\(\int x (a+b \text {sech}(c+d \sqrt {x}))^2 \, dx\) [39]

Optimal result

Integrand size = 18, antiderivative size = 319 \[ \int x \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\frac {2 b^2 x^{3/2}}{d}+\frac {a^2 x^2}{2}+\frac {8 a b x^{3/2} \arctan \left (e^{c+d \sqrt {x}}\right )}{d}-\frac {6 b^2 x \log \left (1+e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 i a b x \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {12 i a b x \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}-\frac {6 b^2 \sqrt {x} \operatorname {PolyLog}\left (2,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 i a b \sqrt {x} \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {24 i a b \sqrt {x} \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}+\frac {3 b^2 \operatorname {PolyLog}\left (3,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {24 i a b \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {24 i a b \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {2 b^2 x^{3/2} \tanh \left (c+d \sqrt {x}\right )}{d} \]

[Out]

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*polylog(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
*polylog(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

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {5544, 4275, 4265, 2611, 6744, 2320, 6724, 4269, 3799, 2221} \[ \int x \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\frac {a^2 x^2}{2}+\frac {8 a b x^{3/2} \arctan \left (e^{c+d \sqrt {x}}\right )}{d}-\frac {24 i a b \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {24 i a b \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {24 i a b \sqrt {x} \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {24 i a b \sqrt {x} \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {12 i a b x \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {12 i a b x \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {3 b^2 \operatorname {PolyLog}\left (3,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {6 b^2 \sqrt {x} \operatorname {PolyLog}\left (2,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {6 b^2 x \log \left (e^{2 \left (c+d \sqrt {x}\right )}+1\right )}{d^2}+\frac {2 b^2 x^{3/2} \tanh \left (c+d \sqrt {x}\right )}{d}+\frac {2 b^2 x^{3/2}}{d} \]

[In]

Int[x*(a + b*Sech[c + d*Sqrt[x]])^2,x]

[Out]

(2*b^2*x^(3/2))/d + (a^2*x^2)/2 + (8*a*b*x^(3/2)*ArcTan[E^(c + d*Sqrt[x])])/d - (6*b^2*x*Log[1 + E^(2*(c + d*S
qrt[x]))])/d^2 - ((12*I)*a*b*x*PolyLog[2, (-I)*E^(c + d*Sqrt[x])])/d^2 + ((12*I)*a*b*x*PolyLog[2, I*E^(c + d*S
qrt[x])])/d^2 - (6*b^2*Sqrt[x]*PolyLog[2, -E^(2*(c + d*Sqrt[x]))])/d^3 + ((24*I)*a*b*Sqrt[x]*PolyLog[3, (-I)*E
^(c + d*Sqrt[x])])/d^3 - ((24*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]))])/d^4 - ((24*I)*a*b*PolyLog[4, (-I)*E^(c + d*Sqrt[x])])/d^4 + ((24*I)*a*b*PolyLog[4, I*E^(c + d
*Sqrt[x])])/d^4 + (2*b^2*x^(3/2)*Tanh[c + d*Sqrt[x]])/d

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 3799

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> Simp[(-I)*((c + d*x)^(m
+ 1)/(d*(m + 1))), x] + Dist[2*I, Int[(c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*fz*x)))), x]
, x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]

Rule 4265

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c +
d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)/E^(I*k*Pi)]/(f*fz*I)), x] + (-Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*
Log[1 - E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*Log[1 + E^((-I)*e
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 4269

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(c + d*x)^m)*(Cot[e + f*x]/f), x
] + Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 4275

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]

Rule 5544

Int[(x_)^(m_.)*((a_.) + (b_.)*Sech[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(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[Simplif
y[(m + 1)/n], 0] && IntegerQ[p]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 6744

