\(\int \frac {a+b \arcsin (c x)}{\sqrt {d+c d x} (f-c f x)^{5/2}} \, dx\) [66]

Optimal result

Integrand size = 30, antiderivative size = 192 \[ \int \frac {a+b \arcsin (c x)}{\sqrt {d+c d x} (f-c f x)^{5/2}} \, dx=-\frac {b \sqrt {1-c^2 x^2}}{3 c f^2 (1-c x) \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {\sqrt {d+c d x} (a+b \arcsin (c x))}{3 c d f (f-c f x)^{3/2}}+\frac {\sqrt {d+c d x} (a+b \arcsin (c x))}{3 c d f^2 \sqrt {f-c f x}}+\frac {b \sqrt {1-c^2 x^2} \log (1-c x)}{3 c f^2 \sqrt {d+c d x} \sqrt {f-c f x}} \] Output:

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

Mathematica [A] (verified)

Time = 0.83 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.68 \[ \int \frac {a+b \arcsin (c x)}{\sqrt {d+c d x} (f-c f x)^{5/2}} \, dx=\frac {\sqrt {d+c d x} \sqrt {f-c f x} \left (-\left ((-2+c x) \left (-b+b c x+a \sqrt {1-c^2 x^2}\right )\right )-b (-2+c x) \sqrt {1-c^2 x^2} \arcsin (c x)+b (-1+c x)^2 \log (f-c f x)\right )}{3 c d f^3 (-1+c x)^2 \sqrt {1-c^2 x^2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.83, 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.

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

\(\Big \downarrow \) 5178

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 5260

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(d^2*(1 - c^2*x^2)^(5/2)*((2*(1 + c*x)*(a + b*ArcSin[c*x]))/(3*c*(1 - c^2* 
x^2)^(3/2)) + (x*(a + b*ArcSin[c*x]))/(3*Sqrt[1 - c^2*x^2]) - b*c*((1 + c* 
x)/(3*c^2*(1 - c^2*x^2)) + ArcTanh[c*x]/(3*c^2) - Log[1 - c^2*x^2]/(6*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 = 1.92 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.44

method	result	size
default	\(a \left (\frac {\sqrt {c d x +d}}{3 f d c \left (-c f x +f \right )^{\frac {3}{2}}}+\frac {\sqrt {c d x +d}}{3 c d \,f^{2} \sqrt {-c f x +f}}\right )+\frac {b \sqrt {-f \left (c x -1\right )}\, \sqrt {d \left (c x +1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \left (i \arcsin \left (c x \right ) x^{2} c^{2}-2 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right ) x^{2} c^{2}+\arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, c x -2 i \arcsin \left (c x \right ) x c +4 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right ) x c -2 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}+i \arcsin \left (c x \right )-c x -2 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right )+1\right )}{3 f^{3} d c \left (c^{4} x^{4}-2 c^{3} x^{3}+2 c x -1\right )}\)	\(276\)
parts	\(a \left (\frac {\sqrt {c d x +d}}{3 f d c \left (-c f x +f \right )^{\frac {3}{2}}}+\frac {\sqrt {c d x +d}}{3 c d \,f^{2} \sqrt {-c f x +f}}\right )+\frac {b \sqrt {-f \left (c x -1\right )}\, \sqrt {d \left (c x +1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \left (i \arcsin \left (c x \right ) x^{2} c^{2}-2 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right ) x^{2} c^{2}+\arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, c x -2 i \arcsin \left (c x \right ) x c +4 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right ) x c -2 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}+i \arcsin \left (c x \right )-c x -2 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right )+1\right )}{3 f^{3} d c \left (c^{4} x^{4}-2 c^{3} x^{3}+2 c x -1\right )}\)	\(276\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 527, normalized size of antiderivative = 2.74 \[ \int \frac {a+b \arcsin (c x)}{\sqrt {d+c d x} (f-c f x)^{5/2}} \, dx=\left [\frac {{\left (b c^{3} x^{3} - b c^{2} x^{2} - b c x + b\right )} \sqrt {d f} \log \left (\frac {c^{6} d f x^{6} - 4 \, c^{5} d f x^{5} + 5 \, c^{4} d f x^{4} - 4 \, c^{2} d f x^{2} + 4 \, c d f x - {\left (c^{4} x^{4} - 4 \, c^{3} x^{3} + 6 \, c^{2} x^{2} - 4 \, c x\right )} \sqrt {-c^{2} x^{2} + 1} \sqrt {c d x + d} \sqrt {-c f x + f} \sqrt {d f} - 2 \, d f}{c^{4} x^{4} - 2 \, c^{3} x^{3} + 2 \, c x - 1}\right ) - 2 \, {\left (a c^{2} x^{2} + \sqrt {-c^{2} x^{2} + 1} b c x - a c x + {\left (b c^{2} x^{2} - b c x - 2 \, b\right )} \arcsin \left (c x\right ) - 2 \, a\right )} \sqrt {c d x + d} \sqrt {-c f x + f}}{6 \, {\left (c^{4} d f^{3} x^{3} - c^{3} d f^{3} x^{2} - c^{2} d f^{3} x + c d f^{3}\right )}}, \frac {{\left (b c^{3} x^{3} - b c^{2} x^{2} - b c x + b\right )} \sqrt {-d f} \arctan \left (\frac {{\left (c^{2} x^{2} - 2 \, c x + 2\right )} \sqrt {-c^{2} x^{2} + 1} \sqrt {c d x + d} \sqrt {-c f x + f} \sqrt {-d f}}{c^{4} d f x^{4} - 2 \, c^{3} d f x^{3} - c^{2} d f x^{2} + 2 \, c d f x}\right ) - {\left (a c^{2} x^{2} + \sqrt {-c^{2} x^{2} + 1} b c x - a c x + {\left (b c^{2} x^{2} - b c x - 2 \, b\right )} \arcsin \left (c x\right ) - 2 \, a\right )} \sqrt {c d x + d} \sqrt {-c f x + f}}{3 \, {\left (c^{4} d f^{3} x^{3} - c^{3} d f^{3} x^{2} - c^{2} d f^{3} x + c d f^{3}\right )}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.18 \[ \int \frac {a+b \arcsin (c x)}{\sqrt {d+c d x} (f-c f x)^{5/2}} \, dx=\frac {1}{3} \, b c {\left (\frac {1}{c^{3} \sqrt {d} f^{\frac {5}{2}} x - c^{2} \sqrt {d} f^{\frac {5}{2}}} + \frac {\log \left (c x - 1\right )}{c^{2} \sqrt {d} f^{\frac {5}{2}}}\right )} + \frac {1}{3} \, b {\left (\frac {\sqrt {-c^{2} d f x^{2} + d f}}{c^{3} d f^{3} x^{2} - 2 \, c^{2} d f^{3} x + c d f^{3}} - \frac {\sqrt {-c^{2} d f x^{2} + d f}}{c^{2} d f^{3} x - c d f^{3}}\right )} \arcsin \left (c x\right ) + \frac {1}{3} \, a {\left (\frac {\sqrt {-c^{2} d f x^{2} + d f}}{c^{3} d f^{3} x^{2} - 2 \, c^{2} d f^{3} x + c d f^{3}} - \frac {\sqrt {-c^{2} d f x^{2} + d f}}{c^{2} d f^{3} x - c d f^{3}}\right )} \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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