\(\int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx\) [47]

Optimal result

Integrand size = 30, antiderivative size = 150 \[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=-\frac {2 b f \sqrt {1-c^2 x^2}}{3 c d^2 (1+c x) \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {(f-c f x)^{3/2} (a+b \arcsin (c x))}{3 c d f (d+c d x)^{3/2}}-\frac {b f \sqrt {1-c^2 x^2} \log (1+c x)}{3 c d^2 \sqrt {d+c d x} \sqrt {f-c f x}} \] Output:

-2/3*b*f*(-c^2*x^2+1)^(1/2)/c/d^2/(c*x+1)/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2) 
-1/3*(-c*f*x+f)^(3/2)*(a+b*arcsin(c*x))/c/d/f/(c*d*x+d)^(3/2)-1/3*b*f*(-c^ 
2*x^2+1)^(1/2)*ln(c*x+1)/c/d^2/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)
 

Mathematica [A] (verified)

Time = 1.41 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=-\frac {f \sqrt {d+c d x} \left ((-1+c x) \left (-a+a c x-b \sqrt {1-c^2 x^2}\right )+b (-1+c x)^2 \arcsin (c x)+b (1+c x) \sqrt {1-c^2 x^2} \log (-f (1+c x))\right )}{3 c d^3 (1+c x)^2 \sqrt {f-c f x}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.71, 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, 27, 456, 49, 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.

\(\displaystyle \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(c d x+d)^{5/2}} \, dx\)

\(\Big \downarrow \) 5178

\(\displaystyle \frac {\left (1-c^2 x^2\right )^{5/2} \int \frac {f^3 (1-c x)^3 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {f^3 \left (1-c^2 x^2\right )^{5/2} \int \frac {(1-c x)^3 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\)

\(\Big \downarrow \) 5260

\(\displaystyle \frac {f^3 \left (1-c^2 x^2\right )^{5/2} \left (-b c \int -\frac {(1-c x)^3}{3 c \left (1-c^2 x^2\right )^2}dx-\frac {(1-c x)^3 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {f^3 \left (1-c^2 x^2\right )^{5/2} \left (\frac {1}{3} b \int \frac {(1-c x)^3}{\left (1-c^2 x^2\right )^2}dx-\frac {(1-c x)^3 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\)

\(\Big \downarrow \) 456

\(\displaystyle \frac {f^3 \left (1-c^2 x^2\right )^{5/2} \left (\frac {1}{3} b \int \frac {1-c x}{(c x+1)^2}dx-\frac {(1-c x)^3 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\)

\(\Big \downarrow \) 49

\(\displaystyle \frac {f^3 \left (1-c^2 x^2\right )^{5/2} \left (\frac {1}{3} b \int \left (\frac {2}{(c x+1)^2}+\frac {1}{-c x-1}\right )dx-\frac {(1-c x)^3 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {f^3 \left (1-c^2 x^2\right )^{5/2} \left (\frac {1}{3} b \left (-\frac {2}{c (c x+1)}-\frac {\log (c x+1)}{c}\right )-\frac {(1-c x)^3 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\)

Input:

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

Output:

(f^3*(1 - c^2*x^2)^(5/2)*(-1/3*((1 - c*x)^3*(a + b*ArcSin[c*x]))/(c*(1 - c 
^2*x^2)^(3/2)) + (b*(-2/(c*(1 + c*x)) - Log[1 + c*x]/c))/3))/((d + c*d*x)^ 
(5/2)*(f - c*f*x)^(5/2))
 

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 49

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

rule 456

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ 
(c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; FreeQ[{a, b, c, d, n, p}, x] && 
EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] &&  !Integ 
erQ[n]))
 

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 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 = 2.63 (sec) , antiderivative size = 449, normalized size of antiderivative = 2.99

