\(\int \frac {(d+c d x)^{5/2} (a+b \arccos (c x))^2}{(e-c e x)^{5/2}} \, dx\) [572]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [F]
Sympy [F(-1)]
Maxima [F(-2)]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 32, antiderivative size = 730 \[ \int \frac {(d+c d x)^{5/2} (a+b \arccos (c x))^2}{(e-c e x)^{5/2}} \, dx=\frac {2 a b d^5 x \left (1-c^2 x^2\right )^{5/2}}{(d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {2 b^2 d^5 \left (1-c^2 x^2\right )^3}{c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {2 b^2 d^5 x \left (1-c^2 x^2\right )^{5/2} \arccos (c x)}{(d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {28 i d^5 \left (1-c^2 x^2\right )^{5/2} (a+b \arccos (c x))^2}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {d^5 \left (1-c^2 x^2\right )^3 (a+b \arccos (c x))^2}{c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {5 d^5 \left (1-c^2 x^2\right )^{5/2} (a+b \arccos (c x))^3}{3 b c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {112 b d^5 \left (1-c^2 x^2\right )^{5/2} (a+b \arccos (c x)) \log \left (1-i e^{-i \arccos (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {112 i b^2 d^5 \left (1-c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,i e^{-i \arccos (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {8 b d^5 \left (1-c^2 x^2\right )^{5/2} (a+b \arccos (c x)) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} \arccos (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {16 b^2 d^5 \left (1-c^2 x^2\right )^{5/2} \tan \left (\frac {\pi }{4}+\frac {1}{2} \arccos (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {28 d^5 \left (1-c^2 x^2\right )^{5/2} (a+b \arccos (c x))^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} \arccos (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {4 d^5 \left (1-c^2 x^2\right )^{5/2} (a+b \arccos (c x))^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} \arccos (c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} \arccos (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}} \] Output:

2*a*b*d^5*x*(-c^2*x^2+1)^(5/2)/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)+2*b^2*d^5* 
(-c^2*x^2+1)^3/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)+2*b^2*d^5*x*(-c^2*x^2+1) 
^(5/2)*arccos(c*x)/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)-28/3*I*d^5*(-c^2*x^2+1 
)^(5/2)*(a+b*arccos(c*x))^2/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)-d^5*(-c^2*x 
^2+1)^3*(a+b*arccos(c*x))^2/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)+5/3*d^5*(-c 
^2*x^2+1)^(5/2)*(a+b*arccos(c*x))^3/b/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)-1 
12/3*b*d^5*(-c^2*x^2+1)^(5/2)*(a+b*arccos(c*x))*ln(1-I/(c*x+I*(-c^2*x^2+1) 
^(1/2)))/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)-112/3*I*b^2*d^5*(-c^2*x^2+1)^( 
5/2)*polylog(2,I/(c*x+I*(-c^2*x^2+1)^(1/2)))/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^ 
(5/2)-8/3*b*d^5*(-c^2*x^2+1)^(5/2)*(a+b*arccos(c*x))*sec(1/4*Pi+1/2*arccos 
(c*x))^2/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)+16/3*b^2*d^5*(-c^2*x^2+1)^(5/2 
)*tan(1/4*Pi+1/2*arccos(c*x))/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)-28/3*d^5* 
(-c^2*x^2+1)^(5/2)*(a+b*arccos(c*x))^2*tan(1/4*Pi+1/2*arccos(c*x))/c/(c*d* 
x+d)^(5/2)/(-c*e*x+e)^(5/2)+4/3*d^5*(-c^2*x^2+1)^(5/2)*(a+b*arccos(c*x))^2 
*sec(1/4*Pi+1/2*arccos(c*x))^2*tan(1/4*Pi+1/2*arccos(c*x))/c/(c*d*x+d)^(5/ 
2)/(-c*e*x+e)^(5/2)
 

Mathematica [A] (verified)

