3.1.33 \(\int x^2 (a+b \text {sech}(c+d \sqrt {x})) \, dx\) [33]

Optimal. Leaf size=310 \[ \frac {a x^3}{3}+\frac {4 b x^{5/2} \text {ArcTan}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {10 i b x^2 \text {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 i b x^2 \text {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {40 i b x^{3/2} \text {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {40 i b x^{3/2} \text {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {120 i b x \text {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {120 i b x \text {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 i b \sqrt {x} \text {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 i b \sqrt {x} \text {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 i b \text {PolyLog}\left (6,-i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {240 i b \text {PolyLog}\left (6,i e^{c+d \sqrt {x}}\right )}{d^6} \]

[Out]

1/3*a*x^3+4*b*x^(5/2)*arctan(exp(c+d*x^(1/2)))/d-10*I*b*x^2*polylog(2,-I*exp(c+d*x^(1/2)))/d^2+10*I*b*x^2*poly
log(2,I*exp(c+d*x^(1/2)))/d^2+40*I*b*x^(3/2)*polylog(3,-I*exp(c+d*x^(1/2)))/d^3-40*I*b*x^(3/2)*polylog(3,I*exp
(c+d*x^(1/2)))/d^3-120*I*b*x*polylog(4,-I*exp(c+d*x^(1/2)))/d^4+120*I*b*x*polylog(4,I*exp(c+d*x^(1/2)))/d^4-24
0*I*b*polylog(6,-I*exp(c+d*x^(1/2)))/d^6+240*I*b*polylog(6,I*exp(c+d*x^(1/2)))/d^6+240*I*b*polylog(5,-I*exp(c+
d*x^(1/2)))*x^(1/2)/d^5-240*I*b*polylog(5,I*exp(c+d*x^(1/2)))*x^(1/2)/d^5

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Rubi [A]
time = 0.20, antiderivative size = 310, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {14, 5544, 4265, 2611, 6744, 2320, 6724} \begin {gather*} \frac {a x^3}{3}+\frac {4 b x^{5/2} \text {ArcTan}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {240 i b \text {Li}_6\left (-i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {240 i b \text {Li}_6\left (i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {240 i b \sqrt {x} \text {Li}_5\left (-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 i b \sqrt {x} \text {Li}_5\left (i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {120 i b x \text {Li}_4\left (-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {120 i b x \text {Li}_4\left (i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {40 i b x^{3/2} \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {40 i b x^{3/2} \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {10 i b x^2 \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 i b x^2 \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x^3)/3 + (4*b*x^(5/2)*ArcTan[E^(c + d*Sqrt[x])])/d - ((10*I)*b*x^2*PolyLog[2, (-I)*E^(c + d*Sqrt[x])])/d^2
+ ((10*I)*b*x^2*PolyLog[2, I*E^(c + d*Sqrt[x])])/d^2 + ((40*I)*b*x^(3/2)*PolyLog[3, (-I)*E^(c + d*Sqrt[x])])/d
^3 - ((40*I)*b*x^(3/2)*PolyLog[3, I*E^(c + d*Sqrt[x])])/d^3 - ((120*I)*b*x*PolyLog[4, (-I)*E^(c + d*Sqrt[x])])
/d^4 + ((120*I)*b*x*PolyLog[4, I*E^(c + d*Sqrt[x])])/d^4 + ((240*I)*b*Sqrt[x]*PolyLog[5, (-I)*E^(c + d*Sqrt[x]
)])/d^5 - ((240*I)*b*Sqrt[x]*PolyLog[5, I*E^(c + d*Sqrt[x])])/d^5 - ((240*I)*b*PolyLog[6, (-I)*E^(c + d*Sqrt[x
])])/d^6 + ((240*I)*b*PolyLog[6, I*E^(c + d*Sqrt[x])])/d^6

