\(\int x (a+b \arcsin (c x))^{5/2} \, dx\) [152]

Optimal result

Integrand size = 14, antiderivative size = 216 \[ \int x (a+b \arcsin (c x))^{5/2} \, dx=\frac {15 b^2 \sqrt {a+b \arcsin (c x)}}{64 c^2}-\frac {15}{32} b^2 x^2 \sqrt {a+b \arcsin (c x)}+\frac {5 b x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^{3/2}}{8 c}-\frac {(a+b \arcsin (c x))^{5/2}}{4 c^2}+\frac {1}{2} x^2 (a+b \arcsin (c x))^{5/2}-\frac {15 b^{5/2} \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {a+b \arcsin (c x)}}{\sqrt {b} \sqrt {\pi }}\right )}{128 c^2}-\frac {15 b^{5/2} \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {a+b \arcsin (c x)}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {2 a}{b}\right )}{128 c^2} \] Output:

15/64*b^2*(a+b*arcsin(c*x))^(1/2)/c^2-15/32*b^2*x^2*(a+b*arcsin(c*x))^(1/2 
)+5/8*b*x*(-c^2*x^2+1)^(1/2)*(a+b*arcsin(c*x))^(3/2)/c-1/4*(a+b*arcsin(c*x 
))^(5/2)/c^2+1/2*x^2*(a+b*arcsin(c*x))^(5/2)-15/128*b^(5/2)*Pi^(1/2)*cos(2 
*a/b)*FresnelC(2*(a+b*arcsin(c*x))^(1/2)/b^(1/2)/Pi^(1/2))/c^2-15/128*b^(5 
/2)*Pi^(1/2)*FresnelS(2*(a+b*arcsin(c*x))^(1/2)/b^(1/2)/Pi^(1/2))*sin(2*a/ 
b)/c^2
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.05 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.60 \[ \int x (a+b \arcsin (c x))^{5/2} \, dx=\frac {i b^3 e^{-\frac {2 i a}{b}} \left (\sqrt {-\frac {i (a+b \arcsin (c x))}{b}} \Gamma \left (\frac {7}{2},-\frac {2 i (a+b \arcsin (c x))}{b}\right )-e^{\frac {4 i a}{b}} \sqrt {\frac {i (a+b \arcsin (c x))}{b}} \Gamma \left (\frac {7}{2},\frac {2 i (a+b \arcsin (c x))}{b}\right )\right )}{32 \sqrt {2} c^2 \sqrt {a+b \arcsin (c x)}} \] Input:

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

Output:

((I/32)*b^3*(Sqrt[((-I)*(a + b*ArcSin[c*x]))/b]*Gamma[7/2, ((-2*I)*(a + b* 
ArcSin[c*x]))/b] - E^(((4*I)*a)/b)*Sqrt[(I*(a + b*ArcSin[c*x]))/b]*Gamma[7 
/2, ((2*I)*(a + b*ArcSin[c*x]))/b]))/(Sqrt[2]*c^2*E^(((2*I)*a)/b)*Sqrt[a + 
 b*ArcSin[c*x]])
 

Rubi [A] (verified)

Time = 1.66 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {5140, 5210, 5140, 5152, 5224, 3042, 3793, 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.

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

\(\Big \downarrow \) 5140

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

\(\Big \downarrow \) 5210

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

\(\Big \downarrow \) 5140

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

\(\Big \downarrow \) 5152

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

\(\Big \downarrow \) 5224

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3793

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} x^2 (a+b \arcsin (c x))^{5/2}-\frac {5}{4} b c \left (\frac {(a+b \arcsin (c x))^{5/2}}{5 b c^3}+\frac {3 b \left (\frac {1}{2} x^2 \sqrt {a+b \arcsin (c x)}-\frac {-\frac {1}{2} \sqrt {\pi } \sqrt {b} \cos \left (\frac {2 a}{b}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {a+b \arcsin (c x)}}{\sqrt {b} \sqrt {\pi }}\right )-\frac {1}{2} \sqrt {\pi } \sqrt {b} \sin \left (\frac {2 a}{b}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {a+b \arcsin (c x)}}{\sqrt {b} \sqrt {\pi }}\right )+\sqrt {a+b \arcsin (c x)}}{4 c^2}\right )}{4 c}-\frac {x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^{3/2}}{2 c^2}\right )\)

Input:

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

Output:

