\(\int \frac {(f-c f x)^{5/2} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx\) [59]

Optimal result

Integrand size = 30, antiderivative size = 432 \[ \int \frac {(f-c f x)^{5/2} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=-\frac {b f^3 x \sqrt {1-c^2 x^2}}{d^2 \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {8 b f^3 \sqrt {1-c^2 x^2}}{3 c d^2 (1+c x) \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {5 b f^3 \sqrt {1-c^2 x^2} \arcsin (c x)^2}{2 c d^2 \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {20 f^3 (1-c x) (a+b \arcsin (c x))}{3 c d^2 \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {2 f^3 (1-c x)^4 (a+b \arcsin (c x))}{3 c d^2 \sqrt {d+c d x} \sqrt {f-c f x} \left (1-c^2 x^2\right )}+\frac {5 f^3 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{3 c d^2 \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {5 f^3 \sqrt {1-c^2 x^2} \arcsin (c x) (a+b \arcsin (c x))}{c d^2 \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {28 b f^3 \sqrt {1-c^2 x^2} \log (1+c x)}{3 c d^2 \sqrt {d+c d x} \sqrt {f-c f x}} \] Output:

-b*f^3*x*(-c^2*x^2+1)^(1/2)/d^2/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)-8/3*b*f^3 
*(-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)-5/2*b*f 
^3*(-c^2*x^2+1)^(1/2)*arcsin(c*x)^2/c/d^2/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2) 
+20/3*f^3*(-c*x+1)*(a+b*arcsin(c*x))/c/d^2/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2 
)-2/3*f^3*(-c*x+1)^4*(a+b*arcsin(c*x))/c/d^2/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1 
/2)/(-c^2*x^2+1)+5/3*f^3*(-c^2*x^2+1)*(a+b*arcsin(c*x))/c/d^2/(c*d*x+d)^(1 
/2)/(-c*f*x+f)^(1/2)+5*f^3*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*(a+b*arcsin(c*x) 
)/c/d^2/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)-28/3*b*f^3*(-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 = 9.05 (sec) , antiderivative size = 847, normalized size of antiderivative = 1.96 \[ \int \frac {(f-c f x)^{5/2} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(f^2*((4*a*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(23 + 34*c*x + 3*c^2*x^2))/(1 + 
 c*x)^2 - 60*a*Sqrt[d]*Sqrt[f]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x] 
)/(Sqrt[d]*Sqrt[f]*(-1 + c^2*x^2))] + (2*b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x] 
*(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2])*(Cos[ArcSin[c*x]/2]*(-8 + 6*Arc 
Sin[c*x] + 9*ArcSin[c*x]^2 - 84*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2 
]]) + Cos[(3*ArcSin[c*x])/2]*((14 - 3*ArcSin[c*x])*ArcSin[c*x] + 28*Log[Co 
s[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]]) + 2*(-4 + 2*(2 + 7*Sqrt[1 - c^2*x^ 
2])*ArcSin[c*x] + 3*(2 + Sqrt[1 - c^2*x^2])*ArcSin[c*x]^2 - 28*(2 + Sqrt[1 
 - c^2*x^2])*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]])*Sin[ArcSin[c*x] 
/2]))/((1 - c*x)*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^4) + (2*b*Sqrt[ 
d + c*d*x]*Sqrt[f - c*f*x]*(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2])*(Cos[ 
(3*ArcSin[c*x])/2]*(ArcSin[c*x] + 2*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c* 
x]/2]]) - Cos[ArcSin[c*x]/2]*(4 + 3*ArcSin[c*x] + 6*Log[Cos[ArcSin[c*x]/2] 
 + Sin[ArcSin[c*x]/2]]) + 2*(-2 + (2 + Sqrt[1 - c^2*x^2])*ArcSin[c*x] - 2* 
(2 + Sqrt[1 - c^2*x^2])*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]])*Sin[ 
ArcSin[c*x]/2]))/((1 - c*x)*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^4) + 
 (b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/ 
2])*(2*(4 + 6*c*x + 6*c^2*x^2 + 52*(1 + c*x)*Log[Cos[ArcSin[c*x]/2] + Sin[ 
ArcSin[c*x]/2]])*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]) - 18*ArcSin[c*x 
]^2*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^3 + ArcSin[c*x]*(-24*Cos[...
 

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.48, 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.

