Integrand size = 18, antiderivative size = 319 \[ \int x \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\frac {2 b^2 x^{3/2}}{d}+\frac {a^2 x^2}{2}+\frac {8 a b x^{3/2} \arctan \left (e^{c+d \sqrt {x}}\right )}{d}-\frac {6 b^2 x \log \left (1+e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 i a b x \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {12 i a b x \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}-\frac {6 b^2 \sqrt {x} \operatorname {PolyLog}\left (2,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 i a b \sqrt {x} \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {24 i a b \sqrt {x} \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}+\frac {3 b^2 \operatorname {PolyLog}\left (3,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {24 i a b \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {24 i a b \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {2 b^2 x^{3/2} \tanh \left (c+d \sqrt {x}\right )}{d} \]
2*b^2*x^(3/2)/d+1/2*a^2*x^2+8*a*b*x^(3/2)*arctan(exp(c+d*x^(1/2)))/d-6*b^2 *x*ln(1+exp(2*c+2*d*x^(1/2)))/d^2-12*I*a*b*x*polylog(2,-I*exp(c+d*x^(1/2)) )/d^2+12*I*a*b*x*polylog(2,I*exp(c+d*x^(1/2)))/d^2+3*b^2*polylog(3,-exp(2* c+2*d*x^(1/2)))/d^4-24*I*a*b*polylog(4,-I*exp(c+d*x^(1/2)))/d^4+24*I*a*b*p olylog(4,I*exp(c+d*x^(1/2)))/d^4-6*b^2*polylog(2,-exp(2*c+2*d*x^(1/2)))*x^ (1/2)/d^3+24*I*a*b*polylog(3,-I*exp(c+d*x^(1/2)))*x^(1/2)/d^3-24*I*a*b*pol ylog(3,I*exp(c+d*x^(1/2)))*x^(1/2)/d^3+2*b^2*x^(3/2)*tanh(c+d*x^(1/2))/d
Time = 5.82 (sec) , antiderivative size = 466, normalized size of antiderivative = 1.46 \[ \int x \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\frac {\cosh \left (c+d \sqrt {x}\right ) \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \left (a^2 x^2 \cosh \left (c+d \sqrt {x}\right )+\frac {2 b \cosh \left (c+d \sqrt {x}\right ) \left (4 b e^{2 c} x^{3/2}+\frac {i \left (1+e^{2 c}\right ) \left (12 i b d^2 x \log \left (1-i e^{c+d \sqrt {x}}\right )+4 a d^3 x^{3/2} \log \left (1-i e^{c+d \sqrt {x}}\right )+12 i b d^2 x \log \left (1+i e^{c+d \sqrt {x}}\right )-4 a d^3 x^{3/2} \log \left (1+i e^{c+d \sqrt {x}}\right )-6 i b d^2 x \log \left (1+e^{2 \left (c+d \sqrt {x}\right )}\right )-12 \left (-i b d \sqrt {x}+a d^2 x\right ) \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )+12 \left (i b d \sqrt {x}+a d^2 x\right ) \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )+24 a d \sqrt {x} \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )-24 a d \sqrt {x} \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )-3 i b \operatorname {PolyLog}\left (3,-e^{2 \left (c+d \sqrt {x}\right )}\right )-24 a \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )+24 a \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )\right )}{d^3}\right )}{d \left (1+e^{2 c}\right )}+\frac {4 b^2 x^{3/2} \text {sech}(c) \sinh \left (d \sqrt {x}\right )}{d}\right )}{2 \left (b+a \cosh \left (c+d \sqrt {x}\right )\right )^2} \]
(Cosh[c + d*Sqrt[x]]*(a + b*Sech[c + d*Sqrt[x]])^2*(a^2*x^2*Cosh[c + d*Sqr t[x]] + (2*b*Cosh[c + d*Sqrt[x]]*(4*b*E^(2*c)*x^(3/2) + (I*(1 + E^(2*c))*( (12*I)*b*d^2*x*Log[1 - I*E^(c + d*Sqrt[x])] + 4*a*d^3*x^(3/2)*Log[1 - I*E^ (c + d*Sqrt[x])] + (12*I)*b*d^2*x*Log[1 + I*E^(c + d*Sqrt[x])] - 4*a*d^3*x ^(3/2)*Log[1 + I*E^(c + d*Sqrt[x])] - (6*I)*b*d^2*x*Log[1 + E^(2*(c + d*Sq rt[x]))] - 12*((-I)*b*d*Sqrt[x] + a*d^2*x)*PolyLog[2, (-I)*E^(c + d*Sqrt[x ])] + 12*(I*b*d*Sqrt[x] + a*d^2*x)*PolyLog[2, I*E^(c + d*Sqrt[x])] + 24*a* d*Sqrt[x]*PolyLog[3, (-I)*E^(c + d*Sqrt[x])] - 24*a*d*Sqrt[x]*PolyLog[3, I *E^(c + d*Sqrt[x])] - (3*I)*b*PolyLog[3, -E^(2*(c + d*Sqrt[x]))] - 24*a*Po lyLog[4, (-I)*E^(c + d*Sqrt[x])] + 24*a*PolyLog[4, I*E^(c + d*Sqrt[x])]))/ d^3))/(d*(1 + E^(2*c))) + (4*b^2*x^(3/2)*Sech[c]*Sinh[d*Sqrt[x]])/d))/(2*( b + a*Cosh[c + d*Sqrt[x]])^2)
Time = 0.69 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5959, 3042, 4678, 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.
