3.582 \(\int \text {sech}^{\frac {23}{4}}(x) \sinh ^5(x) \, dx\)

Optimal. Leaf size=31 \[ -\frac {4}{19} \text {sech}^{\frac {19}{4}}(x)+\frac {8}{11} \text {sech}^{\frac {11}{4}}(x)-\frac {4}{3} \text {sech}^{\frac {3}{4}}(x) \]

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Rubi [A]  time = 0.03, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2622, 270} \[ -\frac {4}{19} \text {sech}^{\frac {19}{4}}(x)+\frac {8}{11} \text {sech}^{\frac {11}{4}}(x)-\frac {4}{3} \text {sech}^{\frac {3}{4}}(x) \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^(23/4)*Sinh[x]^5,x]

[Out]

(-4*Sech[x]^(3/4))/3 + (8*Sech[x]^(11/4))/11 - (4*Sech[x]^(19/4))/19

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rubi steps

\begin {align*} \int \text {sech}^{\frac {23}{4}}(x) \sinh ^5(x) \, dx &=-\operatorname {Subst}\left (\int \frac {\left (-1+x^2\right )^2}{\sqrt [4]{x}} \, dx,x,\text {sech}(x)\right )\\ &=-\operatorname {Subst}\left (\int \left (\frac {1}{\sqrt [4]{x}}-2 x^{7/4}+x^{15/4}\right ) \, dx,x,\text {sech}(x)\right )\\ &=-\frac {4}{3} \text {sech}^{\frac {3}{4}}(x)+\frac {8}{11} \text {sech}^{\frac {11}{4}}(x)-\frac {4}{19} \text {sech}^{\frac {19}{4}}(x)\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 27, normalized size = 0.87 \[ \text {sech}^{\frac {3}{4}}(x) \left (-\frac {4}{19} \text {sech}^4(x)+\frac {8 \text {sech}^2(x)}{11}-\frac {4}{3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^(23/4)*Sinh[x]^5,x]

[Out]

Sech[x]^(3/4)*(-4/3 + (8*Sech[x]^2)/11 - (4*Sech[x]^4)/19)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \text {sech}^{\frac {23}{4}}(x) \sinh ^5(x) \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[Sech[x]^(23/4)*Sinh[x]^5,x]

[Out]

Could not integrate

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fricas [B]  time = 0.67, size = 359, normalized size = 11.58 \[ -\frac {4 \cdot 2^{\frac {3}{4}} {\left (209 \, \cosh \relax (x)^{8} + 1672 \, \cosh \relax (x) \sinh \relax (x)^{7} + 209 \, \sinh \relax (x)^{8} + 76 \, {\left (77 \, \cosh \relax (x)^{2} + 5\right )} \sinh \relax (x)^{6} + 380 \, \cosh \relax (x)^{6} + 152 \, {\left (77 \, \cosh \relax (x)^{3} + 15 \, \cosh \relax (x)\right )} \sinh \relax (x)^{5} + 10 \, {\left (1463 \, \cosh \relax (x)^{4} + 570 \, \cosh \relax (x)^{2} + 87\right )} \sinh \relax (x)^{4} + 870 \, \cosh \relax (x)^{4} + 8 \, {\left (1463 \, \cosh \relax (x)^{5} + 950 \, \cosh \relax (x)^{3} + 435 \, \cosh \relax (x)\right )} \sinh \relax (x)^{3} + 4 \, {\left (1463 \, \cosh \relax (x)^{6} + 1425 \, \cosh \relax (x)^{4} + 1305 \, \cosh \relax (x)^{2} + 95\right )} \sinh \relax (x)^{2} + 380 \, \cosh \relax (x)^{2} + 8 \, {\left (209 \, \cosh \relax (x)^{7} + 285 \, \cosh \relax (x)^{5} + 435 \, \cosh \relax (x)^{3} + 95 \, \cosh \relax (x)\right )} \sinh \relax (x) + 209\right )} \left (\frac {\cosh \relax (x) + \sinh \relax (x)}{\cosh \relax (x)^{2} + 2 \, \cosh \relax (x) \sinh \relax (x) + \sinh \relax (x)^{2} + 1}\right )^{\frac {3}{4}}}{627 \, {\left (\cosh \relax (x)^{8} + 8 \, \cosh \relax (x) \sinh \relax (x)^{7} + \sinh \relax (x)^{8} + 4 \, {\left (7 \, \cosh \relax (x)^{2} + 1\right )} \sinh \relax (x)^{6} + 4 \, \cosh \relax (x)^{6} + 8 \, {\left (7 \, \cosh \relax (x)^{3} + 3 \, \cosh \relax (x)\right )} \sinh \relax (x)^{5} + 2 \, {\left (35 \, \cosh \relax (x)^{4} + 30 \, \cosh \relax (x)^{2} + 3\right )} \sinh \relax (x)^{4} + 6 \, \cosh \relax (x)^{4} + 8 \, {\left (7 \, \cosh \relax (x)^{5} + 10 \, \cosh \relax (x)^{3} + 3 \, \cosh \relax (x)\right )} \sinh \relax (x)^{3} + 4 \, {\left (7 \, \cosh \relax (x)^{6} + 15 \, \cosh \relax (x)^{4} + 9 \, \cosh \relax (x)^{2} + 1\right )} \sinh \relax (x)^{2} + 4 \, \cosh \relax (x)^{2} + 8 \, {\left (\cosh \relax (x)^{7} + 3 \, \cosh \relax (x)^{5} + 3 \, \cosh \relax (x)^{3} + \cosh \relax (x)\right )} \sinh \relax (x) + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^(3/4)*tanh(x)^5,x, algorithm="fricas")

