\(\int \frac {x}{a+b \csc (c+d \sqrt {x})} \, dx\) [43]

Optimal result

Integrand size = 18, antiderivative size = 539 \[ \int \frac {x}{a+b \csc \left (c+d \sqrt {x}\right )} \, dx=\frac {x^2}{2 a}+\frac {2 i b x^{3/2} \log \left (1-\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}-\frac {2 i b x^{3/2} \log \left (1-\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}+\frac {6 b x \operatorname {PolyLog}\left (2,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}-\frac {6 b x \operatorname {PolyLog}\left (2,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {12 i b \sqrt {x} \operatorname {PolyLog}\left (3,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3}-\frac {12 i b \sqrt {x} \operatorname {PolyLog}\left (3,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3}-\frac {12 b \operatorname {PolyLog}\left (4,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^4}+\frac {12 b \operatorname {PolyLog}\left (4,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^4} \] Output:

1/2*x^2/a+2*I*b*x^(3/2)*ln(1-I*a*exp(I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^(1/2)) 
)/a/(-a^2+b^2)^(1/2)/d-2*I*b*x^(3/2)*ln(1-I*a*exp(I*(c+d*x^(1/2)))/(b+(-a^ 
2+b^2)^(1/2)))/a/(-a^2+b^2)^(1/2)/d+6*b*x*polylog(2,I*a*exp(I*(c+d*x^(1/2) 
))/(b-(-a^2+b^2)^(1/2)))/a/(-a^2+b^2)^(1/2)/d^2-6*b*x*polylog(2,I*a*exp(I* 
(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/2)))/a/(-a^2+b^2)^(1/2)/d^2+12*I*b*x^(1/2) 
*polylog(3,I*a*exp(I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^(1/2)))/a/(-a^2+b^2)^(1/ 
2)/d^3-12*I*b*x^(1/2)*polylog(3,I*a*exp(I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/ 
2)))/a/(-a^2+b^2)^(1/2)/d^3-12*b*polylog(4,I*a*exp(I*(c+d*x^(1/2)))/(b-(-a 
^2+b^2)^(1/2)))/a/(-a^2+b^2)^(1/2)/d^4+12*b*polylog(4,I*a*exp(I*(c+d*x^(1/ 
2)))/(b+(-a^2+b^2)^(1/2)))/a/(-a^2+b^2)^(1/2)/d^4
 

Mathematica [A] (verified)

Time = 0.99 (sec) , antiderivative size = 438, normalized size of antiderivative = 0.81 \[ \int \frac {x}{a+b \csc \left (c+d \sqrt {x}\right )} \, dx=\frac {\sqrt {a^2-b^2} d^4 x^2-4 b d^3 x^{3/2} \log \left (1-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{-i b+\sqrt {a^2-b^2}}\right )+4 b d^3 x^{3/2} \log \left (1+\frac {a e^{i \left (c+d \sqrt {x}\right )}}{i b+\sqrt {a^2-b^2}}\right )+12 i b d^2 x \operatorname {PolyLog}\left (2,\frac {a e^{i \left (c+d \sqrt {x}\right )}}{-i b+\sqrt {a^2-b^2}}\right )-12 i b d^2 x \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{i b+\sqrt {a^2-b^2}}\right )-24 b d \sqrt {x} \operatorname {PolyLog}\left (3,\frac {a e^{i \left (c+d \sqrt {x}\right )}}{-i b+\sqrt {a^2-b^2}}\right )+24 b d \sqrt {x} \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{i b+\sqrt {a^2-b^2}}\right )-24 i b \operatorname {PolyLog}\left (4,\frac {a e^{i \left (c+d \sqrt {x}\right )}}{-i b+\sqrt {a^2-b^2}}\right )+24 i b \operatorname {PolyLog}\left (4,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{i b+\sqrt {a^2-b^2}}\right )}{2 a \sqrt {a^2-b^2} d^4} \] Input:

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

Output:

(Sqrt[a^2 - b^2]*d^4*x^2 - 4*b*d^3*x^(3/2)*Log[1 - (a*E^(I*(c + d*Sqrt[x]) 
))/((-I)*b + Sqrt[a^2 - b^2])] + 4*b*d^3*x^(3/2)*Log[1 + (a*E^(I*(c + d*Sq 
rt[x])))/(I*b + Sqrt[a^2 - b^2])] + (12*I)*b*d^2*x*PolyLog[2, (a*E^(I*(c + 
 d*Sqrt[x])))/((-I)*b + Sqrt[a^2 - b^2])] - (12*I)*b*d^2*x*PolyLog[2, -((a 
*E^(I*(c + d*Sqrt[x])))/(I*b + Sqrt[a^2 - b^2]))] - 24*b*d*Sqrt[x]*PolyLog 
[3, (a*E^(I*(c + d*Sqrt[x])))/((-I)*b + Sqrt[a^2 - b^2])] + 24*b*d*Sqrt[x] 
*PolyLog[3, -((a*E^(I*(c + d*Sqrt[x])))/(I*b + Sqrt[a^2 - b^2]))] - (24*I) 
*b*PolyLog[4, (a*E^(I*(c + d*Sqrt[x])))/((-I)*b + Sqrt[a^2 - b^2])] + (24* 
I)*b*PolyLog[4, -((a*E^(I*(c + d*Sqrt[x])))/(I*b + Sqrt[a^2 - b^2]))])/(2* 
a*Sqrt[a^2 - b^2]*d^4)
 

Rubi [A] (verified)

Time = 1.15 (sec) , antiderivative size = 541, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4693, 3042, 4679, 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.

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

\(\Big \downarrow \) 4693

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4679

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

\(\Big \downarrow \) 2009

\(\displaystyle 2 \left (-\frac {6 b \operatorname {PolyLog}\left (4,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^4 \sqrt {b^2-a^2}}+\frac {6 b \operatorname {PolyLog}\left (4,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right )}{a d^4 \sqrt {b^2-a^2}}+\frac {6 i b \sqrt {x} \operatorname {PolyLog}\left (3,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}-\frac {6 i b \sqrt {x} \operatorname {PolyLog}\left (3,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}+\frac {3 b x \operatorname {PolyLog}\left (2,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}-\frac {3 b x \operatorname {PolyLog}\left (2,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}+\frac {i b x^{3/2} \log \left (1-\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d \sqrt {b^2-a^2}}-\frac {i b x^{3/2} \log \left (1-\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{\sqrt {b^2-a^2}+b}\right )}{a d \sqrt {b^2-a^2}}+\frac {x^2}{4 a}\right )\)

Input:

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

Output:

2*(x^2/(4*a) + (I*b*x^(3/2)*Log[1 - (I*a*E^(I*(c + d*Sqrt[x])))/(b - Sqrt[ 
-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d) - (I*b*x^(3/2)*Log[1 - (I*a*E^(I*(c 
+ d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d) + (3*b*x*Po 
lyLog[2, (I*a*E^(I*(c + d*Sqrt[x])))/(b - Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 
 + b^2]*d^2) - (3*b*x*PolyLog[2, (I*a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^ 
2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d^2) + ((6*I)*b*Sqrt[x]*PolyLog[3, (I*a*E^ 
(I*(c + d*Sqrt[x])))/(b - Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d^3) - ( 
(6*I)*b*Sqrt[x]*PolyLog[3, (I*a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^ 
2])])/(a*Sqrt[-a^2 + b^2]*d^3) - (6*b*PolyLog[4, (I*a*E^(I*(c + d*Sqrt[x]) 
))/(b - Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d^4) + (6*b*PolyLog[4, (I* 
a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d^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 4679

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Si 
n[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] && IGt 
Q[m, 0]
 

rule 4693

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

Maple [F]

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

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]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {x}{a+b \csc \left (c+d \sqrt {x}\right )} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f 
or more de
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

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