\(\int \frac {\sqrt {f-c f x} (a+b \arccos (c x))}{(d+c d x)^{3/2}} \, dx\) [510]

Optimal result

Integrand size = 30, antiderivative size = 162 \[ \int \frac {\sqrt {f-c f x} (a+b \arccos (c x))}{(d+c d x)^{3/2}} \, dx=-\frac {2 f^2 (1-c x) \left (1-c^2 x^2\right ) (a+b \arccos (c x))}{c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {f^2 \left (1-c^2 x^2\right )^{3/2} (a+b \arccos (c x))^2}{2 b c (d+c d x)^{3/2} (f-c f x)^{3/2}}+\frac {2 b f^2 \left (1-c^2 x^2\right )^{3/2} \log (1+c x)}{c (d+c d x)^{3/2} (f-c f x)^{3/2}} \] Output:

-2*f^2*(-c*x+1)*(-c^2*x^2+1)*(a+b*arccos(c*x))/c/(c*d*x+d)^(3/2)/(-c*f*x+f 
)^(3/2)-1/2*f^2*(-c^2*x^2+1)^(3/2)*(a+b*arccos(c*x))^2/b/c/(c*d*x+d)^(3/2) 
/(-c*f*x+f)^(3/2)+2*b*f^2*(-c^2*x^2+1)^(3/2)*ln(c*x+1)/c/(c*d*x+d)^(3/2)/( 
-c*f*x+f)^(3/2)
 

Mathematica [A] (verified)

Time = 2.37 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.20 \[ \int \frac {\sqrt {f-c f x} (a+b \arccos (c x))}{(d+c d x)^{3/2}} \, dx=-\frac {\frac {4 a \sqrt {d+c d x} \sqrt {f-c f x}}{1+c x}-2 a \sqrt {d} \sqrt {f} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (-1+c^2 x^2\right )}\right )+\frac {b (-1+c x) \sqrt {d+c d x} \sqrt {f-c f x} \cot \left (\frac {1}{2} \arccos (c x)\right ) \left (-4 \arccos (c x)+\cot \left (\frac {1}{2} \arccos (c x)\right ) \left (\arccos (c x)^2-8 \log \left (\cos \left (\frac {1}{2} \arccos (c x)\right )\right )\right )\right )}{(1+c x) \sqrt {1-c^2 x^2}}}{2 c d^2} \] Input:

Integrate[(Sqrt[f - c*f*x]*(a + b*ArcCos[c*x]))/(d + c*d*x)^(3/2),x]
 

Output:

-1/2*((4*a*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(1 + c*x) - 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))] + (b*(-1 + c*x)*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*Cot[ArcCos[c*x]/2]* 
(-4*ArcCos[c*x] + Cot[ArcCos[c*x]/2]*(ArcCos[c*x]^2 - 8*Log[Cos[ArcCos[c*x 
]/2]])))/((1 + c*x)*Sqrt[1 - c^2*x^2]))/(c*d^2)
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.65, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, 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.

\(\displaystyle \int \frac {\sqrt {f-c f x} (a+b \arccos (c x))}{(c d x+d)^{3/2}} \, dx\)

\(\Big \downarrow \) 5179

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 5275

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {f^2 \left (1-c^2 x^2\right )^{3/2} \left (-\frac {2 (1-c x) (a+b \arccos (c x))}{c \sqrt {1-c^2 x^2}}+\frac {(a+b \arccos (c x))^2}{2 b c}-\frac {2 b \log (c x+1)}{c}\right )}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\)

Input:

Int[(Sqrt[f - c*f*x]*(a + b*ArcCos[c*x]))/(d + c*d*x)^(3/2),x]
 

Output:

(f^2*(1 - c^2*x^2)^(3/2)*((-2*(1 - c*x)*(a + b*ArcCos[c*x]))/(c*Sqrt[1 - c 
^2*x^2]) + (a + b*ArcCos[c*x])^2/(2*b*c) - (2*b*Log[1 + c*x])/c))/((d + c* 
d*x)^(3/2)*(f - c*f*x)^(3/2))
 

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 [C] (verified)

Result contains complex when optimal does not.

Time = 9.10 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.99

method	result	size
default	\(-\frac {\sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \sqrt {-c^{2} x^{2}+1}\, b \arccos \left (c x \right )^{2}}{2 \left (c x +1\right ) d^{2} \left (c x -1\right ) c}-\frac {\sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \sqrt {-c^{2} x^{2}+1}\, a \arccos \left (c x \right )}{\left (c x +1\right ) d^{2} \left (c x -1\right ) c}-\frac {4 i \sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \sqrt {-c^{2} x^{2}+1}\, b \arccos \left (c x \right )}{\left (c x +1\right ) d^{2} \left (c x -1\right ) c}-\frac {2 \sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \left (-i \sqrt {-c^{2} x^{2}+1}+c x -1\right ) \left (a +b \arccos \left (c x \right )\right )}{\left (c x +1\right ) d^{2} \left (c x -1\right ) c}+\frac {4 \sqrt {d \left (c x +1\right )}\, \sqrt {-f \left (c x -1\right )}\, \sqrt {-c^{2} x^{2}+1}\, b \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )}{\left (c x +1\right ) d^{2} \left (c x -1\right ) c}\)	\(323\)

Input:

int((-c*f*x+f)^(1/2)*(a+b*arccos(c*x))/(c*d*x+d)^(3/2),x,method=_RETURNVER 
BOSE)
 

Output:

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

Fricas [F]

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

integrate((-c*f*x+f)^(1/2)*(a+b*arccos(c*x))/(c*d*x+d)^(3/2),x, algorithm= 
"fricas")
 

Output:

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

Sympy [F]

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

integrate((-c*f*x+f)**(1/2)*(a+b*acos(c*x))/(c*d*x+d)**(3/2),x)
 

Output:

Integral(sqrt(-f*(c*x - 1))*(a + b*acos(c*x))/(d*(c*x + 1))**(3/2), x)
 

Maxima [F]

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

integrate((-c*f*x+f)^(1/2)*(a+b*arccos(c*x))/(c*d*x+d)^(3/2),x, algorithm= 
"maxima")
 

Output:

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

Giac [F]

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

integrate((-c*f*x+f)^(1/2)*(a+b*arccos(c*x))/(c*d*x+d)^(3/2),x, algorithm= 
"giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {f-c f x} (a+b \arccos (c x))}{(d+c d x)^{3/2}} \, dx=\int \frac {\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )\,\sqrt {f-c\,f\,x}}{{\left (d+c\,d\,x\right )}^{3/2}} \,d x \] Input:

int(((a + b*acos(c*x))*(f - c*f*x)^(1/2))/(d + c*d*x)^(3/2),x)
 

Output:

int(((a + b*acos(c*x))*(f - c*f*x)^(1/2))/(d + c*d*x)^(3/2), x)
 

Reduce [F]

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

int((-c*f*x+f)^(1/2)*(a+b*acos(c*x))/(c*d*x+d)^(3/2),x)
 

Output:

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