3.2.39 \(\int \frac {a+b \text {ArcSin}(c x)}{x^4 (d-c^2 d x^2)^{5/2}} \, dx\) [139]

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

[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 = 0.18, antiderivative size = 310, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {277, 198, 197, 4779, 12, 1813, 1634} \begin {gather*} -\frac {2 c^2 (a+b \text {ArcSin}(c x))}{d x \left (d-c^2 d x^2\right )^{3/2}}-\frac {a+b \text {ArcSin}(c x)}{3 d x^3 \left (d-c^2 d x^2\right )^{3/2}}+\frac {16 c^4 x (a+b \text {ArcSin}(c x))}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {8 c^4 x (a+b \text {ArcSin}(c x))}{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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p +
1], 0] && NeQ[p, -1]

Rule 277

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^(p + 1)/(a*(m + 1))), x]
- Dist[b*((m + n*(p + 1) + 1)/(a*(m + 1))), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 1634

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rule 1813

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*SubstFor[x^2,
Pq, x]*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && IntegerQ[(m - 1)/2]

Rule 4779

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[x^
m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSin[c*x], u, x] - Dist[b*c*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]], Int[Si
mplifyIntegrand[u/Sqrt[d + e*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IntegerQ[p
- 1/2] && NeQ[p, -2^(-1)] && (IGtQ[(m + 1)/2, 0] || ILtQ[(m + 2*p + 3)/2, 0])

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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 = 0.21, size = 213, normalized size = 0.69 \begin {gather*} -\frac {\sqrt {d-c^2 d x^2} \left (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 \left (1+6 c^2 x^2-24 c^4 x^4+16 c^6 x^6\right ) \text {ArcSin}(c x)-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 )+8 b c^5 x^5 \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )\right )}{6 d^3 x^3 \left (-1+c^2 x^2\right )^2} \end {gather*}

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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Maple [C] Result contains complex when optimal does not.
time = 0.39, size = 1877, normalized size = 6.05

method	result	size
default	\(\text {Expression too large to display}\)	\(1877\)

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,method=_RETURNVERBOSE)

[Out]

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+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^14-344/3*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^2*x^2+1)^(1/2)*c^5+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+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^2-176/3*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+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+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^2*x^2+1)^(1/2)*c^3+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)+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+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^3-64*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^10-448/3*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^12-280/3*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^4-16/3*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+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+a*(-1/3/d/x^3/(-c^2*d*x^2+d)^(3/2)+2*c^2*(-1/d/x/(-c^2*d*x^2
+d)^(3/2)+4*c^2*(1/3*x/d/(-c^2*d*x^2+d)^(3/2)+2/3/d^2*x/(-c^2*d*x^2+d)^(1/2))))+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^8-40/3*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^6-8/3*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

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Maxima [A]
time = 0.52, size = 255, normalized size = 0.82 \begin {gather*} \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 \left (x\right )}{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 \end {gather*}

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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {5}{2}}}\, dx \end {gather*}

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]

Integral((a + b*asin(c*x))/(x**4*(-d*(c*x - 1)*(c*x + 1))**(5/2)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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