3.1.67 \(\int \frac {\sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))}{x^5} \, dx\) [67]

Optimal. Leaf size=301 \[ -\frac {b c \sqrt {d-c^2 d x^2}}{12 x^3 \sqrt {1-c^2 x^2}}+\frac {b c^3 \sqrt {d-c^2 d x^2}}{8 x \sqrt {1-c^2 x^2}}-\frac {\sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))}{4 x^4}+\frac {c^2 \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))}{8 x^2}+\frac {c^4 \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x)) \tanh ^{-1}\left (e^{i \text {ArcSin}(c x)}\right )}{4 \sqrt {1-c^2 x^2}}-\frac {i b c^4 \sqrt {d-c^2 d x^2} \text {PolyLog}\left (2,-e^{i \text {ArcSin}(c x)}\right )}{8 \sqrt {1-c^2 x^2}}+\frac {i b c^4 \sqrt {d-c^2 d x^2} \text {PolyLog}\left (2,e^{i \text {ArcSin}(c x)}\right )}{8 \sqrt {1-c^2 x^2}} \]

[Out]

-1/4*(-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))/x^4+1/8*c^2*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)/x^2-1/12*b*c*(-
c^2*d*x^2+d)^(1/2)/x^3/(-c^2*x^2+1)^(1/2)+1/8*b*c^3*(-c^2*d*x^2+d)^(1/2)/x/(-c^2*x^2+1)^(1/2)+1/4*c^4*(a+b*arc
sin(c*x))*arctanh(I*c*x+(-c^2*x^2+1)^(1/2))*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)-1/8*I*b*c^4*polylog(2,-I*c
*x-(-c^2*x^2+1)^(1/2))*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)+1/8*I*b*c^4*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2))
*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.20, antiderivative size = 301, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {4781, 30, 4789, 4803, 4268, 2317, 2438} \begin {gather*} \frac {c^2 \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))}{8 x^2}-\frac {\sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))}{4 x^4}+\frac {c^4 \sqrt {d-c^2 d x^2} \tanh ^{-1}\left (e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{4 \sqrt {1-c^2 x^2}}-\frac {i b c^4 \sqrt {d-c^2 d x^2} \text {Li}_2\left (-e^{i \text {ArcSin}(c x)}\right )}{8 \sqrt {1-c^2 x^2}}+\frac {i b c^4 \sqrt {d-c^2 d x^2} \text {Li}_2\left (e^{i \text {ArcSin}(c x)}\right )}{8 \sqrt {1-c^2 x^2}}-\frac {b c \sqrt {d-c^2 d x^2}}{12 x^3 \sqrt {1-c^2 x^2}}+\frac {b c^3 \sqrt {d-c^2 d x^2}}{8 x \sqrt {1-c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(b*c*Sqrt[d - c^2*d*x^2])/(x^3*Sqrt[1 - c^2*x^2]) + (b*c^3*Sqrt[d - c^2*d*x^2])/(8*x*Sqrt[1 - c^2*x^2])
- (Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(4*x^4) + (c^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(8*x^2) +
(c^4*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])*ArcTanh[E^(I*ArcSin[c*x])])/(4*Sqrt[1 - c^2*x^2]) - ((I/8)*b*c^4*
Sqrt[d - c^2*d*x^2]*PolyLog[2, -E^(I*ArcSin[c*x])])/Sqrt[1 - c^2*x^2] + ((I/8)*b*c^4*Sqrt[d - c^2*d*x^2]*PolyL
og[2, E^(I*ArcSin[c*x])])/Sqrt[1 - c^2*x^2]

Rule 30

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

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 4268

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^(I*(e + f*
x))]/f), x] + (-Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[d*(m/f), Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IGtQ[m, 0]

Rule 4781

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(f
*x)^(m + 1)*Sqrt[d + e*x^2]*((a + b*ArcSin[c*x])^n/(f*(m + 1))), x] + (-Dist[b*c*(n/(f*(m + 1)))*Simp[Sqrt[d +
e*x^2]/Sqrt[1 - c^2*x^2]], Int[(f*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1), x], x] + Dist[(c^2/(f^2*(m + 1)))*S
imp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]], Int[(f*x)^(m + 2)*((a + b*ArcSin[c*x])^n/Sqrt[1 - c^2*x^2]), x], x]) /
; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[m, -1]

Rule 4789

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/(f*(m + 1)))*Simp[(d + e*x
^2)^p/(1 - c^2*x^2)^p], Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; Free
Q[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && ILtQ[m, -1]

