Integrand size = 30, antiderivative size = 411 \[ \int (d+c d x)^{3/2} (f-c f x)^{5/2} (a+b \arccos (c x)) \, dx=-\frac {b f x (d+c d x)^{3/2} (f-c f x)^{3/2}}{5 \left (1-c^2 x^2\right )^{3/2}}-\frac {3 b c f x^2 (d+c d x)^{3/2} (f-c f x)^{3/2}}{16 \left (1-c^2 x^2\right )^{3/2}}+\frac {2 b c^2 f x^3 (d+c d x)^{3/2} (f-c f x)^{3/2}}{15 \left (1-c^2 x^2\right )^{3/2}}-\frac {b c^4 f x^5 (d+c d x)^{3/2} (f-c f x)^{3/2}}{25 \left (1-c^2 x^2\right )^{3/2}}+\frac {b f (d+c d x)^{3/2} (f-c f x)^{3/2} \sqrt {1-c^2 x^2}}{16 c}+\frac {1}{4} f x (d+c d x)^{3/2} (f-c f x)^{3/2} (a+b \arccos (c x))+\frac {3 f x (d+c d x)^{3/2} (f-c f x)^{3/2} (a+b \arccos (c x))}{8 \left (1-c^2 x^2\right )}+\frac {f (d+c d x)^{3/2} (f-c f x)^{3/2} \left (1-c^2 x^2\right ) (a+b \arccos (c x))}{5 c}+\frac {3 f (d+c d x)^{3/2} (f-c f x)^{3/2} (a+b \arccos (c x))^2}{16 b c \left (1-c^2 x^2\right )^{3/2}} \] Output:
-1/5*b*f*x*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)/(-c^2*x^2+1)^(3/2)-3/16*b*c*f* x^2*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)/(-c^2*x^2+1)^(3/2)+2/15*b*c^2*f*x^3*( c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)/(-c^2*x^2+1)^(3/2)-1/25*b*c^4*f*x^5*(c*d*x +d)^(3/2)*(-c*f*x+f)^(3/2)/(-c^2*x^2+1)^(3/2)+1/16*b*f*(c*d*x+d)^(3/2)*(-c *f*x+f)^(3/2)*(-c^2*x^2+1)^(1/2)/c+1/4*f*x*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2 )*(a+b*arccos(c*x))+3*f*x*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)*(a+b*arccos(c*x ))/(-8*c^2*x^2+8)+1/5*f*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)*(-c^2*x^2+1)*(a+b *arccos(c*x))/c+3/16*f*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)*(a+b*arccos(c*x))^ 2/b/c/(-c^2*x^2+1)^(3/2)
Time = 2.45 (sec) , antiderivative size = 305, normalized size of antiderivative = 0.74 \[ \int (d+c d x)^{3/2} (f-c f x)^{5/2} (a+b \arccos (c x)) \, dx=\frac {d f^2 \left (-1800 b \sqrt {d+c d x} \sqrt {f-c f x} \arccos (c x)^2-3600 a \sqrt {d} \sqrt {f} \sqrt {1-c^2 x^2} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (-1+c^2 x^2\right )}\right )+\sqrt {d+c d x} \sqrt {f-c f x} \left (128 b c x \left (15-10 c^2 x^2+3 c^4 x^4\right )+240 a \sqrt {1-c^2 x^2} \left (8+25 c x-16 c^2 x^2-10 c^3 x^3+8 c^4 x^4\right )+1200 b \cos (2 \arccos (c x))-75 b \cos (4 \arccos (c x))\right )+60 b \sqrt {d+c d x} \sqrt {f-c f x} \arccos (c x) \left (32 \left (1-c^2 x^2\right )^{5/2}+40 \sin (2 \arccos (c x))-5 \sin (4 \arccos (c x))\right )\right )}{9600 c \sqrt {1-c^2 x^2}} \] Input:
Integrate[(d + c*d*x)^(3/2)*(f - c*f*x)^(5/2)*(a + b*ArcCos[c*x]),x]
Output:
(d*f^2*(-1800*b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcCos[c*x]^2 - 3600*a*Sqr t[d]*Sqrt[f]*Sqrt[1 - c^2*x^2]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x] )/(Sqrt[d]*Sqrt[f]*(-1 + c^2*x^2))] + Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(128 *b*c*x*(15 - 10*c^2*x^2 + 3*c^4*x^4) + 240*a*Sqrt[1 - c^2*x^2]*(8 + 25*c*x - 16*c^2*x^2 - 10*c^3*x^3 + 8*c^4*x^4) + 1200*b*Cos[2*ArcCos[c*x]] - 75*b *Cos[4*ArcCos[c*x]]) + 60*b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcCos[c*x]*(3 2*(1 - c^2*x^2)^(5/2) + 40*Sin[2*ArcCos[c*x]] - 5*Sin[4*ArcCos[c*x]])))/(9 600*c*Sqrt[1 - c^2*x^2])
Time = 0.66 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.46, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, 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.
