\(\int x^{3/2} (a+b \csc (c+d \sqrt {x})) \, dx\) [51]

Optimal result

Integrand size = 20, antiderivative size = 258 \[ \int x^{3/2} \left (a+b \csc \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {2}{5} a x^{5/2}-\frac {4 b x^2 \text {arctanh}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {8 i b x^{3/2} \operatorname {PolyLog}\left (2,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i b x^{3/2} \operatorname {PolyLog}\left (2,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {24 b x \operatorname {PolyLog}\left (3,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 b x \operatorname {PolyLog}\left (3,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {48 i b \sqrt {x} \operatorname {PolyLog}\left (4,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {48 i b \sqrt {x} \operatorname {PolyLog}\left (4,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {48 b \operatorname {PolyLog}\left (5,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {48 b \operatorname {PolyLog}\left (5,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5} \] Output:

2/5*a*x^(5/2)-4*b*x^2*arctanh(exp(I*(c+d*x^(1/2))))/d+8*I*b*x^(3/2)*polylo 
g(2,-exp(I*(c+d*x^(1/2))))/d^2-8*I*b*x^(3/2)*polylog(2,exp(I*(c+d*x^(1/2)) 
))/d^2-24*b*x*polylog(3,-exp(I*(c+d*x^(1/2))))/d^3+24*b*x*polylog(3,exp(I* 
(c+d*x^(1/2))))/d^3-48*I*b*x^(1/2)*polylog(4,-exp(I*(c+d*x^(1/2))))/d^4+48 
*I*b*x^(1/2)*polylog(4,exp(I*(c+d*x^(1/2))))/d^4+48*b*polylog(5,-exp(I*(c+ 
d*x^(1/2))))/d^5-48*b*polylog(5,exp(I*(c+d*x^(1/2))))/d^5
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.11 \[ \int x^{3/2} \left (a+b \csc \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {2 \left (a d^5 x^{5/2}+5 b d^4 x^2 \log \left (1-e^{i \left (c+d \sqrt {x}\right )}\right )-5 b d^4 x^2 \log \left (1+e^{i \left (c+d \sqrt {x}\right )}\right )+20 i b d^3 x^{3/2} \operatorname {PolyLog}\left (2,-e^{i \left (c+d \sqrt {x}\right )}\right )-20 i b d^3 x^{3/2} \operatorname {PolyLog}\left (2,e^{i \left (c+d \sqrt {x}\right )}\right )-60 b d^2 x \operatorname {PolyLog}\left (3,-e^{i \left (c+d \sqrt {x}\right )}\right )+60 b d^2 x \operatorname {PolyLog}\left (3,e^{i \left (c+d \sqrt {x}\right )}\right )-120 i b d \sqrt {x} \operatorname {PolyLog}\left (4,-e^{i \left (c+d \sqrt {x}\right )}\right )+120 i b d \sqrt {x} \operatorname {PolyLog}\left (4,e^{i \left (c+d \sqrt {x}\right )}\right )+120 b \operatorname {PolyLog}\left (5,-e^{i \left (c+d \sqrt {x}\right )}\right )-120 b \operatorname {PolyLog}\left (5,e^{i \left (c+d \sqrt {x}\right )}\right )\right )}{5 d^5} \] Input:

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

Output:

(2*(a*d^5*x^(5/2) + 5*b*d^4*x^2*Log[1 - E^(I*(c + d*Sqrt[x]))] - 5*b*d^4*x 
^2*Log[1 + E^(I*(c + d*Sqrt[x]))] + (20*I)*b*d^3*x^(3/2)*PolyLog[2, -E^(I* 
(c + d*Sqrt[x]))] - (20*I)*b*d^3*x^(3/2)*PolyLog[2, E^(I*(c + d*Sqrt[x]))] 
 - 60*b*d^2*x*PolyLog[3, -E^(I*(c + d*Sqrt[x]))] + 60*b*d^2*x*PolyLog[3, E 
^(I*(c + d*Sqrt[x]))] - (120*I)*b*d*Sqrt[x]*PolyLog[4, -E^(I*(c + d*Sqrt[x 
]))] + (120*I)*b*d*Sqrt[x]*PolyLog[4, E^(I*(c + d*Sqrt[x]))] + 120*b*PolyL 
og[5, -E^(I*(c + d*Sqrt[x]))] - 120*b*PolyLog[5, E^(I*(c + d*Sqrt[x]))]))/ 
(5*d^5)
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2010, 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.

\(\displaystyle \int x^{3/2} \left (a+b \csc \left (c+d \sqrt {x}\right )\right ) \, dx\)

\(\Big \downarrow \) 2010

\(\displaystyle \int \left (a x^{3/2}+b x^{3/2} \csc \left (c+d \sqrt {x}\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2}{5} a x^{5/2}-\frac {4 b x^2 \text {arctanh}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {48 b \operatorname {PolyLog}\left (5,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {48 b \operatorname {PolyLog}\left (5,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {48 i b \sqrt {x} \operatorname {PolyLog}\left (4,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {48 i b \sqrt {x} \operatorname {PolyLog}\left (4,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {24 b x \operatorname {PolyLog}\left (3,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 b x \operatorname {PolyLog}\left (3,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {8 i b x^{3/2} \operatorname {PolyLog}\left (2,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i b x^{3/2} \operatorname {PolyLog}\left (2,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}\)

Input:

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

Output:

