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

Optimal. Leaf size=154 \[ \frac{2 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt{d-c^2 d x^2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{b}{6 c d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}+\frac{b \sqrt{1-c^2 x^2} \log \left (1-c^2 x^2\right )}{3 c d^2 \sqrt{d-c^2 d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.078979, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {4655, 4653, 260, 261} \[ \frac{2 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt{d-c^2 d x^2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{b}{6 c d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}+\frac{b \sqrt{1-c^2 x^2} \log \left (1-c^2 x^2\right )}{3 c d^2 \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.238211, size = 113, normalized size = 0.73 \[ -\frac{\sqrt{d-c^2 d x^2} \left (4 a c^3 x^3-6 a c x+b \sqrt{1-c^2 x^2}-2 b \left (1-c^2 x^2\right )^{3/2} \log \left (c^2 x^2-1\right )+2 b c x \left (2 c^2 x^2-3\right ) \sin ^{-1}(c x)\right )}{6 c d^3 \left (c^2 x^2-1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.125, size = 1071, normalized size = 7. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a/d*x/(-c^2*d*x^2+d)^(3/2)+2/3*a/d^2*x/(-c^2*d*x^2+d)^(1/2)+2/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-
10*c^4*x^4+11*c^2*x^2-4)*c^6*x^7-5/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^2*(-
c^2*x^2+1)*x^3+2/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^4*(-c^2*x^2+1)*x^5+I*b
*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*(-c^2*x^2+1)*x+14/3*I*b*(-d*(c^2*x^2-1))^(1/2)
/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x^2-2*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(
3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^4*arcsin(c*x)*x^5-I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11
*c^2*x^2-4)*x-1/2*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c*(-c^2*x^2+1)^(1/2)*x^2+4/
3*I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c/d^3/(c^2*x^2-1)*arcsin(c*x)-7/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d
^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^4*x^5+17/3*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*
x^2-4)*c^2*arcsin(c*x)*x^3-8/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)/c*arcsin(c*x
)*(-c^2*x^2+1)^(1/2)+2/3*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)/c*(-c^2*x^2+1)^(1/2)
-2*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^3*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x^4+8
/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^2*x^3-4*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(
3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*arcsin(c*x)*x-2/3*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c/d^3/(c^2*x^
2-1)*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)

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Maxima [A]  time = 1.76874, size = 190, normalized size = 1.23 \begin{align*} \frac{1}{6} \, b c{\left (\frac{1}{c^{4} d^{\frac{5}{2}} x^{2} - c^{2} d^{\frac{5}{2}}} + \frac{2 \, \log \left (c x + 1\right )}{c^{2} d^{\frac{5}{2}}} + \frac{2 \, \log \left (c x - 1\right )}{c^{2} d^{\frac{5}{2}}}\right )} + \frac{1}{3} \, b{\left (\frac{2 \, x}{\sqrt{-c^{2} d x^{2} + d} d^{2}} + \frac{x}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}} d}\right )} \arcsin \left (c x\right ) + \frac{1}{3} \, a{\left (\frac{2 \, x}{\sqrt{-c^{2} d x^{2} + d} d^{2}} + \frac{x}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}} d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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^{6} - 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} - d^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \arcsin \left (c x\right ) + a}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/(-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)