\(\int \frac {(f+g x) (a+b \arccos (c x))}{\sqrt {d-c^2 d x^2}} \, dx\) [16]

Optimal result

Integrand size = 29, antiderivative size = 120 \[ \int \frac {(f+g x) (a+b \arccos (c x))}{\sqrt {d-c^2 d x^2}} \, dx=-\frac {b g x \sqrt {1-c^2 x^2}}{c \sqrt {d-c^2 d x^2}}-\frac {g \sqrt {d-c^2 d x^2} (a+b \arccos (c x))}{c^2 d}-\frac {f \sqrt {1-c^2 x^2} (a+b \arccos (c x))^2}{2 b c \sqrt {d-c^2 d x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.43 \[ \int \frac {(f+g x) (a+b \arccos (c x))}{\sqrt {d-c^2 d x^2}} \, dx=\frac {-2 \sqrt {d} g \left (a-a c^2 x^2+b c x \sqrt {1-c^2 x^2}\right )+2 b \sqrt {d} g \left (-1+c^2 x^2\right ) \arccos (c x)-b c \sqrt {d} f \sqrt {1-c^2 x^2} \arccos (c x)^2-2 a c f \sqrt {d-c^2 d x^2} \arctan \left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (-1+c^2 x^2\right )}\right )}{2 c^2 \sqrt {d} \sqrt {d-c^2 d x^2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.73, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {5277, 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.

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

\(\Big \downarrow \) 5277

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

\(\Big \downarrow \) 5263

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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))
 

rule 5277

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_ 
) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[Simp[(d + e*x^2)^p/(1 - c^2*x^2)^ 
p]   Int[(f + g*x)^m*(1 - c^2*x^2)^p*(a + b*ArcCos[c*x])^n, x], x] /; FreeQ 
[{a, b, c, d, e, f, g, n}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && Intege 
rQ[p - 1/2] &&  !GtQ[d, 0]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.64 (sec) , antiderivative size = 247, normalized size of antiderivative = 2.06

method	result	size
default	\(\frac {a f \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{\sqrt {c^{2} d}}-\frac {a g \sqrt {-c^{2} d \,x^{2}+d}}{c^{2} d}+b \left (\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right )^{2} f}{2 \left (c^{2} x^{2}-1\right ) c d}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, x c +c^{2} x^{2}-1\right ) g \left (\arccos \left (c x \right )+i\right )}{2 c^{2} d \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (-i \sqrt {-c^{2} x^{2}+1}\, x c +c^{2} x^{2}-1\right ) g \left (\arccos \left (c x \right )-i\right )}{2 c^{2} d \left (c^{2} x^{2}-1\right )}\right )\)	\(247\)
parts	\(\frac {a f \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{\sqrt {c^{2} d}}-\frac {a g \sqrt {-c^{2} d \,x^{2}+d}}{c^{2} d}+b \left (\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right )^{2} f}{2 \left (c^{2} x^{2}-1\right ) c d}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, x c +c^{2} x^{2}-1\right ) g \left (\arccos \left (c x \right )+i\right )}{2 c^{2} d \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (-i \sqrt {-c^{2} x^{2}+1}\, x c +c^{2} x^{2}-1\right ) g \left (\arccos \left (c x \right )-i\right )}{2 c^{2} d \left (c^{2} x^{2}-1\right )}\right )\)	\(247\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-2)]

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

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

Output:

Exception raised: TypeError >> Invalid comparison of non-real zoo
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.90 \[ \int \frac {(f+g x) (a+b \arccos (c x))}{\sqrt {d-c^2 d x^2}} \, dx=\frac {b f \arccos \left (c x\right ) \arcsin \left (c x\right )}{c \sqrt {d}} + \frac {b f \arcsin \left (c x\right )^{2}}{2 \, c \sqrt {d}} - \frac {b g x}{c \sqrt {d}} + \frac {a f \arcsin \left (c x\right )}{c \sqrt {d}} - \frac {\sqrt {-c^{2} d x^{2} + d} b g \arccos \left (c x\right )}{c^{2} d} - \frac {\sqrt {-c^{2} d x^{2} + d} a g}{c^{2} d} \] Input:

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

Output:

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

Giac [F(-2)]

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

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

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const ve 
cteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.59 \[ \int \frac {(f+g x) (a+b \arccos (c x))}{\sqrt {d-c^2 d x^2}} \, dx=\frac {\sqrt {d}\, \left (-\mathit {acos} \left (c x \right )^{2} b c f -2 \sqrt {-c^{2} x^{2}+1}\, \mathit {acos} \left (c x \right ) b g +2 \mathit {asin} \left (c x \right ) a c f -2 \sqrt {-c^{2} x^{2}+1}\, a g -2 b c g x \right )}{2 c^{2} d} \] Input:

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

Output:

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