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

Optimal. Leaf size=150 \[ \frac {2 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{d \sqrt {d-c^2 d x^2}}-\frac {a+b \sin ^{-1}(c x)}{d x \sqrt {d-c^2 d x^2}}+\frac {b c \log (x) \sqrt {d-c^2 d x^2}}{d^2 \sqrt {1-c^2 x^2}}+\frac {b c \sqrt {d-c^2 d x^2} \log \left (1-c^2 x^2\right )}{2 d^2 \sqrt {1-c^2 x^2}} \]

[Out]

(-a-b*arcsin(c*x))/d/x/(-c^2*d*x^2+d)^(1/2)+2*c^2*x*(a+b*arcsin(c*x))/d/(-c^2*d*x^2+d)^(1/2)+b*c*ln(x)*(-c^2*d
*x^2+d)^(1/2)/d^2/(-c^2*x^2+1)^(1/2)+1/2*b*c*ln(-c^2*x^2+1)*(-c^2*d*x^2+d)^(1/2)/d^2/(-c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {4701, 4653, 260, 266, 36, 29, 31} \[ \frac {2 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{d \sqrt {d-c^2 d x^2}}-\frac {a+b \sin ^{-1}(c x)}{d x \sqrt {d-c^2 d x^2}}+\frac {b c \sqrt {1-c^2 x^2} \log (x)}{d \sqrt {d-c^2 d x^2}}+\frac {b c \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{2 d \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*ArcSin[c*x])/(d*x*Sqrt[d - c^2*d*x^2])) + (2*c^2*x*(a + b*ArcSin[c*x]))/(d*Sqrt[d - c^2*d*x^2]) + (b*
c*Sqrt[1 - c^2*x^2]*Log[x])/(d*Sqrt[d - c^2*d*x^2]) + (b*c*Sqrt[1 - c^2*x^2]*Log[1 - c^2*x^2])/(2*d*Sqrt[d - c
^2*d*x^2])

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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 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 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 )^{3/2}} \, dx &=-\frac {a+b \sin ^{-1}(c x)}{d x \sqrt {d-c^2 d x^2}}+\left (2 c^2\right ) \int \frac {a+b \sin ^{-1}(c x)}{\left (d-c^2 d x^2\right )^{3/2}} \, dx+\frac {\left (b c \sqrt {1-c^2 x^2}\right ) \int \frac {1}{x \left (1-c^2 x^2\right )} \, dx}{d \sqrt {d-c^2 d x^2}}\\ &=-\frac {a+b \sin ^{-1}(c x)}{d x \sqrt {d-c^2 d x^2}}+\frac {2 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{d \sqrt {d-c^2 d x^2}}+\frac {\left (b c \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )}{2 d \sqrt {d-c^2 d x^2}}-\frac {\left (2 b c^3 \sqrt {1-c^2 x^2}\right ) \int \frac {x}{1-c^2 x^2} \, dx}{d \sqrt {d-c^2 d x^2}}\\ &=-\frac {a+b \sin ^{-1}(c x)}{d x \sqrt {d-c^2 d x^2}}+\frac {2 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{d \sqrt {d-c^2 d x^2}}+\frac {b c \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{d \sqrt {d-c^2 d x^2}}+\frac {\left (b c \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 d \sqrt {d-c^2 d x^2}}+\frac {\left (b c^3 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-c^2 x} \, dx,x,x^2\right )}{2 d \sqrt {d-c^2 d x^2}}\\ &=-\frac {a+b \sin ^{-1}(c x)}{d x \sqrt {d-c^2 d x^2}}+\frac {2 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{d \sqrt {d-c^2 d x^2}}+\frac {b c \sqrt {1-c^2 x^2} \log (x)}{d \sqrt {d-c^2 d x^2}}+\frac {b c \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{2 d \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 117, normalized size = 0.78 \[ -\frac {\sqrt {d-c^2 d x^2} \left (4 a c^2 x^2-2 a+b c x \sqrt {1-c^2 x^2} \log \left (x^2\right )+b c x \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )+2 b \left (2 c^2 x^2-1\right ) \sin ^{-1}(c x)\right )}{2 d^2 x \left (c^2 x^2-1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 15.03, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b \arcsin \left (c x\right ) + a\right )}}{c^{4} d^{2} x^{6} - 2 \, c^{2} d^{2} x^{4} + d^{2} x^{2}}, x\right ) \]

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)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \arcsin \left (c x\right ) + a}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x^{2}}\,{d x} \]

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)^(3/2),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.26, size = 239, normalized size = 1.59 \[ -\frac {a}{d x \sqrt {-c^{2} d \,x^{2}+d}}+\frac {2 a \,c^{2} x}{d \sqrt {-c^{2} d \,x^{2}+d}}+\frac {2 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right ) c}{d^{2} \left (c^{2} x^{2}-1\right )}-\frac {2 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x \,c^{2}}{\left (c^{2} x^{2}-1\right ) d^{2}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )}{\left (c^{2} x^{2}-1\right ) d^{2} x}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{4}-1\right ) c}{d^{2} \left (c^{2} x^{2}-1\right )} \]

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)^(3/2),x)

[Out]

-a/d/x/(-c^2*d*x^2+d)^(1/2)+2*a*c^2/d*x/(-c^2*d*x^2+d)^(1/2)+2*I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d
^2/(c^2*x^2-1)*arcsin(c*x)*c-2*b*(-d*(c^2*x^2-1))^(1/2)*arcsin(c*x)/(c^2*x^2-1)/d^2*x*c^2+b*(-d*(c^2*x^2-1))^(
1/2)*arcsin(c*x)/(c^2*x^2-1)/d^2/x-b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^2/(c^2*x^2-1)*ln((I*c*x+(-c^2
*x^2+1)^(1/2))^4-1)*c

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maxima [A]  time = 0.43, size = 129, normalized size = 0.86 \[ \frac {1}{2} \, b c {\left (\frac {\log \left (c x + 1\right )}{d^{\frac {3}{2}}} + \frac {\log \left (c x - 1\right )}{d^{\frac {3}{2}}} + \frac {2 \, \log \relax (x)}{d^{\frac {3}{2}}}\right )} + {\left (\frac {2 \, c^{2} x}{\sqrt {-c^{2} d x^{2} + d} d} - \frac {1}{\sqrt {-c^{2} d x^{2} + d} d x}\right )} b \arcsin \left (c x\right ) + {\left (\frac {2 \, c^{2} x}{\sqrt {-c^{2} d x^{2} + d} d} - \frac {1}{\sqrt {-c^{2} d x^{2} + d} d x}\right )} a \]

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)^(3/2),x, algorithm="maxima")

[Out]

1/2*b*c*(log(c*x + 1)/d^(3/2) + log(c*x - 1)/d^(3/2) + 2*log(x)/d^(3/2)) + (2*c^2*x/(sqrt(-c^2*d*x^2 + d)*d) -
1/(sqrt(-c^2*d*x^2 + d)*d*x))*b*arcsin(c*x) + (2*c^2*x/(sqrt(-c^2*d*x^2 + d)*d) - 1/(sqrt(-c^2*d*x^2 + d)*d*x
))*a

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{x^2\,{\left (d-c^2\,d\,x^2\right )}^{3/2}} \,d x \]

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)^(3/2)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {asin}{\left (c x \right )}}{x^{2} \left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

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)**(3/2),x)

[Out]

Integral((a + b*asin(c*x))/(x**2*(-d*(c*x - 1)*(c*x + 1))**(3/2)), x)

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