\(\int (d+c d x)^{5/2} (f-c f x)^{3/2} (a+b \arcsin (c x)) \, dx\) [48]

Optimal result

Integrand size = 30, antiderivative size = 438 \[ \int (d+c d x)^{5/2} (f-c f x)^{3/2} (a+b \arcsin (c x)) \, dx=\frac {b d^2 f x \sqrt {d+c d x} \sqrt {f-c f x}}{5 \sqrt {1-c^2 x^2}}-\frac {3 b c d^2 f x^2 \sqrt {d+c d x} \sqrt {f-c f x}}{16 \sqrt {1-c^2 x^2}}-\frac {2 b c^2 d^2 f x^3 \sqrt {d+c d x} \sqrt {f-c f x}}{15 \sqrt {1-c^2 x^2}}+\frac {b c^4 d^2 f x^5 \sqrt {d+c d x} \sqrt {f-c f x}}{25 \sqrt {1-c^2 x^2}}+\frac {b d^2 f \sqrt {d+c d x} \sqrt {f-c f x} \left (1-c^2 x^2\right )^{3/2}}{16 c}+\frac {3}{8} d^2 f x \sqrt {d+c d x} \sqrt {f-c f x} (a+b \arcsin (c x))+\frac {1}{4} d^2 f x \sqrt {d+c d x} \sqrt {f-c f x} \left (1-c^2 x^2\right ) (a+b \arcsin (c x))-\frac {d^2 f \sqrt {d+c d x} \sqrt {f-c f x} \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))}{5 c}+\frac {3 d^2 f \sqrt {d+c d x} \sqrt {f-c f x} (a+b \arcsin (c x))^2}{16 b c \sqrt {1-c^2 x^2}} \] Output:

1/5*b*d^2*f*x*(c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)/(-c^2*x^2+1)^(1/2)-3/16*b*c 
*d^2*f*x^2*(c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)/(-c^2*x^2+1)^(1/2)-2/15*b*c^2* 
d^2*f*x^3*(c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)/(-c^2*x^2+1)^(1/2)+1/25*b*c^4*d 
^2*f*x^5*(c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)/(-c^2*x^2+1)^(1/2)+1/16*b*d^2*f* 
(c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)*(-c^2*x^2+1)^(3/2)/c+3/8*d^2*f*x*(c*d*x+d 
)^(1/2)*(-c*f*x+f)^(1/2)*(a+b*arcsin(c*x))+1/4*d^2*f*x*(c*d*x+d)^(1/2)*(-c 
*f*x+f)^(1/2)*(-c^2*x^2+1)*(a+b*arcsin(c*x))-1/5*d^2*f*(c*d*x+d)^(1/2)*(-c 
*f*x+f)^(1/2)*(-c^2*x^2+1)^2*(a+b*arcsin(c*x))/c+3/16*d^2*f*(c*d*x+d)^(1/2 
)*(-c*f*x+f)^(1/2)*(a+b*arcsin(c*x))^2/b/c/(-c^2*x^2+1)^(1/2)
 

Mathematica [A] (verified)

Time = 2.47 (sec) , antiderivative size = 305, normalized size of antiderivative = 0.70 \[ \int (d+c d x)^{5/2} (f-c f x)^{3/2} (a+b \arcsin (c x)) \, dx=\frac {d^2 f \left (1800 b \sqrt {d+c d x} \sqrt {f-c f x} \arcsin (c x)^2-3600 a \sqrt {d} \sqrt {f} \sqrt {1-c^2 x^2} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (-1+c^2 x^2\right )}\right )+\sqrt {d+c d x} \sqrt {f-c f x} \left (128 b c x \left (15-10 c^2 x^2+3 c^4 x^4\right )-240 a \sqrt {1-c^2 x^2} \left (8-25 c x-16 c^2 x^2+10 c^3 x^3+8 c^4 x^4\right )+1200 b \cos (2 \arcsin (c x))+75 b \cos (4 \arcsin (c x))\right )-60 b \sqrt {d+c d x} \sqrt {f-c f x} \arcsin (c x) \left (32 \left (1-c^2 x^2\right )^{5/2}-40 \sin (2 \arcsin (c x))-5 \sin (4 \arcsin (c x))\right )\right )}{9600 c \sqrt {1-c^2 x^2}} \] Input:

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

Output:

(d^2*f*(1800*b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]^2 - 3600*a*Sqrt 
[d]*Sqrt[f]*Sqrt[1 - c^2*x^2]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]) 
/(Sqrt[d]*Sqrt[f]*(-1 + c^2*x^2))] + Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(128* 
b*c*x*(15 - 10*c^2*x^2 + 3*c^4*x^4) - 240*a*Sqrt[1 - c^2*x^2]*(8 - 25*c*x 
- 16*c^2*x^2 + 10*c^3*x^3 + 8*c^4*x^4) + 1200*b*Cos[2*ArcSin[c*x]] + 75*b* 
Cos[4*ArcSin[c*x]]) - 60*b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]*(32 
*(1 - c^2*x^2)^(5/2) - 40*Sin[2*ArcSin[c*x]] - 5*Sin[4*ArcSin[c*x]])))/(96 
00*c*Sqrt[1 - c^2*x^2])
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.43, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {5178, 27, 5262, 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.

