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

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

[Out]

-(b*c*Sqrt[d - c^2*d*x^2])/(12*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*Sq
rt[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]*PolyLog
[2, E^(I*ArcSin[c*x])])/Sqrt[1 - c^2*x^2]

________________________________________________________________________________________

Rubi [A]  time = 0.294422, antiderivative size = 301, normalized size of antiderivative = 1., 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 = {4693, 30, 4701, 4709, 4183, 2279, 2391} \[ -\frac{i b c^4 \sqrt{d-c^2 d x^2} \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )}{8 \sqrt{1-c^2 x^2}}+\frac{i b c^4 \sqrt{d-c^2 d x^2} \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )}{8 \sqrt{1-c^2 x^2}}+\frac{c^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 x^2}-\frac{\sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{4 x^4}+\frac{c^4 \sqrt{d-c^2 d x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 \sqrt{1-c^2 x^2}}+\frac{b c^3 \sqrt{d-c^2 d x^2}}{8 x \sqrt{1-c^2 x^2}}-\frac{b c \sqrt{d-c^2 d x^2}}{12 x^3 \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*Sqrt[d - c^2*d*x^2])/(12*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*Sq
rt[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]*PolyLog
[2, E^(I*ArcSin[c*x])])/Sqrt[1 - c^2*x^2]

Rule 4693

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*Sqrt[d + e*x^2])/(f*(m + 1
)*Sqrt[1 - c^2*x^2]), Int[(f*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1), x], x] + Dist[(c^2*Sqrt[d + e*x^2])/(f^2*
(m + 1)*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 30

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

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]

Rule 4709

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

Rule 4183

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 2279

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 2391

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]

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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 3.93246, size = 321, normalized size = 1.07 \[ \frac{b c^4 \sqrt{d-c^2 d x^2} \left (-24 i \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )+24 i \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )-\frac{16 \sin ^4\left (\frac{1}{2} \sin ^{-1}(c x)\right )}{c^3 x^3}-24 \sin ^{-1}(c x) \log \left (1-e^{i \sin ^{-1}(c x)}\right )+24 \sin ^{-1}(c x) \log \left (1+e^{i \sin ^{-1}(c x)}\right )+8 \tan \left (\frac{1}{2} \sin ^{-1}(c x)\right )+8 \cot \left (\frac{1}{2} \sin ^{-1}(c x)\right )-c x \csc ^4\left (\frac{1}{2} \sin ^{-1}(c x)\right )-3 \sin ^{-1}(c x) \csc ^4\left (\frac{1}{2} \sin ^{-1}(c x)\right )+6 \sin ^{-1}(c x) \csc ^2\left (\frac{1}{2} \sin ^{-1}(c x)\right )+3 \sin ^{-1}(c x) \sec ^4\left (\frac{1}{2} \sin ^{-1}(c x)\right )-6 \sin ^{-1}(c x) \sec ^2\left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )}{192 \sqrt{1-c^2 x^2}}+\frac{a \left (c^2 x^2-2\right ) \sqrt{d-c^2 d x^2}}{8 x^4}+\frac{1}{8} a c^4 \sqrt{d} \log \left (\sqrt{d} \sqrt{d-c^2 d x^2}+d\right )-\frac{1}{8} a c^4 \sqrt{d} \log (x) \]

Warning: Unable to verify antiderivative.

[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 = 0.293, size = 571, normalized size = 1.9 \begin{align*} -{\frac{a}{4\,d{x}^{4}} \left ( -{c}^{2}d{x}^{2}+d \right ) ^{{\frac{3}{2}}}}-{\frac{a{c}^{2}}{8\,d{x}^{2}} \left ( -{c}^{2}d{x}^{2}+d \right ) ^{{\frac{3}{2}}}}+{\frac{a{c}^{4}}{8}\sqrt{d}\ln \left ({\frac{1}{x} \left ( 2\,d+2\,\sqrt{d}\sqrt{-{c}^{2}d{x}^{2}+d} \right ) } \right ) }-{\frac{a{c}^{4}}{8}\sqrt{-{c}^{2}d{x}^{2}+d}}+{\frac{b\arcsin \left ( cx \right ){c}^{4}}{8\,{c}^{2}{x}^{2}-8}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}-{\frac{b{c}^{3}}{ \left ( 8\,{c}^{2}{x}^{2}-8 \right ) x}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{3\,b\arcsin \left ( cx \right ){c}^{2}}{ \left ( 8\,{c}^{2}{x}^{2}-8 \right ){x}^{2}}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{bc}{ \left ( 12\,{c}^{2}{x}^{2}-12 \right ){x}^{3}}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{b\arcsin \left ( cx \right ) }{ \left ( 4\,{c}^{2}{x}^{2}-4 \right ){x}^{4}}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}-{\frac{b\arcsin \left ( cx \right ){c}^{4}}{8\,{c}^{2}{x}^{2}-8}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}\ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) }+{\frac{b\arcsin \left ( cx \right ){c}^{4}}{8\,{c}^{2}{x}^{2}-8}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}\ln \left ( 1-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) }+{\frac{ib{c}^{4}}{8\,{c}^{2}{x}^{2}-8}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}{\it polylog} \left ( 2,-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) }-{\frac{ib{c}^{4}}{8\,{c}^{2}{x}^{2}-8}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}{\it polylog} \left ( 2,icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) } \end{align*}

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)

[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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-c^{2} d x^{2} + d}{\left (b \arcsin \left (c x\right ) + a\right )}}{x^{5}}, x\right ) \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \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{align*}

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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-c^{2} d x^{2} + d}{\left (b \arcsin \left (c x\right ) + a\right )}}{x^{5}}\,{d x} \end{align*}

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]

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