3.6.16 \(\int (d+c d x)^{5/2} (f-c f x)^{5/2} (a+b \arcsin (c x)) \, dx\) [516]

3.6.16.1 Optimal result

Integrand size = 30, antiderivative size = 315 \[ \int (d+c d x)^{5/2} (f-c f x)^{5/2} (a+b \arcsin (c x)) \, dx=-\frac {25 b c x^2 (d+c d x)^{5/2} (f-c f x)^{5/2}}{96 \left (1-c^2 x^2\right )^{5/2}}+\frac {5 b c^3 x^4 (d+c d x)^{5/2} (f-c f x)^{5/2}}{96 \left (1-c^2 x^2\right )^{5/2}}+\frac {b (d+c d x)^{5/2} (f-c f x)^{5/2} \sqrt {1-c^2 x^2}}{36 c}+\frac {1}{6} x (d+c d x)^{5/2} (f-c f x)^{5/2} (a+b \arcsin (c x))+\frac {5 x (d+c d x)^{5/2} (f-c f x)^{5/2} (a+b \arcsin (c x))}{16 \left (1-c^2 x^2\right )^2}+\frac {5 x (d+c d x)^{5/2} (f-c f x)^{5/2} (a+b \arcsin (c x))}{24 \left (1-c^2 x^2\right )}+\frac {5 (d+c d x)^{5/2} (f-c f x)^{5/2} (a+b \arcsin (c x))^2}{32 b c \left (1-c^2 x^2\right )^{5/2}} \]

output

-25/96*b*c*x^2*(c*d*x+d)^(5/2)*(-c*f*x+f)^(5/2)/(-c^2*x^2+1)^(5/2)+5/96*b* 
c^3*x^4*(c*d*x+d)^(5/2)*(-c*f*x+f)^(5/2)/(-c^2*x^2+1)^(5/2)+1/6*x*(c*d*x+d 
)^(5/2)*(-c*f*x+f)^(5/2)*(a+b*arcsin(c*x))+5/16*x*(c*d*x+d)^(5/2)*(-c*f*x+ 
f)^(5/2)*(a+b*arcsin(c*x))/(-c^2*x^2+1)^2+5/24*x*(c*d*x+d)^(5/2)*(-c*f*x+f 
)^(5/2)*(a+b*arcsin(c*x))/(-c^2*x^2+1)+5/32*(c*d*x+d)^(5/2)*(-c*f*x+f)^(5/ 
2)*(a+b*arcsin(c*x))^2/b/c/(-c^2*x^2+1)^(5/2)+1/36*b*(c*d*x+d)^(5/2)*(-c*f 
*x+f)^(5/2)*(-c^2*x^2+1)^(1/2)/c
 

3.6.16.2 Mathematica [A] (verified)

Time = 3.06 (sec) , antiderivative size = 303, normalized size of antiderivative = 0.96 \[ \int (d+c d x)^{5/2} (f-c f x)^{5/2} (a+b \arcsin (c x)) \, dx=\frac {d^2 f^2 \left (360 b \sqrt {d+c d x} \sqrt {f-c f x} \arcsin (c x)^2-720 a \sqrt {d} \sqrt {f} \sqrt {1-c^2 x^2} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (-1+c^2 x^2\right )}\right )+\sqrt {d+c d x} \sqrt {f-c f x} \left (1584 a c x \sqrt {1-c^2 x^2}-1248 a c^3 x^3 \sqrt {1-c^2 x^2}+384 a c^5 x^5 \sqrt {1-c^2 x^2}+270 b \cos (2 \arcsin (c x))+27 b \cos (4 \arcsin (c x))+2 b \cos (6 \arcsin (c x))\right )+12 b \sqrt {d+c d x} \sqrt {f-c f x} \arcsin (c x) (45 \sin (2 \arcsin (c x))+9 \sin (4 \arcsin (c x))+\sin (6 \arcsin (c x)))\right )}{2304 c \sqrt {1-c^2 x^2}} \]

input

Integrate[(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)*(a + b*ArcSin[c*x]),x]
 

output

(d^2*f^2*(360*b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]^2 - 720*a*Sqrt 
[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]*(1584 
*a*c*x*Sqrt[1 - c^2*x^2] - 1248*a*c^3*x^3*Sqrt[1 - c^2*x^2] + 384*a*c^5*x^ 
5*Sqrt[1 - c^2*x^2] + 270*b*Cos[2*ArcSin[c*x]] + 27*b*Cos[4*ArcSin[c*x]] + 
 2*b*Cos[6*ArcSin[c*x]]) + 12*b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x 
]*(45*Sin[2*ArcSin[c*x]] + 9*Sin[4*ArcSin[c*x]] + Sin[6*ArcSin[c*x]])))/(2 
304*c*Sqrt[1 - c^2*x^2])
 

3.6.16.3 Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.64, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {5178, 5158, 241, 5158, 244, 2009, 5156, 15, 5152}

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.