(x^2*(a + b*ArcSin[c*x])^(5/2))/2 - (5*b*c*(-1/2*(x*Sqrt[1 - c^2*x^2]*(a + 
 b*ArcSin[c*x])^(3/2))/c^2 + (a + b*ArcSin[c*x])^(5/2)/(5*b*c^3) + (3*b*(( 
x^2*Sqrt[a + b*ArcSin[c*x]])/2 - (Sqrt[a + b*ArcSin[c*x]] - (Sqrt[b]*Sqrt[ 
Pi]*Cos[(2*a)/b]*FresnelC[(2*Sqrt[a + b*ArcSin[c*x]])/(Sqrt[b]*Sqrt[Pi])]) 
/2 - (Sqrt[b]*Sqrt[Pi]*FresnelS[(2*Sqrt[a + b*ArcSin[c*x]])/(Sqrt[b]*Sqrt[ 
Pi])]*Sin[(2*a)/b])/2)/(4*c^2)))/(4*c)))/4
 

Defintions of rubi rules used

rule 2009

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

rule 3042

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

rule 3793

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In 
t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f 
, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
 

rule 5140

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[x 
^(m + 1)*((a + b*ArcSin[c*x])^n/(m + 1)), x] - Simp[b*c*(n/(m + 1))   Int[x 
^(m + 1)*((a + b*ArcSin[c*x])^(n - 1)/Sqrt[1 - c^2*x^2]), x], x] /; FreeQ[{ 
a, b, c}, x] && IGtQ[m, 0] && GtQ[n, 0]
 

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 5210

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

rule 5224

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(407\) vs. \(2(170)=340\).

Time = 0.10 (sec) , antiderivative size = 408, normalized size of antiderivative = 1.89

method	result	size
default	\(-\frac {15 \sqrt {-\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {a +b \arcsin \left (c x \right )}\, \cos \left (\frac {2 a}{b}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) b^{3}-15 \sqrt {-\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {a +b \arcsin \left (c x \right )}\, \sin \left (\frac {2 a}{b}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) b^{3}+32 \arcsin \left (c x \right )^{3} \cos \left (-\frac {2 \left (a +b \arcsin \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) b^{3}+96 \arcsin \left (c x \right )^{2} \cos \left (-\frac {2 \left (a +b \arcsin \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) a \,b^{2}+40 \arcsin \left (c x \right )^{2} \sin \left (-\frac {2 \left (a +b \arcsin \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) b^{3}+96 \arcsin \left (c x \right ) \cos \left (-\frac {2 \left (a +b \arcsin \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) a^{2} b -30 \arcsin \left (c x \right ) \cos \left (-\frac {2 \left (a +b \arcsin \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) b^{3}+80 \arcsin \left (c x \right ) \sin \left (-\frac {2 \left (a +b \arcsin \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) a \,b^{2}+32 \cos \left (-\frac {2 \left (a +b \arcsin \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) a^{3}-30 \cos \left (-\frac {2 \left (a +b \arcsin \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) a \,b^{2}+40 \sin \left (-\frac {2 \left (a +b \arcsin \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) a^{2} b}{128 c^{2} \sqrt {a +b \arcsin \left (c x \right )}}\)	\(408\)

Input:

int(x*(a+b*arcsin(c*x))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/128/c^2/(a+b*arcsin(c*x))^(1/2)*(15*(-1/b)^(1/2)*Pi^(1/2)*(a+b*arcsin(c 
*x))^(1/2)*cos(2*a/b)*FresnelC(2*2^(1/2)/Pi^(1/2)/(-2/b)^(1/2)*(a+b*arcsin 
(c*x))^(1/2)/b)*b^3-15*(-1/b)^(1/2)*Pi^(1/2)*(a+b*arcsin(c*x))^(1/2)*sin(2 
*a/b)*FresnelS(2*2^(1/2)/Pi^(1/2)/(-2/b)^(1/2)*(a+b*arcsin(c*x))^(1/2)/b)* 
b^3+32*arcsin(c*x)^3*cos(-2*(a+b*arcsin(c*x))/b+2*a/b)*b^3+96*arcsin(c*x)^ 
2*cos(-2*(a+b*arcsin(c*x))/b+2*a/b)*a*b^2+40*arcsin(c*x)^2*sin(-2*(a+b*arc 
sin(c*x))/b+2*a/b)*b^3+96*arcsin(c*x)*cos(-2*(a+b*arcsin(c*x))/b+2*a/b)*a^ 
2*b-30*arcsin(c*x)*cos(-2*(a+b*arcsin(c*x))/b+2*a/b)*b^3+80*arcsin(c*x)*si 
n(-2*(a+b*arcsin(c*x))/b+2*a/b)*a*b^2+32*cos(-2*(a+b*arcsin(c*x))/b+2*a/b) 
*a^3-30*cos(-2*(a+b*arcsin(c*x))/b+2*a/b)*a*b^2+40*sin(-2*(a+b*arcsin(c*x) 
)/b+2*a/b)*a^2*b)
 

Fricas [F(-2)]