Time = 15.52 (sec) , antiderivative size = 568, normalized size of antiderivative = 0.78 \[ \int \frac {(d+c d x)^{5/2} (a+b \arccos (c x))^2}{(e-c e x)^{5/2}} \, dx=\frac {d^2 \sqrt {1-c^2 x^2} \left (15 a^2 \sqrt {d} \sqrt {e} (-1+c x)^3 \sqrt {1-c^2 x^2} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {e-c e x}}{\sqrt {d} \sqrt {e} \left (-1+c^2 x^2\right )}\right )-896 i b^2 \sqrt {d+c d x} \sqrt {e-c e x} \operatorname {PolyLog}\left (2,e^{i \arccos (c x)}\right ) \sin ^6\left (\frac {1}{2} \arccos (c x)\right )+\sqrt {d+c d x} \sqrt {e-c e x} \left (a^2 \sqrt {1-c^2 x^2} \left (-23+57 c x-37 c^2 x^2+3 c^3 x^3\right )-2 b \left (15 a (-1+c x)^2+b \left (28 i+23 \sqrt {1-c^2 x^2}+c^2 x^2 \left (28 i+3 \sqrt {1-c^2 x^2}\right )-2 c x \left (28 i+17 \sqrt {1-c^2 x^2}\right )\right )\right ) \arccos (c x)^2 \sin ^2\left (\frac {1}{2} \arccos (c x)\right )+64 b \left (a+b \sqrt {1-c^2 x^2}\right ) \sin ^4\left (\frac {1}{2} \arccos (c x)\right )-40 b^2 \arccos (c x)^3 \sin ^6\left (\frac {1}{2} \arccos (c x)\right )+16 b \left (-3 a c x+3 b \sqrt {1-c^2 x^2}+56 a \log \left (\sin \left (\frac {1}{2} \arccos (c x)\right )\right )\right ) \sin ^6\left (\frac {1}{2} \arccos (c x)\right )-16 b \arccos (c x) \sin ^2\left (\frac {1}{2} \arccos (c x)\right ) \left (-2 a \sqrt {1-c^2 x^2}+\left (-4 b+14 a \sqrt {1-c^2 x^2}\right ) \sin ^2\left (\frac {1}{2} \arccos (c x)\right )+\left (3 b c x+3 a \sqrt {1-c^2 x^2}-56 b \log \left (1-e^{i \arccos (c x)}\right )\right ) \sin ^4\left (\frac {1}{2} \arccos (c x)\right )\right )\right )\right )}{3 c e^3 (-1+c x)^4 (1+c x)} \] Input:

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

Output:

(d^2*Sqrt[1 - c^2*x^2]*(15*a^2*Sqrt[d]*Sqrt[e]*(-1 + c*x)^3*Sqrt[1 - c^2*x 
^2]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])/(Sqrt[d]*Sqrt[e]*(-1 + c^ 
2*x^2))] - (896*I)*b^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*PolyLog[2, E^(I*Arc 
Cos[c*x])]*Sin[ArcCos[c*x]/2]^6 + Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(a^2*Sqr 
t[1 - c^2*x^2]*(-23 + 57*c*x - 37*c^2*x^2 + 3*c^3*x^3) - 2*b*(15*a*(-1 + c 
*x)^2 + b*(28*I + 23*Sqrt[1 - c^2*x^2] + c^2*x^2*(28*I + 3*Sqrt[1 - c^2*x^ 
2]) - 2*c*x*(28*I + 17*Sqrt[1 - c^2*x^2])))*ArcCos[c*x]^2*Sin[ArcCos[c*x]/ 
2]^2 + 64*b*(a + b*Sqrt[1 - c^2*x^2])*Sin[ArcCos[c*x]/2]^4 - 40*b^2*ArcCos 
[c*x]^3*Sin[ArcCos[c*x]/2]^6 + 16*b*(-3*a*c*x + 3*b*Sqrt[1 - c^2*x^2] + 56 
*a*Log[Sin[ArcCos[c*x]/2]])*Sin[ArcCos[c*x]/2]^6 - 16*b*ArcCos[c*x]*Sin[Ar 
cCos[c*x]/2]^2*(-2*a*Sqrt[1 - c^2*x^2] + (-4*b + 14*a*Sqrt[1 - c^2*x^2])*S 
in[ArcCos[c*x]/2]^2 + (3*b*c*x + 3*a*Sqrt[1 - c^2*x^2] - 56*b*Log[1 - E^(I 
*ArcCos[c*x])])*Sin[ArcCos[c*x]/2]^4))))/(3*c*e^3*(-1 + c*x)^4*(1 + c*x))
 

Rubi [A] (verified)

Time = 1.55 (sec) , antiderivative size = 309, normalized size of antiderivative = 0.42, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5179, 27, 5275, 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.

