3.59 \(\int \frac{\sqrt{d-c^2 d x^2} (a+b \sin ^{-1}(c x))}{x^4} \, dx\)

Optimal. Leaf size=111 \[ -\frac{\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 d x^3}-\frac{b c \sqrt{d-c^2 d x^2}}{6 x^2 \sqrt{1-c^2 x^2}}-\frac{b c^3 \log (x) \sqrt{d-c^2 d x^2}}{3 \sqrt{1-c^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0925604, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {4681, 14} \[ -\frac{\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 d x^3}-\frac{b c \sqrt{d-c^2 d x^2}}{6 x^2 \sqrt{1-c^2 x^2}}-\frac{b c^3 \log (x) \sqrt{d-c^2 d x^2}}{3 \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4681

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[(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*ArcSi
n[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && EqQ[m + 2*p
+ 3, 0] && 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{\sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{x^4} \, dx &=-\frac{\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 d x^3}+\frac{\left (b c \sqrt{d-c^2 d x^2}\right ) \int \frac{1-c^2 x^2}{x^3} \, dx}{3 \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 )}{3 d x^3}+\frac{\left (b c \sqrt{d-c^2 d x^2}\right ) \int \left (\frac{1}{x^3}-\frac{c^2}{x}\right ) \, dx}{3 \sqrt{1-c^2 x^2}}\\ &=-\frac{b c \sqrt{d-c^2 d x^2}}{6 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 )}{3 d x^3}-\frac{b c^3 \sqrt{d-c^2 d x^2} \log (x)}{3 \sqrt{1-c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.133413, size = 134, normalized size = 1.21 \[ \frac{\sqrt{d-c^2 d x^2} \left (2 a \left (c^2 x^2-1\right )^2+b c x \left (1-3 c^2 x^2\right ) \sqrt{1-c^2 x^2}+2 b \left (c^2 x^2-1\right )^2 \sin ^{-1}(c x)\right )}{6 x^3 \left (c^2 x^2-1\right )}-\frac{b c^3 \log (x) \sqrt{d-c^2 d x^2}}{3 \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.243, size = 1117, normalized size = 10.1 \begin{align*} \text{result too large to display} \end{align*}

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^4,x)

[Out]

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

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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)^(1/2)*(a+b*arcsin(c*x))/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.60023, size = 883, normalized size = 7.95 \begin{align*} \left [\frac{{\left (b c^{5} x^{5} - b c^{3} x^{3}\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 ) - \sqrt{-c^{2} d x^{2} + d}{\left (b c x^{3} - b c x\right )} \sqrt{-c^{2} x^{2} + 1} + 2 \,{\left (a c^{4} x^{4} - 2 \, a c^{2} x^{2} +{\left (b c^{4} x^{4} - 2 \, b c^{2} x^{2} + b\right )} \arcsin \left (c x\right ) + a\right )} \sqrt{-c^{2} d x^{2} + d}}{6 \,{\left (c^{2} x^{5} - x^{3}\right )}}, -\frac{2 \,{\left (b c^{5} x^{5} - b c^{3} x^{3}\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 ) + \sqrt{-c^{2} d x^{2} + d}{\left (b c x^{3} - b c x\right )} \sqrt{-c^{2} x^{2} + 1} - 2 \,{\left (a c^{4} x^{4} - 2 \, a c^{2} x^{2} +{\left (b c^{4} x^{4} - 2 \, b c^{2} x^{2} + b\right )} \arcsin \left (c x\right ) + a\right )} \sqrt{-c^{2} d x^{2} + d}}{6 \,{\left (c^{2} x^{5} - x^{3}\right )}}\right ] \end{align*}

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^4,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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^{4}}\, dx \end{align*}

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**4,x)

[Out]

Integral(sqrt(-d*(c*x - 1)*(c*x + 1))*(a + b*asin(c*x))/x**4, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-c^{2} d x^{2} + d}{\left (b \arcsin \left (c x\right ) + a\right )}}{x^{4}}\,{d x} \end{align*}

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^4,x, algorithm="giac")

[Out]

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