Exception generated. \[ \int x (a+b \arcsin (c x))^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (constant residues)
 

Sympy [F]

\[ \int x (a+b \arcsin (c x))^{5/2} \, dx=\int x \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{\frac {5}{2}}\, dx \] Input:

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

Output:

Integral(x*(a + b*asin(c*x))**(5/2), x)
 

Maxima [F]

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

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

Output:

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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.13 (sec) , antiderivative size = 1521, normalized size of antiderivative = 7.04 \[ \int x (a+b \arcsin (c x))^{5/2} \, dx=\text {Too large to display} \] Input:

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

Output:

-3/16*sqrt(pi)*a^2*b^(9/2)*erf(-sqrt(b*arcsin(c*x) + a)/sqrt(b) - I*sqrt(b 
*arcsin(c*x) + a)*sqrt(b)/abs(b))*e^(2*I*a/b)/((b^4 + I*b^5/abs(b))*c^2) - 
 3/16*sqrt(pi)*a^2*b^(9/2)*erf(-sqrt(b*arcsin(c*x) + a)/sqrt(b) + I*sqrt(b 
*arcsin(c*x) + a)*sqrt(b)/abs(b))*e^(-2*I*a/b)/((b^4 - I*b^5/abs(b))*c^2) 
- 9/64*I*sqrt(pi)*a*b^(9/2)*erf(-sqrt(b*arcsin(c*x) + a)/sqrt(b) - I*sqrt( 
b*arcsin(c*x) + a)*sqrt(b)/abs(b))*e^(2*I*a/b)/((b^3 + I*b^4/abs(b))*c^2) 
+ 15/256*sqrt(pi)*b^(11/2)*erf(-sqrt(b*arcsin(c*x) + a)/sqrt(b) - I*sqrt(b 
*arcsin(c*x) + a)*sqrt(b)/abs(b))*e^(2*I*a/b)/((b^3 + I*b^4/abs(b))*c^2) + 
 9/64*I*sqrt(pi)*a*b^(9/2)*erf(-sqrt(b*arcsin(c*x) + a)/sqrt(b) + I*sqrt(b 
*arcsin(c*x) + a)*sqrt(b)/abs(b))*e^(-2*I*a/b)/((b^3 - I*b^4/abs(b))*c^2) 
+ 15/256*sqrt(pi)*b^(11/2)*erf(-sqrt(b*arcsin(c*x) + a)/sqrt(b) + I*sqrt(b 
*arcsin(c*x) + a)*sqrt(b)/abs(b))*e^(-2*I*a/b)/((b^3 - I*b^4/abs(b))*c^2) 
+ 3/4*I*sqrt(pi)*a^3*b^(3/2)*erf(-sqrt(b*arcsin(c*x) + a)/sqrt(b) - I*sqrt 
(b*arcsin(c*x) + a)*sqrt(b)/abs(b))*e^(2*I*a/b)/((b^2 + I*b^3/abs(b))*c^2) 
 - 3/16*sqrt(pi)*a^2*b^(5/2)*erf(-sqrt(b*arcsin(c*x) + a)/sqrt(b) - I*sqrt 
(b*arcsin(c*x) + a)*sqrt(b)/abs(b))*e^(2*I*a/b)/((b^2 + I*b^3/abs(b))*c^2) 
 - 3/4*I*sqrt(pi)*a^3*b^(3/2)*erf(-sqrt(b*arcsin(c*x) + a)/sqrt(b) + I*sqr 
t(b*arcsin(c*x) + a)*sqrt(b)/abs(b))*e^(-2*I*a/b)/((b^2 - I*b^3/abs(b))*c^ 
2) - 3/16*sqrt(pi)*a^2*b^(5/2)*erf(-sqrt(b*arcsin(c*x) + a)/sqrt(b) + I*sq 
rt(b*arcsin(c*x) + a)*sqrt(b)/abs(b))*e^(-2*I*a/b)/((b^2 - I*b^3/abs(b)...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x (a+b \arcsin (c x))^{5/2} \, dx=2 \left (\int \sqrt {\mathit {asin} \left (c x \right ) b +a}\, \mathit {asin} \left (c x \right ) x d x \right ) a b +\left (\int \sqrt {\mathit {asin} \left (c x \right ) b +a}\, \mathit {asin} \left (c x \right )^{2} x d x \right ) b^{2}+\left (\int \sqrt {\mathit {asin} \left (c x \right ) b +a}\, x d x \right ) a^{2} \] Input:

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

Output:

2*int(sqrt(asin(c*x)*b + a)*asin(c*x)*x,x)*a*b + int(sqrt(asin(c*x)*b + a) 
*asin(c*x)**2*x,x)*b**2 + int(sqrt(asin(c*x)*b + a)*x,x)*a**2