3.173 \(\int \frac {x^4}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx\)

Optimal. Leaf size=92 \[ \frac {3 x}{16 \left (c^4+\frac {1}{x^4}\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {3 \tanh ^{-1}\left (\sqrt {\frac {1}{c^4 x^4}+1}\right )}{16 c^8 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {x^5}{8 \text {sech}^{\frac {3}{2}}(2 \log (c x))} \]

[Out]

3/16*x/(c^4+1/x^4)/sech(2*ln(c*x))^(3/2)+1/8*x^5/sech(2*ln(c*x))^(3/2)+3/16*arctanh((1+1/c^4/x^4)^(1/2))/c^8/(
1+1/c^4/x^4)^(3/2)/x^3/sech(2*ln(c*x))^(3/2)

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Rubi [A]  time = 0.07, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5551, 5549, 266, 47, 63, 207} \[ \frac {3 x}{16 \left (c^4+\frac {1}{x^4}\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {3 \tanh ^{-1}\left (\sqrt {\frac {1}{c^4 x^4}+1}\right )}{16 c^8 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {x^5}{8 \text {sech}^{\frac {3}{2}}(2 \log (c x))} \]

Antiderivative was successfully verified.

[In]

Int[x^4/Sech[2*Log[c*x]]^(3/2),x]

[Out]

(3*x)/(16*(c^4 + x^(-4))*Sech[2*Log[c*x]]^(3/2)) + x^5/(8*Sech[2*Log[c*x]]^(3/2)) + (3*ArcTanh[Sqrt[1 + 1/(c^4
*x^4)]])/(16*c^8*(1 + 1/(c^4*x^4))^(3/2)*x^3*Sech[2*Log[c*x]]^(3/2))

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 5549

Int[((e_.)*(x_))^(m_.)*Sech[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[(Sech[d*(a + b*Log[x])]^p*
(1 + 1/(E^(2*a*d)*x^(2*b*d)))^p)/x^(-(b*d*p)), Int[(e*x)^m/(x^(b*d*p)*(1 + 1/(E^(2*a*d)*x^(2*b*d)))^p), x], x]
/; FreeQ[{a, b, d, e, m, p}, x] &&  !IntegerQ[p]

Rule 5551

Int[((e_.)*(x_))^(m_.)*Sech[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[(e*x)^(m + 1
)/(e*n*(c*x^n)^((m + 1)/n)), Subst[Int[x^((m + 1)/n - 1)*Sech[d*(a + b*Log[x])]^p, x], x, c*x^n], x] /; FreeQ[
{a, b, c, d, e, m, n, p}, x] && (NeQ[c, 1] || NeQ[n, 1])

Rubi steps

\begin {align*} \int \frac {x^4}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^4}{\text {sech}^{\frac {3}{2}}(2 \log (x))} \, dx,x,c x\right )}{c^5}\\ &=\frac {\operatorname {Subst}\left (\int \left (1+\frac {1}{x^4}\right )^{3/2} x^7 \, dx,x,c x\right )}{c^8 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {(1+x)^{3/2}}{x^3} \, dx,x,\frac {1}{c^4 x^4}\right )}{4 c^8 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))}\\ &=\frac {x^5}{8 \text {sech}^{\frac {3}{2}}(2 \log (c x))}-\frac {3 \operatorname {Subst}\left (\int \frac {\sqrt {1+x}}{x^2} \, dx,x,\frac {1}{c^4 x^4}\right )}{16 c^8 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))}\\ &=\frac {3 x}{16 \left (c^4+\frac {1}{x^4}\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {x^5}{8 \text {sech}^{\frac {3}{2}}(2 \log (c x))}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1+x}} \, dx,x,\frac {1}{c^4 x^4}\right )}{32 c^8 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))}\\ &=\frac {3 x}{16 \left (c^4+\frac {1}{x^4}\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {x^5}{8 \text {sech}^{\frac {3}{2}}(2 \log (c x))}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {1+\frac {1}{c^4 x^4}}\right )}{16 c^8 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))}\\ &=\frac {3 x}{16 \left (c^4+\frac {1}{x^4}\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {x^5}{8 \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {3 \tanh ^{-1}\left (\sqrt {1+\frac {1}{c^4 x^4}}\right )}{16 c^8 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 90, normalized size = 0.98 \[ \frac {3 c x \sinh ^{-1}\left (c^2 x^2\right )+c^3 x^3 \sqrt {c^4 x^4+1} \left (2 c^4 x^4+5\right )}{32 \sqrt {2} c^5 \sqrt {\frac {c^2 x^2}{c^4 x^4+1}} \sqrt {c^4 x^4+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/Sech[2*Log[c*x]]^(3/2),x]

