\(\int \sqrt {d+c d x} \sqrt {f-c f x} (a+b \arccos (c x)) \, dx\) [508]

Optimal result

Integrand size = 30, antiderivative size = 134 \[ \int \sqrt {d+c d x} \sqrt {f-c f x} (a+b \arccos (c x)) \, dx=-\frac {b c x^2 \sqrt {d+c d x} \sqrt {f-c f x}}{4 \sqrt {1-c^2 x^2}}+\frac {1}{2} x \sqrt {d+c d x} \sqrt {f-c f x} (a+b \arccos (c x))+\frac {\sqrt {d+c d x} \sqrt {f-c f x} (a+b \arccos (c x))^2}{4 b c \sqrt {1-c^2 x^2}} \] Output:

-1/4*b*c*x^2*(c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)/(-c^2*x^2+1)^(1/2)+1/2*x*(c* 
d*x+d)^(1/2)*(-c*f*x+f)^(1/2)*(a+b*arccos(c*x))+1/4*(c*d*x+d)^(1/2)*(-c*f* 
x+f)^(1/2)*(a+b*arccos(c*x))^2/b/c/(-c^2*x^2+1)^(1/2)
 

Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.56 \[ \int \sqrt {d+c d x} \sqrt {f-c f x} (a+b \arccos (c x)) \, dx=\frac {1}{2} a x \sqrt {-f (-1+c x)} \sqrt {d (1+c x)}-\frac {a \sqrt {d} \sqrt {f} \arctan \left (\frac {c x \sqrt {-f (-1+c x)} \sqrt {d (1+c x)}}{\sqrt {d} \sqrt {f} (-1+c x) (1+c x)}\right )}{2 c}+\frac {b \sqrt {d+c d x} \sqrt {f-c f x} \sqrt {-d f \left (1-c^2 x^2\right )} (\cos (2 \arccos (c x))+2 \arccos (c x) (-\arccos (c x)+\sin (2 \arccos (c x))))}{8 c \sqrt {(-d-c d x) (f-c f x)} \sqrt {1-c^2 x^2}} \] Input:

Integrate[Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(a + b*ArcCos[c*x]),x]
 

Output:

(a*x*Sqrt[-(f*(-1 + c*x))]*Sqrt[d*(1 + c*x)])/2 - (a*Sqrt[d]*Sqrt[f]*ArcTa 
n[(c*x*Sqrt[-(f*(-1 + c*x))]*Sqrt[d*(1 + c*x)])/(Sqrt[d]*Sqrt[f]*(-1 + c*x 
)*(1 + c*x))])/(2*c) + (b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*Sqrt[-(d*f*(1 - 
c^2*x^2))]*(Cos[2*ArcCos[c*x]] + 2*ArcCos[c*x]*(-ArcCos[c*x] + Sin[2*ArcCo 
s[c*x]])))/(8*c*Sqrt[(-d - c*d*x)*(f - c*f*x)]*Sqrt[1 - c^2*x^2])
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.69, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {5179, 5157, 15, 5153}

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.

\(\displaystyle \int \sqrt {c d x+d} \sqrt {f-c f x} (a+b \arccos (c x)) \, dx\)

\(\Big \downarrow \) 5179

\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {f-c f x} \int \sqrt {1-c^2 x^2} (a+b \arccos (c x))dx}{\sqrt {1-c^2 x^2}}\)

\(\Big \downarrow \) 5157

\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {f-c f x} \left (\frac {1}{2} \int \frac {a+b \arccos (c x)}{\sqrt {1-c^2 x^2}}dx+\frac {1}{2} b c \int xdx+\frac {1}{2} x \sqrt {1-c^2 x^2} (a+b \arccos (c x))\right )}{\sqrt {1-c^2 x^2}}\)