(2*a*x^(5/2))/5 - (4*b*x^2*ArcTanh[E^(I*(c + d*Sqrt[x]))])/d + ((8*I)*b*x^ 
(3/2)*PolyLog[2, -E^(I*(c + d*Sqrt[x]))])/d^2 - ((8*I)*b*x^(3/2)*PolyLog[2 
, E^(I*(c + d*Sqrt[x]))])/d^2 - (24*b*x*PolyLog[3, -E^(I*(c + d*Sqrt[x]))] 
)/d^3 + (24*b*x*PolyLog[3, E^(I*(c + d*Sqrt[x]))])/d^3 - ((48*I)*b*Sqrt[x] 
*PolyLog[4, -E^(I*(c + d*Sqrt[x]))])/d^4 + ((48*I)*b*Sqrt[x]*PolyLog[4, E^ 
(I*(c + d*Sqrt[x]))])/d^4 + (48*b*PolyLog[5, -E^(I*(c + d*Sqrt[x]))])/d^5 
- (48*b*PolyLog[5, E^(I*(c + d*Sqrt[x]))])/d^5
 

Defintions of rubi rules used

rule 2009

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

rule 2010

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] 
, x] /; FreeQ[{c, m}, x] && SumQ[u] &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) 
+ (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(b*x^(3/2)*csc(d*sqrt(x) + c) + a*x^(3/2), x)
 

Sympy [F]

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

integrate(x**(3/2)*(a+b*csc(c+d*x**(1/2))),x)
 

Output:

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 730 vs. \(2 (200) = 400\).

Time = 0.12 (sec) , antiderivative size = 730, normalized size of antiderivative = 2.83 \[ \int x^{3/2} \left (a+b \csc \left (c+d \sqrt {x}\right )\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

1/5*(2*(d*sqrt(x) + c)^5*a - 10*(d*sqrt(x) + c)^4*a*c + 20*(d*sqrt(x) + c) 
^3*a*c^2 - 20*(d*sqrt(x) + c)^2*a*c^3 + 10*(d*sqrt(x) + c)*a*c^4 - 10*b*c^ 
4*log(cot(d*sqrt(x) + c) + csc(d*sqrt(x) + c)) + 10*(-I*(d*sqrt(x) + c)^4* 
b + 4*I*(d*sqrt(x) + c)^3*b*c - 6*I*(d*sqrt(x) + c)^2*b*c^2 + 4*I*(d*sqrt( 
x) + c)*b*c^3)*arctan2(sin(d*sqrt(x) + c), cos(d*sqrt(x) + c) + 1) + 10*(- 
I*(d*sqrt(x) + c)^4*b + 4*I*(d*sqrt(x) + c)^3*b*c - 6*I*(d*sqrt(x) + c)^2* 
b*c^2 + 4*I*(d*sqrt(x) + c)*b*c^3)*arctan2(sin(d*sqrt(x) + c), -cos(d*sqrt 
(x) + c) + 1) + 40*(I*(d*sqrt(x) + c)^3*b - 3*I*(d*sqrt(x) + c)^2*b*c + 3* 
I*(d*sqrt(x) + c)*b*c^2 - I*b*c^3)*dilog(-e^(I*d*sqrt(x) + I*c)) + 40*(-I* 
(d*sqrt(x) + c)^3*b + 3*I*(d*sqrt(x) + c)^2*b*c - 3*I*(d*sqrt(x) + c)*b*c^ 
2 + I*b*c^3)*dilog(e^(I*d*sqrt(x) + I*c)) - 5*((d*sqrt(x) + c)^4*b - 4*(d* 
sqrt(x) + c)^3*b*c + 6*(d*sqrt(x) + c)^2*b*c^2 - 4*(d*sqrt(x) + c)*b*c^3)* 
log(cos(d*sqrt(x) + c)^2 + sin(d*sqrt(x) + c)^2 + 2*cos(d*sqrt(x) + c) + 1 
) + 5*((d*sqrt(x) + c)^4*b - 4*(d*sqrt(x) + c)^3*b*c + 6*(d*sqrt(x) + c)^2 
*b*c^2 - 4*(d*sqrt(x) + c)*b*c^3)*log(cos(d*sqrt(x) + c)^2 + sin(d*sqrt(x) 
 + c)^2 - 2*cos(d*sqrt(x) + c) + 1) + 240*b*polylog(5, -e^(I*d*sqrt(x) + I 
*c)) - 240*b*polylog(5, e^(I*d*sqrt(x) + I*c)) + 240*(-I*(d*sqrt(x) + c)*b 
 + I*b*c)*polylog(4, -e^(I*d*sqrt(x) + I*c)) + 240*(I*(d*sqrt(x) + c)*b - 
I*b*c)*polylog(4, e^(I*d*sqrt(x) + I*c)) - 120*((d*sqrt(x) + c)^2*b - 2*(d 
*sqrt(x) + c)*b*c + b*c^2)*polylog(3, -e^(I*d*sqrt(x) + I*c)) + 120*((d...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x^{3/2} \left (a+b \csc \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {2 \sqrt {x}\, a \,x^{2}}{5}+\left (\int \sqrt {x}\, \csc \left (\sqrt {x}\, d +c \right ) x d x \right ) b \] Input:

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

Output:

(2*sqrt(x)*a*x**2 + 5*int(sqrt(x)*csc(sqrt(x)*d + c)*x,x)*b)/5