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

\(\Big \downarrow \) 5179

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 5275

\(\displaystyle \frac {d^5 \left (1-c^2 x^2\right )^{5/2} \int \left (\frac {c x (a+b \arccos (c x))^2}{\sqrt {1-c^2 x^2}}+\frac {12 (a+b \arccos (c x))^2}{(c x-1) \sqrt {1-c^2 x^2}}+\frac {8 (a+b \arccos (c x))^2}{(c x-1)^2 \sqrt {1-c^2 x^2}}+\frac {5 (a+b \arccos (c x))^2}{\sqrt {1-c^2 x^2}}\right )dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {d^5 \left (1-c^2 x^2\right )^{5/2} \left (-\frac {\sqrt {1-c^2 x^2} (a+b \arccos (c x))^2}{c}-\frac {5 (a+b \arccos (c x))^3}{3 b c}-\frac {28 i (a+b \arccos (c x))^2}{3 c}+\frac {112 b \log \left (1-e^{i \arccos (c x)}\right ) (a+b \arccos (c x))}{3 c}-\frac {28 \cot \left (\frac {1}{2} \arccos (c x)\right ) (a+b \arccos (c x))^2}{3 c}+\frac {8 b \csc ^2\left (\frac {1}{2} \arccos (c x)\right ) (a+b \arccos (c x))}{3 c}+\frac {4 \cot \left (\frac {1}{2} \arccos (c x)\right ) \csc ^2\left (\frac {1}{2} \arccos (c x)\right ) (a+b \arccos (c x))^2}{3 c}-2 a b x-\frac {112 i b^2 \operatorname {PolyLog}\left (2,e^{i \arccos (c x)}\right )}{3 c}-2 b^2 x \arccos (c x)+\frac {16 b^2 \cot \left (\frac {1}{2} \arccos (c x)\right )}{3 c}+\frac {2 b^2 \sqrt {1-c^2 x^2}}{c}\right )}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\)

Input:

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

Output:

(d^5*(1 - c^2*x^2)^(5/2)*(-2*a*b*x + (2*b^2*Sqrt[1 - c^2*x^2])/c - 2*b^2*x 
*ArcCos[c*x] - (((28*I)/3)*(a + b*ArcCos[c*x])^2)/c - (Sqrt[1 - c^2*x^2]*( 
a + b*ArcCos[c*x])^2)/c - (5*(a + b*ArcCos[c*x])^3)/(3*b*c) + (16*b^2*Cot[ 
ArcCos[c*x]/2])/(3*c) - (28*(a + b*ArcCos[c*x])^2*Cot[ArcCos[c*x]/2])/(3*c 
) + (8*b*(a + b*ArcCos[c*x])*Csc[ArcCos[c*x]/2]^2)/(3*c) + (4*(a + b*ArcCo 
s[c*x])^2*Cot[ArcCos[c*x]/2]*Csc[ArcCos[c*x]/2]^2)/(3*c) + (112*b*(a + b*A 
rcCos[c*x])*Log[1 - E^(I*ArcCos[c*x])])/(3*c) - (((112*I)/3)*b^2*PolyLog[2 
, E^(I*ArcCos[c*x])])/c))/((d + c*d*x)^(5/2)*(e - c*e*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 5179
Int[((a_.) + ArcCos[(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*ArcCos[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 5275
Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_ 
) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcCos[c*x] 
)^n/Sqrt[d + e*x^2], (f + g*x)^m*(d + e*x^2)^(p + 1/2), x], x] /; FreeQ[{a, 
 b, c, d, e, f, g}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && ILtQ[p + 1/2, 
 0] && GtQ[d, 0] && IGtQ[n, 0]
 
