3.137 \(\int \frac {a+b \sin ^{-1}(c x)}{x^2 (d-c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=224 \[ \frac {8 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {4 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {a+b \sin ^{-1}(c x)}{d x \left (d-c^2 d x^2\right )^{3/2}}-\frac {b c \sqrt {d-c^2 d x^2}}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {b c \log (x) \sqrt {d-c^2 d x^2}}{d^3 \sqrt {1-c^2 x^2}}+\frac {5 b c \sqrt {d-c^2 d x^2} \log \left (1-c^2 x^2\right )}{6 d^3 \sqrt {1-c^2 x^2}} \]

[Out]

(-a-b*arcsin(c*x))/d/x/(-c^2*d*x^2+d)^(3/2)+4/3*c^2*x*(a+b*arcsin(c*x))/d/(-c^2*d*x^2+d)^(3/2)+8/3*c^2*x*(a+b*
arcsin(c*x))/d^2/(-c^2*d*x^2+d)^(1/2)-1/6*b*c*(-c^2*d*x^2+d)^(1/2)/d^3/(-c^2*x^2+1)^(3/2)+b*c*ln(x)*(-c^2*d*x^
2+d)^(1/2)/d^3/(-c^2*x^2+1)^(1/2)+5/6*b*c*ln(-c^2*x^2+1)*(-c^2*d*x^2+d)^(1/2)/d^3/(-c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {4701, 4655, 4653, 260, 261, 266, 44} \[ \frac {8 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {4 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {a+b \sin ^{-1}(c x)}{d x \left (d-c^2 d x^2\right )^{3/2}}-\frac {b c}{6 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {b c \sqrt {1-c^2 x^2} \log (x)}{d^2 \sqrt {d-c^2 d x^2}}+\frac {5 b c \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{6 d^2 \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])/(x^2*(d - c^2*d*x^2)^(5/2)),x]

[Out]

-(b*c)/(6*d^2*Sqrt[1 - c^2*x^2]*Sqrt[d - c^2*d*x^2]) - (a + b*ArcSin[c*x])/(d*x*(d - c^2*d*x^2)^(3/2)) + (4*c^
2*x*(a + b*ArcSin[c*x]))/(3*d*(d - c^2*d*x^2)^(3/2)) + (8*c^2*x*(a + b*ArcSin[c*x]))/(3*d^2*Sqrt[d - c^2*d*x^2
]) + (b*c*Sqrt[1 - c^2*x^2]*Log[x])/(d^2*Sqrt[d - c^2*d*x^2]) + (5*b*c*Sqrt[1 - c^2*x^2]*Log[1 - c^2*x^2])/(6*
d^2*Sqrt[d - c^2*d*x^2])

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 261

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

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 4653

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(x*(a + b*ArcSin[c
*x])^n)/(d*Sqrt[d + e*x^2]), x] - Dist[(b*c*n*Sqrt[1 - c^2*x^2])/(d*Sqrt[d + e*x^2]), Int[(x*(a + b*ArcSin[c*x
])^(n - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0]

Rule 4655

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(p
+ 1)*(a + b*ArcSin[c*x])^n)/(2*d*(p + 1)), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p + 1)*(a + b*
ArcSin[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*(p + 1)*(1 - c^2*x^2)^FracPart[p
]), Int[x*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2
*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 4701

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(
(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n)/(d*f*(m + 1)), x] + (Dist[(c^2*(m + 2*p + 3))/(f^2*(m
+ 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^F
racPart[p])/(f*(m + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x
])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[m, -1] && Inte
gerQ[m]