[Out]

(c^3*x^3*Sqrt[1 + c^4*x^4]*(5 + 2*c^4*x^4) + 3*c*x*ArcSinh[c^2*x^2])/(32*Sqrt[2]*c^5*Sqrt[(c^2*x^2)/(1 + c^4*x
^4)]*Sqrt[1 + c^4*x^4])

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fricas [A]  time = 0.44, size = 101, normalized size = 1.10 \[ \frac {2 \, \sqrt {2} {\left (2 \, c^{9} x^{9} + 7 \, c^{5} x^{5} + 5 \, c x\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}} + 3 \, \sqrt {2} \log \left (-2 \, c^{4} x^{4} - 2 \, {\left (c^{5} x^{5} + c x\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}} - 1\right )}{128 \, c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/sech(2*log(c*x))^(3/2),x, algorithm="fricas")

[Out]

1/128*(2*sqrt(2)*(2*c^9*x^9 + 7*c^5*x^5 + 5*c*x)*sqrt(c^2*x^2/(c^4*x^4 + 1)) + 3*sqrt(2)*log(-2*c^4*x^4 - 2*(c
^5*x^5 + c*x)*sqrt(c^2*x^2/(c^4*x^4 + 1)) - 1))/c^5

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/sech(2*log(c*x))^(3/2),x, algorithm="giac")

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Warning, integrat
ion of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [abs(t_nostep)]U
nable to cancel step at 0 of 1/2/c^6*c^4*(1/2*ln(sqrt(c^4*t_nostep^4+1)-1)-1/2*ln(sqrt(c^4*t_nostep^4+1)+1)+sq
rt(c^4*t_nostep^4+1))-1/2/c^6*c^4*(-1/2*ln(sqrt(c^4*t_nostep^4+1)-1)+1/2*ln(sqrt(c^4*t_nostep^4+1)+1)-sqrt(c^4
*t_nostep^4+1))Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument
is real):Check [abs(t_nostep)]Unable to cancel step at 0 of 1/2/c^6*c^4*(1/2*ln(sqrt(c^4*t_nostep^4+1)-1)-1/2
*ln(sqrt(c^4*t_nostep^4+1)+1)+sqrt(c^4*t_nostep^4+1))-1/2/c^6*c^4*(-1/2*ln(sqrt(c^4*t_nostep^4+1)-1)+1/2*ln(sq
rt(c^4*t_nostep^4+1)+1)-sqrt(c^4*t_nostep^4+1))Unable to divide, perhaps due to rounding error%%%{1,[6,4,1,0]%
%%}+%%%{1,[2,0,1,0]%%%} / %%%{1,[0,2,0,1]%%%} Error: Bad Argument Value

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maple [A]  time = 0.28, size = 113, normalized size = 1.23 \[ \frac {x^{3} \left (2 c^{4} x^{4}+5\right ) \sqrt {2}}{64 c^{2} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}+\frac {3 \ln \left (\frac {c^{4} x^{2}}{\sqrt {c^{4}}}+\sqrt {c^{4} x^{4}+1}\right ) \sqrt {2}\, x}{64 \sqrt {c^{4}}\, c^{2} \sqrt {c^{4} x^{4}+1}\, \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/sech(2*ln(c*x))^(3/2),x)

[Out]

1/64*x^3*(2*c^4*x^4+5)*2^(1/2)/c^2/(c^2*x^2/(c^4*x^4+1))^(1/2)+3/64*ln(c^4*x^2/(c^4)^(1/2)+(c^4*x^4+1)^(1/2))/
(c^4)^(1/2)*2^(1/2)/c^2*x/(c^4*x^4+1)^(1/2)/(c^2*x^2/(c^4*x^4+1))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\operatorname {sech}\left (2 \, \log \left (c x\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/sech(2*log(c*x))^(3/2),x, algorithm="maxima")

[Out]

integrate(x^4/sech(2*log(c*x))^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^4}{{\left (\frac {1}{\mathrm {cosh}\left (2\,\ln \left (c\,x\right )\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(1/cosh(2*log(c*x)))^(3/2),x)

[Out]

int(x^4/(1/cosh(2*log(c*x)))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\operatorname {sech}^{\frac {3}{2}}{\left (2 \log {\left (c x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/sech(2*ln(c*x))**(3/2),x)

[Out]

Integral(x**4/sech(2*log(c*x))**(3/2), x)

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