3.2.36 \(\int \frac {a+b \text {ArcSin}(c x)}{x (d-c^2 d x^2)^{5/2}} \, dx\) [136]

Optimal. Leaf size=291 \[ -\frac {b c x}{6 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {a+b \text {ArcSin}(c x)}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {a+b \text {ArcSin}(c x)}{d^2 \sqrt {d-c^2 d x^2}}-\frac {2 \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x)) \tanh ^{-1}\left (e^{i \text {ArcSin}(c x)}\right )}{d^2 \sqrt {d-c^2 d x^2}}-\frac {7 b \sqrt {1-c^2 x^2} \tanh ^{-1}(c x)}{6 d^2 \sqrt {d-c^2 d x^2}}+\frac {i b \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,-e^{i \text {ArcSin}(c x)}\right )}{d^2 \sqrt {d-c^2 d x^2}}-\frac {i b \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,e^{i \text {ArcSin}(c x)}\right )}{d^2 \sqrt {d-c^2 d x^2}} \]

[Out]

1/3*(a+b*arcsin(c*x))/d/(-c^2*d*x^2+d)^(3/2)+(a+b*arcsin(c*x))/d^2/(-c^2*d*x^2+d)^(1/2)-1/6*b*c*x/d^2/(-c^2*x^
2+1)^(1/2)/(-c^2*d*x^2+d)^(1/2)-2*(a+b*arcsin(c*x))*arctanh(I*c*x+(-c^2*x^2+1)^(1/2))*(-c^2*x^2+1)^(1/2)/d^2/(
-c^2*d*x^2+d)^(1/2)-7/6*b*arctanh(c*x)*(-c^2*x^2+1)^(1/2)/d^2/(-c^2*d*x^2+d)^(1/2)+I*b*polylog(2,-I*c*x-(-c^2*
x^2+1)^(1/2))*(-c^2*x^2+1)^(1/2)/d^2/(-c^2*d*x^2+d)^(1/2)-I*b*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2))*(-c^2*x^2+1)
^(1/2)/d^2/(-c^2*d*x^2+d)^(1/2)

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Rubi [A]
time = 0.28, antiderivative size = 291, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {4793, 4803, 4268, 2317, 2438, 212, 205} \begin {gather*} \frac {a+b \text {ArcSin}(c x)}{d^2 \sqrt {d-c^2 d x^2}}-\frac {2 \sqrt {1-c^2 x^2} \tanh ^{-1}\left (e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{d^2 \sqrt {d-c^2 d x^2}}+\frac {a+b \text {ArcSin}(c x)}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {i b \sqrt {1-c^2 x^2} \text {Li}_2\left (-e^{i \text {ArcSin}(c x)}\right )}{d^2 \sqrt {d-c^2 d x^2}}-\frac {i b \sqrt {1-c^2 x^2} \text {Li}_2\left (e^{i \text {ArcSin}(c x)}\right )}{d^2 \sqrt {d-c^2 d x^2}}-\frac {b c x}{6 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}-\frac {7 b \sqrt {1-c^2 x^2} \tanh ^{-1}(c x)}{6 d^2 \sqrt {d-c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(b*c*x)/(d^2*Sqrt[1 - c^2*x^2]*Sqrt[d - c^2*d*x^2]) + (a + b*ArcSin[c*x])/(3*d*(d - c^2*d*x^2)^(3/2)) + (
a + b*ArcSin[c*x])/(d^2*Sqrt[d - c^2*d*x^2]) - (2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])*ArcTanh[E^(I*ArcSin[c*
x])])/(d^2*Sqrt[d - c^2*d*x^2]) - (7*b*Sqrt[1 - c^2*x^2]*ArcTanh[c*x])/(6*d^2*Sqrt[d - c^2*d*x^2]) + (I*b*Sqrt
[1 - c^2*x^2]*PolyLog[2, -E^(I*ArcSin[c*x])])/(d^2*Sqrt[d - c^2*d*x^2]) - (I*b*Sqrt[1 - c^2*x^2]*PolyLog[2, E^
(I*ArcSin[c*x])])/(d^2*Sqrt[d - c^2*d*x^2])

Rule 205

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}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 4793