Rubi steps

\begin {align*} \int \frac {a+b \sin ^{-1}(c x)}{x^2 \left (d-c^2 d x^2\right )^{5/2}} \, dx &=-\frac {a+b \sin ^{-1}(c x)}{d x \left (d-c^2 d x^2\right )^{3/2}}+\left (4 c^2\right ) \int \frac {a+b \sin ^{-1}(c x)}{\left (d-c^2 d x^2\right )^{5/2}} \, dx+\frac {\left (b c \sqrt {1-c^2 x^2}\right ) \int \frac {1}{x \left (1-c^2 x^2\right )^2} \, dx}{d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {a+b \sin ^{-1}(c x)}{d x \left (d-c^2 d x^2\right )^{3/2}}+\frac {4 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {\left (8 c^2\right ) \int \frac {a+b \sin ^{-1}(c x)}{\left (d-c^2 d x^2\right )^{3/2}} \, dx}{3 d}+\frac {\left (b c \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1-c^2 x\right )^2} \, dx,x,x^2\right )}{2 d^2 \sqrt {d-c^2 d x^2}}-\frac {\left (4 b c^3 \sqrt {1-c^2 x^2}\right ) \int \frac {x}{\left (1-c^2 x^2\right )^2} \, dx}{3 d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {2 b c}{3 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}-\frac {a+b \sin ^{-1}(c x)}{d x \left (d-c^2 d x^2\right )^{3/2}}+\frac {4 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {8 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (b c \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{x}+\frac {c^2}{\left (-1+c^2 x\right )^2}-\frac {c^2}{-1+c^2 x}\right ) \, dx,x,x^2\right )}{2 d^2 \sqrt {d-c^2 d x^2}}-\frac {\left (8 b c^3 \sqrt {1-c^2 x^2}\right ) \int \frac {x}{1-c^2 x^2} \, dx}{3 d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {b c}{6 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}-\frac {a+b \sin ^{-1}(c x)}{d x \left (d-c^2 d x^2\right )^{3/2}}+\frac {4 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {8 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {b c \sqrt {1-c^2 x^2} \log (x)}{d^2 \sqrt {d-c^2 d x^2}}+\frac {5 b c \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{6 d^2 \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 188, normalized size = 0.84 \[ -\frac {\sqrt {d-c^2 d x^2} \left (16 a c^4 x^4-24 a c^2 x^2+6 a+b c x \sqrt {1-c^2 x^2}-3 b c x \left (1-c^2 x^2\right )^{3/2} \log \left (x^2\right )-5 b c x \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )+2 b \left (8 c^4 x^4-12 c^2 x^2+3\right ) \sin ^{-1}(c x)+5 b c^3 x^3 \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )\right )}{6 d^3 x \left (c^2 x^2-1\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c*x])/(x^2*(d - c^2*d*x^2)^(5/2)),x]

[Out]

-1/6*(Sqrt[d - c^2*d*x^2]*(6*a - 24*a*c^2*x^2 + 16*a*c^4*x^4 + b*c*x*Sqrt[1 - c^2*x^2] + 2*b*(3 - 12*c^2*x^2 +
8*c^4*x^4)*ArcSin[c*x] - 3*b*c*x*(1 - c^2*x^2)^(3/2)*Log[x^2] - 5*b*c*x*Sqrt[1 - c^2*x^2]*Log[1 - c^2*x^2] +
5*b*c^3*x^3*Sqrt[1 - c^2*x^2]*Log[1 - c^2*x^2]))/(d^3*x*(-1 + c^2*x^2)^2)

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fricas [F]  time = 13.04, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b \arcsin \left (c x\right ) + a\right )}}{c^{6} d^{3} x^{8} - 3 \, c^{4} d^{3} x^{6} + 3 \, c^{2} d^{3} x^{4} - d^{3} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/x^2/(-c^2*d*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-c^2*d*x^2 + d)*(b*arcsin(c*x) + a)/(c^6*d^3*x^8 - 3*c^4*d^3*x^6 + 3*c^2*d^3*x^4 - d^3*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \arcsin \left (c x\right ) + a}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/x^2/(-c^2*d*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x) + a)/((-c^2*d*x^2 + d)^(5/2)*x^2), x)

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maple [C]  time = 0.33, size = 1346, normalized size = 6.01 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(c*x))/x^2/(-c^2*d*x^2+d)^(5/2),x)

[Out]

