3.442 \(\int \frac{\tanh ^4(e+f x)}{\sqrt{a+a \sinh ^2(e+f x)}} \, dx\)
Optimal. Leaf size=91 \[ -\frac{\tanh ^3(e+f x)}{4 f \sqrt{a \cosh ^2(e+f x)}}-\frac{3 \tanh (e+f x)}{8 f \sqrt{a \cosh ^2(e+f x)}}+\frac{3 \cosh (e+f x) \tan ^{-1}(\sinh (e+f x))}{8 f \sqrt{a \cosh ^2(e+f x)}} \]
[Out]
(3*ArcTan[Sinh[e + f*x]]*Cosh[e + f*x])/(8*f*Sqrt[a*Cosh[e + f*x]^2]) - (3*Tanh[e + f*x])/(8*f*Sqrt[a*Cosh[e +
f*x]^2]) - Tanh[e + f*x]^3/(4*f*Sqrt[a*Cosh[e + f*x]^2])
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Rubi [A] time = 0.144237, antiderivative size = 91, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used =
{3176, 3207, 2611, 3770} \[ -\frac{\tanh ^3(e+f x)}{4 f \sqrt{a \cosh ^2(e+f x)}}-\frac{3 \tanh (e+f x)}{8 f \sqrt{a \cosh ^2(e+f x)}}+\frac{3 \cosh (e+f x) \tan ^{-1}(\sinh (e+f x))}{8 f \sqrt{a \cosh ^2(e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[Tanh[e + f*x]^4/Sqrt[a + a*Sinh[e + f*x]^2],x]
[Out]
(3*ArcTan[Sinh[e + f*x]]*Cosh[e + f*x])/(8*f*Sqrt[a*Cosh[e + f*x]^2]) - (3*Tanh[e + f*x])/(8*f*Sqrt[a*Cosh[e +
f*x]^2]) - Tanh[e + f*x]^3/(4*f*Sqrt[a*Cosh[e + f*x]^2])
Rule 3176
Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*cos[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]
Rule 3207
Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] && !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])
Rule 2611
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]
Rule 3770
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Rubi steps
\begin{align*} \int \frac{\tanh ^4(e+f x)}{\sqrt{a+a \sinh ^2(e+f x)}} \, dx &=\int \frac{\tanh ^4(e+f x)}{\sqrt{a \cosh ^2(e+f x)}} \, dx\\ &=\frac{\cosh (e+f x) \int \text{sech}(e+f x) \tanh ^4(e+f x) \, dx}{\sqrt{a \cosh ^2(e+f x)}}\\ &=-\frac{\tanh ^3(e+f x)}{4 f \sqrt{a \cosh ^2(e+f x)}}+\frac{(3 \cosh (e+f x)) \int \text{sech}(e+f x) \tanh ^2(e+f x) \, dx}{4 \sqrt{a \cosh ^2(e+f x)}}\\ &=-\frac{3 \tanh (e+f x)}{8 f \sqrt{a \cosh ^2(e+f x)}}-\frac{\tanh ^3(e+f x)}{4 f \sqrt{a \cosh ^2(e+f x)}}+\frac{(3 \cosh (e+f x)) \int \text{sech}(e+f x) \, dx}{8 \sqrt{a \cosh ^2(e+f x)}}\\ &=\frac{3 \tan ^{-1}(\sinh (e+f x)) \cosh (e+f x)}{8 f \sqrt{a \cosh ^2(e+f x)}}-\frac{3 \tanh (e+f x)}{8 f \sqrt{a \cosh ^2(e+f x)}}-\frac{\tanh ^3(e+f x)}{4 f \sqrt{a \cosh ^2(e+f x)}}\\ \end{align*}
Mathematica [A] time = 0.132891, size = 66, normalized size = 0.73 \[ \frac{\tanh (e+f x) \left (-8 \tanh ^2(e+f x)-6 \text{sech}^2(e+f x)+3\right )+3 \cosh (e+f x) \tan ^{-1}(\sinh (e+f x))}{8 f \sqrt{a \cosh ^2(e+f x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[Tanh[e + f*x]^4/Sqrt[a + a*Sinh[e + f*x]^2],x]
[Out]
(3*ArcTan[Sinh[e + f*x]]*Cosh[e + f*x] + Tanh[e + f*x]*(3 - 6*Sech[e + f*x]^2 - 8*Tanh[e + f*x]^2))/(8*f*Sqrt[
a*Cosh[e + f*x]^2])