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/(2*d*f*(p + 1))), x] + (Dist[(m + 2*p + 3)/(2*d*(p
+ 1)), Int[(f*x)^m*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n, x], x] + Dist[b*c*(n/(2*f*(p + 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]) /; Fre
eQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] &&  !GtQ[m, 1] && (IntegerQ[m] ||
IntegerQ[p] || EqQ[n, 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 {a+b \sin ^{-1}(c x)}{x \left (d-c^2 d x^2\right )^{5/2}} \, dx &=\frac {a+b \sin ^{-1}(c x)}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {\int \frac {a+b \sin ^{-1}(c x)}{x \left (d-c^2 d x^2\right )^{3/2}} \, dx}{d}-\frac {\left (b c \sqrt {1-c^2 x^2}\right ) \int \frac {1}{\left (1-c^2 x^2\right )^2} \, dx}{3 d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {b c x}{6 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {a+b \sin ^{-1}(c x)}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {a+b \sin ^{-1}(c x)}{d^2 \sqrt {d-c^2 d x^2}}+\frac {\int \frac {a+b \sin ^{-1}(c x)}{x \sqrt {d-c^2 d x^2}} \, dx}{d^2}-\frac {\left (b c \sqrt {1-c^2 x^2}\right ) \int \frac {1}{1-c^2 x^2} \, dx}{6 d^2 \sqrt {d-c^2 d x^2}}-\frac {\left (b c \sqrt {1-c^2 x^2}\right ) \int \frac {1}{1-c^2 x^2} \, dx}{d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {b c x}{6 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {a+b \sin ^{-1}(c x)}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {a+b \sin ^{-1}(c x)}{d^2 \sqrt {d-c^2 d x^2}}-\frac {7 b \sqrt {1-c^2 x^2} \tanh ^{-1}(c x)}{6 d^2 \sqrt {d-c^2 d x^2}}+\frac {\sqrt {1-c^2 x^2} \int \frac {a+b \sin ^{-1}(c x)}{x \sqrt {1-c^2 x^2}} \, dx}{d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {b c x}{6 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {a+b \sin ^{-1}(c x)}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {a+b \sin ^{-1}(c x)}{d^2 \sqrt {d-c^2 d x^2}}-\frac {7 b \sqrt {1-c^2 x^2} \tanh ^{-1}(c x)}{6 d^2 \sqrt {d-c^2 d x^2}}+\frac {\sqrt {1-c^2 x^2} \text {Subst}\left (\int (a+b x) \csc (x) \, dx,x,\sin ^{-1}(c x)\right )}{d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {b c x}{6 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {a+b \sin ^{-1}(c x)}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {a+b \sin ^{-1}(c x)}{d^2 \sqrt {d-c^2 d x^2}}-\frac {2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^2 \sqrt {d-c^2 d x^2}}-\frac {7 b \sqrt {1-c^2 x^2} \tanh ^{-1}(c x)}{6 d^2 \sqrt {d-c^2 d x^2}}-\frac {\left (b \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (b \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {b c x}{6 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {a+b \sin ^{-1}(c x)}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {a+b \sin ^{-1}(c x)}{d^2 \sqrt {d-c^2 d x^2}}-\frac {2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^2 \sqrt {d-c^2 d x^2}}-\frac {7 b \sqrt {1-c^2 x^2} \tanh ^{-1}(c x)}{6 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (i b \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{d^2 \sqrt {d-c^2 d x^2}}-\frac {\left (i b \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {b c x}{6 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {a+b \sin ^{-1}(c x)}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {a+b \sin ^{-1}(c x)}{d^2 \sqrt {d-c^2 d x^2}}-\frac {2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d^2 \sqrt {d-c^2 d x^2}}-\frac {7 b \sqrt {1-c^2 x^2} \tanh ^{-1}(c x)}{6 d^2 \sqrt {d-c^2 d x^2}}+\frac {i b \sqrt {1-c^2 x^2} \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{d^2 \sqrt {d-c^2 d x^2}}-\frac {i b \sqrt {1-c^2 x^2} \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{d^2 \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]
time = 1.16, size = 456, normalized size = 1.57 \begin {gather*} -\frac {a \left (-4+3 c^2 x^2\right ) \sqrt {d-c^2 d x^2}}{3 d^3 \left (-1+c^2 x^2\right )^2}+\frac {a \log (x)}{d^{5/2}}-\frac {a \log \left (d+\sqrt {d} \sqrt {d-c^2 d x^2}\right )}{d^{5/2}}+\frac {b \left (20 \text {ArcSin}(c x)+12 \text {ArcSin}(c x) \cos (2 \text {ArcSin}(c x))+18 \sqrt {1-c^2 x^2} \text {ArcSin}(c x) \log \left (1-e^{i \text {ArcSin}(c x)}\right )+6 \text {ArcSin}(c x) \cos (3 \text {ArcSin}(c x)) \log \left (1-e^{i \text {ArcSin}(c x)}\right )-18 \sqrt {1-c^2 x^2} \text {ArcSin}(c x) \log \left (1+e^{i \text {ArcSin}(c x)}\right )-6 \text {ArcSin}(c x) \cos (3 \text {ArcSin}(c x)) \log \left (1+e^{i \text {ArcSin}(c x)}\right )+21 \sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )-\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )+7 \cos (3 \text {ArcSin}(c x)) \log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )-\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )-21 \sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )+\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )-7 \cos (3 \text {ArcSin}(c x)) \log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )+\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )+24 i \left (1-c^2 x^2\right )^{3/2} \text {PolyLog}\left (2,-e^{i \text {ArcSin}(c x)}\right )-24 i \left (1-c^2 x^2\right )^{3/2} \text {PolyLog}\left (2,e^{i \text {ArcSin}(c x)}\right )-2 \sin (2 \text {ArcSin}(c x))\right )}{24 d \left (d-c^2 d x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(a*(-4 + 3*c^2*x^2)*Sqrt[d - c^2*d*x^2])/(d^3*(-1 + c^2*x^2)^2) + (a*Log[x])/d^(5/2) - (a*Log[d + Sqrt[d]
*Sqrt[d - c^2*d*x^2]])/d^(5/2) + (b*(20*ArcSin[c*x] + 12*ArcSin[c*x]*Cos[2*ArcSin[c*x]] + 18*Sqrt[1 - c^2*x^2]
*ArcSin[c*x]*Log[1 - E^(I*ArcSin[c*x])] + 6*ArcSin[c*x]*Cos[3*ArcSin[c*x]]*Log[1 - E^(I*ArcSin[c*x])] - 18*Sqr
t[1 - c^2*x^2]*ArcSin[c*x]*Log[1 + E^(I*ArcSin[c*x])] - 6*ArcSin[c*x]*Cos[3*ArcSin[c*x]]*Log[1 + E^(I*ArcSin[c
*x])] + 21*Sqrt[1 - c^2*x^2]*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]] + 7*Cos[3*ArcSin[c*x]]*Log[Cos[ArcSi
n[c*x]/2] - Sin[ArcSin[c*x]/2]] - 21*Sqrt[1 - c^2*x^2]*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]] - 7*Cos[3*
ArcSin[c*x]]*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]] + (24*I)*(1 - c^2*x^2)^(3/2)*PolyLog[2, -E^(I*ArcSin
[c*x])] - (24*I)*(1 - c^2*x^2)^(3/2)*PolyLog[2, E^(I*ArcSin[c*x])] - 2*Sin[2*ArcSin[c*x]]))/(24*d*(d - c^2*d*x
^2)^(3/2))

