Integrand size = 22, antiderivative size = 407 \[ \int x^{3/2} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\frac {2 b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}+\frac {8 a b x^2 \arctan \left (e^{c+d \sqrt {x}}\right )}{d}-\frac {8 b^2 x^{3/2} \log \left (1+e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 i a b x^{3/2} \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {16 i a b x^{3/2} \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}-\frac {12 b^2 x \operatorname {PolyLog}\left (2,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 i a b x \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {48 i a b x \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}+\frac {12 b^2 \sqrt {x} \operatorname {PolyLog}\left (3,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {96 i a b \sqrt {x} \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {96 i a b \sqrt {x} \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )}{d^4}-\frac {6 b^2 \operatorname {PolyLog}\left (4,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {96 i a b \operatorname {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {96 i a b \operatorname {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )}{d^5}+\frac {2 b^2 x^2 \tanh \left (c+d \sqrt {x}\right )}{d} \]
2*b^2*x^2/d+2/5*a^2*x^(5/2)+8*a*b*x^2*arctan(exp(c+d*x^(1/2)))/d-8*b^2*x^( 3/2)*ln(1+exp(2*c+2*d*x^(1/2)))/d^2-96*I*a*b*polylog(5,I*exp(c+d*x^(1/2))) /d^5+16*I*a*b*x^(3/2)*polylog(2,I*exp(c+d*x^(1/2)))/d^2-12*b^2*x*polylog(2 ,-exp(2*c+2*d*x^(1/2)))/d^3+96*I*a*b*polylog(5,-I*exp(c+d*x^(1/2)))/d^5-48 *I*a*b*x*polylog(3,I*exp(c+d*x^(1/2)))/d^3-6*b^2*polylog(4,-exp(2*c+2*d*x^ (1/2)))/d^5-16*I*a*b*x^(3/2)*polylog(2,-I*exp(c+d*x^(1/2)))/d^2+48*I*a*b*x *polylog(3,-I*exp(c+d*x^(1/2)))/d^3+12*b^2*polylog(3,-exp(2*c+2*d*x^(1/2)) )*x^(1/2)/d^4+96*I*a*b*polylog(4,I*exp(c+d*x^(1/2)))*x^(1/2)/d^4-96*I*a*b* polylog(4,-I*exp(c+d*x^(1/2)))*x^(1/2)/d^4+2*b^2*x^2*tanh(c+d*x^(1/2))/d
Time = 5.89 (sec) , antiderivative size = 496, normalized size of antiderivative = 1.22 \[ \int x^{3/2} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\frac {2 \cosh \left (c+d \sqrt {x}\right ) \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \left (\frac {10 b^2 e^{2 c} x^2 \cosh \left (c+d \sqrt {x}\right )}{d \left (1+e^{2 c}\right )}+a^2 x^{5/2} \cosh \left (c+d \sqrt {x}\right )+\frac {5 i b \cosh \left (c+d \sqrt {x}\right ) \left (2 a d^4 x^2 \log \left (1-i e^{c+d \sqrt {x}}\right )-2 a d^4 x^2 \log \left (1+i e^{c+d \sqrt {x}}\right )+4 i b d^3 x^{3/2} \log \left (1+e^{2 \left (c+d \sqrt {x}\right )}\right )-8 a d^3 x^{3/2} \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )+8 a d^3 x^{3/2} \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )+6 i b d^2 x \operatorname {PolyLog}\left (2,-e^{2 \left (c+d \sqrt {x}\right )}\right )+24 a d^2 x \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )-24 a d^2 x \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )-6 i b d \sqrt {x} \operatorname {PolyLog}\left (3,-e^{2 \left (c+d \sqrt {x}\right )}\right )-48 a d \sqrt {x} \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )+48 a d \sqrt {x} \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )+3 i b \operatorname {PolyLog}\left (4,-e^{2 \left (c+d \sqrt {x}\right )}\right )+48 a \operatorname {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )-48 a \operatorname {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )\right )}{d^5}+\frac {5 b^2 x^2 \text {sech}(c) \sinh \left (d \sqrt {x}\right )}{d}\right )}{5 \left (b+a \cosh \left (c+d \sqrt {x}\right )\right )^2} \]
(2*Cosh[c + d*Sqrt[x]]*(a + b*Sech[c + d*Sqrt[x]])^2*((10*b^2*E^(2*c)*x^2* Cosh[c + d*Sqrt[x]])/(d*(1 + E^(2*c))) + a^2*x^(5/2)*Cosh[c + d*Sqrt[x]] + ((5*I)*b*Cosh[c + d*Sqrt[x]]*(2*a*d^4*x^2*Log[1 - I*E^(c + d*Sqrt[x])] - 2*a*d^4*x^2*Log[1 + I*E^(c + d*Sqrt[x])] + (4*I)*b*d^3*x^(3/2)*Log[1 + E^( 2*(c + d*Sqrt[x]))] - 8*a*d^3*x^(3/2)*PolyLog[2, (-I)*E^(c + d*Sqrt[x])] + 8*a*d^3*x^(3/2)*PolyLog[2, I*E^(c + d*Sqrt[x])] + (6*I)*b*d^2*x*PolyLog[2 , -E^(2*(c + d*Sqrt[x]))] + 24*a*d^2*x*PolyLog[3, (-I)*E^(c + d*Sqrt[x])] - 24*a*d^2*x*PolyLog[3, I*E^(c + d*Sqrt[x])] - (6*I)*b*d*Sqrt[x]*PolyLog[3 , -E^(2*(c + d*Sqrt[x]))] - 48*a*d*Sqrt[x]*PolyLog[4, (-I)*E^(c + d*Sqrt[x ])] + 48*a*d*Sqrt[x]*PolyLog[4, I*E^(c + d*Sqrt[x])] + (3*I)*b*PolyLog[4, -E^(2*(c + d*Sqrt[x]))] + 48*a*PolyLog[5, (-I)*E^(c + d*Sqrt[x])] - 48*a*P olyLog[5, I*E^(c + d*Sqrt[x])]))/d^5 + (5*b^2*x^2*Sech[c]*Sinh[d*Sqrt[x]]) /d))/(5*(b + a*Cosh[c + d*Sqrt[x]])^2)
Time = 0.78 (sec) , antiderivative size = 407, 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, 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.