method	result	size
default	\(a \left (-\frac {\sqrt {-c f x +f}}{c d \left (c d x +d \right )^{\frac {3}{2}}}-f \left (-\frac {\sqrt {-c f x +f}}{3 f d c \left (c d x +d \right )^{\frac {3}{2}}}-\frac {\sqrt {-c f x +f}}{3 f c \,d^{2} \sqrt {c d x +d}}\right )\right )+\frac {b \left (i \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right ) x^{3} c^{3}+3 i \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right ) c^{2} x^{2}-\sqrt {-c^{2} x^{2}+1}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right ) x^{2} c^{2}+i c^{2} x^{2}+3 c^{2} x^{2} \arcsin \left (c x \right )+3 i \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right ) x c +2 i c x -c x \sqrt {-c^{2} x^{2}+1}+i \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right )+\sqrt {-c^{2} x^{2}+1}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right )+i+\arcsin \left (c x \right )+\sqrt {-c^{2} x^{2}+1}\right ) \left (-i \sqrt {-c^{2} x^{2}+1}\, c x +c^{2} x^{2}-i \sqrt {-c^{2} x^{2}+1}-2 c x +1\right ) \sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}}{3 \left (3 c^{5} x^{5}+3 c^{4} x^{4}-2 c^{3} x^{3}-2 c^{2} x^{2}-c x -1\right ) c \,d^{3}}\)	\(449\)
parts	\(a \left (-\frac {\sqrt {-c f x +f}}{c d \left (c d x +d \right )^{\frac {3}{2}}}-f \left (-\frac {\sqrt {-c f x +f}}{3 f d c \left (c d x +d \right )^{\frac {3}{2}}}-\frac {\sqrt {-c f x +f}}{3 f c \,d^{2} \sqrt {c d x +d}}\right )\right )+\frac {b \left (i \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right ) x^{3} c^{3}+3 i \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right ) c^{2} x^{2}-\sqrt {-c^{2} x^{2}+1}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right ) x^{2} c^{2}+i c^{2} x^{2}+3 c^{2} x^{2} \arcsin \left (c x \right )+3 i \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right ) x c +2 i c x -c x \sqrt {-c^{2} x^{2}+1}+i \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right )+\sqrt {-c^{2} x^{2}+1}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right )+i+\arcsin \left (c x \right )+\sqrt {-c^{2} x^{2}+1}\right ) \left (-i \sqrt {-c^{2} x^{2}+1}\, c x +c^{2} x^{2}-i \sqrt {-c^{2} x^{2}+1}-2 c x +1\right ) \sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}}{3 \left (3 c^{5} x^{5}+3 c^{4} x^{4}-2 c^{3} x^{3}-2 c^{2} x^{2}-c x -1\right ) c \,d^{3}}\)	\(449\)

Input:

int((-c*f*x+f)^(1/2)*(a+b*arcsin(c*x))/(c*d*x+d)^(5/2),x,method=_RETURNVER 
BOSE)
 

Output:

a*(-1/c/d*(-c*f*x+f)^(1/2)/(c*d*x+d)^(3/2)-f*(-1/3/f/d/c/(c*d*x+d)^(3/2)*( 
-c*f*x+f)^(1/2)-1/3/f/c/d^2/(c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)))+1/3*b*(I*ln 
(I*c*x+(-c^2*x^2+1)^(1/2)+I)*x^3*c^3+3*I*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)*c^ 
2*x^2-(-c^2*x^2+1)^(1/2)*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)*x^2*c^2+I*c^2*x^2+ 
3*c^2*x^2*arcsin(c*x)+3*I*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)*x*c+2*I*c*x-c*x*( 
-c^2*x^2+1)^(1/2)+I*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)+(-c^2*x^2+1)^(1/2)*ln(I 
*c*x+(-c^2*x^2+1)^(1/2)+I)+I+arcsin(c*x)+(-c^2*x^2+1)^(1/2))*(-I*c*x*(-c^2 
*x^2+1)^(1/2)+c^2*x^2-I*(-c^2*x^2+1)^(1/2)-2*c*x+1)*(d*(c*x+1))^(1/2)*(-f* 
(c*x-1))^(1/2)/(3*c^5*x^5+3*c^4*x^4-2*c^3*x^3-2*c^2*x^2-c*x-1)/c/d^3
 

Fricas [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 520, normalized size of antiderivative = 3.47 \[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=\left [\frac {{\left (b c^{3} d x^{3} + b c^{2} d x^{2} - b c d x - b d\right )} \sqrt {\frac {f}{d}} \log \left (\frac {c^{6} f x^{6} + 4 \, c^{5} f x^{5} + 5 \, c^{4} f x^{4} - 4 \, c^{2} f x^{2} - 4 \, c f x + {\left (c^{4} x^{4} + 4 \, c^{3} x^{3} + 6 \, c^{2} x^{2} + 4 \, c x\right )} \sqrt {-c^{2} x^{2} + 1} \sqrt {c d x + d} \sqrt {-c f x + f} \sqrt {\frac {f}{d}} - 2 \, f}{c^{4} x^{4} + 2 \, c^{3} x^{3} - 2 \, c x - 1}\right ) + 2 \, {\left (a c^{2} x^{2} - 2 \, \sqrt {-c^{2} x^{2} + 1} b c x - 2 \, a c x + {\left (b c^{2} x^{2} - 2 \, b c x + b\right )} \arcsin \left (c x\right ) + a\right )} \sqrt {c d x + d} \sqrt {-c f x + f}}{6 \, {\left (c^{4} d^{3} x^{3} + c^{3} d^{3} x^{2} - c^{2} d^{3} x - c d^{3}\right )}}, -\frac {{\left (b c^{3} d x^{3} + b c^{2} d x^{2} - b c d x - b d\right )} \sqrt {-\frac {f}{d}} \arctan \left (\frac {{\left (c^{2} x^{2} + 2 \, c x + 2\right )} \sqrt {-c^{2} x^{2} + 1} \sqrt {c d x + d} \sqrt {-c f x + f} \sqrt {-\frac {f}{d}}}{c^{4} f x^{4} + 2 \, c^{3} f x^{3} - c^{2} f x^{2} - 2 \, c f x}\right ) - {\left (a c^{2} x^{2} - 2 \, \sqrt {-c^{2} x^{2} + 1} b c x - 2 \, a c x + {\left (b c^{2} x^{2} - 2 \, b c x + b\right )} \arcsin \left (c x\right ) + a\right )} \sqrt {c d x + d} \sqrt {-c f x + f}}{3 \, {\left (c^{4} d^{3} x^{3} + c^{3} d^{3} x^{2} - c^{2} d^{3} x - c d^{3}\right )}}\right ] \] Input:

integrate((-c*f*x+f)^(1/2)*(a+b*arcsin(c*x))/(c*d*x+d)^(5/2),x, algorithm= 
"fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=\int \frac {\sqrt {- f \left (c x - 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\left (d \left (c x + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((-c*f*x+f)**(1/2)*(a+b*asin(c*x))/(c*d*x+d)**(5/2),x)
 

