\(\int \frac {\sqrt {d+c d x} (a+b \arccos (c x))^2}{(e-c e x)^{5/2}} \, dx\) [574]

Optimal result

Integrand size = 32, antiderivative size = 486 \[ \int \frac {\sqrt {d+c d x} (a+b \arccos (c x))^2}{(e-c e x)^{5/2}} \, dx=-\frac {i d^3 \left (1-c^2 x^2\right )^{5/2} (a+b \arccos (c x))^2}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {4 b d^3 \left (1-c^2 x^2\right )^{5/2} (a+b \arccos (c x)) \log \left (1-i e^{-i \arccos (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {4 i b^2 d^3 \left (1-c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,i e^{-i \arccos (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {2 b d^3 \left (1-c^2 x^2\right )^{5/2} (a+b \arccos (c x)) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} \arccos (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {4 b^2 d^3 \left (1-c^2 x^2\right )^{5/2} \tan \left (\frac {\pi }{4}+\frac {1}{2} \arccos (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {d^3 \left (1-c^2 x^2\right )^{5/2} (a+b \arccos (c x))^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} \arccos (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {d^3 \left (1-c^2 x^2\right )^{5/2} (a+b \arccos (c x))^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} \arccos (c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} \arccos (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}} \] Output:

-1/3*I*d^3*(-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)-4/3*b*d^3*(-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)-4/3*I*b^2*d^3*(-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)-2/3*b*d^3*(-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)+4/3*b^2*d^3*(-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)-1/ 
3*d^3*(-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)+1/3*d^3*(-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 = 4.75 (sec) , antiderivative size = 371, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {d+c d x} (a+b \arccos (c x))^2}{(e-c e x)^{5/2}} \, dx=\frac {\sqrt {d+c d x} \sqrt {e-c e x} \sqrt {1-c^2 x^2} \left (a^2 \sqrt {1-c^2 x^2}-a^2 c^2 x^2 \sqrt {1-c^2 x^2}+2 b^2 \left (-i-i c^2 x^2+\sqrt {1-c^2 x^2}+c x \left (2 i+\sqrt {1-c^2 x^2}\right )\right ) \arccos (c x)^2 \sin ^2\left (\frac {1}{2} \arccos (c x)\right )+16 a b \sin ^4\left (\frac {1}{2} \arccos (c x)\right )+16 b^2 \sqrt {1-c^2 x^2} \sin ^4\left (\frac {1}{2} \arccos (c x)\right )+32 a b \log \left (\sin \left (\frac {1}{2} \arccos (c x)\right )\right ) \sin ^6\left (\frac {1}{2} \arccos (c x)\right )-32 i b^2 \operatorname {PolyLog}\left (2,e^{i \arccos (c x)}\right ) \sin ^6\left (\frac {1}{2} \arccos (c x)\right )+8 b \arccos (c x) \sin ^2\left (\frac {1}{2} \arccos (c x)\right ) \left (a \sqrt {1-c^2 x^2}+\left (2 b-a \sqrt {1-c^2 x^2}\right ) \sin ^2\left (\frac {1}{2} \arccos (c x)\right )+4 b \log \left (1-e^{i \arccos (c x)}\right ) \sin ^4\left (\frac {1}{2} \arccos (c x)\right )\right )\right )}{3 c e^3 (-1+c x)^4 (1+c x)} \] Input:

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

Output:

(Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*Sqrt[1 - c^2*x^2]*(a^2*Sqrt[1 - c^2*x^2] 
- a^2*c^2*x^2*Sqrt[1 - c^2*x^2] + 2*b^2*(-I - I*c^2*x^2 + Sqrt[1 - c^2*x^2 
] + c*x*(2*I + Sqrt[1 - c^2*x^2]))*ArcCos[c*x]^2*Sin[ArcCos[c*x]/2]^2 + 16 
*a*b*Sin[ArcCos[c*x]/2]^4 + 16*b^2*Sqrt[1 - c^2*x^2]*Sin[ArcCos[c*x]/2]^4 
+ 32*a*b*Log[Sin[ArcCos[c*x]/2]]*Sin[ArcCos[c*x]/2]^6 - (32*I)*b^2*PolyLog 
[2, E^(I*ArcCos[c*x])]*Sin[ArcCos[c*x]/2]^6 + 8*b*ArcCos[c*x]*Sin[ArcCos[c 
*x]/2]^2*(a*Sqrt[1 - c^2*x^2] + (2*b - a*Sqrt[1 - c^2*x^2])*Sin[ArcCos[c*x 
]/2]^2 + 4*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 = 1.32 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.46, 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.

