\(\int \frac {a+b \arcsin (c x)}{(d+c d x)^{3/2} \sqrt {f-c f x}} \, dx\) [61]

Optimal result

Integrand size = 30, antiderivative size = 90 \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{3/2} \sqrt {f-c f x}} \, dx=-\frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{c d f \sqrt {d+c d x}}+\frac {b \sqrt {1-c^2 x^2} \log (1+c x)}{c d \sqrt {d+c d x} \sqrt {f-c f x}} \] Output:

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

Mathematica [A] (verified)

Time = 0.73 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.88 \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{3/2} \sqrt {f-c f x}} \, dx=\frac {\sqrt {d+c d x} \left (a (-1+c x)+b (-1+c x) \arcsin (c x)+b \sqrt {1-c^2 x^2} \log (-f (1+c x))\right )}{c d^2 (1+c x) \sqrt {f-c f x}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.91, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {5178, 27, 5260, 25, 27, 451, 16}

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.

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

\(\Big \downarrow \) 5178

\(\displaystyle \frac {\left (1-c^2 x^2\right )^{3/2} \int \frac {f (1-c x) (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {f \left (1-c^2 x^2\right )^{3/2} \int \frac {(1-c x) (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\)

\(\Big \downarrow \) 5260

\(\displaystyle \frac {f \left (1-c^2 x^2\right )^{3/2} \left (-b c \int -\frac {1-c x}{c \left (1-c^2 x^2\right )}dx-\frac {(1-c x) (a+b \arcsin (c x))}{c \sqrt {1-c^2 x^2}}\right )}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {f \left (1-c^2 x^2\right )^{3/2} \left (b c \int \frac {1-c x}{c \left (1-c^2 x^2\right )}dx-\frac {(1-c x) (a+b \arcsin (c x))}{c \sqrt {1-c^2 x^2}}\right )}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {f \left (1-c^2 x^2\right )^{3/2} \left (b \int \frac {1-c x}{1-c^2 x^2}dx-\frac {(1-c x) (a+b \arcsin (c x))}{c \sqrt {1-c^2 x^2}}\right )}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\)

\(\Big \downarrow \) 451

\(\displaystyle \frac {f \left (1-c^2 x^2\right )^{3/2} \left (b \int \frac {1}{c x+1}dx-\frac {(1-c x) (a+b \arcsin (c x))}{c \sqrt {1-c^2 x^2}}\right )}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {f \left (1-c^2 x^2\right )^{3/2} \left (\frac {b \log (c x+1)}{c}-\frac {(1-c x) (a+b \arcsin (c x))}{c \sqrt {1-c^2 x^2}}\right )}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\)

Input:

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

Output:

(f*(1 - c^2*x^2)^(3/2)*(-(((1 - c*x)*(a + b*ArcSin[c*x]))/(c*Sqrt[1 - c^2* 
x^2])) + (b*Log[1 + c*x])/c))/((d + c*d*x)^(3/2)*(f - c*f*x)^(3/2))
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 451

Int[((c_) + (d_.)*(x_))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[c^2/a   In 
t[1/(c - d*x), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0]
 

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 5260

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e 
_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[(f + g*x)^m*(d + e*x^2)^p, 
 x]}, Simp[(a + b*ArcSin[c*x])   u, x] - Simp[b*c   Int[1/Sqrt[1 - c^2*x^2] 
   u, x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[c^2*d + e, 0] && IG 
tQ[m, 0] && ILtQ[p + 1/2, 0] && GtQ[d, 0] && (LtQ[m, -2*p - 1] || GtQ[m, 3] 
)
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.79 (sec) , antiderivative size = 234, normalized size of antiderivative = 2.60

method	result	size
default	\(-\frac {a \sqrt {-c f x +f}}{f d c \sqrt {c d x +d}}+b \left (\frac {2 i \sqrt {-c^{2} x^{2}+1}\, \sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \arcsin \left (c x \right )}{d^{2} f c \left (c^{2} x^{2}-1\right )}-\frac {\arcsin \left (c x \right ) \sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}+c x -1\right )}{\left (c x +1\right ) c \,d^{2} f \left (c x -1\right )}-\frac {2 \sqrt {-f \left (c x -1\right )}\, \sqrt {d \left (c x +1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right )}{d^{2} f c \left (c^{2} x^{2}-1\right )}\right )\)	\(234\)
parts	\(-\frac {a \sqrt {-c f x +f}}{f d c \sqrt {c d x +d}}+b \left (\frac {2 i \sqrt {-c^{2} x^{2}+1}\, \sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \arcsin \left (c x \right )}{d^{2} f c \left (c^{2} x^{2}-1\right )}-\frac {\arcsin \left (c x \right ) \sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}+c x -1\right )}{\left (c x +1\right ) c \,d^{2} f \left (c x -1\right )}-\frac {2 \sqrt {-f \left (c x -1\right )}\, \sqrt {d \left (c x +1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right )}{d^{2} f c \left (c^{2} x^{2}-1\right )}\right )\)	\(234\)

