\(\int \frac {\sqrt {d+c d x} (a+b \arccos (c x))}{\sqrt {f-c f x}} \, dx\) [526]

Optimal result

Integrand size = 30, antiderivative size = 141 \[ \int \frac {\sqrt {d+c d x} (a+b \arccos (c x))}{\sqrt {f-c f x}} \, dx=\frac {b d x \sqrt {1-c^2 x^2}}{\sqrt {d+c d x} \sqrt {f-c f x}}-\frac {d \left (1-c^2 x^2\right ) (a+b \arccos (c x))}{c \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {d \sqrt {1-c^2 x^2} (a+b \arccos (c x))^2}{2 b c \sqrt {d+c d x} \sqrt {f-c f x}} \] Output:

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

Mathematica [A] (verified)

Time = 1.45 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.41 \[ \int \frac {\sqrt {d+c d x} (a+b \arccos (c x))}{\sqrt {f-c f x}} \, dx=-\frac {\frac {2 \sqrt {d+c d x} \sqrt {f-c f x} \left (b c x+a \sqrt {1-c^2 x^2}\right )}{\sqrt {1-c^2 x^2}}+2 b \sqrt {d+c d x} \sqrt {f-c f x} \arccos (c x)+\frac {b \sqrt {d+c d x} \sqrt {f-c f x} \arccos (c x)^2}{\sqrt {1-c^2 x^2}}+2 a \sqrt {d} \sqrt {f} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (-1+c^2 x^2\right )}\right )}{2 c f} \] Input:

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

Output:

-1/2*((2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(b*c*x + a*Sqrt[1 - c^2*x^2]))/Sq 
rt[1 - c^2*x^2] + 2*b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcCos[c*x] + (b*Sqr 
t[d + c*d*x]*Sqrt[f - c*f*x]*ArcCos[c*x]^2)/Sqrt[1 - c^2*x^2] + 2*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))])/(c*f)
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.63, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {5179, 27, 5263, 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.

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

\(\Big \downarrow \) 5179

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 5263

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 5263

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_ 
) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^p*(a + 
 b*ArcCos[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 = 1.01 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.19

method	result	size
default	\(-\frac {a \sqrt {c d x +d}\, \sqrt {-c f x +f}}{c f}+\frac {a d \sqrt {\left (-c f x +f \right ) \left (c d x +d \right )}\, \arctan \left (\frac {\sqrt {c^{2} d f}\, x}{\sqrt {-c^{2} d f \,x^{2}+d f}}\right )}{\sqrt {c d x +d}\, \sqrt {-c f x +f}\, \sqrt {c^{2} d f}}+b \left (\frac {\sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right )^{2}}{2 \left (c x -1\right ) f c \left (c x +1\right )}-\frac {\sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, x c +c^{2} x^{2}-1\right ) \left (\arccos \left (c x \right )+i\right )}{2 \left (c x -1\right ) f c \left (c x +1\right )}-\frac {\sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \left (-i \sqrt {-c^{2} x^{2}+1}\, x c +c^{2} x^{2}-1\right ) \left (\arccos \left (c x \right )-i\right )}{2 \left (c x -1\right ) f c \left (c x +1\right )}\right )\)	\(309\)
parts	\(-\frac {a \sqrt {c d x +d}\, \sqrt {-c f x +f}}{c f}+\frac {a d \sqrt {\left (-c f x +f \right ) \left (c d x +d \right )}\, \arctan \left (\frac {\sqrt {c^{2} d f}\, x}{\sqrt {-c^{2} d f \,x^{2}+d f}}\right )}{\sqrt {c d x +d}\, \sqrt {-c f x +f}\, \sqrt {c^{2} d f}}+b \left (\frac {\sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right )^{2}}{2 \left (c x -1\right ) f c \left (c x +1\right )}-\frac {\sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, x c +c^{2} x^{2}-1\right ) \left (\arccos \left (c x \right )+i\right )}{2 \left (c x -1\right ) f c \left (c x +1\right )}-\frac {\sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \left (-i \sqrt {-c^{2} x^{2}+1}\, x c +c^{2} x^{2}-1\right ) \left (\arccos \left (c x \right )-i\right )}{2 \left (c x -1\right ) f c \left (c x +1\right )}\right )\)	\(309\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

a*(d*arcsin(c*x)/(c*f*sqrt(d/f)) - sqrt(-c^2*d*f*x^2 + d*f)/(c*f)) + b*sqr 
t(d)*integrate(sqrt(c*x + 1)*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x)/sq 
rt(-c*x + 1), x)/sqrt(f)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

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