| \(\displaystyle \int x \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx\) |
| \(\Big \downarrow \) 5959 |
| \(\displaystyle 2 \int x^{3/2} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2d\sqrt {x}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 2 \int x^{3/2} \left (a+b \csc \left (i c+i d \sqrt {x}+\frac {\pi }{2}\right )\right )^2d\sqrt {x}\) |
| \(\Big \downarrow \) 4678 |
| \(\displaystyle 2 \int \left (x^{3/2} a^2+2 b x^{3/2} \text {sech}\left (c+d \sqrt {x}\right ) a+b^2 x^{3/2} \text {sech}^2\left (c+d \sqrt {x}\right )\right )d\sqrt {x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 \left (\frac {a^2 x^2}{4}+\frac {4 a b x^{3/2} \arctan \left (e^{c+d \sqrt {x}}\right )}{d}-\frac {12 i a b \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {12 i a b \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {12 i a b \sqrt {x} \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {12 i a b \sqrt {x} \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {6 i a b x \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {6 i a b x \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {3 b^2 \operatorname {PolyLog}\left (3,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{2 d^4}-\frac {3 b^2 \sqrt {x} \operatorname {PolyLog}\left (2,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {3 b^2 x \log \left (e^{2 \left (c+d \sqrt {x}\right )}+1\right )}{d^2}+\frac {b^2 x^{3/2} \tanh \left (c+d \sqrt {x}\right )}{d}+\frac {b^2 x^{3/2}}{d}\right )\) |
2*((b^2*x^(3/2))/d + (a^2*x^2)/4 + (4*a*b*x^(3/2)*ArcTan[E^(c + d*Sqrt[x]) ])/d - (3*b^2*x*Log[1 + E^(2*(c + d*Sqrt[x]))])/d^2 - ((6*I)*a*b*x*PolyLog [2, (-I)*E^(c + d*Sqrt[x])])/d^2 + ((6*I)*a*b*x*PolyLog[2, I*E^(c + d*Sqrt [x])])/d^2 - (3*b^2*Sqrt[x]*PolyLog[2, -E^(2*(c + d*Sqrt[x]))])/d^3 + ((12 *I)*a*b*Sqrt[x]*PolyLog[3, (-I)*E^(c + d*Sqrt[x])])/d^3 - ((12*I)*a*b*Sqrt [x]*PolyLog[3, I*E^(c + d*Sqrt[x])])/d^3 + (3*b^2*PolyLog[3, -E^(2*(c + d* Sqrt[x]))])/(2*d^4) - ((12*I)*a*b*PolyLog[4, (-I)*E^(c + d*Sqrt[x])])/d^4 + ((12*I)*a*b*PolyLog[4, I*E^(c + d*Sqrt[x])])/d^4 + (b^2*x^(3/2)*Tanh[c + d*Sqrt[x]])/d)
3.1.39.3.1 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, (a + b*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 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]
\[\int x \left (a +b \,\operatorname {sech}\left (c +d \sqrt {x}\right )\right )^{2}d x\]
\[ \int x \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\int { {\left (b \operatorname {sech}\left (d \sqrt {x} + c\right ) + a\right )}^{2} x \,d x } \]
\[ \int x \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\int x \left (a + b \operatorname {sech}{\left (c + d \sqrt {x} \right )}\right )^{2}\, dx \]
\[ \int x \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\int { {\left (b \operatorname {sech}\left (d \sqrt {x} + c\right ) + a\right )}^{2} x \,d x } \]
1/2*(a^2*d*x^2*e^(2*d*sqrt(x) + 2*c) + a^2*d*x^2 - 8*b^2*x^(3/2))/(d*e^(2* d*sqrt(x) + 2*c) + d) + integrate(2*(2*a*b*d*x*e^(d*sqrt(x) + c) + 3*b^2*s qrt(x))/(d*e^(2*d*sqrt(x) + 2*c) + d), x)
\[ \int x \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\int { {\left (b \operatorname {sech}\left (d \sqrt {x} + c\right ) + a\right )}^{2} x \,d x } \]
Timed out. \[ \int x \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\int x\,{\left (a+\frac {b}{\mathrm {cosh}\left (c+d\,\sqrt {x}\right )}\right )}^2 \,d x \]