\(\int x (a+b \csc (c+d \sqrt {x})) \, dx\) [33]

Optimal result

Integrand size = 16, antiderivative size = 200 \[ \int x \left (a+b \csc \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^2}{2}-\frac {4 b x^{3/2} \text {arctanh}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {6 i b x \operatorname {PolyLog}\left (2,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {6 i b x \operatorname {PolyLog}\left (2,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 b \sqrt {x} \operatorname {PolyLog}\left (3,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {12 b \sqrt {x} \operatorname {PolyLog}\left (3,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {12 i b \operatorname {PolyLog}\left (4,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {12 i b \operatorname {PolyLog}\left (4,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4} \] Output:

1/2*a*x^2-4*b*x^(3/2)*arctanh(exp(I*(c+d*x^(1/2))))/d+6*I*b*x*polylog(2,-e 
xp(I*(c+d*x^(1/2))))/d^2-6*I*b*x*polylog(2,exp(I*(c+d*x^(1/2))))/d^2-12*b* 
x^(1/2)*polylog(3,-exp(I*(c+d*x^(1/2))))/d^3+12*b*x^(1/2)*polylog(3,exp(I* 
(c+d*x^(1/2))))/d^3-12*I*b*polylog(4,-exp(I*(c+d*x^(1/2))))/d^4+12*I*b*pol 
ylog(4,exp(I*(c+d*x^(1/2))))/d^4
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.30 \[ \int x \left (a+b \csc \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^2}{2}-\frac {2 b \left (2 d^3 x^{3/2} \text {arctanh}\left (\cos \left (c+d \sqrt {x}\right )+i \sin \left (c+d \sqrt {x}\right )\right )-3 i d^2 x \operatorname {PolyLog}\left (2,-\cos \left (c+d \sqrt {x}\right )-i \sin \left (c+d \sqrt {x}\right )\right )+3 i d^2 x \operatorname {PolyLog}\left (2,\cos \left (c+d \sqrt {x}\right )+i \sin \left (c+d \sqrt {x}\right )\right )+6 d \sqrt {x} \operatorname {PolyLog}\left (3,-\cos \left (c+d \sqrt {x}\right )-i \sin \left (c+d \sqrt {x}\right )\right )-6 d \sqrt {x} \operatorname {PolyLog}\left (3,\cos \left (c+d \sqrt {x}\right )+i \sin \left (c+d \sqrt {x}\right )\right )+6 i \operatorname {PolyLog}\left (4,-\cos \left (c+d \sqrt {x}\right )-i \sin \left (c+d \sqrt {x}\right )\right )-6 i \operatorname {PolyLog}\left (4,\cos \left (c+d \sqrt {x}\right )+i \sin \left (c+d \sqrt {x}\right )\right )\right )}{d^4} \] Input:

Integrate[x*(a + b*Csc[c + d*Sqrt[x]]),x]
 

Output:

(a*x^2)/2 - (2*b*(2*d^3*x^(3/2)*ArcTanh[Cos[c + d*Sqrt[x]] + I*Sin[c + d*S 
qrt[x]]] - (3*I)*d^2*x*PolyLog[2, -Cos[c + d*Sqrt[x]] - I*Sin[c + d*Sqrt[x 
]]] + (3*I)*d^2*x*PolyLog[2, Cos[c + d*Sqrt[x]] + I*Sin[c + d*Sqrt[x]]] + 
6*d*Sqrt[x]*PolyLog[3, -Cos[c + d*Sqrt[x]] - I*Sin[c + d*Sqrt[x]]] - 6*d*S 
qrt[x]*PolyLog[3, Cos[c + d*Sqrt[x]] + I*Sin[c + d*Sqrt[x]]] + (6*I)*PolyL 
og[4, -Cos[c + d*Sqrt[x]] - I*Sin[c + d*Sqrt[x]]] - (6*I)*PolyLog[4, Cos[c 
 + d*Sqrt[x]] + I*Sin[c + d*Sqrt[x]]]))/d^4
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, 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.

