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

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 238, normalized size of antiderivative = 1.00, 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 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 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 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])

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 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 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]

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*}

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Mathematica [A]  time = 0.35, size = 162, normalized size = 0.68 \[ \frac {\sqrt {d-c^2 d x^2} \left (-16 a c^4 x^4+8 a c^2 x^2+2 a+b c x \sqrt {1-c^2 x^2}+2 b \left (-8 c^4 x^4+4 c^2 x^2+1\right ) \sin ^{-1}(c x)-5 b c^3 x^3 \sqrt {1-c^2 x^2} \log \left (x^2\right )-3 b c^3 x^3 \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )\right )}{6 d^2 x^3 \left (c^2 x^2-1\right )} \]

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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fricas [F]  time = 19.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b \arcsin \left (c x\right ) + a\right )}}{c^{4} d^{2} x^{8} - 2 \, c^{2} d^{2} x^{6} + d^{2} x^{4}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \arcsin \left (c x\right ) + a}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x^{4}}\,{d x} \]

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)

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maple [C]  time = 0.50, size = 1045, normalized size = 4.39 \[ -\frac {a}{3 d \,x^{3} \sqrt {-c^{2} d \,x^{2}+d}}-\frac {4 a \,c^{2}}{3 d x \sqrt {-c^{2} d \,x^{2}+d}}+\frac {8 a \,c^{4} x}{3 d \sqrt {-c^{2} d \,x^{2}+d}}+\frac {4 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x \,c^{4}}{3 \left (8 c^{4} x^{4}-7 c^{2} x^{2}-1\right ) d^{2}}+\frac {4 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x^{3} c^{6}}{\left (8 c^{4} x^{4}-7 c^{2} x^{2}-1\right ) d^{2}}+\frac {16 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right ) c^{3}}{3 d^{2} \left (c^{2} x^{2}-1\right )}-\frac {8 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right ) c^{3}}{3 \left (8 c^{4} x^{4}-7 c^{2} x^{2}-1\right ) d^{2}}-\frac {64 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x^{2} \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right ) c^{5}}{3 \left (8 c^{4} x^{4}-7 c^{2} x^{2}-1\right ) d^{2}}+\frac {32 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x^{7} c^{10}}{3 \left (8 c^{4} x^{4}-7 c^{2} x^{2}-1\right ) d^{2}}+\frac {32 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x^{5} \left (-c^{2} x^{2}+1\right ) c^{8}}{3 \left (8 c^{4} x^{4}-7 c^{2} x^{2}-1\right ) d^{2}}-\frac {64 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x^{3} \arcsin \left (c x \right ) c^{6}}{3 \left (8 c^{4} x^{4}-7 c^{2} x^{2}-1\right ) d^{2}}-\frac {4 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x \left (-c^{2} x^{2}+1\right ) c^{4}}{3 \left (8 c^{4} x^{4}-7 c^{2} x^{2}-1\right ) d^{2}}-\frac {16 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x^{5} c^{8}}{\left (8 c^{4} x^{4}-7 c^{2} x^{2}-1\right ) d^{2}}-\frac {16 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x^{3} \left (-c^{2} x^{2}+1\right ) c^{6}}{3 \left (8 c^{4} x^{4}-7 c^{2} x^{2}-1\right ) d^{2}}+\frac {8 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x \arcsin \left (c x \right ) c^{4}}{\left (8 c^{4} x^{4}-7 c^{2} x^{2}-1\right ) d^{2}}+\frac {4 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{3} \sqrt {-c^{2} x^{2}+1}}{3 \left (8 c^{4} x^{4}-7 c^{2} x^{2}-1\right ) d^{2}}+\frac {4 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) c^{2}}{\left (8 c^{4} x^{4}-7 c^{2} x^{2}-1\right ) d^{2} x}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c}{6 \left (8 c^{4} x^{4}-7 c^{2} x^{2}-1\right ) d^{2} x^{2}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )}{3 \left (8 c^{4} x^{4}-7 c^{2} x^{2}-1\right ) d^{2} x^{3}}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right ) c^{3}}{d^{2} \left (c^{2} x^{2}-1\right )}-\frac {5 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}-1\right ) c^{3}}{3 d^{2} \left (c^{2} x^{2}-1\right )} \]

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)+4/3*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+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^3-8/3*I*b*(-d*(
c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2-1)/d^2*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*c^3-64/3*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*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*c^5+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-
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^6-4/3*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^4-16*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^
8-16/3*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+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+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]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, {\left (\frac {8 \, c^{4} x}{\sqrt {-c^{2} d x^{2} + d} d} - \frac {4 \, c^{2}}{\sqrt {-c^{2} d x^{2} + d} d x} - \frac {1}{\sqrt {-c^{2} d x^{2} + d} d x^{3}}\right )} a - \frac {-\frac {1}{6} \, b {\left (\frac {3 \, c^{3} \log \left (c x + 1\right ) + 3 \, c^{3} \log \left (c x - 1\right ) + 10 \, c^{3} \log \relax (x) - \frac {c}{x^{2}}}{d} + \frac {2 \, {\left (8 \, c^{4} x^{4} - 4 \, c^{2} x^{2} - 1\right )} \arctan \left (\frac {c x}{\sqrt {c x + 1} \sqrt {-c x + 1}}\right )}{\sqrt {c x + 1} \sqrt {-c x + 1} d x^{3}}\right )}}{\sqrt {d}} \]

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]

1/3*(8*c^4*x/(sqrt(-c^2*d*x^2 + d)*d) - 4*c^2/(sqrt(-c^2*d*x^2 + d)*d*x) - 1/(sqrt(-c^2*d*x^2 + d)*d*x^3))*a -
b*integrate(arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/((c^2*d*x^6 - d*x^4)*sqrt(c*x + 1)*sqrt(-c*x + 1)), x)
/sqrt(d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{x^4\,{\left (d-c^2\,d\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {asin}{\left (c x \right )}}{x^{4} \left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

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]

Integral((a + b*asin(c*x))/(x**4*(-d*(c*x - 1)*(c*x + 1))**(3/2)), x)

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