\(\int \frac {a+b \arcsin (c x)}{(d+e x)^{3/2} \sqrt {f+g x}} \, dx\) [99]

Optimal result

Integrand size = 27, antiderivative size = 258 \[ \int \frac {a+b \arcsin (c x)}{(d+e x)^{3/2} \sqrt {f+g x}} \, dx=-\frac {2 \sqrt {f+g x} (a+b \arcsin (c x))}{(e f-d g) \sqrt {d+e x}}+\frac {4 b c \sqrt {-c (c d+e)} \sqrt {\frac {(e f-d g) (1-c x)}{(c d+e) (f+g x)}} \sqrt {-\frac {(e f-d g) (1+c x)}{(c d-e) (f+g x)}} (f+g x) \operatorname {EllipticPi}\left (\frac {(c d+e) g}{e (c f+g)},\arcsin \left (\frac {\sqrt {-c (c f+g)} \sqrt {d+e x}}{\sqrt {-c (c d+e)} \sqrt {f+g x}}\right ),\frac {(c d+e) (c f-g)}{(c d-e) (c f+g)}\right )}{e \sqrt {-c (c f+g)} (e f-d g) \sqrt {1-c^2 x^2}} \] Output:

-2*(g*x+f)^(1/2)*(a+b*arcsin(c*x))/(-d*g+e*f)/(e*x+d)^(1/2)+4*b*c*(-c*(c*d 
+e))^(1/2)*((-d*g+e*f)*(-c*x+1)/(c*d+e)/(g*x+f))^(1/2)*(-(-d*g+e*f)*(c*x+1 
)/(c*d-e)/(g*x+f))^(1/2)*(g*x+f)*EllipticPi((-c*(c*f+g))^(1/2)*(e*x+d)^(1/ 
2)/(-c*(c*d+e))^(1/2)/(g*x+f)^(1/2),(c*d+e)*g/e/(c*f+g),((c*d+e)*(c*f-g)/( 
c*d-e)/(c*f+g))^(1/2))/e/(-c*(c*f+g))^(1/2)/(-d*g+e*f)/(-c^2*x^2+1)^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 5.39 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.53 \[ \int \frac {a+b \arcsin (c x)}{(d+e x)^{3/2} \sqrt {f+g x}} \, dx=\frac {2 \sqrt {f+g x} \left (a+b \arcsin (c x)-\frac {2 b c (d+e x) \left (-\frac {(c d-e) e (-e f+d g) (-1+c x) \sqrt {\frac {(c d+e) (1+c x)}{c (d+e x)}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(-c d+e) (-1+c x)}{c (d+e x)}}}{\sqrt {2}}\right ),\frac {2 c (-e f+d g)}{(c d-e) (c f+g)}\right )}{d+e x}-e (c d+e) g \sqrt {\frac {(-c d+e) (-1+c x)}{c (d+e x)}} \sqrt {\frac {\left (-c^2 d^2+e^2\right ) \left (-1+c^2 x^2\right )}{c^2 (d+e x)^2}} \operatorname {EllipticPi}\left (-\frac {2 e}{c d-e},\arcsin \left (\frac {\sqrt {\frac {(-c d+e) (-1+c x)}{c (d+e x)}}}{\sqrt {2}}\right ),\frac {2 c (-e f+d g)}{(c d-e) (c f+g)}\right )\right )}{(c d-e) e^2 (c f+g) \sqrt {\frac {(-c d+e) (-1+c x)}{c (d+e x)}} \sqrt {\frac {(c d+e) (f+g x)}{(c f+g) (d+e x)}} \sqrt {1-c^2 x^2}}\right )}{(-e f+d g) \sqrt {d+e x}} \] Input:

Integrate[(a + b*ArcSin[c*x])/((d + e*x)^(3/2)*Sqrt[f + g*x]),x]
 

Output:

