \sin ^{-1}(c x)\right ) \int \frac {\sqrt {d-c^2 d x^2}}{x^6} \, dx\\ &=-\frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{5 d x^5}-\frac {\left (b c \sqrt {d-c^2 d x^2}\right ) \int \frac {-3+c^2 x^2+2 c^4 x^4}{x^5} \, dx}{15 \sqrt {1-c^2 x^2}}+\frac {1}{5} \left (2 c^2 \left (a+b \sin ^{-1}(c x)\right )\right ) \int \frac {\sqrt {d-c^2 d x^2}}{x^4} \, dx\\ &=-\frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{5 d x^5}-\frac {2 c^2 \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{15 d x^3}-\frac {\left (b c \sqrt {d-c^2 d x^2}\right ) \int \left (-\frac {3}{x^5}+\frac {c^2}{x^3}+\frac {2 c^4}{x}\right ) \, dx}{15 \sqrt {1-c^2 x^2}}\\ &=-\frac {b c \sqrt {d-c^2 d x^2}}{20 x^4 \sqrt {1-c^2 x^2}}+\frac {b c^3 \sqrt {d-c^2 d x^2}}{30 x^2 \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{5 d x^5}-\frac {2 c^2 \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{15 d x^3}-\frac {2 b c^5 \sqrt {d-c^2 d x^2} \log (x)}{15 \sqrt {1-c^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 162, normalized size = 0.87 \begin {gather*} \frac {\sqrt {d-c^2 d x^2} \left (12 a \left (-1+c^2 x^2\right )^2 \left (3+2 c^2 x^2\right )+b c x \sqrt {1-c^2 x^2} \left (9-6 c^2 x^2-50 c^4 x^4\right )+12 b \left (-1+c^2 x^2\right )^2 \left (3+2 c^2 x^2\right ) \text {ArcSin}(c x)\right )}{180 x^5 \left (-1+c^2 x^2\right )}-\frac {2 b c^5 \sqrt {d-c^2 d x^2} \log (x)}{15 \sqrt {1-c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/x^6,x]

[Out]

(Sqrt[d - c^2*d*x^2]*(12*a*(-1 + c^2*x^2)^2*(3 + 2*c^2*x^2) + b*c*x*Sqrt[1 - c^2*x^2]*(9 - 6*c^2*x^2 - 50*c^4*
x^4) + 12*b*(-1 + c^2*x^2)^2*(3 + 2*c^2*x^2)*ArcSin[c*x]))/(180*x^5*(-1 + c^2*x^2)) - (2*b*c^5*Sqrt[d - c^2*d*
x^2]*Log[x])/(15*Sqrt[1 - c^2*x^2])

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Maple [C] Result contains complex when optimal does not.
time = 0.34, size = 1903, normalized size = 10.18

method	result	size
default	\(\text {Expression too large to display}\)	\(1903\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))/x^6,x,method=_RETURNVERBOSE)

[Out]

a*(-1/5/d/x^5*(-c^2*d*x^2+d)^(3/2)-2/15*c^2/d/x^3*(-c^2*d*x^2+d)^(3/2))+1/4*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x
^6-5*c^4*x^4-15*c^2*x^2+9)/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*c^5+9/5*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x
^4-15*c^2*x^2+9)/x^5/(c^2*x^2-1)*arcsin(c*x)+2/15*I*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+
9)*x^7/(c^2*x^2-1)*(-c^2*x^2+1)*c^12-2/15*I*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)*x^5/(
c^2*x^2-1)*(-c^2*x^2+1)*c^10-3/10*I*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)*x^3/(c^2*x^2-
1)*(-c^2*x^2+1)*c^8+3/10*I*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)*x/(c^2*x^2-1)*(-c^2*x^
2+1)*c^6+6/5*I*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)/(c^2*x^2-1)*arcsin(c*x)*(-c^2*x^2+
1)^(1/2)*c^5+2*I*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)*x^6/(c^2*x^2-1)*arcsin(c*x)*(-c^
2*x^2+1)^(1/2)*c^11-2/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)*x^4/(c^2*x^2-1)*arcsin(
c*x)*(-c^2*x^2+1)^(1/2)*c^9-2*I*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)*x^2/(c^2*x^2-1)*a
rcsin(c*x)*(-c^2*x^2+1)^(1/2)*c^7+2/15*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/(c^2*x^2-1)*ln((I*c*x+(-c^2
*x^2+1)^(1/2))^2-1)*c^5+12/5*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)/x/(c^2*x^2-1)*arcsin
(c*x)*c^4-21/20*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)/x^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2
)*c^3-27/5*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)/x^3/(c^2*x^2-1)*arcsin(c*x)*c^2+9/20*b
*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)/x^4/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*c+2*b*(-d*(c^2*
x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)*x^7/(c^2*x^2-1)*arcsin(c*x)*c^12-5/3*b*(-d*(c^2*x^2-1))^(1/2
)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)*x^5/(c^2*x^2-1)*arcsin(c*x)*c^10-1/2*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^
6-5*c^4*x^4-15*c^2*x^2+9)*x^4/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*c^9-17/3*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c
^4*x^4-15*c^2*x^2+9)*x^3/(c^2*x^2-1)*arcsin(c*x)*c^8+11/12*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c
^2*x^2+9)*x^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*c^7+98/15*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x
^2+9)*x/(c^2*x^2-1)*arcsin(c*x)*c^6-3/10*I*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)*x/(c^2
*x^2-1)*c^6-4*I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*c^5/(15*c^2*x^2-15)+2/15*I*b*(-d*(c^2*
x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)*x^9/(c^2*x^2-1)*c^14-4/15*I*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6
*x^6-5*c^4*x^4-15*c^2*x^2+9)*x^7/(c^2*x^2-1)*c^12-1/6*I*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*
x^2+9)*x^5/(c^2*x^2-1)*c^10+3/5*I*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)*x^3/(c^2*x^2-1)
*c^8