Rule 4803

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[(1/c^(m
+ 1))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]], Subst[Int[(a + b*x)^n*Sin[x]^m, x], x, ArcSin[c*x]], x] /; Free
Q[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{x^5} \, dx &=-\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}+\frac {\left (b c \sqrt {d-c^2 d x^2}\right ) \int \frac {1}{x^4} \, dx}{4 \sqrt {1-c^2 x^2}}-\frac {\left (c^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{x^3 \sqrt {1-c^2 x^2}} \, dx}{4 \sqrt {1-c^2 x^2}}\\ &=-\frac {b c \sqrt {d-c^2 d x^2}}{12 x^3 \sqrt {1-c^2 x^2}}-\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}+\frac {c^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 x^2}-\frac {\left (b c^3 \sqrt {d-c^2 d x^2}\right ) \int \frac {1}{x^2} \, dx}{8 \sqrt {1-c^2 x^2}}-\frac {\left (c^4 \sqrt {d-c^2 d x^2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{x \sqrt {1-c^2 x^2}} \, dx}{8 \sqrt {1-c^2 x^2}}\\ &=-\frac {b c \sqrt {d-c^2 d x^2}}{12 x^3 \sqrt {1-c^2 x^2}}+\frac {b c^3 \sqrt {d-c^2 d x^2}}{8 x \sqrt {1-c^2 x^2}}-\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}+\frac {c^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 x^2}-\frac {\left (c^4 \sqrt {d-c^2 d x^2}\right ) \text {Subst}\left (\int (a+b x) \csc (x) \, dx,x,\sin ^{-1}(c x)\right )}{8 \sqrt {1-c^2 x^2}}\\ &=-\frac {b c \sqrt {d-c^2 d x^2}}{12 x^3 \sqrt {1-c^2 x^2}}+\frac {b c^3 \sqrt {d-c^2 d x^2}}{8 x \sqrt {1-c^2 x^2}}-\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}+\frac {c^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 x^2}+\frac {c^4 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 \sqrt {1-c^2 x^2}}+\frac {\left (b c^4 \sqrt {d-c^2 d x^2}\right ) \text {Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{8 \sqrt {1-c^2 x^2}}-\frac {\left (b c^4 \sqrt {d-c^2 d x^2}\right ) \text {Subst}\left (\int \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{8 \sqrt {1-c^2 x^2}}\\ &=-\frac {b c \sqrt {d-c^2 d x^2}}{12 x^3 \sqrt {1-c^2 x^2}}+\frac {b c^3 \sqrt {d-c^2 d x^2}}{8 x \sqrt {1-c^2 x^2}}-\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}+\frac {c^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 x^2}+\frac {c^4 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 \sqrt {1-c^2 x^2}}-\frac {\left (i b c^4 \sqrt {d-c^2 d x^2}\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{8 \sqrt {1-c^2 x^2}}+\frac {\left (i b c^4 \sqrt {d-c^2 d x^2}\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{8 \sqrt {1-c^2 x^2}}\\ &=-\frac {b c \sqrt {d-c^2 d x^2}}{12 x^3 \sqrt {1-c^2 x^2}}+\frac {b c^3 \sqrt {d-c^2 d x^2}}{8 x \sqrt {1-c^2 x^2}}-\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}+\frac {c^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 x^2}+\frac {c^4 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 \sqrt {1-c^2 x^2}}-\frac {i b c^4 \sqrt {d-c^2 d x^2} \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{8 \sqrt {1-c^2 x^2}}+\frac {i b c^4 \sqrt {d-c^2 d x^2} \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{8 \sqrt {1-c^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 2.88, size = 321, normalized size = 1.07 \begin {gather*} \frac {a \left (-2+c^2 x^2\right ) \sqrt {d-c^2 d x^2}}{8 x^4}-\frac {1}{8} a c^4 \sqrt {d} \log (x)+\frac {1}{8} a c^4 \sqrt {d} \log \left (d+\sqrt {d} \sqrt {d-c^2 d x^2}\right )+\frac {b c^4 \sqrt {d-c^2 d x^2} \left (8 \cot \left (\frac {1}{2} \text {ArcSin}(c x)\right )+6 \text {ArcSin}(c x) \csc ^2\left (\frac {1}{2} \text {ArcSin}(c x)\right )-c x \csc ^4\left (\frac {1}{2} \text {ArcSin}(c x)\right )-3 \text {ArcSin}(c x) \csc ^4\left (\frac {1}{2} \text {ArcSin}(c x)\right )-24 \text {ArcSin}(c x) \log \left (1-e^{i \text {ArcSin}(c x)}\right )+24 \text {ArcSin}(c x) \log \left (1+e^{i \text {ArcSin}(c x)}\right )-24 i \text {PolyLog}\left (2,-e^{i \text {ArcSin}(c x)}\right )+24 i \text {PolyLog}\left (2,e^{i \text {ArcSin}(c x)}\right )-6 \text {ArcSin}(c x) \sec ^2\left (\frac {1}{2} \text {ArcSin}(c x)\right )+3 \text {ArcSin}(c x) \sec ^4\left (\frac {1}{2} \text {ArcSin}(c x)\right )-\frac {16 \sin ^4\left (\frac {1}{2} \text {ArcSin}(c x)\right )}{c^3 x^3}+8 \tan \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )}{192 \sqrt {1-c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(-2 + c^2*x^2)*Sqrt[d - c^2*d*x^2])/(8*x^4) - (a*c^4*Sqrt[d]*Log[x])/8 + (a*c^4*Sqrt[d]*Log[d + Sqrt[d]*Sqr
t[d - c^2*d*x^2]])/8 + (b*c^4*Sqrt[d - c^2*d*x^2]*(8*Cot[ArcSin[c*x]/2] + 6*ArcSin[c*x]*Csc[ArcSin[c*x]/2]^2 -
c*x*Csc[ArcSin[c*x]/2]^4 - 3*ArcSin[c*x]*Csc[ArcSin[c*x]/2]^4 - 24*ArcSin[c*x]*Log[1 - E^(I*ArcSin[c*x])] + 2
4*ArcSin[c*x]*Log[1 + E^(I*ArcSin[c*x])] - (24*I)*PolyLog[2, -E^(I*ArcSin[c*x])] + (24*I)*PolyLog[2, E^(I*ArcS
in[c*x])] - 6*ArcSin[c*x]*Sec[ArcSin[c*x]/2]^2 + 3*ArcSin[c*x]*Sec[ArcSin[c*x]/2]^4 - (16*Sin[ArcSin[c*x]/2]^4
)/(c^3*x^3) + 8*Tan[ArcSin[c*x]/2]))/(192*Sqrt[1 - c^2*x^2])

