\(\int \frac {a+b \arcsin (c x)}{(d+c d x)^{5/2} (f-c f x)^{3/2}} \, dx\) [65]

Optimal result

Integrand size = 30, antiderivative size = 252 \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{5/2} (f-c f x)^{3/2}} \, dx=-\frac {b \sqrt {1-c^2 x^2}}{6 c d^2 f (1+c x) \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {a+b \arcsin (c x)}{3 c d f (d+c d x)^{3/2} \sqrt {f-c f x}}+\frac {2 x (a+b \arcsin (c x))}{3 d^2 f \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {b \sqrt {1-c^2 x^2} \text {arctanh}(c x)}{6 c d^2 f \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {b \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{3 c d^2 f \sqrt {d+c d x} \sqrt {f-c f x}} \] Output:

-1/6*b*(-c^2*x^2+1)^(1/2)/c/d^2/f/(c*x+1)/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2) 
-1/3*(a+b*arcsin(c*x))/c/d/f/(c*d*x+d)^(3/2)/(-c*f*x+f)^(1/2)+2/3*x*(a+b*a 
rcsin(c*x))/d^2/f/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)+1/6*b*(-c^2*x^2+1)^(1/2 
)*arctanh(c*x)/c/d^2/f/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)+1/3*b*(-c^2*x^2+1) 
^(1/2)*ln(-c^2*x^2+1)/c/d^2/f/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)
 

Mathematica [A] (verified)

Time = 1.11 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.71 \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{5/2} (f-c f x)^{3/2}} \, dx=\frac {\sqrt {d+c d x} \left (-4 a+8 a c x+8 a c^2 x^2-2 b \sqrt {1-c^2 x^2}+4 b \left (-1+2 c x+2 c^2 x^2\right ) \arcsin (c x)+5 b (1+c x) \sqrt {1-c^2 x^2} \log (-f (1+c x))+3 b \sqrt {1-c^2 x^2} \log (f-c f x)+3 b c x \sqrt {1-c^2 x^2} \log (f-c f x)\right )}{12 c d^3 f (1+c x)^2 \sqrt {f-c f x}} \] Input:

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

Output:

(Sqrt[d + c*d*x]*(-4*a + 8*a*c*x + 8*a*c^2*x^2 - 2*b*Sqrt[1 - c^2*x^2] + 4 
*b*(-1 + 2*c*x + 2*c^2*x^2)*ArcSin[c*x] + 5*b*(1 + c*x)*Sqrt[1 - c^2*x^2]* 
Log[-(f*(1 + c*x))] + 3*b*Sqrt[1 - c^2*x^2]*Log[f - c*f*x] + 3*b*c*x*Sqrt[ 
1 - c^2*x^2]*Log[f - c*f*x]))/(12*c*d^3*f*(1 + c*x)^2*Sqrt[f - c*f*x])
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.63, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {5178, 27, 5260, 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.

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

\(\Big \downarrow \) 5178

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 5260

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {f \left (1-c^2 x^2\right )^{5/2} \left (\frac {2 x (a+b \arcsin (c x))}{3 \sqrt {1-c^2 x^2}}-\frac {(1-c x) (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}-b c \left (-\frac {\text {arctanh}(c x)}{6 c^2}+\frac {1-c x}{6 c^2 \left (1-c^2 x^2\right )}-\frac {\log \left (1-c^2 x^2\right )}{3 c^2}\right )\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\)

Input:

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

Output:

(f*(1 - c^2*x^2)^(5/2)*(-1/3*((1 - c*x)*(a + b*ArcSin[c*x]))/(c*(1 - c^2*x 
^2)^(3/2)) + (2*x*(a + b*ArcSin[c*x]))/(3*Sqrt[1 - c^2*x^2]) - b*c*((1 - c 
*x)/(6*c^2*(1 - c^2*x^2)) - ArcTanh[c*x]/(6*c^2) - Log[1 - c^2*x^2]/(3*c^2 
))))/((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 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 = 2.03 (sec) , antiderivative size = 499, normalized size of antiderivative = 1.98