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

\(\Big \downarrow \) 5179

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 5275

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(d^3*(1 - c^2*x^2)^(5/2)*(((-1/3*I)*(a + b*ArcCos[c*x])^2)/c + (4*b^2*Cot[ 
ArcCos[c*x]/2])/(3*c) - ((a + b*ArcCos[c*x])^2*Cot[ArcCos[c*x]/2])/(3*c) + 
 (2*b*(a + b*ArcCos[c*x])*Csc[ArcCos[c*x]/2]^2)/(3*c) + ((a + b*ArcCos[c*x 
])^2*Cot[ArcCos[c*x]/2]*Csc[ArcCos[c*x]/2]^2)/(3*c) + (4*b*(a + b*ArcCos[c 
*x])*Log[1 - E^(I*ArcCos[c*x])])/(3*c) - (((4*I)/3)*b^2*PolyLog[2, E^(I*Ar 
cCos[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 [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2994 vs. \(2 (428 ) = 856\).

Time = 4.38 (sec) , antiderivative size = 2995, normalized size of antiderivative = 6.16

method	result	size
default	\(\text {Expression too large to display}\)	\(2995\)
parts	\(\text {Expression too large to display}\)	\(2995\)

Input:

int((c*d*x+d)^(1/2)*(a+b*arccos(c*x))^2/(-c*e*x+e)^(5/2),x,method=_RETURNV 
ERBOSE)
 

Output:

a^2*(1/c/e*(c*d*x+d)^(1/2)/(-c*e*x+e)^(3/2)-d*(1/3/d/c/e/(-c*e*x+e)^(3/2)* 
(c*d*x+d)^(1/2)+1/3/d/c/e^2/(-c*e*x+e)^(1/2)*(c*d*x+d)^(1/2)))+4/3*b^2*(d* 
(c*x+1))^(1/2)*(-e*(c*x-1))^(1/2)/(3*c^4*x^4-6*c^3*x^3+4*c^2*x^2-2*c*x+1)/ 
e^3*c/(c*x+1)*(-c^2*x^2+1)*x^2+4/3*b^2*(d*(c*x+1))^(1/2)*(-e*(c*x-1))^(1/2 
)/(3*c^4*x^4-6*c^3*x^3+4*c^2*x^2-2*c*x+1)/e^3/c/(c*x+1)*(-c^2*x^2+1)^(1/2) 
*arccos(c*x)+b^2*(d*(c*x+1))^(1/2)*(-e*(c*x-1))^(1/2)/(3*c^4*x^4-6*c^3*x^3 
+4*c^2*x^2-2*c*x+1)/e^3*c^3/(c*x+1)*arccos(c*x)^2*x^4+2*b^2*(d*(c*x+1))^(1 
/2)*(-e*(c*x-1))^(1/2)/(3*c^4*x^4-6*c^3*x^3+4*c^2*x^2-2*c*x+1)/e^3*c^2/(c* 
x+1)*arccos(c*x)^2*x^3+4/3*b^2*(d*(c*x+1))^(1/2)*(-e*(c*x-1))^(1/2)/(3*c^4 
*x^4-6*c^3*x^3+4*c^2*x^2-2*c*x+1)/e^3*c/(c*x+1)*arccos(c*x)^2*x^2-2/3*I*b^ 
2*(d*(c*x+1))^(1/2)*(-e*(c*x-1))^(1/2)/(3*c^4*x^4-6*c^3*x^3+4*c^2*x^2-2*c* 
x+1)/e^3/c/(c*x+1)*arccos(c*x)+4/3*I*b^2*(d*(c*x+1))^(1/2)*(-e*(c*x-1))^(1 
/2)/(3*c^4*x^4-6*c^3*x^3+4*c^2*x^2-2*c*x+1)/e^3/(c*x+1)*(-c^2*x^2+1)^(1/2) 
*x-4/3*I*b^2*(d*(c*x+1))^(1/2)*(-e*(c*x-1))^(1/2)/(3*c^4*x^4-6*c^3*x^3+4*c 
^2*x^2-2*c*x+1)/e^3/c/(c*x+1)*(-c^2*x^2+1)^(1/2)+2/3*b^2*(d*(c*x+1))^(1/2) 
*(-e*(c*x-1))^(1/2)/(3*c^4*x^4-6*c^3*x^3+4*c^2*x^2-2*c*x+1)/e^3/(c*x+1)*ar 
ccos(c*x)^2*x-8/3*b^2*(d*(c*x+1))^(1/2)*(-e*(c*x-1))^(1/2)/(3*c^4*x^4-6*c^ 
3*x^3+4*c^2*x^2-2*c*x+1)/e^3*c^3/(c*x+1)*x^4+4/3*b^2*(d*(c*x+1))^(1/2)*(-e 
*(c*x-1))^(1/2)/(3*c^4*x^4-6*c^3*x^3+4*c^2*x^2-2*c*x+1)/e^3/c/(c*x+1)*(-c^ 
2*x^2+1)-8/3*b^2*(d*(c*x+1))^(1/2)*(-e*(c*x-1))^(1/2)/(3*c^4*x^4-6*c^3*...
 

