3.5.42 \(\int \frac {\tanh ^4(e+f x)}{\sqrt {a+a \sinh ^2(e+f x)}} \, dx\) [442]

Optimal. Leaf size=91 \[ \frac {3 \text {ArcTan}(\sinh (e+f x)) \cosh (e+f x)}{8 f \sqrt {a \cosh ^2(e+f x)}}-\frac {3 \tanh (e+f x)}{8 f \sqrt {a \cosh ^2(e+f x)}}-\frac {\tanh ^3(e+f x)}{4 f \sqrt {a \cosh ^2(e+f x)}} \]

[Out]

3/8*arctan(sinh(f*x+e))*cosh(f*x+e)/f/(a*cosh(f*x+e)^2)^(1/2)-3/8*tanh(f*x+e)/f/(a*cosh(f*x+e)^2)^(1/2)-1/4*ta
nh(f*x+e)^3/f/(a*cosh(f*x+e)^2)^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3255, 3286, 2691, 3855} \begin {gather*} \frac {3 \cosh (e+f x) \text {ArcTan}(\sinh (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 \tanh (e+f x)}{8 f \sqrt {a \cosh ^2(e+f x)}} \end {gather*}

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 2691

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 3255

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 3286

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 3855

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*}

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Mathematica [A]
time = 0.10, size = 66, normalized size = 0.73 \begin {gather*} \frac {3 \text {ArcTan}(\sinh (e+f x)) \cosh (e+f x)+\tanh (e+f x) \left (3-6 \text {sech}^2(e+f x)-8 \tanh ^2(e+f x)\right )}{8 f \sqrt {a \cosh ^2(e+f x)}} \end {gather*}

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 = 1.44, size = 68, normalized size = 0.75

method	result	size
default	\(\frac {3 \arctan \left (\sinh \left (f x +e \right )\right ) \left (\cosh ^{4}\left (f x +e \right )\right )-5 \left (\cosh ^{2}\left (f x +e \right )\right ) \sinh \left (f x +e \right )+2 \sinh \left (f x +e \right )}{8 \cosh \left (f x +e \right )^{3} \sqrt {a \left (\cosh ^{2}\left (f x +e \right )\right )}\, f}\)	\(68\)
risch	\(-\frac {5 \,{\mathrm e}^{6 f x +6 e}-3 \,{\mathrm e}^{4 f x +4 e}+3 \,{\mathrm e}^{2 f x +2 e}-5}{4 \sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}\, \left ({\mathrm e}^{2 f x +2 e}+1\right )^{3} f}+\frac {3 i \ln \left ({\mathrm e}^{f x}+i {\mathrm e}^{-e}\right ) \left ({\mathrm e}^{2 f x +2 e}+1\right ) {\mathrm e}^{-f x -e}}{8 f \sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}}-\frac {3 i \ln \left ({\mathrm e}^{f x}-i {\mathrm e}^{-e}\right ) \left ({\mathrm e}^{2 f x +2 e}+1\right ) {\mathrm e}^{-f x -e}}{8 f \sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}}\)	\(211\)

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,method=_RETURNVERBOSE)