method	result	size
default	\(a \left (-\frac {1}{3 f d c \left (c d x +d \right )^{\frac {3}{2}} \sqrt {-c f x +f}}+\frac {-\frac {2}{3 f d c \sqrt {c d x +d}\, \sqrt {-c f x +f}}+\frac {2 \sqrt {c d x +d}}{3 f c \,d^{2} \sqrt {-c f x +f}}}{d}\right )+\frac {b \left (-4 i \arcsin \left (c x \right ) x c -5 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right ) x^{3} c^{3}-3 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right ) x^{3} c^{3}+4 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}+4 i \arcsin \left (c x \right ) x^{3} c^{3}-5 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right ) c^{2} x^{2}-3 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right ) x^{2} c^{2}+4 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, c x +4 i \arcsin \left (c x \right ) x^{2} c^{2}+c^{2} x^{2}+5 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right ) x c +3 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right ) x c -2 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}-4 i \arcsin \left (c x \right )+5 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right )+3 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right )-1\right ) \sqrt {-f \left (c x -1\right )}\, \sqrt {d \left (c x +1\right )}\, \sqrt {-c^{2} x^{2}+1}}{6 \left (c^{5} x^{5}+c^{4} x^{4}-2 c^{3} x^{3}-2 c^{2} x^{2}+c x +1\right ) c \,d^{3} f^{2}}\)	\(499\)
parts	\(a \left (-\frac {1}{3 f d c \left (c d x +d \right )^{\frac {3}{2}} \sqrt {-c f x +f}}+\frac {-\frac {2}{3 f d c \sqrt {c d x +d}\, \sqrt {-c f x +f}}+\frac {2 \sqrt {c d x +d}}{3 f c \,d^{2} \sqrt {-c f x +f}}}{d}\right )+\frac {b \left (-4 i \arcsin \left (c x \right ) x c -5 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right ) x^{3} c^{3}-3 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right ) x^{3} c^{3}+4 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}+4 i \arcsin \left (c x \right ) x^{3} c^{3}-5 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right ) c^{2} x^{2}-3 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right ) x^{2} c^{2}+4 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, c x +4 i \arcsin \left (c x \right ) x^{2} c^{2}+c^{2} x^{2}+5 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right ) x c +3 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right ) x c -2 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}-4 i \arcsin \left (c x \right )+5 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right )+3 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right )-1\right ) \sqrt {-f \left (c x -1\right )}\, \sqrt {d \left (c x +1\right )}\, \sqrt {-c^{2} x^{2}+1}}{6 \left (c^{5} x^{5}+c^{4} x^{4}-2 c^{3} x^{3}-2 c^{2} x^{2}+c x +1\right ) c \,d^{3} f^{2}}\)	\(499\)

Input:

int((a+b*arcsin(c*x))/(c*d*x+d)^(5/2)/(-c*f*x+f)^(3/2),x,method=_RETURNVER 
BOSE)
 

Output:

a*(-1/3/f/d/c/(c*d*x+d)^(3/2)/(-c*f*x+f)^(1/2)+2/3/d*(-1/f/d/c/(c*d*x+d)^( 
1/2)/(-c*f*x+f)^(1/2)+1/f/c/d^2/(-c*f*x+f)^(1/2)*(c*d*x+d)^(1/2)))+1/6*b*( 
-4*I*arcsin(c*x)*c*x-5*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)*x^3*c^3-3*ln(I*c*x+( 
-c^2*x^2+1)^(1/2)-I)*x^3*c^3+4*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*c^2*x^2+4*I* 
arcsin(c*x)*c^3*x^3-5*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)*c^2*x^2-3*ln(I*c*x+(- 
c^2*x^2+1)^(1/2)-I)*x^2*c^2+4*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*c*x+4*I*arcsi 
n(c*x)*c^2*x^2+c^2*x^2+5*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)*x*c+3*ln(I*c*x+(-c 
^2*x^2+1)^(1/2)-I)*x*c-2*arcsin(c*x)*(-c^2*x^2+1)^(1/2)-4*I*arcsin(c*x)+5* 
ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)+3*ln(I*c*x+(-c^2*x^2+1)^(1/2)-I)-1)*(-f*(c* 
x-1))^(1/2)*(d*(c*x+1))^(1/2)*(-c^2*x^2+1)^(1/2)/(c^5*x^5+c^4*x^4-2*c^3*x^ 
3-2*c^2*x^2+c*x+1)/c/d^3/f^2
 

