3.128 \(\int \frac{a+b \sin ^{-1}(c x)}{x^4 (d-c^2 d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=238 \[ \frac{8 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{3 d \sqrt{d-c^2 d x^2}}-\frac{4 c^2 \left (a+b \sin ^{-1}(c x)\right )}{3 d x \sqrt{d-c^2 d x^2}}-\frac{a+b \sin ^{-1}(c x)}{3 d x^3 \sqrt{d-c^2 d x^2}}-\frac{b c \sqrt{d-c^2 d x^2}}{6 d^2 x^2 \sqrt{1-c^2 x^2}}+\frac{5 b c^3 \log (x) \sqrt{d-c^2 d x^2}}{3 d^2 \sqrt{1-c^2 x^2}}+\frac{b c^3 \sqrt{d-c^2 d x^2} \log \left (1-c^2 x^2\right )}{2 d^2 \sqrt{1-c^2 x^2}} \]

[Out]

-(b*c*Sqrt[d - c^2*d*x^2])/(6*d^2*x^2*Sqrt[1 - c^2*x^2]) - (a + b*ArcSin[c*x])/(3*d*x^3*Sqrt[d - c^2*d*x^2]) -
(4*c^2*(a + b*ArcSin[c*x]))/(3*d*x*Sqrt[d - c^2*d*x^2]) + (8*c^4*x*(a + b*ArcSin[c*x]))/(3*d*Sqrt[d - c^2*d*x
^2]) + (5*b*c^3*Sqrt[d - c^2*d*x^2]*Log[x])/(3*d^2*Sqrt[1 - c^2*x^2]) + (b*c^3*Sqrt[d - c^2*d*x^2]*Log[1 - c^2
*x^2])/(2*d^2*Sqrt[1 - c^2*x^2])

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Rubi [A]  time = 0.292013, antiderivative size = 238, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {4701, 4653, 260, 266, 36, 29, 31, 44} \[ \frac{8 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{3 d \sqrt{d-c^2 d x^2}}-\frac{4 c^2 \left (a+b \sin ^{-1}(c x)\right )}{3 d x \sqrt{d-c^2 d x^2}}-\frac{a+b \sin ^{-1}(c x)}{3 d x^3 \sqrt{d-c^2 d x^2}}-\frac{b c \sqrt{1-c^2 x^2}}{6 d x^2 \sqrt{d-c^2 d x^2}}+\frac{5 b c^3 \sqrt{1-c^2 x^2} \log (x)}{3 d \sqrt{d-c^2 d x^2}}+\frac{b c^3 \sqrt{1-c^2 x^2} \log \left (1-c^2 x^2\right )}{2 d \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])/(x^4*(d - c^2*d*x^2)^(3/2)),x]

[Out]

-(b*c*Sqrt[1 - c^2*x^2])/(6*d*x^2*Sqrt[d - c^2*d*x^2]) - (a + b*ArcSin[c*x])/(3*d*x^3*Sqrt[d - c^2*d*x^2]) - (
4*c^2*(a + b*ArcSin[c*x]))/(3*d*x*Sqrt[d - c^2*d*x^2]) + (8*c^4*x*(a + b*ArcSin[c*x]))/(3*d*Sqrt[d - c^2*d*x^2
]) + (5*b*c^3*Sqrt[1 - c^2*x^2]*Log[x])/(3*d*Sqrt[d - c^2*d*x^2]) + (b*c^3*Sqrt[1 - c^2*x^2]*Log[1 - c^2*x^2])
/(2*d*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 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 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 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 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 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{a+b \sin ^{-1}(c x)}{x^4 \left (d-c^2 d x^2\right )^{3/2}} \, dx &=-\frac{a+b \sin ^{-1}(c x)}{3 d x^3 \sqrt{d-c^2 d x^2}}+\frac{1}{3} \left (4 c^2\right ) \int \frac{a+b \sin ^{-1}(c x)}{x^2 \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^3 \left (1-c^2 x^2\right )} \, dx}{3 d \sqrt{d-c^2 d x^2}}\\ &=-\frac{a+b \sin ^{-1}(c x)}{3 d x^3 \sqrt{d-c^2 d x^2}}-\frac{4 c^2 \left (a+b \sin ^{-1}(c x)\right )}{3 d x \sqrt{d-c^2 d x^2}}+\frac{1}{3} \left (8 c^4\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 ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1-c^2 x\right )} \, dx,x,x^2\right )}{6 d \sqrt{d-c^2 d x^2}}+\frac{\left (4 b c^3 \sqrt{1-c^2 x^2}\right ) \int \frac{1}{x \left (1-c^2 x^2\right )} \, dx}{3 d \sqrt{d-c^2 d x^2}}\\ &=-\frac{a+b \sin ^{-1}(c x)}{3 d x^3 \sqrt{d-c^2 d x^2}}-\frac{4 c^2 \left (a+b \sin ^{-1}(c x)\right )}{3 d x \sqrt{d-c^2 d x^2}}+\frac{8 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{3 d \sqrt{d-c^2 d x^2}}+\frac{\left (b c \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{x^2}+\frac{c^2}{x}-\frac{c^4}{-1+c^2 x}\right ) \, dx,x,x^2\right )}{6 d \sqrt{d-c^2 d x^2}}+\frac{\left (2 b c^3 \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 )}{3 d \sqrt{d-c^2 d x^2}}-\frac{\left (8 b c^5 \sqrt{1-c^2 x^2}\right ) \int \frac{x}{1-c^2 x^2} \, dx}{3 d \sqrt{d-c^2 d x^2}}\\ &=-\frac{b c \sqrt{1-c^2 x^2}}{6 d x^2 \sqrt{d-c^2 d x^2}}-\frac{a+b \sin ^{-1}(c x)}{3 d x^3 \sqrt{d-c^2 d x^2}}-\frac{4 c^2 \left (a+b \sin ^{-1}(c x)\right )}{3 d x \sqrt{d-c^2 d x^2}}+\frac{8 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{3 d \sqrt{d-c^2 d x^2}}+\frac{b c^3 \sqrt{1-c^2 x^2} \log (x)}{3 d \sqrt{d-c^2 d x^2}}+\frac{7 b c^3 \sqrt{1-c^2 x^2} \log \left (1-c^2 x^2\right )}{6 d \sqrt{d-c^2 d x^2}}+\frac{\left (2 b c^3 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )}{3 d \sqrt{d-c^2 d x^2}}+\frac{\left (2 b c^5 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{1-c^2 x} \, dx,x,x^2\right )}{3 d \sqrt{d-c^2 d x^2}}\\ &=-\frac{b c \sqrt{1-c^2 x^2}}{6 d x^2 \sqrt{d-c^2 d x^2}}-\frac{a+b \sin ^{-1}(c x)}{3 d x^3 \sqrt{d-c^2 d x^2}}-\frac{4 c^2 \left (a+b \sin ^{-1}(c x)\right )}{3 d x \sqrt{d-c^2 d x^2}}+\frac{8 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{3 d \sqrt{d-c^2 d x^2}}+\frac{5 b c^3 \sqrt{1-c^2 x^2} \log (x)}{3 d \sqrt{d-c^2 d x^2}}+\frac{b c^3 \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*}