[Out]

-4/627*2^(3/4)*(209*cosh(x)^8 + 1672*cosh(x)*sinh(x)^7 + 209*sinh(x)^8 + 76*(77*cosh(x)^2 + 5)*sinh(x)^6 + 380
*cosh(x)^6 + 152*(77*cosh(x)^3 + 15*cosh(x))*sinh(x)^5 + 10*(1463*cosh(x)^4 + 570*cosh(x)^2 + 87)*sinh(x)^4 +
870*cosh(x)^4 + 8*(1463*cosh(x)^5 + 950*cosh(x)^3 + 435*cosh(x))*sinh(x)^3 + 4*(1463*cosh(x)^6 + 1425*cosh(x)^
4 + 1305*cosh(x)^2 + 95)*sinh(x)^2 + 380*cosh(x)^2 + 8*(209*cosh(x)^7 + 285*cosh(x)^5 + 435*cosh(x)^3 + 95*cos
h(x))*sinh(x) + 209)*((cosh(x) + sinh(x))/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 1))^(3/4)/(cosh(x)^8 +
8*cosh(x)*sinh(x)^7 + sinh(x)^8 + 4*(7*cosh(x)^2 + 1)*sinh(x)^6 + 4*cosh(x)^6 + 8*(7*cosh(x)^3 + 3*cosh(x))*si
nh(x)^5 + 2*(35*cosh(x)^4 + 30*cosh(x)^2 + 3)*sinh(x)^4 + 6*cosh(x)^4 + 8*(7*cosh(x)^5 + 10*cosh(x)^3 + 3*cosh
(x))*sinh(x)^3 + 4*(7*cosh(x)^6 + 15*cosh(x)^4 + 9*cosh(x)^2 + 1)*sinh(x)^2 + 4*cosh(x)^2 + 8*(cosh(x)^7 + 3*c
osh(x)^5 + 3*cosh(x)^3 + cosh(x))*sinh(x) + 1)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {sech}\relax (x)^{\frac {3}{4}} \tanh \relax (x)^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^(3/4)*tanh(x)^5,x, algorithm="giac")

[Out]

integrate(sech(x)^(3/4)*tanh(x)^5, x)

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[\int \mathrm {sech}\relax (x )^{\frac {3}{4}} \left (\tanh ^{5}\relax (x )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^(3/4)*tanh(x)^5,x)

[Out]

int(sech(x)^(3/4)*tanh(x)^5,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {sech}\relax (x)^{\frac {3}{4}} \tanh \relax (x)^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^(3/4)*tanh(x)^5,x, algorithm="maxima")

[Out]

integrate(sech(x)^(3/4)*tanh(x)^5, x)

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mupad [B]  time = 0.17, size = 120, normalized size = 3.87 \[ \frac {32\,{\left (\frac {1}{\frac {{\mathrm {e}}^{-x}}{2}+\frac {{\mathrm {e}}^x}{2}}\right )}^{3/4}}{11\,\left ({\mathrm {e}}^{2\,x}+1\right )}-\frac {1312\,{\left (\frac {1}{\frac {{\mathrm {e}}^{-x}}{2}+\frac {{\mathrm {e}}^x}{2}}\right )}^{3/4}}{209\,{\left ({\mathrm {e}}^{2\,x}+1\right )}^2}+\frac {128\,{\left (\frac {1}{\frac {{\mathrm {e}}^{-x}}{2}+\frac {{\mathrm {e}}^x}{2}}\right )}^{3/4}}{19\,{\left ({\mathrm {e}}^{2\,x}+1\right )}^3}-\frac {64\,{\left (\frac {1}{\frac {{\mathrm {e}}^{-x}}{2}+\frac {{\mathrm {e}}^x}{2}}\right )}^{3/4}}{19\,{\left ({\mathrm {e}}^{2\,x}+1\right )}^4}-\frac {4\,{\left (\frac {1}{\frac {{\mathrm {e}}^{-x}}{2}+\frac {{\mathrm {e}}^x}{2}}\right )}^{3/4}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^5*(1/cosh(x))^(3/4),x)

[Out]

(32*(1/(exp(-x)/2 + exp(x)/2))^(3/4))/(11*(exp(2*x) + 1)) - (1312*(1/(exp(-x)/2 + exp(x)/2))^(3/4))/(209*(exp(
2*x) + 1)^2) + (128*(1/(exp(-x)/2 + exp(x)/2))^(3/4))/(19*(exp(2*x) + 1)^3) - (64*(1/(exp(-x)/2 + exp(x)/2))^(
3/4))/(19*(exp(2*x) + 1)^4) - (4*(1/(exp(-x)/2 + exp(x)/2))^(3/4))/3

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**(3/4)*tanh(x)**5,x)

[Out]

Timed out

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