Maple [A] (verified)

Time = 3.62 (sec) , antiderivative size = 974, normalized size of antiderivative = 1.33

method result size
default \(\frac {5 \sqrt {d \left (c x +1\right )}\, \sqrt {-e \left (c x -1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \left (a +b \arccos \left (c x \right )\right )^{3} d^{2}}{3 \left (c x -1\right ) e^{3} c \left (c x +1\right ) b}-\frac {\sqrt {-e \left (c x -1\right )}\, \sqrt {d \left (c x +1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, x c +c^{2} x^{2}-1\right ) \left (\arccos \left (c x \right )^{2} b^{2}+2 \arccos \left (c x \right ) a b +a^{2}-2 b^{2}+2 i \arccos \left (c x \right ) b^{2}+2 i a b \right ) d^{2}}{2 \left (c x -1\right ) e^{3} c \left (c x +1\right )}-\frac {\sqrt {-e \left (c x -1\right )}\, \sqrt {d \left (c x +1\right )}\, \left (-i \sqrt {-c^{2} x^{2}+1}\, x c +c^{2} x^{2}-1\right ) \left (\arccos \left (c x \right )^{2} b^{2}+2 \arccos \left (c x \right ) a b +a^{2}-2 b^{2}-2 i \arccos \left (c x \right ) b^{2}-2 i a b \right ) d^{2}}{2 \left (c x -1\right ) e^{3} c \left (c x +1\right )}+\frac {4 \sqrt {-e \left (c x -1\right )}\, \sqrt {d \left (c x +1\right )}\, \left (7 c^{2} x^{2}+2 c x -5-7 i \sqrt {-c^{2} x^{2}+1}\, x c +7 i \sqrt {-c^{2} x^{2}+1}\right ) \left (63 b^{2} c^{2} x^{2} \arccos \left (c x \right )^{2}+28 i a b c x +126 a b \,c^{2} x^{2} \arccos \left (c x \right )+4 i \sqrt {-c^{2} x^{2}+1}\, b^{2} c x -96 \arccos \left (c x \right )^{2} b^{2} c x +14 \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right ) b^{2} c x -14 i a b -14 i \arccos \left (c x \right ) b^{2}+63 a^{2} c^{2} x^{2}-32 x^{2} c^{2} b^{2}-192 \arccos \left (c x \right ) a b c x +14 \sqrt {-c^{2} x^{2}+1}\, a b c x +28 i \arccos \left (c x \right ) b^{2} c x +37 \arccos \left (c x \right )^{2} b^{2}-10 \arccos \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, b^{2}-14 i \arccos \left (c x \right ) b^{2} c^{2} x^{2}-14 i a b \,c^{2} x^{2}-96 a^{2} c x +56 c x \,b^{2}+74 \arccos \left (c x \right ) a b -10 \sqrt {-c^{2} x^{2}+1}\, a b -4 i \sqrt {-c^{2} x^{2}+1}\, b^{2}+37 a^{2}-24 b^{2}\right ) d^{2}}{3 \left (63 c^{4} x^{4}-222 c^{3} x^{3}+292 c^{2} x^{2}-170 c x +37\right ) e^{3} c \left (c x +1\right )}+\frac {56 i \sqrt {-c^{2} x^{2}+1}\, \sqrt {d \left (c x +1\right )}\, \sqrt {-e \left (c x -1\right )}\, b \left (\arccos \left (c x \right )^{2} b +2 i \arccos \left (c x \right ) \ln \left (\sqrt {c x +i \sqrt {-c^{2} x^{2}+1}}+1\right ) b +2 i \arccos \left (c x \right ) \ln \left (1-\sqrt {c x +i \sqrt {-c^{2} x^{2}+1}}\right ) b +4 \operatorname {polylog}\left (2, \sqrt {c x +i \sqrt {-c^{2} x^{2}+1}}\right ) b +4 \operatorname {polylog}\left (2, -\sqrt {c x +i \sqrt {-c^{2} x^{2}+1}}\right ) b -2 i \ln \left (c x +i \sqrt {-c^{2} x^{2}+1}\right ) a +2 i \ln \left (\sqrt {c x +i \sqrt {-c^{2} x^{2}+1}}-1\right ) a +2 i \ln \left (\sqrt {c x +i \sqrt {-c^{2} x^{2}+1}}+1\right ) a \right ) d^{2}}{3 \left (c x -1\right ) e^{3} c \left (c x +1\right )}\) \(974\)