\(\displaystyle \int (c d x+d)^{5/2} (f-c f x)^{5/2} (a+b \arcsin (c x)) \, dx\)

\(\Big \downarrow \) 5178

\(\displaystyle \frac {(c d x+d)^{5/2} (f-c f x)^{5/2} \int \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))dx}{\left (1-c^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 5158

\(\displaystyle \frac {(c d x+d)^{5/2} (f-c f x)^{5/2} \left (\frac {5}{6} \int \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))dx-\frac {1}{6} b c \int x \left (1-c^2 x^2\right )^2dx+\frac {1}{6} x \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))\right )}{\left (1-c^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 241

\(\displaystyle \frac {(c d x+d)^{5/2} (f-c f x)^{5/2} \left (\frac {5}{6} \int \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))dx+\frac {1}{6} x \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))+\frac {b \left (1-c^2 x^2\right )^3}{36 c}\right )}{\left (1-c^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 5158

\(\displaystyle \frac {(c d x+d)^{5/2} (f-c f x)^{5/2} \left (\frac {5}{6} \left (\frac {3}{4} \int \sqrt {1-c^2 x^2} (a+b \arcsin (c x))dx-\frac {1}{4} b c \int x \left (1-c^2 x^2\right )dx+\frac {1}{4} x \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))\right )+\frac {1}{6} x \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))+\frac {b \left (1-c^2 x^2\right )^3}{36 c}\right )}{\left (1-c^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 244

\(\displaystyle \frac {(c d x+d)^{5/2} (f-c f x)^{5/2} \left (\frac {5}{6} \left (\frac {3}{4} \int \sqrt {1-c^2 x^2} (a+b \arcsin (c x))dx-\frac {1}{4} b c \int \left (x-c^2 x^3\right )dx+\frac {1}{4} x \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))\right )+\frac {1}{6} x \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))+\frac {b \left (1-c^2 x^2\right )^3}{36 c}\right )}{\left (1-c^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {(c d x+d)^{5/2} (f-c f x)^{5/2} \left (\frac {5}{6} \left (\frac {3}{4} \int \sqrt {1-c^2 x^2} (a+b \arcsin (c x))dx+\frac {1}{4} x \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))-\frac {1}{4} b c \left (\frac {x^2}{2}-\frac {c^2 x^4}{4}\right )\right )+\frac {1}{6} x \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))+\frac {b \left (1-c^2 x^2\right )^3}{36 c}\right )}{\left (1-c^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 5156

\(\displaystyle \frac {(c d x+d)^{5/2} (f-c f x)^{5/2} \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {a+b \arcsin (c x)}{\sqrt {1-c^2 x^2}}dx-\frac {1}{2} b c \int xdx+\frac {1}{2} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))\right )+\frac {1}{4} x \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))-\frac {1}{4} b c \left (\frac {x^2}{2}-\frac {c^2 x^4}{4}\right )\right )+\frac {1}{6} x \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))+\frac {b \left (1-c^2 x^2\right )^3}{36 c}\right )}{\left (1-c^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {(c d x+d)^{5/2} (f-c f x)^{5/2} \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {a+b \arcsin (c x)}{\sqrt {1-c^2 x^2}}dx+\frac {1}{2} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))-\frac {1}{4} b c x^2\right )+\frac {1}{4} x \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))-\frac {1}{4} b c \left (\frac {x^2}{2}-\frac {c^2 x^4}{4}\right )\right )+\frac {1}{6} x \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))+\frac {b \left (1-c^2 x^2\right )^3}{36 c}\right )}{\left (1-c^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 5152

\(\displaystyle \frac {(c d x+d)^{5/2} (f-c f x)^{5/2} \left (\frac {1}{6} x \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))+\frac {5}{6} \left (\frac {1}{4} x \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))+\frac {3}{4} \left (\frac {1}{2} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))+\frac {(a+b \arcsin (c x))^2}{4 b c}-\frac {1}{4} b c x^2\right )-\frac {1}{4} b c \left (\frac {x^2}{2}-\frac {c^2 x^4}{4}\right )\right )+\frac {b \left (1-c^2 x^2\right )^3}{36 c}\right )}{\left (1-c^2 x^2\right )^{5/2}}\)

input

Int[(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)*(a + b*ArcSin[c*x]),x]
 

output

((d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)*((b*(1 - c^2*x^2)^3)/(36*c) + (x*(1 - 
 c^2*x^2)^(5/2)*(a + b*ArcSin[c*x]))/6 + (5*(-1/4*(b*c*(x^2/2 - (c^2*x^4)/ 
4)) + (x*(1 - c^2*x^2)^(3/2)*(a + b*ArcSin[c*x]))/4 + (3*(-1/4*(b*c*x^2) + 
 (x*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/2 + (a + b*ArcSin[c*x])^2/(4*b* 
c)))/4))/6))/(1 - c^2*x^2)^(5/2)
 