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

\(\Big \downarrow \) 5178

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 5262

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {d (c d x+d)^{3/2} (f-c f x)^{3/2} \left (\frac {1}{4} x \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))+\frac {3}{8} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))-\frac {\left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))}{5 c}+\frac {3 (a+b \arcsin (c x))^2}{16 b c}+\frac {1}{25} b c^4 x^5+\frac {1}{16} b c^3 x^4-\frac {2}{15} b c^2 x^3-\frac {5}{16} b c x^2+\frac {b x}{5}\right )}{\left (1-c^2 x^2\right )^{3/2}}\)

Input:

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

Output:

(d*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)*((b*x)/5 - (5*b*c*x^2)/16 - (2*b*c^ 
2*x^3)/15 + (b*c^3*x^4)/16 + (b*c^4*x^5)/25 + (3*x*Sqrt[1 - c^2*x^2]*(a + 
b*ArcSin[c*x]))/8 + (x*(1 - c^2*x^2)^(3/2)*(a + b*ArcSin[c*x]))/4 - ((1 - 
c^2*x^2)^(5/2)*(a + b*ArcSin[c*x]))/(5*c) + (3*(a + b*ArcSin[c*x])^2)/(16* 
b*c)))/(1 - c^2*x^2)^(3/2)
 

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 5262

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_ 
) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^p*(a + 
 b*ArcSin[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] & 
& EqQ[c^2*d + e, 0] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ 
[n, 0] && (m == 1 || p > 0 || (n == 1 && p > -1) || (m == 2 && p < -2))
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 2.91 (sec) , antiderivative size = 1194, normalized size of antiderivative = 2.73

method	result	size
default	\(\text {Expression too large to display}\)	\(1194\)
parts	\(\text {Expression too large to display}\)	\(1194\)

Input:

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

Output:

-1/5*a/f/c*(c*d*x+d)^(5/2)*(-c*f*x+f)^(5/2)-1/4*a*d/f/c*(c*d*x+d)^(3/2)*(- 
c*f*x+f)^(5/2)-1/4*a*d^2/f/c*(c*d*x+d)^(1/2)*(-c*f*x+f)^(5/2)+1/8*a*d^2/c* 
(-c*f*x+f)^(3/2)*(c*d*x+d)^(1/2)+3/8*a*d^2*f/c*(-c*f*x+f)^(1/2)*(c*d*x+d)^ 
(1/2)+3/8*a*d^3*f^2*((-c*f*x+f)*(c*d*x+d))^(1/2)/(-c*f*x+f)^(1/2)/(c*d*x+d 
)^(1/2)/(c^2*d*f)^(1/2)*arctan((c^2*d*f)^(1/2)*x/(-c^2*d*f*x^2+d*f)^(1/2)) 
+b*(-3/16*(-f*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)*(-c^2*x^2+1)^(1/2)/c/(c^2*x 
^2-1)*arcsin(c*x)^2*d^2*f-1/800*(d*(c*x+1))^(1/2)*(-f*(c*x-1))^(1/2)*(16*c 
^6*x^6-28*c^4*x^4-16*I*(-c^2*x^2+1)^(1/2)*x^5*c^5+13*c^2*x^2+20*I*(-c^2*x^ 
2+1)^(1/2)*x^3*c^3-5*I*(-c^2*x^2+1)^(1/2)*x*c-1)*(I+5*arcsin(c*x))*d^2*f/c 
/(c^2*x^2-1)-1/256*(d*(c*x+1))^(1/2)*(-f*(c*x-1))^(1/2)*(-8*I*(-c^2*x^2+1) 
^(1/2)*x^4*c^4+8*c^5*x^5+8*I*(-c^2*x^2+1)^(1/2)*x^2*c^2-12*c^3*x^3-I*(-c^2 
*x^2+1)^(1/2)+4*c*x)*(I+4*arcsin(c*x))*d^2*f/c/(c^2*x^2-1)-1/16*(d*(c*x+1) 
)^(1/2)*(-f*(c*x-1))^(1/2)*(I*(-c^2*x^2+1)^(1/2)*c*x+c^2*x^2-1)*(arcsin(c* 
x)-I)*d^2*f/c/(c^2*x^2-1)+1/16*(d*(c*x+1))^(1/2)*(-f*(c*x-1))^(1/2)*(2*I*( 
-c^2*x^2+1)^(1/2)*x^2*c^2+2*c^3*x^3-I*(-c^2*x^2+1)^(1/2)-2*c*x)*(-I+2*arcs 
in(c*x))*d^2*f/c/(c^2*x^2-1)-1/1200*(d*(c*x+1))^(1/2)*(-f*(c*x-1))^(1/2)*( 
I*(-c^2*x^2+1)^(1/2)*c*x+c^2*x^2-1)*(11*I+45*arcsin(c*x))*cos(4*arcsin(c*x 
))*d^2*f/c/(c^2*x^2-1)-1/600*(d*(c*x+1))^(1/2)*(-f*(c*x-1))^(1/2)*(I*c^2*x 
^2-c*x*(-c^2*x^2+1)^(1/2)-I)*(7*I+15*arcsin(c*x))*sin(4*arcsin(c*x))*d^2*f 
/c/(c^2*x^2-1)-1/256*(d*(c*x+1))^(1/2)*(-f*(c*x-1))^(1/2)*(I*c^2*x^2-c*...
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

