3.2.76 \(\int \sqrt {a+b \text {sech}^2(x)} \tanh ^5(x) \, dx\) [176]
Optimal. Leaf size=83 \[ \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a}}\right )-\sqrt {a+b \text {sech}^2(x)}+\frac {(a+2 b) \left (a+b \text {sech}^2(x)\right )^{3/2}}{3 b^2}-\frac {\left (a+b \text {sech}^2(x)\right )^{5/2}}{5 b^2} \]
[Out]
1/3*(a+2*b)*(a+b*sech(x)^2)^(3/2)/b^2-1/5*(a+b*sech(x)^2)^(5/2)/b^2+arctanh((a+b*sech(x)^2)^(1/2)/a^(1/2))*a^(
1/2)-(a+b*sech(x)^2)^(1/2)
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Rubi [A]
time = 0.10, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {4224, 457, 90,
52, 65, 214} \begin {gather*} -\frac {\left (a+b \text {sech}^2(x)\right )^{5/2}}{5 b^2}+\frac {(a+2 b) \left (a+b \text {sech}^2(x)\right )^{3/2}}{3 b^2}-\sqrt {a+b \text {sech}^2(x)}+\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a}}\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Sqrt[a + b*Sech[x]^2]*Tanh[x]^5,x]
[Out]
Sqrt[a]*ArcTanh[Sqrt[a + b*Sech[x]^2]/Sqrt[a]] - Sqrt[a + b*Sech[x]^2] + ((a + 2*b)*(a + b*Sech[x]^2)^(3/2))/(
3*b^2) - (a + b*Sech[x]^2)^(5/2)/(5*b^2)
Rule 52
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]
Rule 65
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 90
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))
Rule 214
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]
Rule 457
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Rule 4224
Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Sec[e + f*x], x]}, Dist[1/f, Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p/x)
, x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[
n, 2] || EqQ[n, 4] || IGtQ[p, 0] || IntegersQ[2*n, p])
Rubi steps
\begin {align*} \int \sqrt {a+b \text {sech}^2(x)} \tanh ^5(x) \, dx &=-\text {Subst}\left (\int \frac {\left (-1+x^2\right )^2 \sqrt {a+b x^2}}{x} \, dx,x,\text {sech}(x)\right )\\ &=-\left (\frac {1}{2} \text {Subst}\left (\int \frac {(-1+x)^2 \sqrt {a+b x}}{x} \, dx,x,\text {sech}^2(x)\right )\right )\\ &=-\left (\frac {1}{2} \text {Subst}\left (\int \left (\frac {(-a-2 b) \sqrt {a+b x}}{b}+\frac {\sqrt {a+b x}}{x}+\frac {(a+b x)^{3/2}}{b}\right ) \, dx,x,\text {sech}^2(x)\right )\right )\\ &=\frac {(a+2 b) \left (a+b \text {sech}^2(x)\right )^{3/2}}{3 b^2}-\frac {\left (a+b \text {sech}^2(x)\right )^{5/2}}{5 b^2}-\frac {1}{2} \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,\text {sech}^2(x)\right )\\ &=-\sqrt {a+b \text {sech}^2(x)}+\frac {(a+2 b) \left (a+b \text {sech}^2(x)\right )^{3/2}}{3 b^2}-\frac {\left (a+b \text {sech}^2(x)\right )^{5/2}}{5 b^2}-\frac {1}{2} a \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\text {sech}^2(x)\right )\\ &=-\sqrt {a+b \text {sech}^2(x)}+\frac {(a+2 b) \left (a+b \text {sech}^2(x)\right )^{3/2}}{3 b^2}-\frac {\left (a+b \text {sech}^2(x)\right )^{5/2}}{5 b^2}-\frac {a \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \text {sech}^2(x)}\right )}{b}\\ &=\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a}}\right )-\sqrt {a+b \text {sech}^2(x)}+\frac {(a+2 b) \left (a+b \text {sech}^2(x)\right )^{3/2}}{3 b^2}-\frac {\left (a+b \text {sech}^2(x)\right )^{5/2}}{5 b^2}\\ \end {align*}
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Mathematica [A]
time = 0.49, size = 114, normalized size = 1.37 \begin {gather*} \frac {1}{15} \cosh (x) \sqrt {a+b \text {sech}^2(x)} \left (\frac {15 \sqrt {2} \sqrt {a} \log \left (\sqrt {2} \sqrt {a} \cosh (x)+\sqrt {a+2 b+a \cosh (2 x)}\right )}{\sqrt {a+2 b+a \cosh (2 x)}}+\left (-15+\frac {2 a^2}{b^2}+\frac {10 a}{b}\right ) \text {sech}(x)+\left (10-\frac {a}{b}\right ) \text {sech}^3(x)-3 \text {sech}^5(x)\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Sqrt[a + b*Sech[x]^2]*Tanh[x]^5,x]
[Out]
(Cosh[x]*Sqrt[a + b*Sech[x]^2]*((15*Sqrt[2]*Sqrt[a]*Log[Sqrt[2]*Sqrt[a]*Cosh[x] + Sqrt[a + 2*b + a*Cosh[2*x]]]
)/Sqrt[a + 2*b + a*Cosh[2*x]] + (-15 + (2*a^2)/b^2 + (10*a)/b)*Sech[x] + (10 - a/b)*Sech[x]^3 - 3*Sech[x]^5))/
15
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Maple [F]
time = 1.94, size = 0, normalized size = 0.00 \[\int \sqrt {a +b \mathrm {sech}\left (x \right )^{2}}\, \left (\tanh ^{5}\left (x \right )\right )\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*sech(x)^2)^(1/2)*tanh(x)^5,x)
[Out]
int((a+b*sech(x)^2)^(1/2)*tanh(x)^5,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((a+b*sech(x)^2)^(1/2)*tanh(x)^5,x, algorithm="maxima")
[Out]
integrate(sqrt(b*sech(x)^2 + a)*tanh(x)^5, x)
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 1929 vs.
\(2 (67) = 134\).
time = 0.65, size = 4594, normalized size = 55.35 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sech(x)^2)^(1/2)*tanh(x)^5,x, algorithm="fricas")
[Out]
[1/60*(15*(b^2*cosh(x)^10 + 10*b^2*cosh(x)*sinh(x)^9 + b^2*sinh(x)^10 + 5*b^2*cosh(x)^8 + 5*(9*b^2*cosh(x)^2 +
b^2)*sinh(x)^8 + 10*b^2*cosh(x)^6 + 40*(3*b^2*cosh(x)^3 + b^2*cosh(x))*sinh(x)^7 + 10*(21*b^2*cosh(x)^4 + 14*
b^2*cosh(x)^2 + b^2)*sinh(x)^6 + 10*b^2*cosh(x)^4 + 4*(63*b^2*cosh(x)^5 + 70*b^2*cosh(x)^3 + 15*b^2*cosh(x))*s
inh(x)^5 + 10*(21*b^2*cosh(x)^6 + 35*b^2*cosh(x)^4 + 15*b^2*cosh(x)^2 + b^2)*sinh(x)^4 + 5*b^2*cosh(x)^2 + 40*
(3*b^2*cosh(x)^7 + 7*b^2*cosh(x)^5 + 5*b^2*cosh(x)^3 + b^2*cosh(x))*sinh(x)^3 + 5*(9*b^2*cosh(x)^8 + 28*b^2*co
