3.6.39 \(\int \frac {a+b \arcsin (c x)}{(d+c d x)^{5/2} (f-c f x)^{5/2}} \, dx\) [539]

3.6.39.1 Optimal result

Integrand size = 30, antiderivative size = 188 \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{5/2} (f-c f x)^{5/2}} \, dx=-\frac {b \left (1-c^2 x^2\right )^{3/2}}{6 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {x \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {2 x \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {b \left (1-c^2 x^2\right )^{5/2} \log \left (1-c^2 x^2\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}} \]

output

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

3.6.39.2 Mathematica [A] (verified)

Time = 1.24 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.95 \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{5/2} (f-c f x)^{5/2}} \, dx=\frac {\sqrt {d+c d x} \left (-6 a c x+4 a c^3 x^3+b \sqrt {1-c^2 x^2}+2 b c x \left (-3+2 c^2 x^2\right ) \arcsin (c x)-2 b \left (1-c^2 x^2\right )^{3/2} \log (-f (1+c x))-2 b \sqrt {1-c^2 x^2} \log (f-c f x)+2 b c^2 x^2 \sqrt {1-c^2 x^2} \log (f-c f x)\right )}{6 c d^3 (-1+c x) \sqrt {f-c f x} (f+c f x)^2} \]

input

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

output

(Sqrt[d + c*d*x]*(-6*a*c*x + 4*a*c^3*x^3 + b*Sqrt[1 - c^2*x^2] + 2*b*c*x*( 
-3 + 2*c^2*x^2)*ArcSin[c*x] - 2*b*(1 - c^2*x^2)^(3/2)*Log[-(f*(1 + c*x))] 
- 2*b*Sqrt[1 - c^2*x^2]*Log[f - c*f*x] + 2*b*c^2*x^2*Sqrt[1 - c^2*x^2]*Log 
[f - c*f*x]))/(6*c*d^3*(-1 + c*x)*Sqrt[f - c*f*x]*(f + c*f*x)^2)
 

3.6.39.3 Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.70, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5178, 5162, 241, 5160, 240}

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.

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

\(\Big \downarrow \) 5178

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

\(\Big \downarrow \) 5162

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

\(\Big \downarrow \) 241

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

\(\Big \downarrow \) 5160

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

\(\Big \downarrow \) 240

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

input

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

output

((1 - c^2*x^2)^(5/2)*(-1/6*b/(c*(1 - c^2*x^2)) + (x*(a + b*ArcSin[c*x]))/( 
3*(1 - c^2*x^2)^(3/2)) + (2*((x*(a + b*ArcSin[c*x]))/Sqrt[1 - c^2*x^2] + ( 
b*Log[1 - c^2*x^2])/(2*c)))/3))/((d + c*d*x)^(5/2)*(f - c*f*x)^(5/2))
 

3.6.39.3.1 Defintions of rubi rules used

rule 240

Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x 
^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
 

rule 241

Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x^2)^(p + 1)/ 
(2*b*(p + 1)), x] /; FreeQ[{a, b, p}, x] && NeQ[p, -1]
 

rule 5160

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x 
_Symbol] :> Simp[x*((a + b*ArcSin[c*x])^n/(d*Sqrt[d + e*x^2])), x] - Simp[b 
*c*(n/d)*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]   Int[x*((a + b*ArcSin[c*x 
])^(n - 1)/(1 - c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d 
 + e, 0] && GtQ[n, 0]
 

rule 5162

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_ 
Symbol] :> Simp[(-x)*(d + e*x^2)^(p + 1)*((a + b*ArcSin[c*x])^n/(2*d*(p + 1 
))), x] + (Simp[(2*p + 3)/(2*d*(p + 1))   Int[(d + e*x^2)^(p + 1)*(a + b*Ar 
cSin[c*x])^n, x], x] + Simp[b*c*(n/(2*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2 
*x^2)^p]   Int[x*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x 
]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, 
 -1] && NeQ[p, -3/2]
 

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]
 

3.6.39.4 Maple [F]

\[\int \frac {a +b \arcsin \left (c x \right )}{\left (c d x +d \right )^{\frac {5}{2}} \left (-c f x +f \right )^{\frac {5}{2}}}d x\]

input

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

output

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

3.6.39.5 Fricas [F]

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

input

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

output

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

3.6.39.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.6.39.7 Maxima [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.94 \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{5/2} (f-c f x)^{5/2}} \, dx=\frac {1}{6} \, b c {\left (\frac {1}{c^{4} d^{\frac {5}{2}} f^{\frac {5}{2}} x^{2} - c^{2} d^{\frac {5}{2}} f^{\frac {5}{2}}} + \frac {2 \, \log \left (c x + 1\right )}{c^{2} d^{\frac {5}{2}} f^{\frac {5}{2}}} + \frac {2 \, \log \left (c x - 1\right )}{c^{2} d^{\frac {5}{2}} f^{\frac {5}{2}}}\right )} + \frac {1}{3} \, b {\left (\frac {x}{{\left (-c^{2} d f x^{2} + d f\right )}^{\frac {3}{2}} d f} + \frac {2 \, x}{\sqrt {-c^{2} d f x^{2} + d f} d^{2} f^{2}}\right )} \arcsin \left (c x\right ) + \frac {1}{3} \, a {\left (\frac {x}{{\left (-c^{2} d f x^{2} + d f\right )}^{\frac {3}{2}} d f} + \frac {2 \, x}{\sqrt {-c^{2} d f x^{2} + d f} d^{2} f^{2}}\right )} \]

input

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

output

1/6*b*c*(1/(c^4*d^(5/2)*f^(5/2)*x^2 - c^2*d^(5/2)*f^(5/2)) + 2*log(c*x + 1 
)/(c^2*d^(5/2)*f^(5/2)) + 2*log(c*x - 1)/(c^2*d^(5/2)*f^(5/2))) + 1/3*b*(x 
/((-c^2*d*f*x^2 + d*f)^(3/2)*d*f) + 2*x/(sqrt(-c^2*d*f*x^2 + d*f)*d^2*f^2) 
)*arcsin(c*x) + 1/3*a*(x/((-c^2*d*f*x^2 + d*f)^(3/2)*d*f) + 2*x/(sqrt(-c^2 
*d*f*x^2 + d*f)*d^2*f^2))
 

3.6.39.8 Giac [F]

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

input

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

output

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

3.6.39.9 Mupad [F(-1)]

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

input

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

output

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