Fricas [F]

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

integrate((c*d*x+d)^(1/2)*(a+b*arccos(c*x))^2/(-c*e*x+e)^(5/2),x, algorith 
m="fricas")
 

Output:

integral(-(b^2*arccos(c*x)^2 + 2*a*b*arccos(c*x) + a^2)*sqrt(c*d*x + d)*sq 
rt(-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]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((c*d*x+d)^(1/2)*(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]

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

integrate((c*d*x+d)^(1/2)*(a+b*arccos(c*x))^2/(-c*e*x+e)^(5/2),x, algorith 
m="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {d+c d x} (a+b \arccos (c x))^2}{(e-c e x)^{5/2}} \, dx=\frac {\sqrt {d}\, \left (6 \sqrt {-c x +1}\, \left (\int \frac {\sqrt {c x +1}\, \mathit {acos} \left (c x \right )}{\sqrt {-c x +1}\, c^{2} x^{2}-2 \sqrt {-c x +1}\, c x +\sqrt {-c x +1}}d x \right ) a b \,c^{2} x -6 \sqrt {-c x +1}\, \left (\int \frac {\sqrt {c x +1}\, \mathit {acos} \left (c x \right )}{\sqrt {-c x +1}\, c^{2} x^{2}-2 \sqrt {-c x +1}\, c x +\sqrt {-c x +1}}d x \right ) a b c +3 \sqrt {-c x +1}\, \left (\int \frac {\sqrt {c x +1}\, \mathit {acos} \left (c x \right )^{2}}{\sqrt {-c x +1}\, c^{2} x^{2}-2 \sqrt {-c x +1}\, c x +\sqrt {-c x +1}}d x \right ) b^{2} c^{2} x -3 \sqrt {-c x +1}\, \left (\int \frac {\sqrt {c x +1}\, \mathit {acos} \left (c x \right )^{2}}{\sqrt {-c x +1}\, c^{2} x^{2}-2 \sqrt {-c x +1}\, c x +\sqrt {-c x +1}}d x \right ) b^{2} c -\sqrt {c x +1}\, a^{2} c x -\sqrt {c x +1}\, a^{2}\right )}{3 \sqrt {e}\, \sqrt {-c x +1}\, c \,e^{2} \left (c x -1\right )} \] Input:

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

Output:

(sqrt(d)*(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)/(sqrt( - c*x + 1)*c**2*x**2 - 2*sqrt( 
 - c*x + 1)*c*x + sqrt( - c*x + 1)),x)*b**2*c**2*x - 3*sqrt( - c*x + 1)*in 
t((sqrt(c*x + 1)*acos(c*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 - sqrt(c*x + 1)*a**2*c*x - sqrt(c* 
x + 1)*a**2))/(3*sqrt(e)*sqrt( - c*x + 1)*c*e**2*(c*x - 1))