Output:

Integral(sqrt(-f*(c*x - 1))*(a + b*asin(c*x))/(d*(c*x + 1))**(5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.43 \[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=-\frac {1}{3} \, b c {\left (\frac {2 \, \sqrt {f}}{c^{3} d^{\frac {5}{2}} x + c^{2} d^{\frac {5}{2}}} + \frac {\sqrt {f} \log \left (c x + 1\right )}{c^{2} d^{\frac {5}{2}}}\right )} - \frac {1}{3} \, b {\left (\frac {2 \, \sqrt {-c^{2} d f x^{2} + d f}}{c^{3} d^{3} x^{2} + 2 \, c^{2} d^{3} x + c d^{3}} - \frac {\sqrt {-c^{2} d f x^{2} + d f}}{c^{2} d^{3} x + c d^{3}}\right )} \arcsin \left (c x\right ) - \frac {1}{3} \, a {\left (\frac {2 \, \sqrt {-c^{2} d f x^{2} + d f}}{c^{3} d^{3} x^{2} + 2 \, c^{2} d^{3} x + c d^{3}} - \frac {\sqrt {-c^{2} d f x^{2} + d f}}{c^{2} d^{3} x + c d^{3}}\right )} \] Input:

integrate((-c*f*x+f)^(1/2)*(a+b*arcsin(c*x))/(c*d*x+d)^(5/2),x, algorithm= 
"maxima")
 

Output:

-1/3*b*c*(2*sqrt(f)/(c^3*d^(5/2)*x + c^2*d^(5/2)) + sqrt(f)*log(c*x + 1)/( 
c^2*d^(5/2))) - 1/3*b*(2*sqrt(-c^2*d*f*x^2 + d*f)/(c^3*d^3*x^2 + 2*c^2*d^3 
*x + c*d^3) - sqrt(-c^2*d*f*x^2 + d*f)/(c^2*d^3*x + c*d^3))*arcsin(c*x) - 
1/3*a*(2*sqrt(-c^2*d*f*x^2 + d*f)/(c^3*d^3*x^2 + 2*c^2*d^3*x + c*d^3) - sq 
rt(-c^2*d*f*x^2 + d*f)/(c^2*d^3*x + c*d^3))
 

Giac [F]

\[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=\int { \frac {\sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (c d x + d\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((-c*f*x+f)^(1/2)*(a+b*arcsin(c*x))/(c*d*x+d)^(5/2),x, algorithm= 
"giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=\int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\sqrt {f-c\,f\,x}}{{\left (d+c\,d\,x\right )}^{5/2}} \,d x \] Input:

int(((a + b*asin(c*x))*(f - c*f*x)^(1/2))/(d + c*d*x)^(5/2),x)
 

Output:

int(((a + b*asin(c*x))*(f - c*f*x)^(1/2))/(d + c*d*x)^(5/2), x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=\frac {\sqrt {f}\, \left (\sqrt {-c x +1}\, a c x -\sqrt {-c x +1}\, a +3 \sqrt {c x +1}\, \left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right )}{\sqrt {c x +1}\, c^{2} x^{2}+2 \sqrt {c x +1}\, c x +\sqrt {c x +1}}d x \right ) b \,c^{2} x +3 \sqrt {c x +1}\, \left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right )}{\sqrt {c x +1}\, c^{2} x^{2}+2 \sqrt {c x +1}\, c x +\sqrt {c x +1}}d x \right ) b c \right )}{3 \sqrt {d}\, \sqrt {c x +1}\, c \,d^{2} \left (c x +1\right )} \] Input:

int((-c*f*x+f)^(1/2)*(a+b*asin(c*x))/(c*d*x+d)^(5/2),x)
 

Output:

(sqrt(f)*(sqrt( - c*x + 1)*a*c*x - sqrt( - c*x + 1)*a + 3*sqrt(c*x + 1)*in 
t((sqrt( - c*x + 1)*asin(c*x))/(sqrt(c*x + 1)*c**2*x**2 + 2*sqrt(c*x + 1)* 
c*x + sqrt(c*x + 1)),x)*b*c**2*x + 3*sqrt(c*x + 1)*int((sqrt( - c*x + 1)*a 
sin(c*x))/(sqrt(c*x + 1)*c**2*x**2 + 2*sqrt(c*x + 1)*c*x + sqrt(c*x + 1)), 
x)*b*c))/(3*sqrt(d)*sqrt(c*x + 1)*c*d**2*(c*x + 1))