(2*Sqrt[f + g*x]*(a + b*ArcSin[c*x] - (2*b*c*(d + e*x)*(-(((c*d - e)*e*(-( 
e*f) + d*g)*(-1 + c*x)*Sqrt[((c*d + e)*(1 + c*x))/(c*(d + e*x))]*EllipticF 
[ArcSin[Sqrt[((-(c*d) + e)*(-1 + c*x))/(c*(d + e*x))]/Sqrt[2]], (2*c*(-(e* 
f) + d*g))/((c*d - e)*(c*f + g))])/(d + e*x)) - e*(c*d + e)*g*Sqrt[((-(c*d 
) + e)*(-1 + c*x))/(c*(d + e*x))]*Sqrt[((-(c^2*d^2) + e^2)*(-1 + c^2*x^2)) 
/(c^2*(d + e*x)^2)]*EllipticPi[(-2*e)/(c*d - e), ArcSin[Sqrt[((-(c*d) + e) 
*(-1 + c*x))/(c*(d + e*x))]/Sqrt[2]], (2*c*(-(e*f) + d*g))/((c*d - e)*(c*f 
 + g))]))/((c*d - e)*e^2*(c*f + g)*Sqrt[((-(c*d) + e)*(-1 + c*x))/(c*(d + 
e*x))]*Sqrt[((c*d + e)*(f + g*x))/((c*f + g)*(d + e*x))]*Sqrt[1 - c^2*x^2] 
)))/((-(e*f) + d*g)*Sqrt[d + e*x])
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {5286, 27, 726}

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.

\(\displaystyle \int \frac {a+b \arcsin (c x)}{(d+e x)^{3/2} \sqrt {f+g x}} \, dx\)

\(\Big \downarrow \) 5286

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 726

\(\displaystyle \frac {4 b c \sqrt {-c (c d+e)} (f+g x) \sqrt {\frac {(1-c x) (e f-d g)}{(c d+e) (f+g x)}} \sqrt {-\frac {(c x+1) (e f-d g)}{(c d-e) (f+g x)}} \operatorname {EllipticPi}\left (\frac {(c d+e) g}{e (c f+g)},\arcsin \left (\frac {\sqrt {-c (c f+g)} \sqrt {d+e x}}{\sqrt {-c (c d+e)} \sqrt {f+g x}}\right ),\frac {(c d+e) (c f-g)}{(c d-e) (c f+g)}\right )}{e \sqrt {1-c^2 x^2} \sqrt {-c (c f+g)} (e f-d g)}-\frac {2 \sqrt {f+g x} (a+b \arcsin (c x))}{\sqrt {d+e x} (e f-d g)}\)

Input:

Int[(a + b*ArcSin[c*x])/((d + e*x)^(3/2)*Sqrt[f + g*x]),x]
 

Output:

(-2*Sqrt[f + g*x]*(a + b*ArcSin[c*x]))/((e*f - d*g)*Sqrt[d + e*x]) + (4*b* 
c*Sqrt[-(c*(c*d + e))]*Sqrt[((e*f - d*g)*(1 - c*x))/((c*d + e)*(f + g*x))] 
*Sqrt[-(((e*f - d*g)*(1 + c*x))/((c*d - e)*(f + g*x)))]*(f + g*x)*Elliptic 
Pi[((c*d + e)*g)/(e*(c*f + g)), ArcSin[(Sqrt[-(c*(c*f + g))]*Sqrt[d + e*x] 
)/(Sqrt[-(c*(c*d + e))]*Sqrt[f + g*x])], ((c*d + e)*(c*f - g))/((c*d - e)* 
(c*f + g))])/(e*Sqrt[-(c*(c*f + g))]*(e*f - d*g)*Sqrt[1 - c^2*x^2])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 726

Int[Sqrt[(d_.) + (e_.)*(x_)]/(Sqrt[(f_.) + (g_.)*(x_)]*Sqrt[(a_) + (c_.)*(x 
_)^2]), x_Symbol] :> With[{q = Rt[-4*a*c, 2]}, Simp[Sqrt[2]*Sqrt[2*c*f - g* 
q]*Sqrt[-q + 2*c*x]*(d + e*x)*Sqrt[(e*f - d*g)*((q + 2*c*x)/((2*c*f - g*q)* 
(d + e*x)))]*(Sqrt[(e*f - d*g)*((2*a + q*x)/((q*f - 2*a*g)*(d + e*x)))]/(g* 
Sqrt[2*c*d - e*q]*Sqrt[2*a*(c/q) + c*x]*Sqrt[a + c*x^2]))*EllipticPi[e*((2* 
c*f - g*q)/(g*(2*c*d - e*q))), ArcSin[Sqrt[2*c*d - e*q]*(Sqrt[f + g*x]/(Sqr 
t[2*c*f - g*q]*Sqrt[d + e*x]))], (q*d - 2*a*e)*((2*c*f - g*q)/((q*f - 2*a*g 
)*(2*c*d - e*q)))], x]] /; FreeQ[{a, c, d, e, f, g}, x]
 

rule 5286

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*(u_), x_Symbol] :> With[{v = IntHide 
[u, x]}, Simp[(a + b*ArcSin[c*x])   v, x] - Simp[b*c   Int[SimplifyIntegran 
d[v/Sqrt[1 - c^2*x^2], x], x], x] /; InverseFunctionFreeQ[v, x]] /; FreeQ[{ 
a, b, c}, x]
 