[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] Leaf count of result is larger than twice the leaf count of optimal. 672 vs. \(2 (86) = 172\).
time = 0.55, size = 672, normalized size = 7.38 \begin {gather*} \frac {\frac {15 \, \arctan \left (e^{\left (-f x - e\right )}\right )}{\sqrt {a}} - \frac {15 \, e^{\left (-f x - e\right )} + 55 \, e^{\left (-3 \, f x - 3 \, e\right )} + 73 \, e^{\left (-5 \, f x - 5 \, e\right )} - 15 \, e^{\left (-7 \, f x - 7 \, e\right )}}{4 \, \sqrt {a} e^{\left (-2 \, f x - 2 \, e\right )} + 6 \, \sqrt {a} e^{\left (-4 \, f x - 4 \, e\right )} + 4 \, \sqrt {a} e^{\left (-6 \, f x - 6 \, e\right )} + \sqrt {a} e^{\left (-8 \, f x - 8 \, e\right )} + \sqrt {a}}}{48 \, f} + \frac {\frac {15 \, \arctan \left (e^{\left (-f x - e\right )}\right )}{\sqrt {a}} - \frac {15 \, e^{\left (-f x - e\right )} - 73 \, e^{\left (-3 \, f x - 3 \, e\right )} - 55 \, e^{\left (-5 \, f x - 5 \, e\right )} - 15 \, e^{\left (-7 \, f x - 7 \, e\right )}}{4 \, \sqrt {a} e^{\left (-2 \, f x - 2 \, e\right )} + 6 \, \sqrt {a} e^{\left (-4 \, f x - 4 \, e\right )} + 4 \, \sqrt {a} e^{\left (-6 \, f x - 6 \, e\right )} + \sqrt {a} e^{\left (-8 \, f x - 8 \, e\right )} + \sqrt {a}}}{48 \, f} - \frac {3 \, {\left (\frac {3 \, \arctan \left (e^{\left (-f x - e\right )}\right )}{\sqrt {a}} - \frac {3 \, e^{\left (-f x - e\right )} + 11 \, e^{\left (-3 \, f x - 3 \, e\right )} - 11 \, e^{\left (-5 \, f x - 5 \, e\right )} - 3 \, e^{\left (-7 \, f x - 7 \, e\right )}}{4 \, \sqrt {a} e^{\left (-2 \, f x - 2 \, e\right )} + 6 \, \sqrt {a} e^{\left (-4 \, f x - 4 \, e\right )} + 4 \, \sqrt {a} e^{\left (-6 \, f x - 6 \, e\right )} + \sqrt {a} e^{\left (-8 \, f x - 8 \, e\right )} + \sqrt {a}}\right )}}{32 \, f} - \frac {35 \, \arctan \left (e^{\left (-f x - e\right )}\right )}{32 \, \sqrt {a} f} - \frac {279 \, e^{\left (-f x - e\right )} + 511 \, e^{\left (-3 \, f x - 3 \, e\right )} + 385 \, e^{\left (-5 \, f x - 5 \, e\right )} + 105 \, e^{\left (-7 \, f x - 7 \, e\right )}}{192 \, {\left (4 \, \sqrt {a} e^{\left (-2 \, f x - 2 \, e\right )} + 6 \, \sqrt {a} e^{\left (-4 \, f x - 4 \, e\right )} + 4 \, \sqrt {a} e^{\left (-6 \, f x - 6 \, e\right )} + \sqrt {a} e^{\left (-8 \, f x - 8 \, e\right )} + \sqrt {a}\right )} f} + \frac {105 \, e^{\left (-f x - e\right )} + 385 \, e^{\left (-3 \, f x - 3 \, e\right )} + 511 \, e^{\left (-5 \, f x - 5 \, e\right )} + 279 \, e^{\left (-7 \, f x - 7 \, e\right )}}{192 \, {\left (4 \, \sqrt {a} e^{\left (-2 \, f x - 2 \, e\right )} + 6 \, \sqrt {a} e^{\left (-4 \, f x - 4 \, e\right )} + 4 \, \sqrt {a} e^{\left (-6 \, f x - 6 \, e\right )} + \sqrt {a} e^{\left (-8 \, f x - 8 \, e\right )} + \sqrt {a}\right )} f} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 1328 vs. \(2 (79) = 158\).
time = 0.41, size = 1328, normalized size = 14.59 \begin {gather*} \text {Too large to display} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tanh ^{4}{\left (e + f x \right )}}{\sqrt {a \left (\sinh ^{2}{\left (e + f x \right )} + 1\right )}}\, dx \end {gather*}

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 [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(tanh(f*x+e)^4/(a+a*sinh(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {tanh}\left (e+f\,x\right )}^4}{\sqrt {a\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(e + f*x)^4/(a + a*sinh(e + f*x)^2)^(1/2),x)

[Out]

int(tanh(e + f*x)^4/(a + a*sinh(e + f*x)^2)^(1/2), x)

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