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Maple [A] time = 0.108, size = 68, normalized size = 0.8 \begin{align*}{\frac{3\,\arctan \left ( \sinh \left ( fx+e \right ) \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{4}-5\, \left ( \cosh \left ( fx+e \right ) \right ) ^{2}\sinh \left ( fx+e \right ) +2\,\sinh \left ( fx+e \right ) }{8\, \left ( \cosh \left ( fx+e \right ) \right ) ^{3}f}{\frac{1}{\sqrt{a \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tanh(f*x+e)^4/(a+a*sinh(f*x+e)^2)^(1/2),x)
[Out]
1/8*(3*arctan(sinh(f*x+e))*cosh(f*x+e)^4-5*cosh(f*x+e)^2*sinh(f*x+e)+2*sinh(f*x+e))/cosh(f*x+e)^3/(a*cosh(f*x+
e)^2)^(1/2)/f
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Maxima [B] time = 1.95509, size = 848, normalized size = 9.32 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(f*x+e)^4/(a+a*sinh(f*x+e)^2)^(1/2),x, algorithm="maxima")
[Out]
1/48*(15*arctan(e^(-f*x - e))/sqrt(a) - (15*e^(-f*x - e) + 55*e^(-3*f*x - 3*e) + 73*e^(-5*f*x - 5*e) - 15*e^(-
7*f*x - 7*e))/(4*sqrt(a)*e^(-2*f*x - 2*e) + 6*sqrt(a)*e^(-4*f*x - 4*e) + 4*sqrt(a)*e^(-6*f*x - 6*e) + sqrt(a)*
e^(-8*f*x - 8*e) + sqrt(a)))/f + 1/48*(15*arctan(e^(-f*x - e))/sqrt(a) - (15*e^(-f*x - e) - 73*e^(-3*f*x - 3*e
) - 55*e^(-5*f*x - 5*e) - 15*e^(-7*f*x - 7*e))/(4*sqrt(a)*e^(-2*f*x - 2*e) + 6*sqrt(a)*e^(-4*f*x - 4*e) + 4*sq
rt(a)*e^(-6*f*x - 6*e) + sqrt(a)*e^(-8*f*x - 8*e) + sqrt(a)))/f - 3/32*(3*arctan(e^(-f*x - e))/sqrt(a) - (3*e^
(-f*x - e) + 11*e^(-3*f*x - 3*e) - 11*e^(-5*f*x - 5*e) - 3*e^(-7*f*x - 7*e))/(4*sqrt(a)*e^(-2*f*x - 2*e) + 6*s
qrt(a)*e^(-4*f*x - 4*e) + 4*sqrt(a)*e^(-6*f*x - 6*e) + sqrt(a)*e^(-8*f*x - 8*e) + sqrt(a)))/f - 35/32*arctan(e
^(-f*x - e))/(sqrt(a)*f) - 1/192*(279*e^(-f*x - e) + 511*e^(-3*f*x - 3*e) + 385*e^(-5*f*x - 5*e) + 105*e^(-7*f
*x - 7*e))/((4*sqrt(a)*e^(-2*f*x - 2*e) + 6*sqrt(a)*e^(-4*f*x - 4*e) + 4*sqrt(a)*e^(-6*f*x - 6*e) + sqrt(a)*e^
(-8*f*x - 8*e) + sqrt(a))*f) + 1/192*(105*e^(-f*x - e) + 385*e^(-3*f*x - 3*e) + 511*e^(-5*f*x - 5*e) + 279*e^(
-7*f*x - 7*e))/((4*sqrt(a)*e^(-2*f*x - 2*e) + 6*sqrt(a)*e^(-4*f*x - 4*e) + 4*sqrt(a)*e^(-6*f*x - 6*e) + sqrt(a
)*e^(-8*f*x - 8*e) + sqrt(a))*f)
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Fricas [B] time = 2.04753, size = 3584, normalized size = 39.38 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(f*x+e)^4/(a+a*sinh(f*x+e)^2)^(1/2),x, algorithm="fricas")
[Out]
-1/4*(35*cosh(f*x + e)*e^(f*x + e)*sinh(f*x + e)^6 + 5*e^(f*x + e)*sinh(f*x + e)^7 + 3*(35*cosh(f*x + e)^2 - 1
)*e^(f*x + e)*sinh(f*x + e)^5 + 5*(35*cosh(f*x + e)^3 - 3*cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e)^4 + (175*co
sh(f*x + e)^4 - 30*cosh(f*x + e)^2 + 3)*e^(f*x + e)*sinh(f*x + e)^3 + 3*(35*cosh(f*x + e)^5 - 10*cosh(f*x + e)
^3 + 3*cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e)^2 + (35*cosh(f*x + e)^6 - 15*cosh(f*x + e)^4 + 9*cosh(f*x + e)
^2 - 5)*e^(f*x + e)*sinh(f*x + e) - 3*(8*cosh(f*x + e)*e^(f*x + e)*sinh(f*x + e)^7 + e^(f*x + e)*sinh(f*x + e)
^8 + 4*(7*cosh(f*x + e)^2 + 1)*e^(f*x + e)*sinh(f*x + e)^6 + 8*(7*cosh(f*x + e)^3 + 3*cosh(f*x + e))*e^(f*x +
e)*sinh(f*x + e)^5 + 2*(35*cosh(f*x + e)^4 + 30*cosh(f*x + e)^2 + 3)*e^(f*x + e)*sinh(f*x + e)^4 + 8*(7*cosh(f
