3.139 \(\int \frac {a+b \sin ^{-1}(c x)}{x^4 (d-c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=310 \[ -\frac {2 c^2 \left (a+b \sin ^{-1}(c x)\right )}{d x \left (d-c^2 d x^2\right )^{3/2}}-\frac {a+b \sin ^{-1}(c x)}{3 d x^3 \left (d-c^2 d x^2\right )^{3/2}}+\frac {16 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {8 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {b c \sqrt {d-c^2 d x^2}}{6 d^3 x^2 \sqrt {1-c^2 x^2}}-\frac {b c^3 \sqrt {d-c^2 d x^2}}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {8 b c^3 \log (x) \sqrt {d-c^2 d x^2}}{3 d^3 \sqrt {1-c^2 x^2}}+\frac {4 b c^3 \sqrt {d-c^2 d x^2} \log \left (1-c^2 x^2\right )}{3 d^3 \sqrt {1-c^2 x^2}} \]

[Out]

1/3*(-a-b*arcsin(c*x))/d/x^3/(-c^2*d*x^2+d)^(3/2)-2*c^2*(a+b*arcsin(c*x))/d/x/(-c^2*d*x^2+d)^(3/2)+8/3*c^4*x*(
a+b*arcsin(c*x))/d/(-c^2*d*x^2+d)^(3/2)+16/3*c^4*x*(a+b*arcsin(c*x))/d^2/(-c^2*d*x^2+d)^(1/2)-1/6*b*c^3*(-c^2*
d*x^2+d)^(1/2)/d^3/(-c^2*x^2+1)^(3/2)-1/6*b*c*(-c^2*d*x^2+d)^(1/2)/d^3/x^2/(-c^2*x^2+1)^(1/2)+8/3*b*c^3*ln(x)*
(-c^2*d*x^2+d)^(1/2)/d^3/(-c^2*x^2+1)^(1/2)+4/3*b*c^3*ln(-c^2*x^2+1)*(-c^2*d*x^2+d)^(1/2)/d^3/(-c^2*x^2+1)^(1/
2)

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Rubi [A]  time = 0.39, antiderivative size = 310, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {4701, 4655, 4653, 260, 261, 266, 44} \[ \frac {16 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {8 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {2 c^2 \left (a+b \sin ^{-1}(c x)\right )}{d x \left (d-c^2 d x^2\right )^{3/2}}-\frac {a+b \sin ^{-1}(c x)}{3 d x^3 \left (d-c^2 d x^2\right )^{3/2}}-\frac {b c^3}{6 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}-\frac {b c \sqrt {1-c^2 x^2}}{6 d^2 x^2 \sqrt {d-c^2 d x^2}}+\frac {8 b c^3 \sqrt {1-c^2 x^2} \log (x)}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {4 b c^3 \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{3 d^2 \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])/(x^4*(d - c^2*d*x^2)^(5/2)),x]

[Out]

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

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 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -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 4655

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(p
+ 1)*(a + b*ArcSin[c*x])^n)/(2*d*(p + 1)), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(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])/(2*(p + 1)*(1 - c^2*x^2)^FracPart[p
]), Int[x*(1 - c^2*x^2)^(p + 1/2)*(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] && LtQ[p, -1] && NeQ[p, -3/2]

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]