\(\displaystyle \int (c d x+d)^{3/2} (f-c f x)^{5/2} (a+b \arccos (c x)) \, dx\) |
\(\Big \downarrow \) 5179 |
\(\displaystyle \frac {(c d x+d)^{3/2} (f-c f x)^{3/2} \int f (1-c x) \left (1-c^2 x^2\right )^{3/2} (a+b \arccos (c x))dx}{\left (1-c^2 x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {f (c d x+d)^{3/2} (f-c f x)^{3/2} \int (1-c x) \left (1-c^2 x^2\right )^{3/2} (a+b \arccos (c x))dx}{\left (1-c^2 x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 5263 |
\(\displaystyle \frac {f (c d x+d)^{3/2} (f-c f x)^{3/2} \int \left (\left (1-c^2 x^2\right )^{3/2} (a+b \arccos (c x))-c x \left (1-c^2 x^2\right )^{3/2} (a+b \arccos (c x))\right )dx}{\left (1-c^2 x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {f (c d x+d)^{3/2} (f-c f x)^{3/2} \left (\frac {1}{4} x \left (1-c^2 x^2\right )^{3/2} (a+b \arccos (c x))+\frac {3}{8} x \sqrt {1-c^2 x^2} (a+b \arccos (c x))+\frac {\left (1-c^2 x^2\right )^{5/2} (a+b \arccos (c x))}{5 c}-\frac {3 (a+b \arccos (c x))^2}{16 b c}+\frac {1}{25} b c^4 x^5-\frac {1}{16} b c^3 x^4-\frac {2}{15} b c^2 x^3+\frac {5}{16} b c x^2+\frac {b x}{5}\right )}{\left (1-c^2 x^2\right )^{3/2}}\) |
Input:
Int[(d + c*d*x)^(3/2)*(f - c*f*x)^(5/2)*(a + b*ArcCos[c*x]),x]
Output:
(f*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)*((b*x)/5 + (5*b*c*x^2)/16 - (2*b*c^ 2*x^3)/15 - (b*c^3*x^4)/16 + (b*c^4*x^5)/25 + (3*x*Sqrt[1 - c^2*x^2]*(a + b*ArcCos[c*x]))/8 + (x*(1 - c^2*x^2)^(3/2)*(a + b*ArcCos[c*x]))/4 + ((1 - c^2*x^2)^(5/2)*(a + b*ArcCos[c*x]))/(5*c) - (3*(a + b*ArcCos[c*x])^2)/(16* b*c)))/(1 - c^2*x^2)^(3/2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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))
Result contains complex when optimal does not.