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 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 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*} \int x^2 \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx &=\int \left (a x^2+b x^2 \text {sech}\left (c+d \sqrt {x}\right )\right ) \, dx\\ &=\frac {a x^3}{3}+b \int x^2 \text {sech}\left (c+d \sqrt {x}\right ) \, dx\\ &=\frac {a x^3}{3}+(2 b) \text {Subst}\left (\int x^5 \text {sech}(c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {a x^3}{3}+\frac {4 b x^{5/2} \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {(10 i b) \text {Subst}\left (\int x^4 \log \left (1-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(10 i b) \text {Subst}\left (\int x^4 \log \left (1+i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d}\\ &=\frac {a x^3}{3}+\frac {4 b x^{5/2} \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {10 i b x^2 \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 i b x^2 \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {(40 i b) \text {Subst}\left (\int x^3 \text {Li}_2\left (-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2}-\frac {(40 i b) \text {Subst}\left (\int x^3 \text {Li}_2\left (i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2}\\ &=\frac {a x^3}{3}+\frac {4 b x^{5/2} \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {10 i b x^2 \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 i b x^2 \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {40 i b x^{3/2} \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {40 i b x^{3/2} \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {(120 i b) \text {Subst}\left (\int x^2 \text {Li}_3\left (-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3}+\frac {(120 i b) \text {Subst}\left (\int x^2 \text {Li}_3\left (i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3}\\ &=\frac {a x^3}{3}+\frac {4 b x^{5/2} \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {10 i b x^2 \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 i b x^2 \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {40 i b x^{3/2} \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {40 i b x^{3/2} \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {120 i b x \text {Li}_4\left (-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {120 i b x \text {Li}_4\left (i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {(240 i b) \text {Subst}\left (\int x \text {Li}_4\left (-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^4}-\frac {(240 i b) \text {Subst}\left (\int x \text {Li}_4\left (i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^4}\\ &=\frac {a x^3}{3}+\frac {4 b x^{5/2} \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {10 i b x^2 \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 i b x^2 \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {40 i b x^{3/2} \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {40 i b x^{3/2} \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {120 i b x \text {Li}_4\left (-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {120 i b x \text {Li}_4\left (i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 i b \sqrt {x} \text {Li}_5\left (-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 i b \sqrt {x} \text {Li}_5\left (i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {(240 i b) \text {Subst}\left (\int \text {Li}_5\left (-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^5}+\frac {(240 i b) \text {Subst}\left (\int \text {Li}_5\left (i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^5}\\ &=\frac {a x^3}{3}+\frac {4 b x^{5/2} \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {10 i b x^2 \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 i b x^2 \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {40 i b x^{3/2} \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {40 i b x^{3/2} \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {120 i b x \text {Li}_4\left (-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {120 i b x \text {Li}_4\left (i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 i b \sqrt {x} \text {Li}_5\left (-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 i b \sqrt {x} \text {Li}_5\left (i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {(240 i b) \text {Subst}\left (\int \frac {\text {Li}_5(-i x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^6}+\frac {(240 i b) \text {Subst}\left (\int \frac {\text {Li}_5(i x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^6}\\ &=\frac {a x^3}{3}+\frac {4 b x^{5/2} \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {10 i b x^2 \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {10 i b x^2 \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {40 i b x^{3/2} \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {40 i b x^{3/2} \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {120 i b x \text {Li}_4\left (-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {120 i b x \text {Li}_4\left (i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 i b \sqrt {x} \text {Li}_5\left (-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 i b \sqrt {x} \text {Li}_5\left (i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 i b \text {Li}_6\left (-i e^{c+d \sqrt {x}}\right )}{d^6}+\frac {240 i b \text {Li}_6\left (i e^{c+d \sqrt {x}}\right )}{d^6}\\ \end {align*}

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Mathematica [A]
time = 1.29, size = 311, normalized size = 1.00 \begin {gather*} \frac {a x^3}{3}+\frac {2 i b \left (d^5 x^{5/2} \log \left (1-i e^{c+d \sqrt {x}}\right )-d^5 x^{5/2} \log \left (1+i e^{c+d \sqrt {x}}\right )-5 d^4 x^2 \text {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )+5 d^4 x^2 \text {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )+20 d^3 x^{3/2} \text {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )-20 d^3 x^{3/2} \text {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )-60 d^2 x \text {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )+60 d^2 x \text {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )+120 d \sqrt {x} \text {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )-120 d \sqrt {x} \text {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )-120 \text {PolyLog}\left (6,-i e^{c+d \sqrt {x}}\right )+120 \text {PolyLog}\left (6,i e^{c+d \sqrt {x}}\right )\right )}{d^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x^3)/3 + ((2*I)*b*(d^5*x^(5/2)*Log[1 - I*E^(c + d*Sqrt[x])] - d^5*x^(5/2)*Log[1 + I*E^(c + d*Sqrt[x])] - 5*
d^4*x^2*PolyLog[2, (-I)*E^(c + d*Sqrt[x])] + 5*d^4*x^2*PolyLog[2, I*E^(c + d*Sqrt[x])] + 20*d^3*x^(3/2)*PolyLo
g[3, (-I)*E^(c + d*Sqrt[x])] - 20*d^3*x^(3/2)*PolyLog[3, I*E^(c + d*Sqrt[x])] - 60*d^2*x*PolyLog[4, (-I)*E^(c
+ d*Sqrt[x])] + 60*d^2*x*PolyLog[4, I*E^(c + d*Sqrt[x])] + 120*d*Sqrt[x]*PolyLog[5, (-I)*E^(c + d*Sqrt[x])] -
120*d*Sqrt[x]*PolyLog[5, I*E^(c + d*Sqrt[x])] - 120*PolyLog[6, (-I)*E^(c + d*Sqrt[x])] + 120*PolyLog[6, I*E^(c
+ d*Sqrt[x])]))/d^6

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Maple [F]
time = 2.07, size = 0, normalized size = 0.00 \[\int x^{2} \left (a +b \,\mathrm {sech}\left (c +d \sqrt {x}\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*x^3 + 2*b*integrate(x^2*e^(d*sqrt(x) + c)/(e^(2*d*sqrt(x) + 2*c) + 1), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \left (a + b \operatorname {sech}{\left (c + d \sqrt {x} \right )}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^2\,\left (a+\frac {b}{\mathrm {cosh}\left (c+d\,\sqrt {x}\right )}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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