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

Optimal. Leaf size=185 \[ -\frac{3}{2} c^2 d x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac{3 c d \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b \sqrt{1-c^2 x^2}}-\frac{\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{x}+\frac{b c^3 d x^2 \sqrt{d-c^2 d x^2}}{4 \sqrt{1-c^2 x^2}}+\frac{b c d \log (x) \sqrt{d-c^2 d x^2}}{\sqrt{1-c^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.167893, antiderivative size = 185, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {4695, 4647, 4641, 30, 14} \[ -\frac{3}{2} c^2 d x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac{3 c d \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b \sqrt{1-c^2 x^2}}-\frac{\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{x}+\frac{b c^3 d x^2 \sqrt{d-c^2 d x^2}}{4 \sqrt{1-c^2 x^2}}+\frac{b c d \log (x) \sqrt{d-c^2 d x^2}}{\sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4695

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*(a + b*ArcSin[c*x])^n)/(f*(m + 1)), x] + (-Dist[(2*e*p)/(f^2*(m + 1)), Int[(f*x)^
(m + 2)*(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])/
(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}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1]

Rule 4647

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(x*Sqrt[d + e*x^2]*(
a + b*ArcSin[c*x])^n)/2, x] + (Dist[Sqrt[d + e*x^2]/(2*Sqrt[1 - c^2*x^2]), Int[(a + b*ArcSin[c*x])^n/Sqrt[1 -
c^2*x^2], x], x] - Dist[(b*c*n*Sqrt[d + e*x^2])/(2*Sqrt[1 - c^2*x^2]), Int[x*(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]

Rule 4641

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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]]

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

Mathematica [A]  time = 0.536894, size = 222, normalized size = 1.2 \[ \frac{3}{2} a c d^{3/2} \tan ^{-1}\left (\frac{c x \sqrt{-d \left (c^2 x^2-1\right )}}{\sqrt{d} \left (c^2 x^2-1\right )}\right )+\sqrt{-d \left (c^2 x^2-1\right )} \left (-\frac{1}{2} a c^2 d x-\frac{a d}{x}\right )-\frac{b c d \sqrt{d \left (1-c^2 x^2\right )} \left (\frac{2 \sqrt{1-c^2 x^2} \sin ^{-1}(c x)}{c x}-2 \log (c x)+\sin ^{-1}(c x)^2\right )}{2 \sqrt{1-c^2 x^2}}-\frac{b c d \sqrt{d \left (1-c^2 x^2\right )} \left (2 \sin ^{-1}(c x) \left (\sin ^{-1}(c x)+\sin \left (2 \sin ^{-1}(c x)\right )\right )+\cos \left (2 \sin ^{-1}(c x)\right )\right )}{8 \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a*d)/x) - (a*c^2*d*x)/2)*Sqrt[-(d*(-1 + c^2*x^2))] + (3*a*c*d^(3/2)*ArcTan[(c*x*Sqrt[-(d*(-1 + c^2*x^2))])
/(Sqrt[d]*(-1 + c^2*x^2))])/2 - (b*c*d*Sqrt[d*(1 - c^2*x^2)]*((2*Sqrt[1 - c^2*x^2]*ArcSin[c*x])/(c*x) + ArcSin
[c*x]^2 - 2*Log[c*x]))/(2*Sqrt[1 - c^2*x^2]) - (b*c*d*Sqrt[d*(1 - c^2*x^2)]*(Cos[2*ArcSin[c*x]] + 2*ArcSin[c*x
]*(ArcSin[c*x] + Sin[2*ArcSin[c*x]])))/(8*Sqrt[1 - c^2*x^2])

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Maple [C]  time = 0.201, size = 464, normalized size = 2.5 \begin{align*} -{\frac{a}{dx} \left ( -{c}^{2}d{x}^{2}+d \right ) ^{{\frac{5}{2}}}}-a{c}^{2}x \left ( -{c}^{2}d{x}^{2}+d \right ) ^{{\frac{3}{2}}}-{\frac{3\,a{c}^{2}dx}{2}\sqrt{-{c}^{2}d{x}^{2}+d}}-{\frac{3\,a{c}^{2}{d}^{2}}{2}\arctan \left ({x\sqrt{{c}^{2}d}{\frac{1}{\sqrt{-{c}^{2}d{x}^{2}+d}}}} \right ){\frac{1}{\sqrt{{c}^{2}d}}}}+{\frac{3\,b \left ( \arcsin \left ( cx \right ) \right ) ^{2}dc}{4\,{c}^{2}{x}^{2}-4}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{bd{c}^{3}{x}^{2}}{4\,{c}^{2}{x}^{2}-4}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{bd{c}^{4}\arcsin \left ( cx \right ){x}^{3}}{2\,{c}^{2}{x}^{2}-2}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{bdc}{8\,{c}^{2}{x}^{2}-8}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{ib\arcsin \left ( cx \right ) dc}{{c}^{2}{x}^{2}-1}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{b{c}^{2}d\arcsin \left ( cx \right ) x}{2\,{c}^{2}{x}^{2}-2}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{bd\arcsin \left ( cx \right ) }{x \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}-{\frac{bdc}{{c}^{2}{x}^{2}-1}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}\ln \left ( \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2}-1 \right ) } \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^2,x)

[Out]

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

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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^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (a c^{2} d x^{2} - a d +{\left (b c^{2} d x^{2} - b d\right )} \arcsin \left (c x\right )\right )} \sqrt{-c^{2} d x^{2} + d}}{x^{2}}, x\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^2,x, algorithm="fricas")

[Out]

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

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

[Out]

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

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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^{2}}\,{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^2,x, algorithm="giac")

[Out]

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