3.6.79 \(\int \frac {\sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2}{x} \, dx\) [579]

3.6.79.1 Optimal result

Integrand size = 35, antiderivative size = 432 \[ \int \frac {\sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2}{x} \, dx=-2 b^2 \sqrt {d+c d x} \sqrt {e-c e x}-\frac {2 a b c x \sqrt {d+c d x} \sqrt {e-c e x}}{\sqrt {1-c^2 x^2}}-\frac {2 b^2 c x \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x)}{\sqrt {1-c^2 x^2}}+\sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2-\frac {2 \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2 \text {arctanh}\left (e^{i \arcsin (c x)}\right )}{\sqrt {1-c^2 x^2}}+\frac {2 i b \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x)) \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right )}{\sqrt {1-c^2 x^2}}-\frac {2 i b \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x)) \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right )}{\sqrt {1-c^2 x^2}}-\frac {2 b^2 \sqrt {d+c d x} \sqrt {e-c e x} \operatorname {PolyLog}\left (3,-e^{i \arcsin (c x)}\right )}{\sqrt {1-c^2 x^2}}+\frac {2 b^2 \sqrt {d+c d x} \sqrt {e-c e x} \operatorname {PolyLog}\left (3,e^{i \arcsin (c x)}\right )}{\sqrt {1-c^2 x^2}} \]

output

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

3.6.79.2 Mathematica [A] (verified)

Time = 2.25 (sec) , antiderivative size = 434, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2}{x} \, dx=a^2 \sqrt {d+c d x} \sqrt {e-c e x}+a^2 \sqrt {d} \sqrt {e} \log (c x)-a^2 \sqrt {d} \sqrt {e} \log \left (d e+\sqrt {d} \sqrt {e} \sqrt {d+c d x} \sqrt {e-c e x}\right )-\frac {2 a b \sqrt {d+c d x} \sqrt {e-c e x} \left (c x-\sqrt {1-c^2 x^2} \arcsin (c x)-\arcsin (c x) \log \left (1-e^{i \arcsin (c x)}\right )+\arcsin (c x) \log \left (1+e^{i \arcsin (c x)}\right )-i \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right )+i \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right )\right )}{\sqrt {1-c^2 x^2}}-\frac {b^2 \sqrt {d+c d x} \sqrt {e-c e x} \left (2 \sqrt {1-c^2 x^2}+2 c x \arcsin (c x)-\sqrt {1-c^2 x^2} \arcsin (c x)^2-\arcsin (c x)^2 \log \left (1-e^{i \arcsin (c x)}\right )+\arcsin (c x)^2 \log \left (1+e^{i \arcsin (c x)}\right )-2 i \arcsin (c x) \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right )+2 i \arcsin (c x) \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right )+2 \operatorname {PolyLog}\left (3,-e^{i \arcsin (c x)}\right )-2 \operatorname {PolyLog}\left (3,e^{i \arcsin (c x)}\right )\right )}{\sqrt {1-c^2 x^2}} \]

input

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

output

a^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x] + a^2*Sqrt[d]*Sqrt[e]*Log[c*x] - a^2*S 
qrt[d]*Sqrt[e]*Log[d*e + Sqrt[d]*Sqrt[e]*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]] 
- (2*a*b*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(c*x - Sqrt[1 - c^2*x^2]*ArcSin[c 
*x] - ArcSin[c*x]*Log[1 - E^(I*ArcSin[c*x])] + ArcSin[c*x]*Log[1 + E^(I*Ar 
cSin[c*x])] - I*PolyLog[2, -E^(I*ArcSin[c*x])] + I*PolyLog[2, E^(I*ArcSin[ 
c*x])]))/Sqrt[1 - c^2*x^2] - (b^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(2*Sqrt[ 
1 - c^2*x^2] + 2*c*x*ArcSin[c*x] - Sqrt[1 - c^2*x^2]*ArcSin[c*x]^2 - ArcSi 
n[c*x]^2*Log[1 - E^(I*ArcSin[c*x])] + ArcSin[c*x]^2*Log[1 + E^(I*ArcSin[c* 
x])] - (2*I)*ArcSin[c*x]*PolyLog[2, -E^(I*ArcSin[c*x])] + (2*I)*ArcSin[c*x 
]*PolyLog[2, E^(I*ArcSin[c*x])] + 2*PolyLog[3, -E^(I*ArcSin[c*x])] - 2*Pol 
yLog[3, E^(I*ArcSin[c*x])]))/Sqrt[1 - c^2*x^2]
 

3.6.79.3 Rubi [A] (verified)

Time = 1.31 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.48, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {5238, 5198, 2009, 5218, 3042, 4671, 3011, 2720, 7143}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sqrt {c d x+d} \sqrt {e-c e x} (a+b \arcsin (c x))^2}{x} \, dx\)

\(\Big \downarrow \) 5238

\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \int \frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{x}dx}{\sqrt {1-c^2 x^2}}\)

\(\Big \downarrow \) 5198

\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (\int \frac {(a+b \arcsin (c x))^2}{x \sqrt {1-c^2 x^2}}dx-2 b c \int (a+b \arcsin (c x))dx+\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2\right )}{\sqrt {1-c^2 x^2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (\int \frac {(a+b \arcsin (c x))^2}{x \sqrt {1-c^2 x^2}}dx+\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2-2 b c \left (a x+b x \arcsin (c x)+\frac {b \sqrt {1-c^2 x^2}}{c}\right )\right )}{\sqrt {1-c^2 x^2}}\)