Rubi steps

\begin {align*} \int \frac {a+b \sin ^{-1}(c x)}{x^4 \left (d-c^2 d x^2\right )^{5/2}} \, dx &=-\frac {a+b \sin ^{-1}(c x)}{3 d x^3 \left (d-c^2 d x^2\right )^{3/2}}+\left (2 c^2\right ) \int \frac {a+b \sin ^{-1}(c x)}{x^2 \left (d-c^2 d x^2\right )^{5/2}} \, dx+\frac {\left (b c \sqrt {1-c^2 x^2}\right ) \int \frac {1}{x^3 \left (1-c^2 x^2\right )^2} \, dx}{3 d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {a+b \sin ^{-1}(c x)}{3 d x^3 \left (d-c^2 d x^2\right )^{3/2}}-\frac {2 c^2 \left (a+b \sin ^{-1}(c x)\right )}{d x \left (d-c^2 d x^2\right )^{3/2}}+\left (8 c^4\right ) \int \frac {a+b \sin ^{-1}(c x)}{\left (d-c^2 d x^2\right )^{5/2}} \, 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 )^2} \, dx,x,x^2\right )}{6 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (2 b c^3 \sqrt {1-c^2 x^2}\right ) \int \frac {1}{x \left (1-c^2 x^2\right )^2} \, dx}{d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {a+b \sin ^{-1}(c x)}{3 d x^3 \left (d-c^2 d x^2\right )^{3/2}}-\frac {2 c^2 \left (a+b \sin ^{-1}(c x)\right )}{d x \left (d-c^2 d x^2\right )^{3/2}}+\frac {8 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {\left (16 c^4\right ) \int \frac {a+b \sin ^{-1}(c x)}{\left (d-c^2 d x^2\right )^{3/2}} \, dx}{3 d}+\frac {\left (b c \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{x^2}+\frac {2 c^2}{x}+\frac {c^4}{\left (-1+c^2 x\right )^2}-\frac {2 c^4}{-1+c^2 x}\right ) \, dx,x,x^2\right )}{6 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (b c^3 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1-c^2 x\right )^2} \, dx,x,x^2\right )}{d^2 \sqrt {d-c^2 d x^2}}-\frac {\left (8 b c^5 \sqrt {1-c^2 x^2}\right ) \int \frac {x}{\left (1-c^2 x^2\right )^2} \, dx}{3 d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {7 b c^3}{6 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}-\frac {b c \sqrt {1-c^2 x^2}}{6 d^2 x^2 \sqrt {d-c^2 d x^2}}-\frac {a+b \sin ^{-1}(c x)}{3 d x^3 \left (d-c^2 d x^2\right )^{3/2}}-\frac {2 c^2 \left (a+b \sin ^{-1}(c x)\right )}{d x \left (d-c^2 d x^2\right )^{3/2}}+\frac {8 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {16 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {2 b c^3 \sqrt {1-c^2 x^2} \log (x)}{3 d^2 \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 )}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (b c^3 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{x}+\frac {c^2}{\left (-1+c^2 x\right )^2}-\frac {c^2}{-1+c^2 x}\right ) \, dx,x,x^2\right )}{d^2 \sqrt {d-c^2 d x^2}}-\frac {\left (16 b c^5 \sqrt {1-c^2 x^2}\right ) \int \frac {x}{1-c^2 x^2} \, dx}{3 d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {b c^3}{6 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}-\frac {b c \sqrt {1-c^2 x^2}}{6 d^2 x^2 \sqrt {d-c^2 d x^2}}-\frac {a+b \sin ^{-1}(c x)}{3 d x^3 \left (d-c^2 d x^2\right )^{3/2}}-\frac {2 c^2 \left (a+b \sin ^{-1}(c x)\right )}{d x \left (d-c^2 d x^2\right )^{3/2}}+\frac {8 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {16 c^4 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {8 b c^3 \sqrt {1-c^2 x^2} \log (x)}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {4 b c^3 \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{3 d^2 \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 213, normalized size = 0.69 \[ -\frac {\sqrt {d-c^2 d x^2} \left (32 a c^6 x^6-48 a c^4 x^4+12 a c^2 x^2+2 a+b c x \sqrt {1-c^2 x^2}+8 b c^5 x^5 \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )-8 b c^3 x^3 \left (1-c^2 x^2\right )^{3/2} \log \left (x^2\right )-8 b c^3 x^3 \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )+2 b \left (16 c^6 x^6-24 c^4 x^4+6 c^2 x^2+1\right ) \sin ^{-1}(c x)\right )}{6 d^3 x^3 \left (c^2 x^2-1\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c*x])/(x^4*(d - c^2*d*x^2)^(5/2)),x]

[Out]

-1/6*(Sqrt[d - c^2*d*x^2]*(2*a + 12*a*c^2*x^2 - 48*a*c^4*x^4 + 32*a*c^6*x^6 + b*c*x*Sqrt[1 - c^2*x^2] + 2*b*(1
+ 6*c^2*x^2 - 24*c^4*x^4 + 16*c^6*x^6)*ArcSin[c*x] - 8*b*c^3*x^3*(1 - c^2*x^2)^(3/2)*Log[x^2] - 8*b*c^3*x^3*S
qrt[1 - c^2*x^2]*Log[1 - c^2*x^2] + 8*b*c^5*x^5*Sqrt[1 - c^2*x^2]*Log[1 - c^2*x^2]))/(d^3*x^3*(-1 + c^2*x^2)^2
)

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fricas [F]  time = 1.19, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b \arcsin \left (c x\right ) + a\right )}}{c^{6} d^{3} x^{10} - 3 \, c^{4} d^{3} x^{8} + 3 \, c^{2} d^{3} x^{6} - d^{3} x^{4}}, x\right ) \]

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)^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-c^2*d*x^2 + d)*(b*arcsin(c*x) + a)/(c^6*d^3*x^10 - 3*c^4*d^3*x^8 + 3*c^2*d^3*x^6 - d^3*x^4), x
)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \arcsin \left (c x\right ) + a}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{4}}\,{d x} \]

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)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x) + a)/((-c^2*d*x^2 + d)^(5/2)*x^4), x)

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maple [C]  time = 0.52, size = 1875, normalized size = 6.05 \[ \text {result too large to display} \]

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)^(5/2),x)

[Out]