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Maple [A]
time = 0.28, size = 571, normalized size = 1.90

method	result	size
default	\(-\frac {a \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{4 d \,x^{4}}-\frac {a \,c^{2} \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{8 d \,x^{2}}+\frac {a \,c^{4} \sqrt {d}\, \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {-c^{2} d \,x^{2}+d}}{x}\right )}{8}-\frac {a \,c^{4} \sqrt {-c^{2} d \,x^{2}+d}}{8}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) c^{4}}{8 c^{2} x^{2}-8}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c^{3}}{8 \left (c^{2} x^{2}-1\right ) x}-\frac {3 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) c^{2}}{8 \left (c^{2} x^{2}-1\right ) x^{2}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c}{12 \left (c^{2} x^{2}-1\right ) x^{3}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )}{4 \left (c^{2} x^{2}-1\right ) x^{4}}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c^{4} \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{8 c^{2} x^{2}-8}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c^{4} \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )}{8 c^{2} x^{2}-8}+\frac {i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c^{4} \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{8 c^{2} x^{2}-8}-\frac {i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c^{4} \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )}{8 c^{2} x^{2}-8}\)	\(571\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))/x^5,x,method=_RETURNVERBOSE)

[Out]

-1/4*a/d/x^4*(-c^2*d*x^2+d)^(3/2)-1/8*a*c^2/d/x^2*(-c^2*d*x^2+d)^(3/2)+1/8*a*c^4*d^(1/2)*ln((2*d+2*d^(1/2)*(-c
^2*d*x^2+d)^(1/2))/x)-1/8*a*c^4*(-c^2*d*x^2+d)^(1/2)+1/8*b*(-d*(c^2*x^2-1))^(1/2)/(c^2*x^2-1)*arcsin(c*x)*c^4-
1/8*b*(-d*(c^2*x^2-1))^(1/2)/(c^2*x^2-1)/x*(-c^2*x^2+1)^(1/2)*c^3-3/8*b*(-d*(c^2*x^2-1))^(1/2)/(c^2*x^2-1)/x^2
*arcsin(c*x)*c^2+1/12*b*(-d*(c^2*x^2-1))^(1/2)/(c^2*x^2-1)/x^3*(-c^2*x^2+1)^(1/2)*c+1/4*b*(-d*(c^2*x^2-1))^(1/
2)/(c^2*x^2-1)/x^4*arcsin(c*x)-b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)*c^4/(8*c^2*x^2-8)*arcsin(c*x)*ln(1+
I*c*x+(-c^2*x^2+1)^(1/2))+b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)*c^4/(8*c^2*x^2-8)*arcsin(c*x)*ln(1-I*c*x
-(-c^2*x^2+1)^(1/2))+I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)*c^4/(8*c^2*x^2-8)*polylog(2,-I*c*x-(-c^2*x^
2+1)^(1/2))-I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)*c^4/(8*c^2*x^2-8)*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2)
)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))/x^5,x, algorithm="maxima")

[Out]

b*sqrt(d)*integrate(sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/x^5, x) + 1/8*(c^4
*sqrt(d)*log(2*sqrt(-c^2*d*x^2 + d)*sqrt(d)/abs(x) + 2*d/abs(x)) - sqrt(-c^2*d*x^2 + d)*c^4 - (-c^2*d*x^2 + d)
^(3/2)*c^2/(d*x^2) - 2*(-c^2*d*x^2 + d)^(3/2)/(d*x^4))*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((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))/x^5,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {- d \left (c x - 1\right ) \left (c x + 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{x^{5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))/x^5,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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