\(\Big \downarrow \) 5218

\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (\int \frac {(a+b \arcsin (c x))^2}{c x}d\arcsin (c x)+\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2-2 b c \left (a x+b x \arcsin (c x)+\frac {b \sqrt {1-c^2 x^2}}{c}\right )\right )}{\sqrt {1-c^2 x^2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (\int (a+b \arcsin (c x))^2 \csc (\arcsin (c x))d\arcsin (c x)+\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2-2 b c \left (a x+b x \arcsin (c x)+\frac {b \sqrt {1-c^2 x^2}}{c}\right )\right )}{\sqrt {1-c^2 x^2}}\)

\(\Big \downarrow \) 4671

\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (-2 b \int (a+b \arcsin (c x)) \log \left (1-e^{i \arcsin (c x)}\right )d\arcsin (c x)+2 b \int (a+b \arcsin (c x)) \log \left (1+e^{i \arcsin (c x)}\right )d\arcsin (c x)-2 \text {arctanh}\left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))^2+\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2-2 b c \left (a x+b x \arcsin (c x)+\frac {b \sqrt {1-c^2 x^2}}{c}\right )\right )}{\sqrt {1-c^2 x^2}}\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (2 b \left (i \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))-i b \int \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right )d\arcsin (c x)\right )-2 b \left (i \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))-i b \int \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right )d\arcsin (c x)\right )-2 \text {arctanh}\left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))^2+\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2-2 b c \left (a x+b x \arcsin (c x)+\frac {b \sqrt {1-c^2 x^2}}{c}\right )\right )}{\sqrt {1-c^2 x^2}}\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (2 b \left (i \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))-b \int e^{-i \arcsin (c x)} \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right )de^{i \arcsin (c x)}\right )-2 b \left (i \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))-b \int e^{-i \arcsin (c x)} \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right )de^{i \arcsin (c x)}\right )-2 \text {arctanh}\left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))^2+\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2-2 b c \left (a x+b x \arcsin (c x)+\frac {b \sqrt {1-c^2 x^2}}{c}\right )\right )}{\sqrt {1-c^2 x^2}}\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (-2 \text {arctanh}\left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))^2+\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2-2 b c \left (a x+b x \arcsin (c x)+\frac {b \sqrt {1-c^2 x^2}}{c}\right )+2 b \left (i \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))-b \operatorname {PolyLog}\left (3,-e^{i \arcsin (c x)}\right )\right )-2 b \left (i \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))-b \operatorname {PolyLog}\left (3,e^{i \arcsin (c x)}\right )\right )\right )}{\sqrt {1-c^2 x^2}}\)

input

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

output

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

3.6.79.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4671

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] + (-Simp[d*(m/f)   Int[(c + 
d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Simp[d*(m/f)   Int[(c + d*x 
)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IG 
tQ[m, 0]
 

rule 5198

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*ArcS 
in[c*x])^n/(f*(m + 2))), x] + (Simp[(1/(m + 2))*Simp[Sqrt[d + e*x^2]/Sqrt[1 
 - c^2*x^2]]   Int[(f*x)^m*((a + b*ArcSin[c*x])^n/Sqrt[1 - c^2*x^2]), x], x 
] - Simp[b*c*(n/(f*(m + 2)))*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]) /; FreeQ[{a, b, c, d, e, 
 f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && (IGtQ[m, -2] || EqQ[n, 1])
 

rule 5218

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

rule 5238

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((h_.)*(x_))^(m_.)*((d_) + (e_ 
.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(q_), x_Symbol] :> Simp[((-d^2)*(g/e))^In 
tPart[q]*(d + e*x)^FracPart[q]*((f + g*x)^FracPart[q]/(1 - c^2*x^2)^FracPar 
t[q])   Int[(h*x)^m*(d + e*x)^(p - q)*(1 - c^2*x^2)^q*(a + b*ArcSin[c*x])^n 
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && EqQ[e*f + d*g, 0] & 
& EqQ[c^2*d^2 - e^2, 0] && HalfIntegerQ[p, q] && GeQ[p - q, 0]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

3.6.79.4 Maple [F]

\[\int \frac {\sqrt {c d x +d}\, \sqrt {-c e x +e}\, \left (a +b \arcsin \left (c x \right )\right )^{2}}{x}d x\]

input

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

output

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

3.6.79.5 Fricas [F]

\[ \int \frac {\sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2}{x} \, dx=\int { \frac {\sqrt {c d x + d} \sqrt {-c e x + e} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{x} \,d x } \]

input

integrate((c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)*(a+b*arcsin(c*x))^2/x,x, algori 
thm="fricas")
 

output

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

3.6.79.6 Sympy [F]

\[ \int \frac {\sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2}{x} \, dx=\int \frac {\sqrt {d \left (c x + 1\right )} \sqrt {- e \left (c x - 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{x}\, dx \]

input

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

output

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

3.6.79.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2}{x} \, dx=\text {Exception raised: ValueError} \]

input

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

output

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(e>0)', see `assume?` for more de 
tails)Is e
 

3.6.79.8 Giac [F]

\[ \int \frac {\sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2}{x} \, dx=\int { \frac {\sqrt {c d x + d} \sqrt {-c e x + e} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{x} \,d x } \]

input

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

output

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

3.6.79.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2}{x} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,\sqrt {d+c\,d\,x}\,\sqrt {e-c\,e\,x}}{x} \,d x \]

input

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

output

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