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

Optimal. Leaf size=147 \[ -\frac{2 c^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 d x}-\frac{\sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 d x^3}-\frac{b c \sqrt{1-c^2 x^2}}{6 x^2 \sqrt{d-c^2 d x^2}}+\frac{2 b c^3 \sqrt{1-c^2 x^2} \log (x)}{3 \sqrt{d-c^2 d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.187793, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {4701, 4681, 29, 30} \[ -\frac{2 c^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 d x}-\frac{\sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 d x^3}-\frac{b c \sqrt{1-c^2 x^2}}{6 x^2 \sqrt{d-c^2 d x^2}}+\frac{2 b c^3 \sqrt{1-c^2 x^2} \log (x)}{3 \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 29

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

Rule 30

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

Rubi steps

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

Mathematica [A]  time = 0.203436, size = 152, normalized size = 1.03 \[ \frac{\sqrt{d-c^2 d x^2} \left (a \left (-4 c^4 x^4+2 c^2 x^2+2\right )+b c x \sqrt{1-c^2 x^2} \left (6 c^2 x^2+1\right )+2 b \left (-2 c^4 x^4+c^2 x^2+1\right ) \sin ^{-1}(c x)\right )}{6 d x^3 \left (c^2 x^2-1\right )}+\frac{2 b c^3 \log (x) \sqrt{d-c^2 d x^2}}{3 d \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.237, size = 849, normalized size = 5.8 \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))/x^4/(-c^2*d*x^2+d)^(1/2),x)

[Out]

-1/3*a/d/x^3*(-c^2*d*x^2+d)^(1/2)-2/3*a*c^2/d/x*(-c^2*d*x^2+d)^(1/2)-2/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4
-2*c^2*x^2-1)/d*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*c^3-2/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-2*c^2*x^2-1)/d*x^
3*(-c^2*x^2+1)*c^6+1/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-2*c^2*x^2-1)/d*x*c^4-1/3*I*b*(-d*(c^2*x^2-1))^(1/
2)/(3*c^4*x^4-2*c^2*x^2-1)/d*x*(-c^2*x^2+1)*c^4-2*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-2*c^2*x^2-1)/d*x^3*arcsi
n(c*x)*c^6-2*I*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-2*c^2*x^2-1)/d*x^2*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*c^5+4/3*I
*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d/(c^2*x^2-1)*arcsin(c*x)*c^3-2/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(3*c
^4*x^4-2*c^2*x^2-1)/d*x^5*c^8+1/3*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-2*c^2*x^2-1)/d*x*arcsin(c*x)*c^4+1/3*I*b
*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-2*c^2*x^2-1)/d*x^3*c^6+1/2*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-2*c^2*x^2-1)
/d*c^3*(-c^2*x^2+1)^(1/2)+4/3*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-2*c^2*x^2-1)/d/x*arcsin(c*x)*c^2+1/6*b*(-d*(
c^2*x^2-1))^(1/2)/(3*c^4*x^4-2*c^2*x^2-1)/d/x^2*(-c^2*x^2+1)^(1/2)*c+1/3*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-2
*c^2*x^2-1)/d/x^3*arcsin(c*x)-2/3*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d/(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((a+b*arcsin(c*x))/x^4/(-c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.25134, size = 895, normalized size = 6.09 \begin{align*} \left [\frac{2 \,{\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 (2 \, a c^{4} x^{4} - a c^{2} x^{2} +{\left (2 \, b c^{4} x^{4} - 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} d x^{5} - d x^{3}\right )}}, \frac{4 \,{\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 (2 \, a c^{4} x^{4} - a c^{2} x^{2} +{\left (2 \, b c^{4} x^{4} - 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} d x^{5} - d x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/x^4/(-c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

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

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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 )}}{x^{4} \sqrt{- d \left (c x - 1\right ) \left (c x + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(c*x))/x**4/(-c**2*d*x**2+d)**(1/2),x)

[Out]

Integral((a + b*asin(c*x))/(x**4*sqrt(-d*(c*x - 1)*(c*x + 1))), 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}{\sqrt{-c^{2} d x^{2} + d} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/x^4/(-c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

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