sh(x)^6 + 30*b^2*cosh(x)^4 + 12*b^2*cosh(x)^2 + b^2)*sinh(x)^2 + b^2 + 10*(b^2*cosh(x)^9 + 4*b^2*cosh(x)^7 + 6
*b^2*cosh(x)^5 + 4*b^2*cosh(x)^3 + b^2*cosh(x))*sinh(x))*sqrt(a)*log(((a^3 + 2*a^2*b + a*b^2)*cosh(x)^8 + 8*(a
^3 + 2*a^2*b + a*b^2)*cosh(x)*sinh(x)^7 + (a^3 + 2*a^2*b + a*b^2)*sinh(x)^8 + 2*(2*a^3 + 5*a^2*b + 4*a*b^2 + b
^3)*cosh(x)^6 + 2*(2*a^3 + 5*a^2*b + 4*a*b^2 + b^3 + 14*(a^3 + 2*a^2*b + a*b^2)*cosh(x)^2)*sinh(x)^6 + 4*(14*(
a^3 + 2*a^2*b + a*b^2)*cosh(x)^3 + 3*(2*a^3 + 5*a^2*b + 4*a*b^2 + b^3)*cosh(x))*sinh(x)^5 + (6*a^3 + 14*a^2*b
+ 9*a*b^2)*cosh(x)^4 + (70*(a^3 + 2*a^2*b + a*b^2)*cosh(x)^4 + 6*a^3 + 14*a^2*b + 9*a*b^2 + 30*(2*a^3 + 5*a^2*
b + 4*a*b^2 + b^3)*cosh(x)^2)*sinh(x)^4 + 4*(14*(a^3 + 2*a^2*b + a*b^2)*cosh(x)^5 + 10*(2*a^3 + 5*a^2*b + 4*a*
b^2 + b^3)*cosh(x)^3 + (6*a^3 + 14*a^2*b + 9*a*b^2)*cosh(x))*sinh(x)^3 + a^3 + 2*(2*a^3 + 3*a^2*b)*cosh(x)^2 +
2*(14*(a^3 + 2*a^2*b + a*b^2)*cosh(x)^6 + 15*(2*a^3 + 5*a^2*b + 4*a*b^2 + b^3)*cosh(x)^4 + 2*a^3 + 3*a^2*b +
3*(6*a^3 + 14*a^2*b + 9*a*b^2)*cosh(x)^2)*sinh(x)^2 + sqrt(2)*((a^2 + 2*a*b + b^2)*cosh(x)^6 + 6*(a^2 + 2*a*b
+ b^2)*cosh(x)*sinh(x)^5 + (a^2 + 2*a*b + b^2)*sinh(x)^6 + 3*(a^2 + 2*a*b + b^2)*cosh(x)^4 + 3*(5*(a^2 + 2*a*b
+ b^2)*cosh(x)^2 + a^2 + 2*a*b + b^2)*sinh(x)^4 + 4*(5*(a^2 + 2*a*b + b^2)*cosh(x)^3 + 3*(a^2 + 2*a*b + b^2)*
cosh(x))*sinh(x)^3 + (3*a^2 + 4*a*b)*cosh(x)^2 + (15*(a^2 + 2*a*b + b^2)*cosh(x)^4 + 18*(a^2 + 2*a*b + b^2)*co
sh(x)^2 + 3*a^2 + 4*a*b)*sinh(x)^2 + a^2 + 2*(3*(a^2 + 2*a*b + b^2)*cosh(x)^5 + 6*(a^2 + 2*a*b + b^2)*cosh(x)^
3 + (3*a^2 + 4*a*b)*cosh(x))*sinh(x))*sqrt(a)*sqrt((a*cosh(x)^2 + a*sinh(x)^2 + a + 2*b)/(cosh(x)^2 - 2*cosh(x
)*sinh(x) + sinh(x)^2)) + 4*(2*(a^3 + 2*a^2*b + a*b^2)*cosh(x)^7 + 3*(2*a^3 + 5*a^2*b + 4*a*b^2 + b^3)*cosh(x)
^5 + (6*a^3 + 14*a^2*b + 9*a*b^2)*cosh(x)^3 + (2*a^3 + 3*a^2*b)*cosh(x))*sinh(x))/(cosh(x)^6 + 6*cosh(x)^5*sin
h(x) + 15*cosh(x)^4*sinh(x)^2 + 20*cosh(x)^3*sinh(x)^3 + 15*cosh(x)^2*sinh(x)^4 + 6*cosh(x)*sinh(x)^5 + sinh(x
)^6)) + 15*(b^2*cosh(x)^10 + 10*b^2*cosh(x)*sinh(x)^9 + b^2*sinh(x)^10 + 5*b^2*cosh(x)^8 + 5*(9*b^2*cosh(x)^2
+ b^2)*sinh(x)^8 + 10*b^2*cosh(x)^6 + 40*(3*b^2*cosh(x)^3 + b^2*cosh(x))*sinh(x)^7 + 10*(21*b^2*cosh(x)^4 + 14
*b^2*cosh(x)^2 + b^2)*sinh(x)^6 + 10*b^2*cosh(x)^4 + 4*(63*b^2*cosh(x)^5 + 70*b^2*cosh(x)^3 + 15*b^2*cosh(x))*
sinh(x)^5 + 10*(21*b^2*cosh(x)^6 + 35*b^2*cosh(x)^4 + 15*b^2*cosh(x)^2 + b^2)*sinh(x)^4 + 5*b^2*cosh(x)^2 + 40
*(3*b^2*cosh(x)^7 + 7*b^2*cosh(x)^5 + 5*b^2*cosh(x)^3 + b^2*cosh(x))*sinh(x)^3 + 5*(9*b^2*cosh(x)^8 + 28*b^2*c
osh(x)^6 + 30*b^2*cosh(x)^4 + 12*b^2*cosh(x)^2 + b^2)*sinh(x)^2 + b^2 + 10*(b^2*cosh(x)^9 + 4*b^2*cosh(x)^7 +