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Maple [A]
time = 0.16, size = 449, normalized size = 1.54

method	result	size
default	\(\frac {a}{3 d \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {a}{d^{2} \sqrt {-c^{2} d \,x^{2}+d}}-\frac {a \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {-c^{2} d \,x^{2}+d}}{x}\right )}{d^{\frac {5}{2}}}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x^{2} c^{2}}{d^{3} \left (c^{2} x^{2}-1\right )^{2}}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, x c}{6 d^{3} \left (c^{2} x^{2}-1\right )^{2}}+\frac {4 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )}{3 d^{3} \left (c^{2} x^{2}-1\right )^{2}}+\frac {b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{3} \left (c^{2} x^{2}-1\right )}-\frac {7 i b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arctan \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{3 d^{3} \left (c^{2} x^{2}-1\right )}-\frac {i b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \dilog \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{3} \left (c^{2} x^{2}-1\right )}-\frac {i b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \dilog \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{3} \left (c^{2} x^{2}-1\right )}\)	\(449\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a/d/(-c^2*d*x^2+d)^(3/2)+a/d^2/(-c^2*d*x^2+d)^(1/2)-a/d^(5/2)*ln((2*d+2*d^(1/2)*(-c^2*d*x^2+d)^(1/2))/x)-b
*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-1)^2*arcsin(c*x)*x^2*c^2-1/6*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-1)^2*(
-c^2*x^2+1)^(1/2)*x*c+4/3*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-1)^2*arcsin(c*x)+b*(-c^2*x^2+1)^(1/2)*(-d*(c^2
*x^2-1))^(1/2)/d^3/(c^2*x^2-1)*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-7/3*I*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*
x^2-1))^(1/2)/d^3/(c^2*x^2-1)*arctan(I*c*x+(-c^2*x^2+1)^(1/2))-I*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d
^3/(c^2*x^2-1)*dilog(1+I*c*x+(-c^2*x^2+1)^(1/2))-I*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-1)
*dilog(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((a+b*arcsin(c*x))/x/(-c^2*d*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

-1/3*a*(3*log(2*sqrt(-c^2*d*x^2 + d)*sqrt(d)/abs(x) + 2*d/abs(x))/d^(5/2) - 3/(sqrt(-c^2*d*x^2 + d)*d^2) - 1/(
(-c^2*d*x^2 + d)^(3/2)*d)) + b*integrate(arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/((c^4*d^2*x^5 - 2*c^2*d^2*
x^3 + d^2*x)*sqrt(c*x + 1)*sqrt(-c*x + 1)), x)/sqrt(d)

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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/(-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^7 - 3*c^4*d^3*x^5 + 3*c^2*d^3*x^3 - d^3*x), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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/(-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), 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\,{\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*(d - c^2*d*x^2)^(5/2)),x)

[Out]

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

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