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 5153

\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {f-c f x} \left (\frac {1}{2} x \sqrt {1-c^2 x^2} (a+b \arccos (c x))-\frac {(a+b \arccos (c x))^2}{4 b c}+\frac {1}{4} b c x^2\right )}{\sqrt {1-c^2 x^2}}\)

Input:

Int[Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(a + b*ArcCos[c*x]),x]
 

Output:

(Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*((b*c*x^2)/4 + (x*Sqrt[1 - c^2*x^2]*(a + 
b*ArcCos[c*x]))/2 - (a + b*ArcCos[c*x])^2/(4*b*c)))/Sqrt[1 - c^2*x^2]
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 5153

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

rule 5157

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

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]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.82 (sec) , antiderivative size = 366, normalized size of antiderivative = 2.73

method	result	size
default	\(-\frac {a \sqrt {c d x +d}\, \left (-c f x +f \right )^{\frac {3}{2}}}{2 c f}+\frac {a \sqrt {-c f x +f}\, \sqrt {c d x +d}}{2 c}+\frac {a d f \sqrt {\left (-c f x +f \right ) \left (c d x +d \right )}\, \arctan \left (\frac {\sqrt {c^{2} d f}\, x}{\sqrt {-c^{2} d f \,x^{2}+d f}}\right )}{2 \sqrt {-c f x +f}\, \sqrt {c d x +d}\, \sqrt {c^{2} d f}}+b \left (\frac {\sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right )^{2}}{4 \left (c^{2} x^{2}-1\right ) c}+\frac {\sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \left (2 c^{3} x^{3}-2 c x +2 i \sqrt {-c^{2} x^{2}+1}\, x^{2} c^{2}-i \sqrt {-c^{2} x^{2}+1}\right ) \left (i+2 \arccos \left (c x \right )\right )}{16 \left (c^{2} x^{2}-1\right ) c}+\frac {\sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \left (-2 i \sqrt {-c^{2} x^{2}+1}\, x^{2} c^{2}+2 c^{3} x^{3}+i \sqrt {-c^{2} x^{2}+1}-2 c x \right ) \left (-i+2 \arccos \left (c x \right )\right )}{16 \left (c^{2} x^{2}-1\right ) c}\right )\)	\(366\)
parts	\(-\frac {a \sqrt {c d x +d}\, \left (-c f x +f \right )^{\frac {3}{2}}}{2 c f}+\frac {a \sqrt {-c f x +f}\, \sqrt {c d x +d}}{2 c}+\frac {a d f \sqrt {\left (-c f x +f \right ) \left (c d x +d \right )}\, \arctan \left (\frac {\sqrt {c^{2} d f}\, x}{\sqrt {-c^{2} d f \,x^{2}+d f}}\right )}{2 \sqrt {-c f x +f}\, \sqrt {c d x +d}\, \sqrt {c^{2} d f}}+b \left (\frac {\sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right )^{2}}{4 \left (c^{2} x^{2}-1\right ) c}+\frac {\sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \left (2 c^{3} x^{3}-2 c x +2 i \sqrt {-c^{2} x^{2}+1}\, x^{2} c^{2}-i \sqrt {-c^{2} x^{2}+1}\right ) \left (i+2 \arccos \left (c x \right )\right )}{16 \left (c^{2} x^{2}-1\right ) c}+\frac {\sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \left (-2 i \sqrt {-c^{2} x^{2}+1}\, x^{2} c^{2}+2 c^{3} x^{3}+i \sqrt {-c^{2} x^{2}+1}-2 c x \right ) \left (-i+2 \arccos \left (c x \right )\right )}{16 \left (c^{2} x^{2}-1\right ) c}\right )\)	\(366\)

Input:

int((c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)*(a+b*arccos(c*x)),x,method=_RETURNVER 
BOSE)
 

Output:

-1/2*a/c/f*(c*d*x+d)^(1/2)*(-c*f*x+f)^(3/2)+1/2*a/c*(-c*f*x+f)^(1/2)*(c*d* 
x+d)^(1/2)+1/2*a*d*f*((-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*(1/4*(d*(c*x+1))^(1/2)*(-f*(c*x-1))^(1/2)*(-c^2*x^2+1)^(1/2)/(c^2*x^2- 
1)/c*arccos(c*x)^2+1/16*(d*(c*x+1))^(1/2)*(-f*(c*x-1))^(1/2)*(2*c^3*x^3-2* 
c*x+2*I*(-c^2*x^2+1)^(1/2)*x^2*c^2-I*(-c^2*x^2+1)^(1/2))*(I+2*arccos(c*x)) 
/(c^2*x^2-1)/c+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*arccos(c*x))/( 
c^2*x^2-1)/c)
 

Fricas [F]

\[ \int \sqrt {d+c d x} \sqrt {f-c f x} (a+b \arccos (c x)) \, dx=\int { \sqrt {c d x + d} \sqrt {-c f x + f} {\left (b \arccos \left (c x\right ) + a\right )} \,d x } \] Input:

integrate((c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)*(a+b*arccos(c*x)),x, algorithm= 
"fricas")
 

Output:

integral(sqrt(c*d*x + d)*sqrt(-c*f*x + f)*(b*arccos(c*x) + a), x)
 

Sympy [F]

\[ \int \sqrt {d+c d x} \sqrt {f-c f x} (a+b \arccos (c x)) \, dx=\int \sqrt {d \left (c x + 1\right )} \sqrt {- f \left (c x - 1\right )} \left (a + b \operatorname {acos}{\left (c x \right )}\right )\, dx \] Input:

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

Output:

Integral(sqrt(d*(c*x + 1))*sqrt(-f*(c*x - 1))*(a + b*acos(c*x)), x)
 

Maxima [F]

\[ \int \sqrt {d+c d x} \sqrt {f-c f x} (a+b \arccos (c x)) \, dx=\int { \sqrt {c d x + d} \sqrt {-c f x + f} {\left (b \arccos \left (c x\right ) + a\right )} \,d x } \] Input:

integrate((c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)*(a+b*arccos(c*x)),x, algorithm= 
"maxima")
 

Output:

b*sqrt(d)*sqrt(f)*integrate(sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(sqrt(c*x 
+ 1)*sqrt(-c*x + 1), c*x), x) + 1/2*(sqrt(-c^2*d*f*x^2 + d*f)*x + d*f*arcs 
in(c*x)/(sqrt(d*f)*c))*a
 

Giac [F]

\[ \int \sqrt {d+c d x} \sqrt {f-c f x} (a+b \arccos (c x)) \, dx=\int { \sqrt {c d x + d} \sqrt {-c f x + f} {\left (b \arccos \left (c x\right ) + a\right )} \,d x } \] Input:

integrate((c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)*(a+b*arccos(c*x)),x, algorithm= 
"giac")
 

Output:

integrate(sqrt(c*d*x + d)*sqrt(-c*f*x + f)*(b*arccos(c*x) + a), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sqrt {d+c d x} \sqrt {f-c f x} (a+b \arccos (c x)) \, dx=\frac {\sqrt {f}\, \sqrt {d}\, \left (-2 \mathit {asin} \left (\frac {\sqrt {-c x +1}}{\sqrt {2}}\right ) a +\sqrt {c x +1}\, \sqrt {-c x +1}\, a c x +2 \left (\int \sqrt {c x +1}\, \sqrt {-c x +1}\, \mathit {acos} \left (c x \right )d x \right ) b c \right )}{2 c} \] Input:

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

Output:

(sqrt(f)*sqrt(d)*( - 2*asin(sqrt( - c*x + 1)/sqrt(2))*a + sqrt(c*x + 1)*sq 
rt( - c*x + 1)*a*c*x + 2*int(sqrt(c*x + 1)*sqrt( - c*x + 1)*acos(c*x),x)*b 
*c))/(2*c)