Input:

int((c*d*x+d)^(5/2)*(a+b*arccos(c*x))^2/(-c*e*x+e)^(5/2),x,method=_RETURNV 
ERBOSE)
 

Output:

5/3*(d*(c*x+1))^(1/2)*(-e*(c*x-1))^(1/2)*(-c^2*x^2+1)^(1/2)/(c*x-1)/e^3/c/ 
(c*x+1)*(a+b*arccos(c*x))^3*d^2/b-1/2*(-e*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2) 
*(I*(-c^2*x^2+1)^(1/2)*c*x+c^2*x^2-1)*(arccos(c*x)^2*b^2+2*arccos(c*x)*a*b 
+a^2-2*b^2+2*I*arccos(c*x)*b^2+2*I*a*b)*d^2/(c*x-1)/e^3/c/(c*x+1)-1/2*(-e* 
(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)*(-I*(-c^2*x^2+1)^(1/2)*x*c+c^2*x^2-1)*(ar 
ccos(c*x)^2*b^2+2*arccos(c*x)*a*b+a^2-2*b^2-2*I*b^2*arccos(c*x)-2*I*a*b)*d 
^2/(c*x-1)/e^3/c/(c*x+1)+4/3*(-e*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)*(7*c^2*x 
^2+2*c*x-5-7*I*(-c^2*x^2+1)^(1/2)*c*x+7*I*(-c^2*x^2+1)^(1/2))*(63*b^2*c^2* 
x^2*arccos(c*x)^2+28*I*a*b*c*x+126*a*b*c^2*x^2*arccos(c*x)+4*I*(-c^2*x^2+1 
)^(1/2)*b^2*c*x-96*arccos(c*x)^2*b^2*c*x+14*(-c^2*x^2+1)^(1/2)*arccos(c*x) 
*b^2*c*x-14*I*a*b-14*I*arccos(c*x)*b^2+63*a^2*c^2*x^2-32*x^2*c^2*b^2-192*a 
rccos(c*x)*a*b*c*x+14*(-c^2*x^2+1)^(1/2)*a*b*c*x+28*I*arccos(c*x)*b^2*c*x+ 
37*arccos(c*x)^2*b^2-10*arccos(c*x)*(-c^2*x^2+1)^(1/2)*b^2-14*I*arccos(c*x 
)*b^2*c^2*x^2-14*I*a*b*c^2*x^2-96*a^2*c*x+56*c*x*b^2+74*arccos(c*x)*a*b-10 
*(-c^2*x^2+1)^(1/2)*a*b-4*I*(-c^2*x^2+1)^(1/2)*b^2+37*a^2-24*b^2)*d^2/(63* 
c^4*x^4-222*c^3*x^3+292*c^2*x^2-170*c*x+37)/e^3/c/(c*x+1)+56/3*I*(-c^2*x^2 
+1)^(1/2)*(d*(c*x+1))^(1/2)*(-e*(c*x-1))^(1/2)/(c*x-1)/e^3/c/(c*x+1)*b*(ar 
ccos(c*x)^2*b+2*I*arccos(c*x)*ln((c*x+I*(-c^2*x^2+1)^(1/2))^(1/2)+1)*b+2*I 
*arccos(c*x)*ln(1-(c*x+I*(-c^2*x^2+1)^(1/2))^(1/2))*b+4*polylog(2,(c*x+I*( 
-c^2*x^2+1)^(1/2))^(1/2))*b+4*polylog(2,-(c*x+I*(-c^2*x^2+1)^(1/2))^(1/...
 