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

\(\Big \downarrow \) 2010

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a x^2}{2}-\frac {4 b x^{3/2} \text {arctanh}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {12 i b \operatorname {PolyLog}\left (4,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {12 i b \operatorname {PolyLog}\left (4,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {12 b \sqrt {x} \operatorname {PolyLog}\left (3,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {12 b \sqrt {x} \operatorname {PolyLog}\left (3,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {6 i b x \operatorname {PolyLog}\left (2,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {6 i b x \operatorname {PolyLog}\left (2,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}\)

Input:

Int[x*(a + b*Csc[c + d*Sqrt[x]]),x]
 

Output:

(a*x^2)/2 - (4*b*x^(3/2)*ArcTanh[E^(I*(c + d*Sqrt[x]))])/d + ((6*I)*b*x*Po 
lyLog[2, -E^(I*(c + d*Sqrt[x]))])/d^2 - ((6*I)*b*x*PolyLog[2, E^(I*(c + d* 
Sqrt[x]))])/d^2 - (12*b*Sqrt[x]*PolyLog[3, -E^(I*(c + d*Sqrt[x]))])/d^3 + 
(12*b*Sqrt[x]*PolyLog[3, E^(I*(c + d*Sqrt[x]))])/d^3 - ((12*I)*b*PolyLog[4 
, -E^(I*(c + d*Sqrt[x]))])/d^4 + ((12*I)*b*PolyLog[4, E^(I*(c + d*Sqrt[x]) 
)])/d^4
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(b*x*csc(d*sqrt(x) + c) + a*x, x)
 

Sympy [F]

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

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

Output:

Integral(x*(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. 534 vs. \(2 (154) = 308\).

Time = 0.11 (sec) , antiderivative size = 534, normalized size of antiderivative = 2.67 \[ \int x \left (a+b \csc \left (c+d \sqrt {x}\right )\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

1/2*((d*sqrt(x) + c)^4*a - 4*(d*sqrt(x) + c)^3*a*c + 6*(d*sqrt(x) + c)^2*a 
*c^2 - 4*(d*sqrt(x) + c)*a*c^3 + 4*b*c^3*log(cot(d*sqrt(x) + c) + csc(d*sq 
rt(x) + c)) + 4*(-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)*arctan2(sin(d*sqrt(x) + c), cos(d*sqrt(x) + c) + 1) 
 + 4*(-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)*arctan2(sin(d*sqrt(x) + c), -cos(d*sqrt(x) + c) + 1) + 12*(I*( 
d*sqrt(x) + c)^2*b - 2*I*(d*sqrt(x) + c)*b*c + I*b*c^2)*dilog(-e^(I*d*sqrt 
(x) + I*c)) + 12*(-I*(d*sqrt(x) + c)^2*b + 2*I*(d*sqrt(x) + c)*b*c - I*b*c 
^2)*dilog(e^(I*d*sqrt(x) + I*c)) - 2*((d*sqrt(x) + c)^3*b - 3*(d*sqrt(x) + 
 c)^2*b*c + 3*(d*sqrt(x) + c)*b*c^2)*log(cos(d*sqrt(x) + c)^2 + sin(d*sqrt 
(x) + c)^2 + 2*cos(d*sqrt(x) + c) + 1) + 2*((d*sqrt(x) + c)^3*b - 3*(d*sqr 
t(x) + c)^2*b*c + 3*(d*sqrt(x) + c)*b*c^2)*log(cos(d*sqrt(x) + c)^2 + sin( 
d*sqrt(x) + c)^2 - 2*cos(d*sqrt(x) + c) + 1) - 24*I*b*polylog(4, -e^(I*d*s 
qrt(x) + I*c)) + 24*I*b*polylog(4, e^(I*d*sqrt(x) + I*c)) - 24*((d*sqrt(x) 
 + c)*b - b*c)*polylog(3, -e^(I*d*sqrt(x) + I*c)) + 24*((d*sqrt(x) + c)*b 
- b*c)*polylog(3, e^(I*d*sqrt(x) + I*c)))/d^4
 

Giac [F]

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

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

Output:

integrate((b*csc(d*sqrt(x) + c) + a)*x, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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