-a/d/x/(-c^2*d*x^2+d)^(3/2)+4/3*a*c^2*x/d/(-c^2*d*x^2+d)^(3/2)+8/3*a*c^2/d^2*x/(-c^2*d*x^2+d)^(1/2)+4*I*b*(-d*
(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*x*c^2-80/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25
*c^4*x^4+26*c^2*x^2-9)/d^3*x^5*(-c^2*x^2+1)*c^6+9*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)
/d^3/x*arcsin(c*x)+3/2*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*(-c^2*x^2+1)^(1/2)*c-b
*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^3/(c^2*x^2-1)*ln((I*c*x+(-c^2*x^2+1)^(1/2))^2-1)*c-64/3*b*(-d*(c^
2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*x^5*arcsin(c*x)*c^6+56*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6
*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*x^3*arcsin(c*x)*c^4-4/3*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^
2*x^2-9)/d^3*x^2*c^3*(-c^2*x^2+1)^(1/2)-5/3*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^3/(c^2*x^2-1)*ln(1+(
I*c*x+(-c^2*x^2+1)^(1/2))^2)*c-44*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*x*arcsin(c*
x)*c^2-64/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*x^4*arcsin(c*x)*(-c^2*x^2+1)^(1
/2)*c^5+140/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*x^5*c^6-24*I*b*(-d*(c^2*x^2-1
))^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*c-112/3*I*b*(-d*(c^2*x^2-1))^(
1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*x^7*c^8+32/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*
c^2*x^2-9)/d^3*x^7*(-c^2*x^2+1)*c^8+136/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*x
^2*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*c^3+16/3*I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^3/(c^2*x^2-1)*arcsi
n(c*x)*c+20*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*x^3*(-c^2*x^2+1)*c^4-24*I*b*(-d
*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*x^3*c^4-4*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25
*c^4*x^4+26*c^2*x^2-9)/d^3*x*(-c^2*x^2+1)*c^2+32/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2
-9)/d^3*x^9*c^10

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, a {\left (\frac {8 \, c^{2} x}{\sqrt {-c^{2} d x^{2} + d} d^{2}} + \frac {4 \, c^{2} x}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d} - \frac {3}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d x}\right )} + \frac {\frac {1}{6} \, b {\left (\frac {c}{c^{2} d^{2} x^{2} - d^{2}} + \frac {2 \, {\left (8 \, c^{4} x^{4} - 12 \, c^{2} x^{2} + 3\right )} \arctan \left (\frac {c x}{\sqrt {c x + 1} \sqrt {-c x + 1}}\right )}{{\left (c^{2} d^{2} x^{3} - d^{2} x\right )} \sqrt {c x + 1} \sqrt {-c x + 1}} + \frac {5 \, c \log \left (c x + 1\right )}{d^{2}} + \frac {5 \, c \log \left (c x - 1\right )}{d^{2}} + \frac {6 \, c \log \relax (x)}{d^{2}}\right )}}{\sqrt {d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/x^2/(-c^2*d*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

1/3*a*(8*c^2*x/(sqrt(-c^2*d*x^2 + d)*d^2) + 4*c^2*x/((-c^2*d*x^2 + d)^(3/2)*d) - 3/((-c^2*d*x^2 + d)^(3/2)*d*x
)) + b*integrate(arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/((c^4*d^2*x^6 - 2*c^2*d^2*x^4 + d^2*x^2)*sqrt(c*x
+ 1)*sqrt(-c*x + 1)), x)/sqrt(d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{x^2\,{\left (d-c^2\,d\,x^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))/(x^2*(d - c^2*d*x^2)^(5/2)),x)

[Out]

int((a + b*asin(c*x))/(x^2*(d - c^2*d*x^2)^(5/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {asin}{\left (c x \right )}}{x^{2} \left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(c*x))/x**2/(-c**2*d*x**2+d)**(5/2),x)

[Out]

Integral((a + b*asin(c*x))/(x**2*(-d*(c*x - 1)*(c*x + 1))**(5/2)), x)

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