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

\(\Big \downarrow \) 5178

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 5260

\(\displaystyle \frac {f^5 \left (1-c^2 x^2\right )^{5/2} \left (-b c \int \left (-\frac {2 (1-c x)^4}{3 c \left (1-c^2 x^2\right )^2}+\frac {20 (1-c x)}{3 c \left (1-c^2 x^2\right )}+\frac {5 \arcsin (c x)}{c \sqrt {1-c^2 x^2}}+\frac {5}{3 c}\right )dx-\frac {2 (1-c x)^4 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}+\frac {20 (1-c x) (a+b \arcsin (c x))}{3 c \sqrt {1-c^2 x^2}}+\frac {5 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{3 c}+\frac {5 \arcsin (c x) (a+b \arcsin (c x))}{c}\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {f^5 \left (1-c^2 x^2\right )^{5/2} \left (-\frac {2 (1-c x)^4 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}+\frac {20 (1-c x) (a+b \arcsin (c x))}{3 c \sqrt {1-c^2 x^2}}+\frac {5 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{3 c}+\frac {5 \arcsin (c x) (a+b \arcsin (c x))}{c}-b c \left (\frac {5 \arcsin (c x)^2}{2 c^2}+\frac {8}{3 c^2 (c x+1)}+\frac {28 \log (c x+1)}{3 c^2}+\frac {x}{c}\right )\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\)

Input:

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

Output:

(f^5*(1 - c^2*x^2)^(5/2)*((-2*(1 - c*x)^4*(a + b*ArcSin[c*x]))/(3*c*(1 - c 
^2*x^2)^(3/2)) + (20*(1 - c*x)*(a + b*ArcSin[c*x]))/(3*c*Sqrt[1 - c^2*x^2] 
) + (5*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(3*c) + (5*ArcSin[c*x]*(a + 
b*ArcSin[c*x]))/c - b*c*(x/c + 8/(3*c^2*(1 + c*x)) + (5*ArcSin[c*x]^2)/(2* 
c^2) + (28*Log[1 + c*x])/(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 = 3.67 (sec) , antiderivative size = 407, normalized size of antiderivative = 0.94

method	result	size
default	\(-\frac {\sqrt {-f \left (c x -1\right )}\, \sqrt {d \left (c x +1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \left (-16 b +60 \arcsin \left (c x \right ) a c x +68 \sqrt {-c^{2} x^{2}+1}\, a c x -22 x b c -6 b \,c^{3} x^{3}+6 a \,c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}+15 \arcsin \left (c x \right )^{2} b \,c^{2} x^{2}+30 \arcsin \left (c x \right ) a \,c^{2} x^{2}-12 x^{2} c^{2} b +6 b \,c^{2} x^{2} \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}+30 b c x \arcsin \left (c x \right )^{2}+46 \sqrt {-c^{2} x^{2}+1}\, a +68 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, b c x +46 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, b +15 \arcsin \left (c x \right )^{2} b +30 \arcsin \left (c x \right ) a +56 i b \arcsin \left (c x \right )-56 i a -112 b \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right )-112 i a c x -224 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right ) b c x +112 i \arcsin \left (c x \right ) b c x +56 i \arcsin \left (c x \right ) b \,c^{2} x^{2}-112 \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right ) b \,c^{2} x^{2}-56 i a \,c^{2} x^{2}\right ) f^{2}}{6 c \,d^{3} \left (c^{4} x^{4}+2 c^{3} x^{3}-2 c x -1\right )}\)	\(407\)

Input:

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

Output:

-1/6*(-f*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)*(-c^2*x^2+1)^(1/2)/c/d^3/(c^4*x^ 
4+2*c^3*x^3-2*c*x-1)*(-16*b+60*arcsin(c*x)*a*c*x+68*(-c^2*x^2+1)^(1/2)*a*c 
*x-22*x*b*c-6*b*c^3*x^3+6*a*c^2*x^2*(-c^2*x^2+1)^(1/2)+15*arcsin(c*x)^2*b* 
c^2*x^2+30*arcsin(c*x)*a*c^2*x^2-12*x^2*c^2*b+6*b*c^2*x^2*arcsin(c*x)*(-c^ 
2*x^2+1)^(1/2)+30*b*c*x*arcsin(c*x)^2+46*(-c^2*x^2+1)^(1/2)*a+68*arcsin(c* 
x)*(-c^2*x^2+1)^(1/2)*b*c*x+46*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*b+15*arcsin( 
c*x)^2*b+30*arcsin(c*x)*a-112*b*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)-224*ln(I*c* 
x+(-c^2*x^2+1)^(1/2)+I)*b*c*x-112*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)*b*c^2*x^2 
+112*I*arcsin(c*x)*b*c*x+56*I*arcsin(c*x)*b*c^2*x^2+56*I*b*arcsin(c*x)-56* 
I*a-112*I*a*c*x-56*I*a*c^2*x^2)*f^2
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