Maple [F]

\[\int \frac {a +b \arcsin \left (c x \right )}{\left (e x +d \right )^{\frac {3}{2}} \sqrt {g x +f}}d x\]

Input:

int((a+b*arcsin(c*x))/(e*x+d)^(3/2)/(g*x+f)^(1/2),x)
 

Output:

int((a+b*arcsin(c*x))/(e*x+d)^(3/2)/(g*x+f)^(1/2),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {a+b \arcsin (c x)}{(d+e x)^{3/2} \sqrt {f+g x}} \, dx=\text {Timed out} \] Input:

integrate((a+b*arcsin(c*x))/(e*x+d)^(3/2)/(g*x+f)^(1/2),x, algorithm="fric 
as")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {a+b \arcsin (c x)}{(d+e x)^{3/2} \sqrt {f+g x}} \, dx=\int \frac {a + b \operatorname {asin}{\left (c x \right )}}{\left (d + e x\right )^{\frac {3}{2}} \sqrt {f + g x}}\, dx \] Input:

integrate((a+b*asin(c*x))/(e*x+d)**(3/2)/(g*x+f)**(1/2),x)
 

Output:

Integral((a + b*asin(c*x))/((d + e*x)**(3/2)*sqrt(f + g*x)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b \arcsin (c x)}{(d+e x)^{3/2} \sqrt {f+g x}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((a+b*arcsin(c*x))/(e*x+d)^(3/2)/(g*x+f)^(1/2),x, algorithm="maxi 
ma")
 

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*(d*g-e*f)>0)', see `assume?` f 
or more de
 

Giac [F]

\[ \int \frac {a+b \arcsin (c x)}{(d+e x)^{3/2} \sqrt {f+g x}} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{{\left (e x + d\right )}^{\frac {3}{2}} \sqrt {g x + f}} \,d x } \] Input:

integrate((a+b*arcsin(c*x))/(e*x+d)^(3/2)/(g*x+f)^(1/2),x, algorithm="giac 
")
 

Output:

integrate((b*arcsin(c*x) + a)/((e*x + d)^(3/2)*sqrt(g*x + f)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \arcsin (c x)}{(d+e x)^{3/2} \sqrt {f+g x}} \, dx=\int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{\sqrt {f+g\,x}\,{\left (d+e\,x\right )}^{3/2}} \,d x \] Input:

int((a + b*asin(c*x))/((f + g*x)^(1/2)*(d + e*x)^(3/2)),x)
 

Output:

int((a + b*asin(c*x))/((f + g*x)^(1/2)*(d + e*x)^(3/2)), x)
 

Reduce [F]

\[ \int \frac {a+b \arcsin (c x)}{(d+e x)^{3/2} \sqrt {f+g x}} \, dx=\frac {2 \sqrt {g}\, \sqrt {e}\, \sqrt {e x +d}\, a +\sqrt {e x +d}\, \left (\int \frac {\mathit {asin} \left (c x \right )}{\sqrt {g x +f}\, \sqrt {e x +d}\, d +\sqrt {g x +f}\, \sqrt {e x +d}\, e x}d x \right ) b d e g -\sqrt {e x +d}\, \left (\int \frac {\mathit {asin} \left (c x \right )}{\sqrt {g x +f}\, \sqrt {e x +d}\, d +\sqrt {g x +f}\, \sqrt {e x +d}\, e x}d x \right ) b \,e^{2} f +2 \sqrt {g x +f}\, a e}{\sqrt {e x +d}\, e \left (d g -e f \right )} \] Input:

int((a+b*asin(c*x))/(e*x+d)^(3/2)/(g*x+f)^(1/2),x)
 

Output:

(2*sqrt(g)*sqrt(e)*sqrt(d + e*x)*a + sqrt(d + e*x)*int(asin(c*x)/(sqrt(f + 
 g*x)*sqrt(d + e*x)*d + sqrt(f + g*x)*sqrt(d + e*x)*e*x),x)*b*d*e*g - sqrt 
(d + e*x)*int(asin(c*x)/(sqrt(f + g*x)*sqrt(d + e*x)*d + sqrt(f + g*x)*sqr 
t(d + e*x)*e*x),x)*b*e**2*f + 2*sqrt(f + g*x)*a*e)/(sqrt(d + e*x)*e*(d*g - 
 e*f))