3.6.93 \(\int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}} \, dx\) [593]

3.6.93.1 Optimal result

Integrand size = 32, antiderivative size = 217 \[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}} \, dx=\frac {x \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {i \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {2 b \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x)) \log \left (1+e^{2 i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {i b^2 \left (1-c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}} \]

output

x*(-c^2*x^2+1)*(a+b*arcsin(c*x))^2/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)-I*(-c^ 
2*x^2+1)^(3/2)*(a+b*arcsin(c*x))^2/c/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)+2*b* 
(-c^2*x^2+1)^(3/2)*(a+b*arcsin(c*x))*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)/c/ 
(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)-I*b^2*(-c^2*x^2+1)^(3/2)*polylog(2,-(I*c* 
x+(-c^2*x^2+1)^(1/2))^2)/c/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)
 

3.6.93.2 Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(550\) vs. \(2(217)=434\).

Time = 0.69 (sec) , antiderivative size = 550, normalized size of antiderivative = 2.53 \[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}} \, dx=\frac {a^2 c x+2 a b c x \arcsin (c x)+2 i b^2 \pi \sqrt {1-c^2 x^2} \arcsin (c x)+b^2 c x \arcsin (c x)^2-i b^2 \sqrt {1-c^2 x^2} \arcsin (c x)^2+4 b^2 \pi \sqrt {1-c^2 x^2} \log \left (1+e^{-i \arcsin (c x)}\right )+b^2 \pi \sqrt {1-c^2 x^2} \log \left (1-i e^{i \arcsin (c x)}\right )+2 b^2 \sqrt {1-c^2 x^2} \arcsin (c x) \log \left (1-i e^{i \arcsin (c x)}\right )-b^2 \pi \sqrt {1-c^2 x^2} \log \left (1+i e^{i \arcsin (c x)}\right )+2 b^2 \sqrt {1-c^2 x^2} \arcsin (c x) \log \left (1+i e^{i \arcsin (c x)}\right )-4 b^2 \pi \sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )\right )+b^2 \pi \sqrt {1-c^2 x^2} \log \left (-\cos \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )+2 a b \sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )+2 a b \sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )-b^2 \pi \sqrt {1-c^2 x^2} \log \left (\sin \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )-2 i b^2 \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )-2 i b^2 \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c d e \sqrt {d+c d x} \sqrt {e-c e x}} \]

input

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

output

(a^2*c*x + 2*a*b*c*x*ArcSin[c*x] + (2*I)*b^2*Pi*Sqrt[1 - c^2*x^2]*ArcSin[c 
*x] + b^2*c*x*ArcSin[c*x]^2 - I*b^2*Sqrt[1 - c^2*x^2]*ArcSin[c*x]^2 + 4*b^ 
2*Pi*Sqrt[1 - c^2*x^2]*Log[1 + E^((-I)*ArcSin[c*x])] + b^2*Pi*Sqrt[1 - c^2 
*x^2]*Log[1 - I*E^(I*ArcSin[c*x])] + 2*b^2*Sqrt[1 - c^2*x^2]*ArcSin[c*x]*L 
og[1 - I*E^(I*ArcSin[c*x])] - b^2*Pi*Sqrt[1 - c^2*x^2]*Log[1 + I*E^(I*ArcS 
in[c*x])] + 2*b^2*Sqrt[1 - c^2*x^2]*ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x] 
)] - 4*b^2*Pi*Sqrt[1 - c^2*x^2]*Log[Cos[ArcSin[c*x]/2]] + b^2*Pi*Sqrt[1 - 
c^2*x^2]*Log[-Cos[(Pi + 2*ArcSin[c*x])/4]] + 2*a*b*Sqrt[1 - c^2*x^2]*Log[C 
os[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]] + 2*a*b*Sqrt[1 - c^2*x^2]*Log[Cos[ 
ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]] - b^2*Pi*Sqrt[1 - c^2*x^2]*Log[Sin[(P 
i + 2*ArcSin[c*x])/4]] - (2*I)*b^2*Sqrt[1 - c^2*x^2]*PolyLog[2, (-I)*E^(I* 
ArcSin[c*x])] - (2*I)*b^2*Sqrt[1 - c^2*x^2]*PolyLog[2, I*E^(I*ArcSin[c*x]) 
])/(c*d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])
 

3.6.93.3 Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.65, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5178, 5160, 5180, 3042, 4202, 2620, 2715, 2838}

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.

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

\(\Big \downarrow \) 5178

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

\(\Big \downarrow \) 5160

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

\(\Big \downarrow \) 5180

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4202

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

\(\displaystyle \frac {\left (1-c^2 x^2\right )^{3/2} \left (\frac {x (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}-\frac {2 b \left (\frac {i (a+b \arcsin (c x))^2}{2 b}-2 i \left (-\frac {1}{2} i \log \left (1+e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))-\frac {1}{4} b \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )\right )\right )}{c}\right )}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\)

input

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

output

((1 - c^2*x^2)^(3/2)*((x*(a + b*ArcSin[c*x])^2)/Sqrt[1 - c^2*x^2] - (2*b*( 
((I/2)*(a + b*ArcSin[c*x])^2)/b - (2*I)*((-1/2*I)*(a + b*ArcSin[c*x])*Log[ 
1 + E^((2*I)*ArcSin[c*x])] - (b*PolyLog[2, -E^((2*I)*ArcSin[c*x])])/4)))/c 
))/((d + c*d*x)^(3/2)*(e - c*e*x)^(3/2))
 

3.6.93.3.1 Defintions of rubi rules used

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4202

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I 
*((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I   Int[(c + d*x)^m*(E^(2*I*( 
e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] && IGt 
Q[m, 0]
 

rule 5160

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x 
_Symbol] :> Simp[x*((a + b*ArcSin[c*x])^n/(d*Sqrt[d + e*x^2])), x] - Simp[b 
*c*(n/d)*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]   Int[x*((a + b*ArcSin[c*x 
])^(n - 1)/(1 - c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d 
 + e, 0] && GtQ[n, 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]
 

rule 5180

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

3.6.93.4 Maple [F]

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

input

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

output

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

3.6.93.5 Fricas [F]

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

input

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

output

integral((b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2)*sqrt(c*d*x + d)*sqr 
t(-c*e*x + e)/(c^4*d^2*e^2*x^4 - 2*c^2*d^2*e^2*x^2 + d^2*e^2), x)
 

3.6.93.6 Sympy [F]

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

input

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

output

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

3.6.93.7 Maxima [F]

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

input

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

output

-b^2*integrate(arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2/((c^2*d*e*x^2 
- d*e)*sqrt(c*x + 1)*sqrt(-c*x + 1)), x)/(sqrt(d)*sqrt(e)) + 2*a*b*x*arcsi 
n(c*x)/(sqrt(-c^2*d*e*x^2 + d*e)*d*e) + a^2*x/(sqrt(-c^2*d*e*x^2 + d*e)*d* 
e) - a*b*sqrt(1/(d*e))*log(x^2 - 1/c^2)/(c*d*e)
 

3.6.93.8 Giac [F]

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

input

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

output

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

3.6.93.9 Mupad [F(-1)]

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

input

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

output

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