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a + b*x)))^p]/(b*c*p*Log[F])), x] - Dist[f*(m/(b*c*p*Log[F])), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int x^3 (a+b \text {sech}(c+d x))^2 \, dx,x,\sqrt {x}\right ) \\ & = 2 \text {Subst}\left (\int \left (a^2 x^3+2 a b x^3 \text {sech}(c+d x)+b^2 x^3 \text {sech}^2(c+d x)\right ) \, dx,x,\sqrt {x}\right ) \\ & = \frac {a^2 x^2}{2}+(4 a b) \text {Subst}\left (\int x^3 \text {sech}(c+d x) \, dx,x,\sqrt {x}\right )+\left (2 b^2\right ) \text {Subst}\left (\int x^3 \text {sech}^2(c+d x) \, dx,x,\sqrt {x}\right ) \\ & = \frac {a^2 x^2}{2}+\frac {8 a b x^{3/2} \arctan \left (e^{c+d \sqrt {x}}\right )}{d}+\frac {2 b^2 x^{3/2} \tanh \left (c+d \sqrt {x}\right )}{d}-\frac {(12 i a b) \text {Subst}\left (\int x^2 \log \left (1-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(12 i a b) \text {Subst}\left (\int x^2 \log \left (1+i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d}-\frac {\left (6 b^2\right ) \text {Subst}\left (\int x^2 \tanh (c+d x) \, dx,x,\sqrt {x}\right )}{d} \\ & = \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 {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 {2 b^2 x^{3/2} \tanh \left (c+d \sqrt {x}\right )}{d}+\frac {(24 i a b) \text {Subst}\left (\int x \operatorname {PolyLog}\left (2,-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2}-\frac {(24 i a b) \text {Subst}\left (\int x \operatorname {PolyLog}\left (2,i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2}-\frac {\left (12 b^2\right ) \text {Subst}\left (\int \frac {e^{2 (c+d x)} x^2}{1+e^{2 (c+d x)}} \, dx,x,\sqrt {x}\right )}{d} \\ & = \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 {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 {2 b^2 x^{3/2} \tanh \left (c+d \sqrt {x}\right )}{d}-\frac {(24 i a b) \text {Subst}\left (\int \operatorname {PolyLog}\left (3,-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3}+\frac {(24 i a b) \text {Subst}\left (\int \operatorname {PolyLog}\left (3,i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3}+\frac {\left (12 b^2\right ) \text {Subst}\left (\int x \log \left (1+e^{2 (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2} \\ & = \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 {2 b^2 x^{3/2} \tanh \left (c+d \sqrt {x}\right )}{d}-\frac {(24 i a b) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(3,-i x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^4}+\frac {(24 i a b) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(3,i x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^4}+\frac {\left (6 b^2\right ) \text {Subst}\left (\int \operatorname {PolyLog}\left (2,-e^{2 (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3} \\ & = \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 {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}+\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,-x)}{x} \, dx,x,e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4} \\ & = \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} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 5.82 (sec) , antiderivative size = 466, normalized size of antiderivative = 1.46 \[ \int x \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\frac {\cosh \left (c+d \sqrt {x}\right ) \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \left (a^2 x^2 \cosh \left (c+d \sqrt {x}\right )+\frac {2 b \cosh \left (c+d \sqrt {x}\right ) \left (4 b e^{2 c} x^{3/2}+\frac {i \left (1+e^{2 c}\right ) \left (12 i b d^2 x \log \left (1-i e^{c+d \sqrt {x}}\right )+4 a d^3 x^{3/2} \log \left (1-i e^{c+d \sqrt {x}}\right )+12 i b d^2 x \log \left (1+i e^{c+d \sqrt {x}}\right )-4 a d^3 x^{3/2} \log \left (1+i e^{c+d \sqrt {x}}\right )-6 i b d^2 x \log \left (1+e^{2 \left (c+d \sqrt {x}\right )}\right )-12 \left (-i b d \sqrt {x}+a d^2 x\right ) \operatorname {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )+12 \left (i b d \sqrt {x}+a d^2 x\right ) \operatorname {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )+24 a d \sqrt {x} \operatorname {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )-24 a d \sqrt {x} \operatorname {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )-3 i b \operatorname {PolyLog}\left (3,-e^{2 \left (c+d \sqrt {x}\right )}\right )-24 a \operatorname {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )+24 a \operatorname {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )\right )}{d^3}\right )}{d \left (1+e^{2 c}\right )}+\frac {4 b^2 x^{3/2} \text {sech}(c) \sinh \left (d \sqrt {x}\right )}{d}\right )}{2 \left (b+a \cosh \left (c+d \sqrt {x}\right )\right )^2} \]

[In]

Integrate[x*(a + b*Sech[c + d*Sqrt[x]])^2,x]

[Out]

(Cosh[c + d*Sqrt[x]]*(a + b*Sech[c + d*Sqrt[x]])^2*(a^2*x^2*Cosh[c + d*Sqrt[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*Sqrt[x]))] - 12*((-I)*b*d*Sqrt[x] + a*d^2*x)*PolyLog[2, (-I)*E^(c + d*Sq
rt[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]))] - 2
4*a*PolyLog[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)

Maple [F]

\[\int x \left (a +b \,\operatorname {sech}\left (c +d \sqrt {x}\right )\right )^{2}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int x \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\int { {\left (b \operatorname {sech}\left (d \sqrt {x} + c\right ) + a\right )}^{2} x \,d x } \]

[In]

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

[Out]

integral(b^2*x*sech(d*sqrt(x) + c)^2 + 2*a*b*x*sech(d*sqrt(x) + c) + a^2*x, x)

Sympy [F]

\[ \int x \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\int x \left (a + b \operatorname {sech}{\left (c + d \sqrt {x} \right )}\right )^{2}\, dx \]

[In]

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

[Out]

Integral(x*(a + b*sech(c + d*sqrt(x)))**2, x)

Maxima [F]

\[ \int x \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\int { {\left (b \operatorname {sech}\left (d \sqrt {x} + c\right ) + a\right )}^{2} x \,d x } \]

[In]

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

[Out]

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*sqrt(x))/(d*e^(2*d*sqrt(x) + 2*c) + d), x)

Giac [F]

\[ \int x \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\int { {\left (b \operatorname {sech}\left (d \sqrt {x} + c\right ) + a\right )}^{2} x \,d x } \]

[In]

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

[Out]

integrate((b*sech(d*sqrt(x) + c) + a)^2*x, x)

Mupad [F(-1)]

Timed out. \[ \int x \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx=\int x\,{\left (a+\frac {b}{\mathrm {cosh}\left (c+d\,\sqrt {x}\right )}\right )}^2 \,d x \]

[In]

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

[Out]

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