Time = 1.86 (sec) , antiderivative size = 1185, normalized size of antiderivative = 2.88
method | result | size |
default | \(\text {Expression too large to display}\) | \(1185\) |
parts | \(\text {Expression too large to display}\) | \(1185\) |
Input:
int((c*d*x+d)^(3/2)*(-c*f*x+f)^(5/2)*(a+b*arccos(c*x)),x,method=_RETURNVER BOSE)
Output:
-1/5*a/c/f*(c*d*x+d)^(3/2)*(-c*f*x+f)^(7/2)-3/20*a*d/c/f*(c*d*x+d)^(1/2)*( -c*f*x+f)^(7/2)+1/20*a*d/c*(-c*f*x+f)^(5/2)*(c*d*x+d)^(1/2)+1/8*a*d*f/c*(- c*f*x+f)^(3/2)*(c*d*x+d)^(1/2)+3/8*a*d*f^2/c*(-c*f*x+f)^(1/2)*(c*d*x+d)^(1 /2)+3/8*a*d^2*f^3*((-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 *(3/16*(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*f^2*d+1/800*(d*(c*x+1))^(1/2)*(-f*(c*x-1))^(1/2)*(16*c^6* x^6-28*c^4*x^4+16*I*(-c^2*x^2+1)^(1/2)*x^5*c^5+13*c^2*x^2-20*I*(-c^2*x^2+1 )^(1/2)*x^3*c^3+5*I*(-c^2*x^2+1)^(1/2)*c*x-1)*(I+5*arccos(c*x))*f^2*d/(c^2 *x^2-1)/c-1/256*(d*(c*x+1))^(1/2)*(-f*(c*x-1))^(1/2)*(8*c^5*x^5-12*c^3*x^3 +8*I*(-c^2*x^2+1)^(1/2)*x^4*c^4+4*c*x-8*I*(-c^2*x^2+1)^(1/2)*x^2*c^2+I*(-c ^2*x^2+1)^(1/2))*(I+4*arccos(c*x))*f^2*d/(c^2*x^2-1)/c+1/16*(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)*f^2*d/(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))*f^2*d/(c^2*x^2-1)/c+1/1200*(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)*(11*I+45*arccos(c*x))*cos(4*arccos(c*x) )*f^2*d/(c^2*x^2-1)/c+1/600*(d*(c*x+1))^(1/2)*(-f*(c*x-1))^(1/2)*(I*c^2*x^ 2+c*x*(-c^2*x^2+1)^(1/2)-I)*(7*I+15*arccos(c*x))*sin(4*arccos(c*x))*f^2*d/ (c^2*x^2-1)/c-3/256*(d*(c*x+1))^(1/2)*(-f*(c*x-1))^(1/2)*(-I*(-c^2*x^2+...
\[ \int (d+c d x)^{3/2} (f-c f x)^{5/2} (a+b \arccos (c x)) \, dx=\int { {\left (c d x + d\right )}^{\frac {3}{2}} {\left (-c f x + f\right )}^{\frac {5}{2}} {\left (b \arccos \left (c x\right ) + a\right )} \,d x } \] Input:
integrate((c*d*x+d)^(3/2)*(-c*f*x+f)^(5/2)*(a+b*arccos(c*x)),x, algorithm= "fricas")
Output:
integral((a*c^3*d*f^2*x^3 - a*c^2*d*f^2*x^2 - a*c*d*f^2*x + a*d*f^2 + (b*c ^3*d*f^2*x^3 - b*c^2*d*f^2*x^2 - b*c*d*f^2*x + b*d*f^2)*arccos(c*x))*sqrt( c*d*x + d)*sqrt(-c*f*x + f), x)
Timed out. \[ \int (d+c d x)^{3/2} (f-c f x)^{5/2} (a+b \arccos (c x)) \, dx=\text {Timed out} \] Input:
integrate((c*d*x+d)**(3/2)*(-c*f*x+f)**(5/2)*(a+b*acos(c*x)),x)
Output:
Timed out
\[ \int (d+c d x)^{3/2} (f-c f x)^{5/2} (a+b \arccos (c x)) \, dx=\int { {\left (c d x + d\right )}^{\frac {3}{2}} {\left (-c f x + f\right )}^{\frac {5}{2}} {\left (b \arccos \left (c x\right ) + a\right )} \,d x } \] Input:
integrate((c*d*x+d)^(3/2)*(-c*f*x+f)^(5/2)*(a+b*arccos(c*x)),x, algorithm= "maxima")