Mathematica [A]  time = 0.311926, size = 162, normalized size = 0.68 \[ \frac{\sqrt{d-c^2 d x^2} \left (-16 a c^4 x^4+8 a c^2 x^2+2 a+b c x \sqrt{1-c^2 x^2}-5 b c^3 x^3 \sqrt{1-c^2 x^2} \log \left (x^2\right )-3 b c^3 x^3 \sqrt{1-c^2 x^2} \log \left (1-c^2 x^2\right )+2 b \left (-8 c^4 x^4+4 c^2 x^2+1\right ) \sin ^{-1}(c x)\right )}{6 d^2 x^3 \left (c^2 x^2-1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c*x])/(x^4*(d - c^2*d*x^2)^(3/2)),x]

[Out]

(Sqrt[d - c^2*d*x^2]*(2*a + 8*a*c^2*x^2 - 16*a*c^4*x^4 + b*c*x*Sqrt[1 - c^2*x^2] + 2*b*(1 + 4*c^2*x^2 - 8*c^4*
x^4)*ArcSin[c*x] - 5*b*c^3*x^3*Sqrt[1 - c^2*x^2]*Log[x^2] - 3*b*c^3*x^3*Sqrt[1 - c^2*x^2]*Log[1 - c^2*x^2]))/(
6*d^2*x^3*(-1 + c^2*x^2))

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Maple [C]  time = 0.25, size = 1045, normalized size = 4.4 \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)^(3/2),x)

[Out]

-1/3*a/d/x^3/(-c^2*d*x^2+d)^(1/2)-4/3*a*c^2/d/x/(-c^2*d*x^2+d)^(1/2)+8/3*a*c^4/d*x/(-c^2*d*x^2+d)^(1/2)+4*I*b*
(-d*(c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2-1)/d^2*x^3*c^6+16/3*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^3-16*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2-1)/d^2*x^5*c^8+32/3*I*b*(-d
*(c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2-1)/d^2*x^7*c^10+32/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2-
1)/d^2*x^5*(-c^2*x^2+1)*c^8-4/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2-1)/d^2*x*(-c^2*x^2+1)*c^4-64/3
*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2-1)/d^2*x^3*arcsin(c*x)*c^6-16/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c
^4*x^4-7*c^2*x^2-1)/d^2*x^3*(-c^2*x^2+1)*c^6+4/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2-1)/d^2*x*c^4-
64/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2-1)/d^2*x^2*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*c^5+8*b*(-d*(c^
2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2-1)/d^2*x*arcsin(c*x)*c^4-8/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*
x^2-1)/d^2*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*c^3+4/3*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2-1)/d^2*c^3*(-c
^2*x^2+1)^(1/2)+4*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2-1)/d^2/x*arcsin(c*x)*c^2+1/6*b*(-d*(c^2*x^2-1)
)^(1/2)/(8*c^4*x^4-7*c^2*x^2-1)/d^2/x^2*(-c^2*x^2+1)^(1/2)*c+1/3*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2
-1)/d^2/x^3*arcsin(c*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(1+(I*c*x+(-c^2*x^2+1)^(
1/2))^2)*c^3-5/3*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))^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)^(3/2),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{\sqrt{-c^{2} d x^{2} + d}{\left (b \arcsin \left (c x\right ) + a\right )}}{c^{4} d^{2} x^{8} - 2 \, c^{2} d^{2} x^{6} + d^{2} x^{4}}, x\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)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(-c^2*d*x^2 + d)*(b*arcsin(c*x) + a)/(c^4*d^2*x^8 - 2*c^2*d^2*x^6 + d^2*x^4), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**(3/2),x)

[Out]

Timed out

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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{3}{2}} 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)^(3/2),x, algorithm="giac")

[Out]

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