\(\int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx\) [45]

Optimal result

Integrand size = 30, antiderivative size = 141 \[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=-\frac {b f x \sqrt {1-c^2 x^2}}{\sqrt {d+c d x} \sqrt {f-c f x}}+\frac {f \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{c \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {f \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{2 b c \sqrt {d+c d x} \sqrt {f-c f x}} \] Output:

-b*f*x*(-c^2*x^2+1)^(1/2)/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)+f*(-c^2*x^2+1)* 
(a+b*arcsin(c*x))/c/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)+1/2*f*(-c^2*x^2+1)^(1 
/2)*(a+b*arcsin(c*x))^2/b/c/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)
 

Mathematica [A] (verified)

Time = 1.38 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.42 \[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=\frac {\frac {2 \sqrt {d+c d x} \sqrt {f-c f x} \left (-b c x+a \sqrt {1-c^2 x^2}\right )}{\sqrt {1-c^2 x^2}}+2 b \sqrt {d+c d x} \sqrt {f-c f x} \arcsin (c x)+\frac {b \sqrt {d+c d x} \sqrt {f-c f x} \arcsin (c x)^2}{\sqrt {1-c^2 x^2}}-2 a \sqrt {d} \sqrt {f} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (-1+c^2 x^2\right )}\right )}{2 c d} \] Input:

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

Output:

((2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(-(b*c*x) + a*Sqrt[1 - c^2*x^2]))/Sqrt 
[1 - c^2*x^2] + 2*b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x] + (b*Sqrt[ 
d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]^2)/Sqrt[1 - c^2*x^2] - 2*a*Sqrt[d]* 
Sqrt[f]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(Sqrt[d]*Sqrt[f]*(-1 
+ c^2*x^2))])/(2*c*d)
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.62, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {5178, 27, 5262, 2009}

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.

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

\(\Big \downarrow \) 5178

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 5262

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(f*Sqrt[1 - c^2*x^2]*(-(b*x) + (Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/c + 
 (a + b*ArcSin[c*x])^2/(2*b*c)))/(Sqrt[d + c*d*x]*Sqrt[f - c*f*x])
 

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 2009

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

rule 5178

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(p_)*((f_) 
 + (g_.)*(x_))^(q_), x_Symbol] :> Simp[(d + e*x)^q*((f + g*x)^q/(1 - c^2*x^ 
2)^q)   Int[(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, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^2 
- e^2, 0] && HalfIntegerQ[p, q] && GeQ[p - q, 0]
 

rule 5262

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_ 
) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^p*(a + 
 b*ArcSin[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] & 
& EqQ[c^2*d + e, 0] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ 
[n, 0] && (m == 1 || p > 0 || (n == 1 && p > -1) || (m == 2 && p < -2))
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.81 (sec) , antiderivative size = 308, normalized size of antiderivative = 2.18