Fricas [F]

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

integrate((a+b*arcsin(c*x))/(c*d*x+d)^(5/2)/(-c*f*x+f)^(3/2),x, algorithm= 
"fricas")
 

Output:

integral(sqrt(c*d*x + d)*sqrt(-c*f*x + f)*(b*arcsin(c*x) + a)/(c^5*d^3*f^2 
*x^5 + c^4*d^3*f^2*x^4 - 2*c^3*d^3*f^2*x^3 - 2*c^2*d^3*f^2*x^2 + c*d^3*f^2 
*x + d^3*f^2), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{5/2} (f-c f x)^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((a+b*asin(c*x))/(c*d*x+d)**(5/2)/(-c*f*x+f)**(3/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.93 \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{5/2} (f-c f x)^{3/2}} \, dx=-\frac {1}{12} \, b c {\left (\frac {2 \, \sqrt {d} \sqrt {f}}{c^{3} d^{3} f^{2} x + c^{2} d^{3} f^{2}} - \frac {5 \, \log \left (c x + 1\right )}{c^{2} d^{\frac {5}{2}} f^{\frac {3}{2}}} - \frac {3 \, \log \left (c x - 1\right )}{c^{2} d^{\frac {5}{2}} f^{\frac {3}{2}}}\right )} - \frac {1}{3} \, b {\left (\frac {1}{\sqrt {-c^{2} d f x^{2} + d f} c^{2} d^{2} f x + \sqrt {-c^{2} d f x^{2} + d f} c d^{2} f} - \frac {2 \, x}{\sqrt {-c^{2} d f x^{2} + d f} d^{2} f}\right )} \arcsin \left (c x\right ) - \frac {1}{3} \, a {\left (\frac {1}{\sqrt {-c^{2} d f x^{2} + d f} c^{2} d^{2} f x + \sqrt {-c^{2} d f x^{2} + d f} c d^{2} f} - \frac {2 \, x}{\sqrt {-c^{2} d f x^{2} + d f} d^{2} f}\right )} \] Input:

integrate((a+b*arcsin(c*x))/(c*d*x+d)^(5/2)/(-c*f*x+f)^(3/2),x, algorithm= 
"maxima")
 

Output:

-1/12*b*c*(2*sqrt(d)*sqrt(f)/(c^3*d^3*f^2*x + c^2*d^3*f^2) - 5*log(c*x + 1 
)/(c^2*d^(5/2)*f^(3/2)) - 3*log(c*x - 1)/(c^2*d^(5/2)*f^(3/2))) - 1/3*b*(1 
/(sqrt(-c^2*d*f*x^2 + d*f)*c^2*d^2*f*x + sqrt(-c^2*d*f*x^2 + d*f)*c*d^2*f) 
 - 2*x/(sqrt(-c^2*d*f*x^2 + d*f)*d^2*f))*arcsin(c*x) - 1/3*a*(1/(sqrt(-c^2 
*d*f*x^2 + d*f)*c^2*d^2*f*x + sqrt(-c^2*d*f*x^2 + d*f)*c*d^2*f) - 2*x/(sqr 
t(-c^2*d*f*x^2 + d*f)*d^2*f))
 

Giac [F]

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

integrate((a+b*arcsin(c*x))/(c*d*x+d)^(5/2)/(-c*f*x+f)^(3/2),x, algorithm= 
"giac")
 

Output:

integrate((b*arcsin(c*x) + a)/((c*d*x + d)^(5/2)*(-c*f*x + f)^(3/2)), x)
 

Mupad [F(-1)]

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

int((a + b*asin(c*x))/((d + c*d*x)^(5/2)*(f - c*f*x)^(3/2)),x)
 

Output:

int((a + b*asin(c*x))/((d + c*d*x)^(5/2)*(f - c*f*x)^(3/2)), x)
 

Reduce [F]

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

int((a+b*asin(c*x))/(c*d*x+d)^(5/2)/(-c*f*x+f)^(3/2),x)
                                                                                    
                                                                                    
 

Output:

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