6*b^2*cosh(x)^5 + 4*b^2*cosh(x)^3 + b^2*cosh(x))*sinh(x))*sqrt(a)*log(-(a*cosh(x)^4 + 4*a*cosh(x)*sinh(x)^3 +
a*sinh(x)^4 + 2*b*cosh(x)^2 + 2*(3*a*cosh(x)^2 + b)*sinh(x)^2 + sqrt(2)*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(
x)^2 - 1)*sqrt(a)*sqrt((a*cosh(x)^2 + a*sinh(x)^2 + a + 2*b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 4*
(a*cosh(x)^3 + b*cosh(x))*sinh(x) + a)/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)) + 4*sqrt(2)*((2*a^2 + 10*a
*b - 15*b^2)*cosh(x)^8 + 8*(2*a^2 + 10*a*b - 15*b^2)*cosh(x)*sinh(x)^7 + (2*a^2 + 10*a*b - 15*b^2)*sinh(x)^8 +
4*(2*a^2 + 9*a*b - 5*b^2)*cosh(x)^6 + 4*(7*(2*a^2 + 10*a*b - 15*b^2)*cosh(x)^2 + 2*a^2 + 9*a*b - 5*b^2)*sinh(
x)^6 + 8*(7*(2*a^2 + 10*a*b - 15*b^2)*cosh(x)^3 + 3*(2*a^2 + 9*a*b - 5*b^2)*cosh(x))*sinh(x)^5 + 2*(6*a^2 + 26
*a*b - 29*b^2)*cosh(x)^4 + 2*(35*(2*a^2 + 10*a*b - 15*b^2)*cosh(x)^4 + 30*(2*a^2 + 9*a*b - 5*b^2)*cosh(x)^2 +
6*a^2 + 26*a*b - 29*b^2)*sinh(x)^4 + 8*(7*(2*a^2 + 10*a*b - 15*b^2)*cosh(x)^5 + 10*(2*a^2 + 9*a*b - 5*b^2)*cos
h(x)^3 + (6*a^2 + 26*a*b - 29*b^2)*cosh(x))*sinh(x)^3 + 4*(2*a^2 + 9*a*b - 5*b^2)*cosh(x)^2 + 4*(7*(2*a^2 + 10
*a*b - 15*b^2)*cosh(x)^6 + 15*(2*a^2 + 9*a*b - 5*b^2)*cosh(x)^4 + 3*(6*a^2 + 26*a*b - 29*b^2)*cosh(x)^2 + 2*a^
2 + 9*a*b - 5*b^2)*sinh(x)^2 + 2*a^2 + 10*a*b - 15*b^2 + 8*((2*a^2 + 10*a*b - 15*b^2)*cosh(x)^7 + 3*(2*a^2 + 9
*a*b - 5*b^2)*cosh(x)^5 + (6*a^2 + 26*a*b - 29*b^2)*cosh(x)^3 + (2*a^2 + 9*a*b - 5*b^2)*cosh(x))*sinh(x))*sqrt
((a*cosh(x)^2 + a*sinh(x)^2 + a + 2*b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)))/(b^2*cosh(x)^10 + 10*b^2*
cosh(x)*sinh(x)^9 + b^2*sinh(x)^10 + 5*b^2*cosh(x)^8 + 5*(9*b^2*cosh(x)^2 + b^2)*sinh(x)^8 + 10*b^2*cosh(x)^6
+ 40*(3*b^2*cosh(x)^3 + b^2*cosh(x))*sinh(x)^7 + 10*(21*b^2*cosh(x)^4 + 14*b^2*cosh(x)^2 + b^2)*sinh(x)^6 + 10
*b^2*cosh(x)^4 + 4*(63*b^2*cosh(x)^5 + 70*b^2*c...
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a + b \operatorname {sech}^{2}{\left (x \right )}} \tanh ^{5}{\left (x \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sech(x)**2)**(1/2)*tanh(x)**5,x)
[Out]
Integral(sqrt(a + b*sech(x)**2)*tanh(x)**5, x)
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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sech(x)^2)^(1/2)*tanh(x)^5,x, algorithm="giac")
[Out]
Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument Type
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {tanh}\left (x\right )}^5\,\sqrt {a+\frac {b}{{\mathrm {cosh}\left (x\right )}^2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tanh(x)^5*(a + b/cosh(x)^2)^(1/2),x)
[Out]
int(tanh(x)^5*(a + b/cosh(x)^2)^(1/2), x)
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