6*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3/x*arcsin(c*x)*c^2+1/6*b*(-d*(c^
2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3/x^2*(-c^2*x^2+1)^(1/2)*c-64*b*(-d*(c^2*x^2
-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x^7*arcsin(c*x)*c^10+160*b*(-d*(c^2*x^2-1))^(1/
2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x^5*arcsin(c*x)*c^8-8/3*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*
x^2+1)^(1/2)/d^3/(c^2*x^2-1)*ln((I*c*x+(-c^2*x^2+1)^(1/2))^4-1)*c^3-344/3*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8
-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x^3*arcsin(c*x)*c^6-2*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6
+35*c^4*x^4-10*c^2*x^2-1)/d^3*x^2*c^5*(-c^2*x^2+1)^(1/2)+12*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35
*c^4*x^4-10*c^2*x^2-1)/d^3*x*arcsin(c*x)*c^4+128/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^
4-10*c^2*x^2-1)/d^3*x^11*c^14-448/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)
/d^3*x^9*c^12+128*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x^4*arcsin(c*
x)*(-c^2*x^2+1)^(1/2)*c^7-176/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3
*x^2*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*c^5-64*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2
*x^2-1)/d^3*x^6*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*c^9+560/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*
c^4*x^4-10*c^2*x^2-1)/d^3*x^7*c^10-280/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x
^2-1)/d^3*x^5*c^8+32/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x^3*c^6+
8/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x*c^4-1/3*a/d/x^3/(-c^2*d*x
^2+d)^(3/2)+128/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x^9*(-c^2*x^2
+1)*c^12-320/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x^7*(-c^2*x^2+1)
*c^10+80*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x^5*(-c^2*x^2+1)*c^8-8
/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x*(-c^2*x^2+1)*c^4+32/3*I*b*
(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^3/(c^2*x^2-1)*arcsin(c*x)*c^3-16/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*
c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*c^3-40/3*I*b*(-d*(c^2*x^2-1))^(
1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x^3*(-c^2*x^2+1)*c^6+2*b*(-d*(c^2*x^2-1))^(1/2)/(12*c
^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*c^3*(-c^2*x^2+1)^(1/2)+1/3*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8
-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3/x^3*arcsin(c*x)+8/3*a*c^4*x/d/(-c^2*d*x^2+d)^(3/2)+16/3*a*c^4/d^2*x/(
-c^2*d*x^2+d)^(1/2)-2*a*c^2/d/x/(-c^2*d*x^2+d)^(3/2)

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maxima [A]  time = 1.33, size = 255, normalized size = 0.82 \[ \frac {1}{6} \, b c {\left (\frac {8 \, c^{2} \log \left (c x + 1\right )}{d^{\frac {5}{2}}} + \frac {8 \, c^{2} \log \left (c x - 1\right )}{d^{\frac {5}{2}}} + \frac {16 \, c^{2} \log \relax (x)}{d^{\frac {5}{2}}} + \frac {1}{c^{2} d^{\frac {5}{2}} x^{4} - d^{\frac {5}{2}} x^{2}}\right )} + \frac {1}{3} \, {\left (\frac {16 \, c^{4} x}{\sqrt {-c^{2} d x^{2} + d} d^{2}} + \frac {8 \, c^{4} x}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d} - \frac {6 \, c^{2}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d x} - \frac {1}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d x^{3}}\right )} b \arcsin \left (c x\right ) + \frac {1}{3} \, {\left (\frac {16 \, c^{4} x}{\sqrt {-c^{2} d x^{2} + d} d^{2}} + \frac {8 \, c^{4} x}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d} - \frac {6 \, c^{2}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d x} - \frac {1}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d x^{3}}\right )} a \]

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)^(5/2),x, algorithm="maxima")

[Out]

1/6*b*c*(8*c^2*log(c*x + 1)/d^(5/2) + 8*c^2*log(c*x - 1)/d^(5/2) + 16*c^2*log(x)/d^(5/2) + 1/(c^2*d^(5/2)*x^4
- d^(5/2)*x^2)) + 1/3*(16*c^4*x/(sqrt(-c^2*d*x^2 + d)*d^2) + 8*c^4*x/((-c^2*d*x^2 + d)^(3/2)*d) - 6*c^2/((-c^2
*d*x^2 + d)^(3/2)*d*x) - 1/((-c^2*d*x^2 + d)^(3/2)*d*x^3))*b*arcsin(c*x) + 1/3*(16*c^4*x/(sqrt(-c^2*d*x^2 + d)
*d^2) + 8*c^4*x/((-c^2*d*x^2 + d)^(3/2)*d) - 6*c^2/((-c^2*d*x^2 + d)^(3/2)*d*x) - 1/((-c^2*d*x^2 + d)^(3/2)*d*
x^3))*a

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{x^4\,{\left (d-c^2\,d\,x^2\right )}^{5/2}} \,d x \]

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)^(5/2)),x)

[Out]

int((a + b*asin(c*x))/(x^4*(d - c^2*d*x^2)^(5/2)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)**(5/2),x)

[Out]

Timed out

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