Fricas [F]

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

integrate((c*d*x+d)^(5/2)*(a+b*arccos(c*x))^2/(-c*e*x+e)^(5/2),x, algorith 
m="fricas")
 

Output:

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

Sympy [F(-1)]

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

integrate((c*d*x+d)**(5/2)*(a+b*acos(c*x))**2/(-c*e*x+e)**(5/2),x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(d+c d x)^{5/2} (a+b \arccos (c x))^2}{(e-c e x)^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((c*d*x+d)^(5/2)*(a+b*arccos(c*x))^2/(-c*e*x+e)^(5/2),x, algorith 
m="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(e>0)', see `assume?` for more de 
tails)Is e
 

Giac [F]

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

integrate((c*d*x+d)^(5/2)*(a+b*arccos(c*x))^2/(-c*e*x+e)^(5/2),x, algorith 
m="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(d+c d x)^{5/2} (a+b \arccos (c x))^2}{(e-c e x)^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(sqrt(d)*d**2*( - 30*sqrt( - c*x + 1)*asin(sqrt( - c*x + 1)/sqrt(2))*a**2* 
c*x + 30*sqrt( - c*x + 1)*asin(sqrt( - c*x + 1)/sqrt(2))*a**2 + 6*sqrt( - 
c*x + 1)*int((sqrt(c*x + 1)*acos(c*x)*x**2)/(sqrt( - c*x + 1)*c**2*x**2 - 
2*sqrt( - c*x + 1)*c*x + sqrt( - c*x + 1)),x)*a*b*c**4*x - 6*sqrt( - c*x + 
 1)*int((sqrt(c*x + 1)*acos(c*x)*x**2)/(sqrt( - c*x + 1)*c**2*x**2 - 2*sqr 
t( - c*x + 1)*c*x + sqrt( - c*x + 1)),x)*a*b*c**3 + 12*sqrt( - c*x + 1)*in 
t((sqrt(c*x + 1)*acos(c*x)*x)/(sqrt( - c*x + 1)*c**2*x**2 - 2*sqrt( - c*x 
+ 1)*c*x + sqrt( - c*x + 1)),x)*a*b*c**3*x - 12*sqrt( - c*x + 1)*int((sqrt 
(c*x + 1)*acos(c*x)*x)/(sqrt( - c*x + 1)*c**2*x**2 - 2*sqrt( - c*x + 1)*c* 
x + sqrt( - c*x + 1)),x)*a*b*c**2 + 6*sqrt( - c*x + 1)*int((sqrt(c*x + 1)* 
acos(c*x))/(sqrt( - c*x + 1)*c**2*x**2 - 2*sqrt( - c*x + 1)*c*x + sqrt( - 
c*x + 1)),x)*a*b*c**2*x - 6*sqrt( - c*x + 1)*int((sqrt(c*x + 1)*acos(c*x)) 
/(sqrt( - c*x + 1)*c**2*x**2 - 2*sqrt( - c*x + 1)*c*x + sqrt( - c*x + 1)), 
x)*a*b*c + 3*sqrt( - c*x + 1)*int((sqrt(c*x + 1)*acos(c*x)**2*x**2)/(sqrt( 
 - c*x + 1)*c**2*x**2 - 2*sqrt( - c*x + 1)*c*x + sqrt( - c*x + 1)),x)*b**2 
*c**4*x - 3*sqrt( - c*x + 1)*int((sqrt(c*x + 1)*acos(c*x)**2*x**2)/(sqrt( 
- c*x + 1)*c**2*x**2 - 2*sqrt( - c*x + 1)*c*x + sqrt( - c*x + 1)),x)*b**2* 
c**3 + 6*sqrt( - c*x + 1)*int((sqrt(c*x + 1)*acos(c*x)**2*x)/(sqrt( - c*x 
+ 1)*c**2*x**2 - 2*sqrt( - c*x + 1)*c*x + sqrt( - c*x + 1)),x)*b**2*c**3*x 
 - 6*sqrt( - c*x + 1)*int((sqrt(c*x + 1)*acos(c*x)**2*x)/(sqrt( - c*x +...