3.6.11 \(\int (d+c d x)^{3/2} (f-c f x)^{3/2} (a+b \arcsin (c x)) \, dx\) [511]

3.6.11.1 Optimal result

Integrand size = 30, antiderivative size = 226 \[ \int (d+c d x)^{3/2} (f-c f x)^{3/2} (a+b \arcsin (c x)) \, dx=-\frac {5 b c x^2 (d+c d x)^{3/2} (f-c f x)^{3/2}}{16 \left (1-c^2 x^2\right )^{3/2}}+\frac {b c^3 x^4 (d+c d x)^{3/2} (f-c f x)^{3/2}}{16 \left (1-c^2 x^2\right )^{3/2}}+\frac {1}{4} x (d+c d x)^{3/2} (f-c f x)^{3/2} (a+b \arcsin (c x))+\frac {3 x (d+c d x)^{3/2} (f-c f x)^{3/2} (a+b \arcsin (c x))}{8 \left (1-c^2 x^2\right )}+\frac {3 (d+c d x)^{3/2} (f-c f x)^{3/2} (a+b \arcsin (c x))^2}{16 b c \left (1-c^2 x^2\right )^{3/2}} \]

output

-5/16*b*c*x^2*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)/(-c^2*x^2+1)^(3/2)+1/16*b*c 
^3*x^4*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)/(-c^2*x^2+1)^(3/2)+1/4*x*(c*d*x+d) 
^(3/2)*(-c*f*x+f)^(3/2)*(a+b*arcsin(c*x))+3/8*x*(c*d*x+d)^(3/2)*(-c*f*x+f) 
^(3/2)*(a+b*arcsin(c*x))/(-c^2*x^2+1)+3/16*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2 
)*(a+b*arcsin(c*x))^2/b/c/(-c^2*x^2+1)^(3/2)
 

3.6.11.2 Mathematica [A] (verified)

Time = 2.28 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.09 \[ \int (d+c d x)^{3/2} (f-c f x)^{3/2} (a+b \arcsin (c x)) \, dx=\frac {24 b d f \sqrt {d+c d x} \sqrt {f-c f x} \arcsin (c x)^2-48 a d^{3/2} f^{3/2} \sqrt {1-c^2 x^2} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (-1+c^2 x^2\right )}\right )+d f \sqrt {d+c d x} \sqrt {f-c f x} \left (16 a c x \left (5-2 c^2 x^2\right ) \sqrt {1-c^2 x^2}+16 b \cos (2 \arcsin (c x))+b \cos (4 \arcsin (c x))\right )+4 b d f \sqrt {d+c d x} \sqrt {f-c f x} \arcsin (c x) (8 \sin (2 \arcsin (c x))+\sin (4 \arcsin (c x)))}{128 c \sqrt {1-c^2 x^2}} \]

input

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

output

(24*b*d*f*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]^2 - 48*a*d^(3/2)*f^( 
3/2)*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))] + d*f*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(16*a*c* 
x*(5 - 2*c^2*x^2)*Sqrt[1 - c^2*x^2] + 16*b*Cos[2*ArcSin[c*x]] + b*Cos[4*Ar 
cSin[c*x]]) + 4*b*d*f*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]*(8*Sin[2 
*ArcSin[c*x]] + Sin[4*ArcSin[c*x]]))/(128*c*Sqrt[1 - c^2*x^2])
 

3.6.11.3 Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.66, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {5178, 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.

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

\(\Big \downarrow \) 5178

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

\(\Big \downarrow \) 5158

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

\(\Big \downarrow \) 244

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 5156

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

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 5152

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

input

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

output

((d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)*(-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*Sqr 
t[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/2 + (a + b*ArcSin[c*x])^2/(4*b*c)))/4) 
)/(1 - c^2*x^2)^(3/2)
 

3.6.11.3.1 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 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]
 

3.6.11.4 Maple [F]

\[\int \left (c d x +d \right )^{\frac {3}{2}} \left (-c f x +f \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )d x\]

input

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

output

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

3.6.11.5 Fricas [F]

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

input

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

output

integral(-(a*c^2*d*f*x^2 - a*d*f + (b*c^2*d*f*x^2 - b*d*f)*arcsin(c*x))*sq 
rt(c*d*x + d)*sqrt(-c*f*x + f), x)
 

3.6.11.6 Sympy [F]

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

input

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

output

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

3.6.11.7 Maxima [F]

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

input

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

output

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

3.6.11.8 Giac [F]

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

input

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

output

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

3.6.11.9 Mupad [F(-1)]

Timed out. \[ \int (d+c d x)^{3/2} (f-c f x)^{3/2} (a+b \arcsin (c x)) \, dx=\int \left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d+c\,d\,x\right )}^{3/2}\,{\left (f-c\,f\,x\right )}^{3/2} \,d x \]

input

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

output

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