Output:
b*sqrt(d)*sqrt(f)*integrate((c^3*d*f^2*x^3 - c^2*d*f^2*x^2 - c*d*f^2*x + d *f^2)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c *x), x) + 1/40*(15*sqrt(-c^2*d*f*x^2 + d*f)*d*f^2*x + 15*d^2*f^3*arcsin(c* x)/(sqrt(d*f)*c) + 10*(-c^2*d*f*x^2 + d*f)^(3/2)*f*x + 8*(-c^2*d*f*x^2 + d *f)^(5/2)/(c*d))*a
\[ \int (d+c d x)^{3/2} (f-c f x)^{5/2} (a+b \arccos (c x)) \, dx=\int { {\left (c d x + d\right )}^{\frac {3}{2}} {\left (-c f x + f\right )}^{\frac {5}{2}} {\left (b \arccos \left (c x\right ) + a\right )} \,d x } \] Input:
integrate((c*d*x+d)^(3/2)*(-c*f*x+f)^(5/2)*(a+b*arccos(c*x)),x, algorithm= "giac")
Output:
integrate((c*d*x + d)^(3/2)*(-c*f*x + f)^(5/2)*(b*arccos(c*x) + a), x)
Timed out. \[ \int (d+c d x)^{3/2} (f-c f x)^{5/2} (a+b \arccos (c x)) \, dx=\int \left (a+b\,\mathrm {acos}\left (c\,x\right )\right )\,{\left (d+c\,d\,x\right )}^{3/2}\,{\left (f-c\,f\,x\right )}^{5/2} \,d x \] Input:
int((a + b*acos(c*x))*(d + c*d*x)^(3/2)*(f - c*f*x)^(5/2),x)
Output:
int((a + b*acos(c*x))*(d + c*d*x)^(3/2)*(f - c*f*x)^(5/2), x)
\[ \int (d+c d x)^{3/2} (f-c f x)^{5/2} (a+b \arccos (c x)) \, dx=\frac {\sqrt {f}\, \sqrt {d}\, d \,f^{2} \left (-30 \mathit {asin} \left (\frac {\sqrt {-c x +1}}{\sqrt {2}}\right ) a +8 \sqrt {c x +1}\, \sqrt {-c x +1}\, a \,c^{4} x^{4}-10 \sqrt {c x +1}\, \sqrt {-c x +1}\, a \,c^{3} x^{3}-16 \sqrt {c x +1}\, \sqrt {-c x +1}\, a \,c^{2} x^{2}+25 \sqrt {c x +1}\, \sqrt {-c x +1}\, a c x +8 \sqrt {c x +1}\, \sqrt {-c x +1}\, a +40 \left (\int \sqrt {c x +1}\, \sqrt {-c x +1}\, \mathit {acos} \left (c x \right ) x^{3}d x \right ) b \,c^{4}-40 \left (\int \sqrt {c x +1}\, \sqrt {-c x +1}\, \mathit {acos} \left (c x \right ) x^{2}d x \right ) b \,c^{3}-40 \left (\int \sqrt {c x +1}\, \sqrt {-c x +1}\, \mathit {acos} \left (c x \right ) x d x \right ) b \,c^{2}+40 \left (\int \sqrt {c x +1}\, \sqrt {-c x +1}\, \mathit {acos} \left (c x \right )d x \right ) b c \right )}{40 c} \] Input:
int((c*d*x+d)^(3/2)*(-c*f*x+f)^(5/2)*(a+b*acos(c*x)),x)
Output:
(sqrt(f)*sqrt(d)*d*f**2*( - 30*asin(sqrt( - c*x + 1)/sqrt(2))*a + 8*sqrt(c *x + 1)*sqrt( - c*x + 1)*a*c**4*x**4 - 10*sqrt(c*x + 1)*sqrt( - c*x + 1)*a *c**3*x**3 - 16*sqrt(c*x + 1)*sqrt( - c*x + 1)*a*c**2*x**2 + 25*sqrt(c*x + 1)*sqrt( - c*x + 1)*a*c*x + 8*sqrt(c*x + 1)*sqrt( - c*x + 1)*a + 40*int(s qrt(c*x + 1)*sqrt( - c*x + 1)*acos(c*x)*x**3,x)*b*c**4 - 40*int(sqrt(c*x + 1)*sqrt( - c*x + 1)*acos(c*x)*x**2,x)*b*c**3 - 40*int(sqrt(c*x + 1)*sqrt( - c*x + 1)*acos(c*x)*x,x)*b*c**2 + 40*int(sqrt(c*x + 1)*sqrt( - c*x + 1)* acos(c*x),x)*b*c))/(40*c)