3.6.30 \(\int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))}{(f-c f x)^{3/2}} \, dx\) [530]

3.6.30.1 Optimal result

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

output

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

3.6.30.2 Mathematica [A] (verified)

Time = 2.36 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.73 \[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))}{(f-c f x)^{3/2}} \, dx=-\frac {\frac {4 a \sqrt {d+c d x} \sqrt {f-c f x}}{-1+c x}-2 a \sqrt {d} \sqrt {f} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (-1+c^2 x^2\right )}\right )+\frac {b (1+c x) \sqrt {d+c d x} \sqrt {f-c f x} \left (\cos \left (\frac {1}{2} \arcsin (c x)\right ) \left ((-4+\arcsin (c x)) \arcsin (c x)-8 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right )-\left (\arcsin (c x) (4+\arcsin (c x))-8 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right ) \sin \left (\frac {1}{2} \arcsin (c x)\right )\right )}{\sqrt {1-c^2 x^2} \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right ) \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )^2}}{2 c f^2} \]

input

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

3.6.30.3 Rubi [A] (verified)

Time = 0.53 (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 = {5178, 27, 5274, 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.

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

\(\Big \downarrow \) 5178

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 5274

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

\(\Big \downarrow \) 2009

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

input

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

output

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

3.6.30.3.1 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 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]
 

rule 5274

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_ 
) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcSin[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]
 

3.6.30.4 Maple [F]

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

input

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

output

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

3.6.30.5 Fricas [F]

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

input

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

output

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

3.6.30.6 Sympy [F]

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

input

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

output

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

3.6.30.7 Maxima [F]

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

input

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

output

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

3.6.30.8 Giac [F]

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

input

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

output

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

3.6.30.9 Mupad [F(-1)]

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

input

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

output

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