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Maxima [A]
time = 0.48, size = 140, normalized size = 0.75 \begin {gather*} -\frac {1}{60} \, {\left (8 \, c^{4} \sqrt {d} \log \left (x\right ) - \frac {2 \, c^{2} \sqrt {d} x^{2} - 3 \, \sqrt {d}}{x^{4}}\right )} b c - \frac {1}{15} \, b {\left (\frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2}}{d x^{3}} + \frac {3 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}}}{d x^{5}}\right )} \arcsin \left (c x\right ) - \frac {1}{15} \, a {\left (\frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2}}{d x^{3}} + \frac {3 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}}}{d x^{5}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))/x^6,x, algorithm="maxima")

[Out]

-1/60*(8*c^4*sqrt(d)*log(x) - (2*c^2*sqrt(d)*x^2 - 3*sqrt(d))/x^4)*b*c - 1/15*b*(2*(-c^2*d*x^2 + d)^(3/2)*c^2/
(d*x^3) + 3*(-c^2*d*x^2 + d)^(3/2)/(d*x^5))*arcsin(c*x) - 1/15*a*(2*(-c^2*d*x^2 + d)^(3/2)*c^2/(d*x^3) + 3*(-c
^2*d*x^2 + d)^(3/2)/(d*x^5))

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Fricas [A]
time = 2.54, size = 501, normalized size = 2.68 \begin {gather*} \left [\frac {4 \, {\left (b c^{7} x^{7} - b c^{5} x^{5}\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 (2 \, b c^{3} x^{3} - {\left (2 \, b c^{3} - 3 \, b c\right )} x^{5} - 3 \, b c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} + 4 \, {\left (2 \, a c^{6} x^{6} - a c^{4} x^{4} - 4 \, a c^{2} x^{2} + {\left (2 \, b c^{6} x^{6} - b c^{4} x^{4} - 4 \, b c^{2} x^{2} + 3 \, b\right )} \arcsin \left (c x\right ) + 3 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{60 \, {\left (c^{2} x^{7} - x^{5}\right )}}, -\frac {8 \, {\left (b c^{7} x^{7} - b c^{5} x^{5}\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 (2 \, b c^{3} x^{3} - {\left (2 \, b c^{3} - 3 \, b c\right )} x^{5} - 3 \, b c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} - 4 \, {\left (2 \, a c^{6} x^{6} - a c^{4} x^{4} - 4 \, a c^{2} x^{2} + {\left (2 \, b c^{6} x^{6} - b c^{4} x^{4} - 4 \, b c^{2} x^{2} + 3 \, b\right )} \arcsin \left (c x\right ) + 3 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{60 \, {\left (c^{2} x^{7} - x^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))/x^6,x, algorithm="fricas")

[Out]

[1/60*(4*(b*c^7*x^7 - b*c^5*x^5)*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)) - (2*b*c^3*x^3 - (2*b*c^3 - 3*b*c)*x^5 - 3*b*c*x)*sqrt(-c^2*d*
x^2 + d)*sqrt(-c^2*x^2 + 1) + 4*(2*a*c^6*x^6 - a*c^4*x^4 - 4*a*c^2*x^2 + (2*b*c^6*x^6 - b*c^4*x^4 - 4*b*c^2*x^
2 + 3*b)*arcsin(c*x) + 3*a)*sqrt(-c^2*d*x^2 + d))/(c^2*x^7 - x^5), -1/60*(8*(b*c^7*x^7 - b*c^5*x^5)*sqrt(-d)*a
rctan(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)) + (2*b*c^3
*x^3 - (2*b*c^3 - 3*b*c)*x^5 - 3*b*c*x)*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1) - 4*(2*a*c^6*x^6 - a*c^4*x^4 -
4*a*c^2*x^2 + (2*b*c^6*x^6 - b*c^4*x^4 - 4*b*c^2*x^2 + 3*b)*arcsin(c*x) + 3*a)*sqrt(-c^2*d*x^2 + d))/(c^2*x^7
- x^5)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {- d \left (c x - 1\right ) \left (c x + 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{x^{6}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c**2*d*x**2+d)**(1/2)*(a+b*asin(c*x))/x**6,x)

[Out]

Integral(sqrt(-d*(c*x - 1)*(c*x + 1))*(a + b*asin(c*x))/x**6, x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))/x^6,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\sqrt {d-c^2\,d\,x^2}}{x^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*asin(c*x))*(d - c^2*d*x^2)^(1/2))/x^6,x)

[Out]

int(((a + b*asin(c*x))*(d - c^2*d*x^2)^(1/2))/x^6, x)

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