Integrand size = 22, antiderivative size = 601 \[ \int \frac {x^{3/2}}{a+b \text {sech}\left (c+d \sqrt {x}\right )} \, dx=\frac {2 x^{5/2}}{5 a}-\frac {2 b x^2 \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}+\frac {2 b x^2 \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d}-\frac {8 b x^{3/2} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {8 b x^{3/2} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {24 b x \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3}-\frac {24 b x \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3}-\frac {48 b \sqrt {x} \operatorname {PolyLog}\left (4,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^4}+\frac {48 b \sqrt {x} \operatorname {PolyLog}\left (4,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^4}+\frac {48 b \operatorname {PolyLog}\left (5,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^5}-\frac {48 b \operatorname {PolyLog}\left (5,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^5} \]
2/5*x^(5/2)/a-2*b*x^2*ln(1+a*exp(c+d*x^(1/2))/(b-(-a^2+b^2)^(1/2)))/a/d/(- a^2+b^2)^(1/2)+2*b*x^2*ln(1+a*exp(c+d*x^(1/2))/(b+(-a^2+b^2)^(1/2)))/a/d/( -a^2+b^2)^(1/2)-8*b*x^(3/2)*polylog(2,-a*exp(c+d*x^(1/2))/(b-(-a^2+b^2)^(1 /2)))/a/d^2/(-a^2+b^2)^(1/2)+8*b*x^(3/2)*polylog(2,-a*exp(c+d*x^(1/2))/(b+ (-a^2+b^2)^(1/2)))/a/d^2/(-a^2+b^2)^(1/2)+24*b*x*polylog(3,-a*exp(c+d*x^(1 /2))/(b-(-a^2+b^2)^(1/2)))/a/d^3/(-a^2+b^2)^(1/2)-24*b*x*polylog(3,-a*exp( c+d*x^(1/2))/(b+(-a^2+b^2)^(1/2)))/a/d^3/(-a^2+b^2)^(1/2)+48*b*polylog(5,- a*exp(c+d*x^(1/2))/(b-(-a^2+b^2)^(1/2)))/a/d^5/(-a^2+b^2)^(1/2)-48*b*polyl og(5,-a*exp(c+d*x^(1/2))/(b+(-a^2+b^2)^(1/2)))/a/d^5/(-a^2+b^2)^(1/2)-48*b *polylog(4,-a*exp(c+d*x^(1/2))/(b-(-a^2+b^2)^(1/2)))*x^(1/2)/a/d^4/(-a^2+b ^2)^(1/2)+48*b*polylog(4,-a*exp(c+d*x^(1/2))/(b+(-a^2+b^2)^(1/2)))*x^(1/2) /a/d^4/(-a^2+b^2)^(1/2)
Time = 1.16 (sec) , antiderivative size = 460, normalized size of antiderivative = 0.77 \[ \int \frac {x^{3/2}}{a+b \text {sech}\left (c+d \sqrt {x}\right )} \, dx=\frac {2 \left (\sqrt {-a^2+b^2} d^5 x^{5/2}-5 b d^4 x^2 \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {-a^2+b^2}}\right )+5 b d^4 x^2 \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )-20 b d^3 x^{3/2} \operatorname {PolyLog}\left (2,\frac {a e^{c+d \sqrt {x}}}{-b+\sqrt {-a^2+b^2}}\right )+20 b d^3 x^{3/2} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )+60 b d^2 x \operatorname {PolyLog}\left (3,\frac {a e^{c+d \sqrt {x}}}{-b+\sqrt {-a^2+b^2}}\right )-60 b d^2 x \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )-120 b d \sqrt {x} \operatorname {PolyLog}\left (4,\frac {a e^{c+d \sqrt {x}}}{-b+\sqrt {-a^2+b^2}}\right )+120 b d \sqrt {x} \operatorname {PolyLog}\left (4,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )+120 b \operatorname {PolyLog}\left (5,\frac {a e^{c+d \sqrt {x}}}{-b+\sqrt {-a^2+b^2}}\right )-120 b \operatorname {PolyLog}\left (5,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {-a^2+b^2}}\right )\right )}{5 a \sqrt {-a^2+b^2} d^5} \]
(2*(Sqrt[-a^2 + b^2]*d^5*x^(5/2) - 5*b*d^4*x^2*Log[1 + (a*E^(c + d*Sqrt[x] ))/(b - Sqrt[-a^2 + b^2])] + 5*b*d^4*x^2*Log[1 + (a*E^(c + d*Sqrt[x]))/(b + Sqrt[-a^2 + b^2])] - 20*b*d^3*x^(3/2)*PolyLog[2, (a*E^(c + d*Sqrt[x]))/( -b + Sqrt[-a^2 + b^2])] + 20*b*d^3*x^(3/2)*PolyLog[2, -((a*E^(c + d*Sqrt[x ]))/(b + Sqrt[-a^2 + b^2]))] + 60*b*d^2*x*PolyLog[3, (a*E^(c + d*Sqrt[x])) /(-b + Sqrt[-a^2 + b^2])] - 60*b*d^2*x*PolyLog[3, -((a*E^(c + d*Sqrt[x]))/ (b + Sqrt[-a^2 + b^2]))] - 120*b*d*Sqrt[x]*PolyLog[4, (a*E^(c + d*Sqrt[x]) )/(-b + Sqrt[-a^2 + b^2])] + 120*b*d*Sqrt[x]*PolyLog[4, -((a*E^(c + d*Sqrt [x]))/(b + Sqrt[-a^2 + b^2]))] + 120*b*PolyLog[5, (a*E^(c + d*Sqrt[x]))/(- b + Sqrt[-a^2 + b^2])] - 120*b*PolyLog[5, -((a*E^(c + d*Sqrt[x]))/(b + Sqr t[-a^2 + b^2]))]))/(5*a*Sqrt[-a^2 + b^2]*d^5)
Time = 1.25 (sec) , antiderivative size = 602, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {5959, 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.
