\(\int \frac {(c e+d e x)^3}{(a+b \arcsin (c+d x))^{3/2}} \, dx\) [145]

Optimal result

Integrand size = 25, antiderivative size = 270 \[ \int \frac {(c e+d e x)^3}{(a+b \arcsin (c+d x))^{3/2}} \, dx=-\frac {2 e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{b d \sqrt {a+b \arcsin (c+d x)}}-\frac {e^3 \sqrt {\frac {\pi }{2}} \cos \left (\frac {4 a}{b}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d}+\frac {e^3 \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b} \sqrt {\pi }}\right )}{b^{3/2} d}+\frac {e^3 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {2 a}{b}\right )}{b^{3/2} d}-\frac {e^3 \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b}}\right ) \sin \left (\frac {4 a}{b}\right )}{b^{3/2} d} \] Output:

-2*e^3*(d*x+c)^3*(1-(d*x+c)^2)^(1/2)/b/d/(a+b*arcsin(d*x+c))^(1/2)-1/2*e^3 
*2^(1/2)*Pi^(1/2)*cos(4*a/b)*FresnelC(2*2^(1/2)/Pi^(1/2)*(a+b*arcsin(d*x+c 
))^(1/2)/b^(1/2))/b^(3/2)/d+e^3*Pi^(1/2)*cos(2*a/b)*FresnelC(2*(a+b*arcsin 
(d*x+c))^(1/2)/b^(1/2)/Pi^(1/2))/b^(3/2)/d+e^3*Pi^(1/2)*FresnelS(2*(a+b*ar 
csin(d*x+c))^(1/2)/b^(1/2)/Pi^(1/2))*sin(2*a/b)/b^(3/2)/d-1/2*e^3*2^(1/2)* 
Pi^(1/2)*FresnelS(2*2^(1/2)/Pi^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b^(1/2))*si 
n(4*a/b)/b^(3/2)/d
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.11 \[ \int \frac {(c e+d e x)^3}{(a+b \arcsin (c+d x))^{3/2}} \, dx=-\frac {i e^3 e^{-\frac {4 i a}{b}} \left (\sqrt {2} e^{\frac {2 i a}{b}} \sqrt {-\frac {i (a+b \arcsin (c+d x))}{b}} \Gamma \left (\frac {1}{2},-\frac {2 i (a+b \arcsin (c+d x))}{b}\right )-\sqrt {2} e^{\frac {6 i a}{b}} \sqrt {\frac {i (a+b \arcsin (c+d x))}{b}} \Gamma \left (\frac {1}{2},\frac {2 i (a+b \arcsin (c+d x))}{b}\right )-\sqrt {-\frac {i (a+b \arcsin (c+d x))}{b}} \Gamma \left (\frac {1}{2},-\frac {4 i (a+b \arcsin (c+d x))}{b}\right )+e^{\frac {8 i a}{b}} \sqrt {\frac {i (a+b \arcsin (c+d x))}{b}} \Gamma \left (\frac {1}{2},\frac {4 i (a+b \arcsin (c+d x))}{b}\right )-2 i e^{\frac {4 i a}{b}} \sin (2 \arcsin (c+d x))+i e^{\frac {4 i a}{b}} \sin (4 \arcsin (c+d x))\right )}{4 b d \sqrt {a+b \arcsin (c+d x)}} \] Input:

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

Output:

((-1/4*I)*e^3*(Sqrt[2]*E^(((2*I)*a)/b)*Sqrt[((-I)*(a + b*ArcSin[c + d*x])) 
/b]*Gamma[1/2, ((-2*I)*(a + b*ArcSin[c + d*x]))/b] - Sqrt[2]*E^(((6*I)*a)/ 
b)*Sqrt[(I*(a + b*ArcSin[c + d*x]))/b]*Gamma[1/2, ((2*I)*(a + b*ArcSin[c + 
 d*x]))/b] - Sqrt[((-I)*(a + b*ArcSin[c + d*x]))/b]*Gamma[1/2, ((-4*I)*(a 
+ b*ArcSin[c + d*x]))/b] + E^(((8*I)*a)/b)*Sqrt[(I*(a + b*ArcSin[c + d*x]) 
)/b]*Gamma[1/2, ((4*I)*(a + b*ArcSin[c + d*x]))/b] - (2*I)*E^(((4*I)*a)/b) 
*Sin[2*ArcSin[c + d*x]] + I*E^(((4*I)*a)/b)*Sin[4*ArcSin[c + d*x]]))/(b*d* 
E^(((4*I)*a)/b)*Sqrt[a + b*ArcSin[c + d*x]])
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {5304, 27, 5142, 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.

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

\(\Big \downarrow \) 5304

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 5142

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 5142

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

rule 5304

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m 
_.), x_Symbol] :> Simp[1/d   Subst[Int[((d*e - c*f)/d + f*(x/d))^m*(a + b*A 
rcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
 

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.22

method	result	size
default	\(\frac {e^{3} \left (2 \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}\, \sin \left (\frac {4 a}{b}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \sqrt {-\frac {1}{b}}-2 \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}\, \cos \left (\frac {4 a}{b}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \sqrt {-\frac {1}{b}}+4 \sqrt {a +b \arcsin \left (d x +c \right )}\, \cos \left (\frac {2 a}{b}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) \sqrt {\pi }\, \sqrt {-\frac {1}{b}}-4 \sqrt {a +b \arcsin \left (d x +c \right )}\, \sin \left (\frac {2 a}{b}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) \sqrt {\pi }\, \sqrt {-\frac {1}{b}}+2 \sin \left (-\frac {2 \left (a +b \arcsin \left (d x +c \right )\right )}{b}+\frac {2 a}{b}\right )-\sin \left (-\frac {4 \left (a +b \arcsin \left (d x +c \right )\right )}{b}+\frac {4 a}{b}\right )\right )}{4 d b \sqrt {a +b \arcsin \left (d x +c \right )}}\)	\(329\)

Input:

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

Output:

1/4/d*e^3/b*(2*Pi^(1/2)*2^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*sin(4*a/b)*Fresn 
elS(2*2^(1/2)/Pi^(1/2)/(-1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*(-1/b)^(1 
/2)-2*Pi^(1/2)*2^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*cos(4*a/b)*FresnelC(2*2^( 
1/2)/Pi^(1/2)/(-1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*(-1/b)^(1/2)+4*(a+ 
b*arcsin(d*x+c))^(1/2)*cos(2*a/b)*FresnelC(2*2^(1/2)/Pi^(1/2)/(-2/b)^(1/2) 
*(a+b*arcsin(d*x+c))^(1/2)/b)*Pi^(1/2)*(-1/b)^(1/2)-4*(a+b*arcsin(d*x+c))^ 
(1/2)*sin(2*a/b)*FresnelS(2*2^(1/2)/Pi^(1/2)/(-2/b)^(1/2)*(a+b*arcsin(d*x+ 
c))^(1/2)/b)*Pi^(1/2)*(-1/b)^(1/2)+2*sin(-2*(a+b*arcsin(d*x+c))/b+2*a/b)-s 
in(-4*(a+b*arcsin(d*x+c))/b+4*a/b))/(a+b*arcsin(d*x+c))^(1/2)
 

Fricas [F(-2)]

Exception generated. \[ \int \frac {(c e+d e x)^3}{(a+b \arcsin (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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