method	result	size
default	\(\frac {a \sqrt {-c f x +f}\, \sqrt {c d x +d}}{c d}+\frac {a f \sqrt {\left (-c f x +f \right ) \left (c d x +d \right )}\, \arctan \left (\frac {\sqrt {c^{2} d f}\, x}{\sqrt {-c^{2} d f \,x^{2}+d f}}\right )}{\sqrt {-c f x +f}\, \sqrt {c d x +d}\, \sqrt {c^{2} d f}}+b \left (-\frac {\sqrt {-f \left (c x -1\right )}\, \sqrt {d \left (c x +1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{2}}{2 \left (c x +1\right ) d c \left (c x -1\right )}+\frac {\sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \left (c^{2} x^{2}-i \sqrt {-c^{2} x^{2}+1}\, c x -1\right ) \left (\arcsin \left (c x \right )+i\right )}{2 \left (c x +1\right ) d c \left (c x -1\right )}+\frac {\left (\arcsin \left (c x \right )-i\right ) \sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, c x +c^{2} x^{2}-1\right )}{2 \left (c x +1\right ) d c \left (c x -1\right )}\right )\)	\(308\)
parts	\(\frac {a \sqrt {-c f x +f}\, \sqrt {c d x +d}}{c d}+\frac {a f \sqrt {\left (-c f x +f \right ) \left (c d x +d \right )}\, \arctan \left (\frac {\sqrt {c^{2} d f}\, x}{\sqrt {-c^{2} d f \,x^{2}+d f}}\right )}{\sqrt {-c f x +f}\, \sqrt {c d x +d}\, \sqrt {c^{2} d f}}+b \left (-\frac {\sqrt {-f \left (c x -1\right )}\, \sqrt {d \left (c x +1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{2}}{2 \left (c x +1\right ) d c \left (c x -1\right )}+\frac {\sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \left (c^{2} x^{2}-i \sqrt {-c^{2} x^{2}+1}\, c x -1\right ) \left (\arcsin \left (c x \right )+i\right )}{2 \left (c x +1\right ) d c \left (c x -1\right )}+\frac {\left (\arcsin \left (c x \right )-i\right ) \sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, c x +c^{2} x^{2}-1\right )}{2 \left (c x +1\right ) d c \left (c x -1\right )}\right )\)	\(308\)

Input:

int((-c*f*x+f)^(1/2)*(a+b*arcsin(c*x))/(c*d*x+d)^(1/2),x,method=_RETURNVER 
BOSE)
 

Output:

a/c/d*(-c*f*x+f)^(1/2)*(c*d*x+d)^(1/2)+a*f*((-c*f*x+f)*(c*d*x+d))^(1/2)/(- 
c*f*x+f)^(1/2)/(c*d*x+d)^(1/2)/(c^2*d*f)^(1/2)*arctan((c^2*d*f)^(1/2)*x/(- 
c^2*d*f*x^2+d*f)^(1/2))+b*(-1/2*(-f*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)*(-c^2 
*x^2+1)^(1/2)/(c*x+1)/d/c/(c*x-1)*arcsin(c*x)^2+1/2*(d*(c*x+1))^(1/2)*(-f* 
(c*x-1))^(1/2)*(c^2*x^2-I*c*x*(-c^2*x^2+1)^(1/2)-1)*(arcsin(c*x)+I)/(c*x+1 
)/d/c/(c*x-1)+1/2*(arcsin(c*x)-I)*(d*(c*x+1))^(1/2)*(-f*(c*x-1))^(1/2)*(I* 
(-c^2*x^2+1)^(1/2)*c*x+c^2*x^2-1)/(c*x+1)/d/c/(c*x-1))
 

Fricas [F]

\[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=\int { \frac {\sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt {c d x + d}} \,d x } \] Input:

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

Output:

integral(sqrt(-c*f*x + f)*(b*arcsin(c*x) + a)/sqrt(c*d*x + d), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

\[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=\int { \frac {\sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt {c d x + d}} \,d x } \] Input:

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

Output:

a*(f*arcsin(c*x)/(c*d*sqrt(f/d)) + sqrt(-c^2*d*f*x^2 + d*f)/(c*d)) + b*sqr 
t(f)*integrate(sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/s 
qrt(c*x + 1), x)/sqrt(d)
 

Giac [F]

\[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=\int { \frac {\sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt {c d x + d}} \,d x } \] Input:

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

Output:

integrate(sqrt(-c*f*x + f)*(b*arcsin(c*x) + a)/sqrt(c*d*x + d), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=\frac {\sqrt {f}\, \left (-2 \mathit {asin} \left (\frac {\sqrt {-c x +1}}{\sqrt {2}}\right ) a +\sqrt {c x +1}\, \sqrt {-c x +1}\, a +\left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right )}{\sqrt {c x +1}}d x \right ) b c \right )}{\sqrt {d}\, c} \] Input:

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

Output:

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