1/3*(3*(-c^2*d*f*x^2 + d*f)^(5/2)/(c^5*d^5*x^4 + 4*c^4*d^5*x^3 + 6*c^3*d^5 
*x^2 + 4*c^2*d^5*x + c*d^5) - 5*(-c^2*d*f*x^2 + d*f)^(3/2)*f/(c^4*d^4*x^3 
+ 3*c^3*d^4*x^2 + 3*c^2*d^4*x + c*d^4) - 10*sqrt(-c^2*d*f*x^2 + d*f)*f^2/( 
c^3*d^3*x^2 + 2*c^2*d^3*x + c*d^3) + 35*sqrt(-c^2*d*f*x^2 + d*f)*f^2/(c^2* 
d^3*x + c*d^3) + 15*f^3*arcsin(c*x)/(c*d^3*sqrt(f/d)))*a + b*sqrt(f)*integ 
rate((c^2*f^2*x^2 - 2*c*f^2*x + f^2)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x 
+ 1)*sqrt(-c*x + 1))/((c^2*d^2*x^2 + 2*c*d^2*x + d^2)*sqrt(c*x + 1)), x)/s 
qrt(d)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(f-c f x)^{5/2} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=\frac {\sqrt {f}\, f^{2} \left (-30 \sqrt {c x +1}\, \mathit {asin} \left (\frac {\sqrt {-c x +1}}{\sqrt {2}}\right ) a c x -30 \sqrt {c x +1}\, \mathit {asin} \left (\frac {\sqrt {-c x +1}}{\sqrt {2}}\right ) a +3 \sqrt {-c x +1}\, a \,c^{2} x^{2}+34 \sqrt {-c x +1}\, a c x +23 \sqrt {-c x +1}\, a +3 \sqrt {c x +1}\, \left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right ) x^{2}}{\sqrt {c x +1}\, c^{2} x^{2}+2 \sqrt {c x +1}\, c x +\sqrt {c x +1}}d x \right ) b \,c^{4} x +3 \sqrt {c x +1}\, \left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right ) x^{2}}{\sqrt {c x +1}\, c^{2} x^{2}+2 \sqrt {c x +1}\, c x +\sqrt {c x +1}}d x \right ) b \,c^{3}-6 \sqrt {c x +1}\, \left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right ) x}{\sqrt {c x +1}\, c^{2} x^{2}+2 \sqrt {c x +1}\, c x +\sqrt {c x +1}}d x \right ) b \,c^{3} x -6 \sqrt {c x +1}\, \left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right ) x}{\sqrt {c x +1}\, c^{2} x^{2}+2 \sqrt {c x +1}\, c x +\sqrt {c x +1}}d x \right ) b \,c^{2}+3 \sqrt {c x +1}\, \left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right )}{\sqrt {c x +1}\, c^{2} x^{2}+2 \sqrt {c x +1}\, c x +\sqrt {c x +1}}d x \right ) b \,c^{2} x +3 \sqrt {c x +1}\, \left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right )}{\sqrt {c x +1}\, c^{2} x^{2}+2 \sqrt {c x +1}\, c x +\sqrt {c x +1}}d x \right ) b c \right )}{3 \sqrt {d}\, \sqrt {c x +1}\, c \,d^{2} \left (c x +1\right )} \] Input:

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

Output:

(sqrt(f)*f**2*( - 30*sqrt(c*x + 1)*asin(sqrt( - c*x + 1)/sqrt(2))*a*c*x - 
30*sqrt(c*x + 1)*asin(sqrt( - c*x + 1)/sqrt(2))*a + 3*sqrt( - c*x + 1)*a*c 
**2*x**2 + 34*sqrt( - c*x + 1)*a*c*x + 23*sqrt( - c*x + 1)*a + 3*sqrt(c*x 
+ 1)*int((sqrt( - c*x + 1)*asin(c*x)*x**2)/(sqrt(c*x + 1)*c**2*x**2 + 2*sq 
rt(c*x + 1)*c*x + sqrt(c*x + 1)),x)*b*c**4*x + 3*sqrt(c*x + 1)*int((sqrt( 
- c*x + 1)*asin(c*x)*x**2)/(sqrt(c*x + 1)*c**2*x**2 + 2*sqrt(c*x + 1)*c*x 
+ sqrt(c*x + 1)),x)*b*c**3 - 6*sqrt(c*x + 1)*int((sqrt( - c*x + 1)*asin(c* 
x)*x)/(sqrt(c*x + 1)*c**2*x**2 + 2*sqrt(c*x + 1)*c*x + sqrt(c*x + 1)),x)*b 
*c**3*x - 6*sqrt(c*x + 1)*int((sqrt( - c*x + 1)*asin(c*x)*x)/(sqrt(c*x + 1 
)*c**2*x**2 + 2*sqrt(c*x + 1)*c*x + sqrt(c*x + 1)),x)*b*c**2 + 3*sqrt(c*x 
+ 1)*int((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)*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))/(3*sqrt(d)*sqrt(c*x + 1)*c*d**2*(c*x + 1))