*x + e)^5 + 10*cosh(f*x + e)^3 + 3*cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e)^3 + 4*(7*cosh(f*x + e)^6 + 15*cosh
(f*x + e)^4 + 9*cosh(f*x + e)^2 + 1)*e^(f*x + e)*sinh(f*x + e)^2 + 8*(cosh(f*x + e)^7 + 3*cosh(f*x + e)^5 + 3*
cosh(f*x + e)^3 + cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e) + (cosh(f*x + e)^8 + 4*cosh(f*x + e)^6 + 6*cosh(f*x
+ e)^4 + 4*cosh(f*x + e)^2 + 1)*e^(f*x + e))*arctan(cosh(f*x + e) + sinh(f*x + e)) + (5*cosh(f*x + e)^7 - 3*c
osh(f*x + e)^5 + 3*cosh(f*x + e)^3 - 5*cosh(f*x + e))*e^(f*x + e))*sqrt(a*e^(4*f*x + 4*e) + 2*a*e^(2*f*x + 2*e
) + a)*e^(-f*x - e)/(a*f*cosh(f*x + e)^8 + (a*f*e^(2*f*x + 2*e) + a*f)*sinh(f*x + e)^8 + 4*a*f*cosh(f*x + e)^6
+ 8*(a*f*cosh(f*x + e)*e^(2*f*x + 2*e) + a*f*cosh(f*x + e))*sinh(f*x + e)^7 + 4*(7*a*f*cosh(f*x + e)^2 + a*f
+ (7*a*f*cosh(f*x + e)^2 + a*f)*e^(2*f*x + 2*e))*sinh(f*x + e)^6 + 6*a*f*cosh(f*x + e)^4 + 8*(7*a*f*cosh(f*x +
e)^3 + 3*a*f*cosh(f*x + e) + (7*a*f*cosh(f*x + e)^3 + 3*a*f*cosh(f*x + e))*e^(2*f*x + 2*e))*sinh(f*x + e)^5 +
2*(35*a*f*cosh(f*x + e)^4 + 30*a*f*cosh(f*x + e)^2 + 3*a*f + (35*a*f*cosh(f*x + e)^4 + 30*a*f*cosh(f*x + e)^2
+ 3*a*f)*e^(2*f*x + 2*e))*sinh(f*x + e)^4 + 4*a*f*cosh(f*x + e)^2 + 8*(7*a*f*cosh(f*x + e)^5 + 10*a*f*cosh(f*
x + e)^3 + 3*a*f*cosh(f*x + e) + (7*a*f*cosh(f*x + e)^5 + 10*a*f*cosh(f*x + e)^3 + 3*a*f*cosh(f*x + e))*e^(2*f
*x + 2*e))*sinh(f*x + e)^3 + 4*(7*a*f*cosh(f*x + e)^6 + 15*a*f*cosh(f*x + e)^4 + 9*a*f*cosh(f*x + e)^2 + a*f +
(7*a*f*cosh(f*x + e)^6 + 15*a*f*cosh(f*x + e)^4 + 9*a*f*cosh(f*x + e)^2 + a*f)*e^(2*f*x + 2*e))*sinh(f*x + e)
^2 + a*f + (a*f*cosh(f*x + e)^8 + 4*a*f*cosh(f*x + e)^6 + 6*a*f*cosh(f*x + e)^4 + 4*a*f*cosh(f*x + e)^2 + a*f)
*e^(2*f*x + 2*e) + 8*(a*f*cosh(f*x + e)^7 + 3*a*f*cosh(f*x + e)^5 + 3*a*f*cosh(f*x + e)^3 + a*f*cosh(f*x + e)
+ (a*f*cosh(f*x + e)^7 + 3*a*f*cosh(f*x + e)^5 + 3*a*f*cosh(f*x + e)^3 + a*f*cosh(f*x + e))*e^(2*f*x + 2*e))*s
inh(f*x + e))
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{4}{\left (e + f x \right )}}{\sqrt{a \left (\sinh ^{2}{\left (e + f x \right )} + 1\right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(f*x+e)**4/(a+a*sinh(f*x+e)**2)**(1/2),x)
[Out]
Integral(tanh(e + f*x)**4/sqrt(a*(sinh(e + f*x)**2 + 1)), x)
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Giac [A] time = 1.37038, size = 130, normalized size = 1.43 \begin{align*} \frac{\frac{3 \, \arctan \left (e^{\left (f x + e\right )}\right )}{\sqrt{a}} - \frac{5 \, \sqrt{a} e^{\left (7 \, f x + 7 \, e\right )} - 3 \, \sqrt{a} e^{\left (5 \, f x + 5 \, e\right )} + 3 \, \sqrt{a} e^{\left (3 \, f x + 3 \, e\right )} - 5 \, \sqrt{a} e^{\left (f x + e\right )}}{a{\left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right )}^{4}}}{4 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(f*x+e)^4/(a+a*sinh(f*x+e)^2)^(1/2),x, algorithm="giac")
[Out]
1/4*(3*arctan(e^(f*x + e))/sqrt(a) - (5*sqrt(a)*e^(7*f*x + 7*e) - 3*sqrt(a)*e^(5*f*x + 5*e) + 3*sqrt(a)*e^(3*f
*x + 3*e) - 5*sqrt(a)*e^(f*x + e))/(a*(e^(2*f*x + 2*e) + 1)^4))/f