\(\displaystyle \int x^{3/2} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx\) |
\(\Big \downarrow \) 5959 |
\(\displaystyle 2 \int x^2 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2d\sqrt {x}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 \int x^2 \left (a+b \csc \left (i c+i d \sqrt {x}+\frac {\pi }{2}\right )\right )^2d\sqrt {x}\) |
\(\Big \downarrow \) 4678 |
\(\displaystyle 2 \int \left (a^2 x^2+b^2 \text {sech}^2\left (c+d \sqrt {x}\right ) x^2+2 a b \text {sech}\left (c+d \sqrt {x}\right ) x^2\right )d\sqrt {x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (\frac {1}{5} a^2 x^{5/2}+\frac {4 a b x^2 \arctan \left (e^{c+d \sqrt {x}}\right )}{d}+\frac {48 i a b \operatorname {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {48 i a b \operatorname {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {48 i a b \sqrt {x} \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {48 i a b \sqrt {x} \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {24 i a b x \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {24 i a b x \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {8 i a b x^{3/2} \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {8 i a b x^{3/2} \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}-\frac {3 b^2 \operatorname {PolyLog}\left (4,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {6 b^2 \sqrt {x} \operatorname {PolyLog}\left (3,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {6 b^2 x \operatorname {PolyLog}\left (2,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {4 b^2 x^{3/2} \log \left (e^{2 \left (c+d \sqrt {x}\right )}+1\right )}{d^2}+\frac {b^2 x^2 \tanh \left (c+d \sqrt {x}\right )}{d}+\frac {b^2 x^2}{d}\right )\) |
2*((b^2*x^2)/d + (a^2*x^(5/2))/5 + (4*a*b*x^2*ArcTan[E^(c + d*Sqrt[x])])/d - (4*b^2*x^(3/2)*Log[1 + E^(2*(c + d*Sqrt[x]))])/d^2 - ((8*I)*a*b*x^(3/2) *PolyLog[2, (-I)*E^(c + d*Sqrt[x])])/d^2 + ((8*I)*a*b*x^(3/2)*PolyLog[2, I *E^(c + d*Sqrt[x])])/d^2 - (6*b^2*x*PolyLog[2, -E^(2*(c + d*Sqrt[x]))])/d^ 3 + ((24*I)*a*b*x*PolyLog[3, (-I)*E^(c + d*Sqrt[x])])/d^3 - ((24*I)*a*b*x* PolyLog[3, I*E^(c + d*Sqrt[x])])/d^3 + (6*b^2*Sqrt[x]*PolyLog[3, -E^(2*(c + d*Sqrt[x]))])/d^4 - ((48*I)*a*b*Sqrt[x]*PolyLog[4, (-I)*E^(c + d*Sqrt[x] )])/d^4 + ((48*I)*a*b*Sqrt[x]*PolyLog[4, I*E^(c + d*Sqrt[x])])/d^4 - (3*b^ 2*PolyLog[4, -E^(2*(c + d*Sqrt[x]))])/d^5 + ((48*I)*a*b*PolyLog[5, (-I)*E^ (c + d*Sqrt[x])])/d^5 - ((48*I)*a*b*PolyLog[5, I*E^(c + d*Sqrt[x])])/d^5 + (b^2*x^2*Tanh[c + d*Sqrt[x]])/d)
3.1.57.3.1 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]
\[\int x^{\frac {3}{2}} \left (a +b \,\operatorname {sech}\left (c +d \sqrt {x}\right )\right )^{2}d x\]
\[ \int x^{3/2} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\int { {\left (b \operatorname {sech}\left (d \sqrt {x} + c\right ) + a\right )}^{2} x^{\frac {3}{2}} \,d x } \]
integral(b^2*x^(3/2)*sech(d*sqrt(x) + c)^2 + 2*a*b*x^(3/2)*sech(d*sqrt(x) + c) + a^2*x^(3/2), x)
\[ \int x^{3/2} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\int x^{\frac {3}{2}} \left (a + b \operatorname {sech}{\left (c + d \sqrt {x} \right )}\right )^{2}\, dx \]
\[ \int x^{3/2} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\int { {\left (b \operatorname {sech}\left (d \sqrt {x} + c\right ) + a\right )}^{2} x^{\frac {3}{2}} \,d x } \]
2/5*(a^2*d*x^(5/2)*e^(2*d*sqrt(x) + 2*c) + a^2*d*x^(5/2) - 10*b^2*x^2)/(d* e^(2*d*sqrt(x) + 2*c) + d) + integrate(4*(a*b*d*x^(5/2)*e^(d*sqrt(x) + c) + 2*b^2*x^2)/(d*x*e^(2*d*sqrt(x) + 2*c) + d*x), x)
\[ \int x^{3/2} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\int { {\left (b \operatorname {sech}\left (d \sqrt {x} + c\right ) + a\right )}^{2} x^{\frac {3}{2}} \,d x } \]
Timed out. \[ \int x^{3/2} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\int x^{3/2}\,{\left (a+\frac {b}{\mathrm {cosh}\left (c+d\,\sqrt {x}\right )}\right )}^2 \,d x \]