| \(\displaystyle \int \frac {x^{3/2}}{a+b \text {sech}\left (c+d \sqrt {x}\right )} \, dx\) |
| \(\Big \downarrow \) 5959 |
| \(\displaystyle 2 \int \frac {x^2}{a+b \text {sech}\left (c+d \sqrt {x}\right )}d\sqrt {x}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 2 \int \frac {x^2}{a+b \csc \left (i c+i d \sqrt {x}+\frac {\pi }{2}\right )}d\sqrt {x}\) |
| \(\Big \downarrow \) 4679 |
| \(\displaystyle 2 \int \left (\frac {x^2}{a}-\frac {b x^2}{a \left (b+a \cosh \left (c+d \sqrt {x}\right )\right )}\right )d\sqrt {x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 \left (\frac {24 b \operatorname {PolyLog}\left (5,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {b^2-a^2}}\right )}{a d^5 \sqrt {b^2-a^2}}-\frac {24 b \operatorname {PolyLog}\left (5,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {b^2-a^2}}\right )}{a d^5 \sqrt {b^2-a^2}}-\frac {24 b \sqrt {x} \operatorname {PolyLog}\left (4,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {b^2-a^2}}\right )}{a d^4 \sqrt {b^2-a^2}}+\frac {24 b \sqrt {x} \operatorname {PolyLog}\left (4,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {b^2-a^2}}\right )}{a d^4 \sqrt {b^2-a^2}}+\frac {12 b x \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}-\frac {12 b x \operatorname {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}-\frac {4 b x^{3/2} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}+\frac {4 b x^{3/2} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}-\frac {b x^2 \log \left (\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {b^2-a^2}}+1\right )}{a d \sqrt {b^2-a^2}}+\frac {b x^2 \log \left (\frac {a e^{c+d \sqrt {x}}}{\sqrt {b^2-a^2}+b}+1\right )}{a d \sqrt {b^2-a^2}}+\frac {x^{5/2}}{5 a}\right )\) |
2*(x^(5/2)/(5*a) - (b*x^2*Log[1 + (a*E^(c + d*Sqrt[x]))/(b - Sqrt[-a^2 + b ^2])])/(a*Sqrt[-a^2 + b^2]*d) + (b*x^2*Log[1 + (a*E^(c + d*Sqrt[x]))/(b + Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d) - (4*b*x^(3/2)*PolyLog[2, -((a* E^(c + d*Sqrt[x]))/(b - Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^2) + (4 *b*x^(3/2)*PolyLog[2, -((a*E^(c + d*Sqrt[x]))/(b + Sqrt[-a^2 + b^2]))])/(a *Sqrt[-a^2 + b^2]*d^2) + (12*b*x*PolyLog[3, -((a*E^(c + d*Sqrt[x]))/(b - S qrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^3) - (12*b*x*PolyLog[3, -((a*E^( c + d*Sqrt[x]))/(b + Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^3) - (24*b *Sqrt[x]*PolyLog[4, -((a*E^(c + d*Sqrt[x]))/(b - Sqrt[-a^2 + b^2]))])/(a*S qrt[-a^2 + b^2]*d^4) + (24*b*Sqrt[x]*PolyLog[4, -((a*E^(c + d*Sqrt[x]))/(b + Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^4) + (24*b*PolyLog[5, -((a*E ^(c + d*Sqrt[x]))/(b - Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^5) - (24 *b*PolyLog[5, -((a*E^(c + d*Sqrt[x]))/(b + Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a ^2 + b^2]*d^5))
3.1.62.3.1 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[(x_)^(m_.)*((a_.) + (b_.)*Sech[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbo l] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sech[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]
\[\int \frac {x^{\frac {3}{2}}}{a +b \,\operatorname {sech}\left (c +d \sqrt {x}\right )}d x\]
\[ \int \frac {x^{3/2}}{a+b \text {sech}\left (c+d \sqrt {x}\right )} \, dx=\int { \frac {x^{\frac {3}{2}}}{b \operatorname {sech}\left (d \sqrt {x} + c\right ) + a} \,d x } \]
\[ \int \frac {x^{3/2}}{a+b \text {sech}\left (c+d \sqrt {x}\right )} \, dx=\int \frac {x^{\frac {3}{2}}}{a + b \operatorname {sech}{\left (c + d \sqrt {x} \right )}}\, dx \]
Exception generated. \[ \int \frac {x^{3/2}}{a+b \text {sech}\left (c+d \sqrt {x}\right )} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a-b>0)', see `assume?` for more details)Is
\[ \int \frac {x^{3/2}}{a+b \text {sech}\left (c+d \sqrt {x}\right )} \, dx=\int { \frac {x^{\frac {3}{2}}}{b \operatorname {sech}\left (d \sqrt {x} + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {x^{3/2}}{a+b \text {sech}\left (c+d \sqrt {x}\right )} \, dx=\int \frac {x^{3/2}}{a+\frac {b}{\mathrm {cosh}\left (c+d\,\sqrt {x}\right )}} \,d x \]