Input:

int((a+b*arcsin(c*x))/(c*d*x+d)^(3/2)/(-c*f*x+f)^(1/2),x,method=_RETURNVER 
BOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 348, normalized size of antiderivative = 3.87 \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{3/2} \sqrt {f-c f x}} \, dx=\left [\frac {{\left (b c x + b\right )} \sqrt {d f} \log \left (\frac {c^{6} d f x^{6} + 4 \, c^{5} d f x^{5} + 5 \, c^{4} d f x^{4} - 4 \, c^{2} d f x^{2} - 4 \, c d f x - {\left (c^{4} x^{4} + 4 \, c^{3} x^{3} + 6 \, c^{2} x^{2} + 4 \, c x\right )} \sqrt {-c^{2} x^{2} + 1} \sqrt {c d x + d} \sqrt {-c f x + f} \sqrt {d f} - 2 \, d f}{c^{4} x^{4} + 2 \, c^{3} x^{3} - 2 \, c x - 1}\right ) - 2 \, \sqrt {c d x + d} \sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )}}{2 \, {\left (c^{2} d^{2} f x + c d^{2} f\right )}}, \frac {{\left (b c x + b\right )} \sqrt {-d f} \arctan \left (\frac {{\left (c^{2} x^{2} + 2 \, c x + 2\right )} \sqrt {-c^{2} x^{2} + 1} \sqrt {c d x + d} \sqrt {-c f x + f} \sqrt {-d f}}{c^{4} d f x^{4} + 2 \, c^{3} d f x^{3} - c^{2} d f x^{2} - 2 \, c d f x}\right ) - \sqrt {c d x + d} \sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )}}{c^{2} d^{2} f x + c d^{2} f}\right ] \] Input:

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

Output:

[1/2*((b*c*x + b)*sqrt(d*f)*log((c^6*d*f*x^6 + 4*c^5*d*f*x^5 + 5*c^4*d*f*x 
^4 - 4*c^2*d*f*x^2 - 4*c*d*f*x - (c^4*x^4 + 4*c^3*x^3 + 6*c^2*x^2 + 4*c*x) 
*sqrt(-c^2*x^2 + 1)*sqrt(c*d*x + d)*sqrt(-c*f*x + f)*sqrt(d*f) - 2*d*f)/(c 
^4*x^4 + 2*c^3*x^3 - 2*c*x - 1)) - 2*sqrt(c*d*x + d)*sqrt(-c*f*x + f)*(b*a 
rcsin(c*x) + a))/(c^2*d^2*f*x + c*d^2*f), ((b*c*x + b)*sqrt(-d*f)*arctan(( 
c^2*x^2 + 2*c*x + 2)*sqrt(-c^2*x^2 + 1)*sqrt(c*d*x + d)*sqrt(-c*f*x + f)*s 
qrt(-d*f)/(c^4*d*f*x^4 + 2*c^3*d*f*x^3 - c^2*d*f*x^2 - 2*c*d*f*x)) - sqrt( 
c*d*x + d)*sqrt(-c*f*x + f)*(b*arcsin(c*x) + a))/(c^2*d^2*f*x + c*d^2*f)]
 

Sympy [F]

\[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{3/2} \sqrt {f-c f x}} \, dx=\int \frac {a + b \operatorname {asin}{\left (c x \right )}}{\left (d \left (c x + 1\right )\right )^{\frac {3}{2}} \sqrt {- f \left (c x - 1\right )}}\, dx \] Input:

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

Output:

Integral((a + b*asin(c*x))/((d*(c*x + 1))**(3/2)*sqrt(-f*(c*x - 1))), x)
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.07 \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{3/2} \sqrt {f-c f x}} \, dx=-\frac {\sqrt {-c^{2} d f x^{2} + d f} b \arcsin \left (c x\right )}{c^{2} d^{2} f x + c d^{2} f} - \frac {\sqrt {-c^{2} d f x^{2} + d f} a}{c^{2} d^{2} f x + c d^{2} f} + \frac {b \log \left (c x + 1\right )}{c d^{\frac {3}{2}} \sqrt {f}} \] Input:

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

Output:

-sqrt(-c^2*d*f*x^2 + d*f)*b*arcsin(c*x)/(c^2*d^2*f*x + c*d^2*f) - sqrt(-c^ 
2*d*f*x^2 + d*f)*a/(c^2*d^2*f*x + c*d^2*f) + b*log(c*x + 1)/(c*d^(3/2)*sqr 
t(f))
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{3/2} \sqrt {f-c f x}} \, dx=\frac {-\sqrt {-c x +1}\, a +\sqrt {c x +1}\, \left (\int \frac {\mathit {asin} \left (c x \right )}{\sqrt {c x +1}\, \sqrt {-c x +1}\, c x +\sqrt {c x +1}\, \sqrt {-c x +1}}d x \right ) b c}{\sqrt {f}\, \sqrt {d}\, \sqrt {c x +1}\, c d} \] Input:

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

Output:

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