b*sqrt(d)*sqrt(f)*integrate(-(c^3*d^2*f*x^3 + c^2*d^2*f*x^2 - c*d^2*f*x - 
d^2*f)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 
 1)), x) + 1/40*(15*sqrt(-c^2*d*f*x^2 + d*f)*d^2*f*x + 15*d^3*f^2*arcsin(c 
*x)/(sqrt(d*f)*c) + 10*(-c^2*d*f*x^2 + d*f)^(3/2)*d*x - 8*(-c^2*d*f*x^2 + 
d*f)^(5/2)/(c*f))*a
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

\[ \int (d+c d x)^{5/2} (f-c f x)^{3/2} (a+b \arcsin (c x)) \, dx=\frac {\sqrt {f}\, \sqrt {d}\, d^{2} f \left (-30 \mathit {asin} \left (\frac {\sqrt {-c x +1}}{\sqrt {2}}\right ) a -8 \sqrt {c x +1}\, \sqrt {-c x +1}\, a \,c^{4} x^{4}-10 \sqrt {c x +1}\, \sqrt {-c x +1}\, a \,c^{3} x^{3}+16 \sqrt {c x +1}\, \sqrt {-c x +1}\, a \,c^{2} x^{2}+25 \sqrt {c x +1}\, \sqrt {-c x +1}\, a c x -8 \sqrt {c x +1}\, \sqrt {-c x +1}\, a -40 \left (\int \sqrt {c x +1}\, \sqrt {-c x +1}\, \mathit {asin} \left (c x \right ) x^{3}d x \right ) b \,c^{4}-40 \left (\int \sqrt {c x +1}\, \sqrt {-c x +1}\, \mathit {asin} \left (c x \right ) x^{2}d x \right ) b \,c^{3}+40 \left (\int \sqrt {c x +1}\, \sqrt {-c x +1}\, \mathit {asin} \left (c x \right ) x d x \right ) b \,c^{2}+40 \left (\int \sqrt {c x +1}\, \sqrt {-c x +1}\, \mathit {asin} \left (c x \right )d x \right ) b c \right )}{40 c} \] Input:

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

Output:

(sqrt(f)*sqrt(d)*d**2*f*( - 30*asin(sqrt( - c*x + 1)/sqrt(2))*a - 8*sqrt(c 
*x + 1)*sqrt( - c*x + 1)*a*c**4*x**4 - 10*sqrt(c*x + 1)*sqrt( - c*x + 1)*a 
*c**3*x**3 + 16*sqrt(c*x + 1)*sqrt( - c*x + 1)*a*c**2*x**2 + 25*sqrt(c*x + 
 1)*sqrt( - c*x + 1)*a*c*x - 8*sqrt(c*x + 1)*sqrt( - c*x + 1)*a - 40*int(s 
qrt(c*x + 1)*sqrt( - c*x + 1)*asin(c*x)*x**3,x)*b*c**4 - 40*int(sqrt(c*x + 
 1)*sqrt( - c*x + 1)*asin(c*x)*x**2,x)*b*c**3 + 40*int(sqrt(c*x + 1)*sqrt( 
 - c*x + 1)*asin(c*x)*x,x)*b*c**2 + 40*int(sqrt(c*x + 1)*sqrt( - c*x + 1)* 
asin(c*x),x)*b*c))/(40*c)