3.6.16.3.1 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 241

Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x^2)^(p + 1)/ 
(2*b*(p + 1)), x] /; FreeQ[{a, b, p}, x] && NeQ[p, -1]
 

rule 244

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand 
Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p 
, 0]
 

rule 2009

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

rule 5152

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

rule 5156

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

rule 5158

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x 
_Symbol] :> Simp[x*(d + e*x^2)^p*((a + b*ArcSin[c*x])^n/(2*p + 1)), x] + (S 
imp[2*d*(p/(2*p + 1))   Int[(d + e*x^2)^(p - 1)*(a + b*ArcSin[c*x])^n, x], 
x] - Simp[b*c*(n/(2*p + 1))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p]   Int[x*(1 
- c^2*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c 
, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0]
 

rule 5178

Int[((a_.) + ArcSin[(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*ArcSin[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]
 

3.6.16.4 Maple [F]

\[\int \left (c d x +d \right )^{\frac {5}{2}} \left (-c f x +f \right )^{\frac {5}{2}} \left (a +b \arcsin \left (c x \right )\right )d x\]

input

int((c*d*x+d)^(5/2)*(-c*f*x+f)^(5/2)*(a+b*arcsin(c*x)),x)
 

output

int((c*d*x+d)^(5/2)*(-c*f*x+f)^(5/2)*(a+b*arcsin(c*x)),x)
 

3.6.16.5 Fricas [F]

\[ \int (d+c d x)^{5/2} (f-c f x)^{5/2} (a+b \arcsin (c x)) \, dx=\int { {\left (c d x + d\right )}^{\frac {5}{2}} {\left (-c f x + f\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \]

input

integrate((c*d*x+d)^(5/2)*(-c*f*x+f)^(5/2)*(a+b*arcsin(c*x)),x, algorithm= 
"fricas")
 

output

integral((a*c^4*d^2*f^2*x^4 - 2*a*c^2*d^2*f^2*x^2 + a*d^2*f^2 + (b*c^4*d^2 
*f^2*x^4 - 2*b*c^2*d^2*f^2*x^2 + b*d^2*f^2)*arcsin(c*x))*sqrt(c*d*x + d)*s 
qrt(-c*f*x + f), x)
 

3.6.16.6 Sympy [F(-1)]

Timed out. \[ \int (d+c d x)^{5/2} (f-c f x)^{5/2} (a+b \arcsin (c x)) \, dx=\text {Timed out} \]

input

integrate((c*d*x+d)**(5/2)*(-c*f*x+f)**(5/2)*(a+b*asin(c*x)),x)
 

output

Timed out
 

3.6.16.7 Maxima [F]

\[ \int (d+c d x)^{5/2} (f-c f x)^{5/2} (a+b \arcsin (c x)) \, dx=\int { {\left (c d x + d\right )}^{\frac {5}{2}} {\left (-c f x + f\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \]

input

integrate((c*d*x+d)^(5/2)*(-c*f*x+f)^(5/2)*(a+b*arcsin(c*x)),x, algorithm= 
"maxima")
 

output

b*sqrt(d)*sqrt(f)*integrate((c^4*d^2*f^2*x^4 - 2*c^2*d^2*f^2*x^2 + d^2*f^2 
)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)), 
 x) + 1/48*(15*sqrt(-c^2*d*f*x^2 + d*f)*d^2*f^2*x + 15*d^3*f^3*arcsin(c*x) 
/(sqrt(d*f)*c) + 10*(-c^2*d*f*x^2 + d*f)^(3/2)*d*f*x + 8*(-c^2*d*f*x^2 + d 
*f)^(5/2)*x)*a
 

3.6.16.8 Giac [F]

\[ \int (d+c d x)^{5/2} (f-c f x)^{5/2} (a+b \arcsin (c x)) \, dx=\int { {\left (c d x + d\right )}^{\frac {5}{2}} {\left (-c f x + f\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \]

input

integrate((c*d*x+d)^(5/2)*(-c*f*x+f)^(5/2)*(a+b*arcsin(c*x)),x, algorithm= 
"giac")
 

output

integrate((c*d*x + d)^(5/2)*(-c*f*x + f)^(5/2)*(b*arcsin(c*x) + a), x)
 

3.6.16.9 Mupad [F(-1)]

Timed out. \[ \int (d+c d x)^{5/2} (f-c f x)^{5/2} (a+b \arcsin (c x)) \, dx=\int \left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d+c\,d\,x\right )}^{5/2}\,{\left (f-c\,f\,x\right )}^{5/2} \,d x \]

input

int((a + b*asin(c*x))*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2),x)
 

output

int((a + b*asin(c*x))*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2), x)