Integrand size = 18, antiderivative size = 539 \[ \int \frac {x}{a+b \csc \left (c+d \sqrt {x}\right )} \, dx=\frac {x^2}{2 a}+\frac {2 i b x^{3/2} \log \left (1-\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}-\frac {2 i b x^{3/2} \log \left (1-\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}+\frac {6 b x \operatorname {PolyLog}\left (2,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}-\frac {6 b x \operatorname {PolyLog}\left (2,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {12 i b \sqrt {x} \operatorname {PolyLog}\left (3,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3}-\frac {12 i b \sqrt {x} \operatorname {PolyLog}\left (3,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3}-\frac {12 b \operatorname {PolyLog}\left (4,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^4}+\frac {12 b \operatorname {PolyLog}\left (4,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^4} \]
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/d/(-a^2+b^2)^(1/2)-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/d/(-a^2+b^2)^(1/2)+6*b*x*polylog(2,I*a*exp(I*(c+d*x^(1/2) ))/(b-(-a^2+b^2)^(1/2)))/a/d^2/(-a^2+b^2)^(1/2)-6*b*x*polylog(2,I*a*exp(I* (c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/2)))/a/d^2/(-a^2+b^2)^(1/2)-12*b*polylog(4 ,I*a*exp(I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^(1/2)))/a/d^4/(-a^2+b^2)^(1/2)+12* b*polylog(4,I*a*exp(I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/2)))/a/d^4/(-a^2+b^2 )^(1/2)+12*I*b*polylog(3,I*a*exp(I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^(1/2)))*x^ (1/2)/a/d^3/(-a^2+b^2)^(1/2)-12*I*b*polylog(3,I*a*exp(I*(c+d*x^(1/2)))/(b+ (-a^2+b^2)^(1/2)))*x^(1/2)/a/d^3/(-a^2+b^2)^(1/2)
Time = 1.50 (sec) , antiderivative size = 438, normalized size of antiderivative = 0.81 \[ \int \frac {x}{a+b \csc \left (c+d \sqrt {x}\right )} \, dx=\frac {\sqrt {a^2-b^2} d^4 x^2-4 b d^3 x^{3/2} \log \left (1-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{-i b+\sqrt {a^2-b^2}}\right )+4 b d^3 x^{3/2} \log \left (1+\frac {a e^{i \left (c+d \sqrt {x}\right )}}{i b+\sqrt {a^2-b^2}}\right )+12 i b d^2 x \operatorname {PolyLog}\left (2,\frac {a e^{i \left (c+d \sqrt {x}\right )}}{-i b+\sqrt {a^2-b^2}}\right )-12 i b d^2 x \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{i b+\sqrt {a^2-b^2}}\right )-24 b d \sqrt {x} \operatorname {PolyLog}\left (3,\frac {a e^{i \left (c+d \sqrt {x}\right )}}{-i b+\sqrt {a^2-b^2}}\right )+24 b d \sqrt {x} \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{i b+\sqrt {a^2-b^2}}\right )-24 i b \operatorname {PolyLog}\left (4,\frac {a e^{i \left (c+d \sqrt {x}\right )}}{-i b+\sqrt {a^2-b^2}}\right )+24 i b \operatorname {PolyLog}\left (4,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{i b+\sqrt {a^2-b^2}}\right )}{2 a \sqrt {a^2-b^2} d^4} \]
(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)
Time = 1.17 (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 |
\(\displaystyle 2 \left (-\frac {6 b \operatorname {PolyLog}\left (4,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^4 \sqrt {b^2-a^2}}+\frac {6 b \operatorname {PolyLog}\left (4,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right )}{a d^4 \sqrt {b^2-a^2}}+\frac {6 i b \sqrt {x} \operatorname {PolyLog}\left (3,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}-\frac {6 i b \sqrt {x} \operatorname {PolyLog}\left (3,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}+\frac {3 b x \operatorname {PolyLog}\left (2,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}-\frac {3 b x \operatorname {PolyLog}\left (2,\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}+\frac {i b x^{3/2} \log \left (1-\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d \sqrt {b^2-a^2}}-\frac {i b x^{3/2} \log \left (1-\frac {i a e^{i \left (c+d \sqrt {x}\right )}}{\sqrt {b^2-a^2}+b}\right )}{a d \sqrt {b^2-a^2}}+\frac {x^2}{4 a}\right )\) |
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) )
3.1.43.3.1 Defintions of rubi rules used
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]
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]
\[\int \frac {x}{a +b \csc \left (c +d \sqrt {x}\right )}d x\]
\[ \int \frac {x}{a+b \csc \left (c+d \sqrt {x}\right )} \, dx=\int { \frac {x}{b \csc \left (d \sqrt {x} + c\right ) + a} \,d x } \]
\[ \int \frac {x}{a+b \csc \left (c+d \sqrt {x}\right )} \, dx=\int \frac {x}{a + b \csc {\left (c + d \sqrt {x} \right )}}\, dx \]
Exception generated. \[ \int \frac {x}{a+b \csc \left (c+d \sqrt {x}\right )} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {x}{a+b \csc \left (c+d \sqrt {x}\right )} \, dx=\int { \frac {x}{b \csc \left (d \sqrt {x} + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {x}{a+b \csc \left (c+d \sqrt {x}\right )} \, dx=\int \frac {x}{a+\frac {b}{\sin \left (c+d\,\sqrt {x}\right )}} \,d x \]