3.5.32 \(\int \sqrt {a+a \sinh ^2(e+f x)} \tanh ^4(e+f x) \, dx\) [432]

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

[Out]

-3/2*arctan(sinh(f*x+e))*sech(f*x+e)*(a*cosh(f*x+e)^2)^(1/2)/f+3/2*(a*cosh(f*x+e)^2)^(1/2)*tanh(f*x+e)/f-1/2*(
a*cosh(f*x+e)^2)^(1/2)*tanh(f*x+e)^3/f

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Rubi [A]
time = 0.09, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3255, 3286, 2672, 294, 327, 209} \begin {gather*} -\frac {3 \text {sech}(e+f x) \sqrt {a \cosh ^2(e+f x)} \text {ArcTan}(\sinh (e+f x))}{2 f}-\frac {\tanh ^3(e+f x) \sqrt {a \cosh ^2(e+f x)}}{2 f}+\frac {3 \tanh (e+f x) \sqrt {a \cosh ^2(e+f x)}}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Sinh[e + f*x]^2]*Tanh[e + f*x]^4,x]

[Out]

(-3*ArcTan[Sinh[e + f*x]]*Sqrt[a*Cosh[e + f*x]^2]*Sech[e + f*x])/(2*f) + (3*Sqrt[a*Cosh[e + f*x]^2]*Tanh[e + f
*x])/(2*f) - (Sqrt[a*Cosh[e + f*x]^2]*Tanh[e + f*x]^3)/(2*f)

Rule 209

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

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2672

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

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
]])

Rubi steps

\begin {align*} \int \sqrt {a+a \sinh ^2(e+f x)} \tanh ^4(e+f x) \, dx &=\int \sqrt {a \cosh ^2(e+f x)} \tanh ^4(e+f x) \, dx\\ &=\left (\sqrt {a \cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \int \sinh (e+f x) \tanh ^3(e+f x) \, dx\\ &=\frac {\left (\sqrt {a \cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \text {Subst}\left (\int \frac {x^4}{\left (1+x^2\right )^2} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=-\frac {\sqrt {a \cosh ^2(e+f x)} \tanh ^3(e+f x)}{2 f}+\frac {\left (3 \sqrt {a \cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \text {Subst}\left (\int \frac {x^2}{1+x^2} \, dx,x,\sinh (e+f x)\right )}{2 f}\\ &=\frac {3 \sqrt {a \cosh ^2(e+f x)} \tanh (e+f x)}{2 f}-\frac {\sqrt {a \cosh ^2(e+f x)} \tanh ^3(e+f x)}{2 f}-\frac {\left (3 \sqrt {a \cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sinh (e+f x)\right )}{2 f}\\ &=-\frac {3 \tan ^{-1}(\sinh (e+f x)) \sqrt {a \cosh ^2(e+f x)} \text {sech}(e+f x)}{2 f}+\frac {3 \sqrt {a \cosh ^2(e+f x)} \tanh (e+f x)}{2 f}-\frac {\sqrt {a \cosh ^2(e+f x)} \tanh ^3(e+f x)}{2 f}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 55, normalized size = 0.60 \begin {gather*} \frac {a (-3 \text {ArcTan}(\sinh (e+f x)) \cosh (e+f x)+(2+\cosh (2 (e+f x))) \tanh (e+f x))}{2 f \sqrt {a \cosh ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + a*Sinh[e + f*x]^2]*Tanh[e + f*x]^4,x]

[Out]

(a*(-3*ArcTan[Sinh[e + f*x]]*Cosh[e + f*x] + (2 + Cosh[2*(e + f*x)])*Tanh[e + f*x]))/(2*f*Sqrt[a*Cosh[e + f*x]
^2])

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Maple [A]
time = 1.24, size = 69, normalized size = 0.76

method	result	size
default	\(-\frac {a \left (3 \arctan \left (\sinh \left (f x +e \right )\right ) \left (\cosh ^{2}\left (f x +e \right )\right )-2 \left (\cosh ^{2}\left (f x +e \right )\right ) \sinh \left (f x +e \right )-\sinh \left (f x +e \right )\right )}{2 \cosh \left (f x +e \right ) \sqrt {a \left (\cosh ^{2}\left (f x +e \right )\right )}\, f}\)	\(69\)
risch	\(\frac {\sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}\, {\mathrm e}^{2 f x +2 e}}{2 f \left ({\mathrm e}^{2 f x +2 e}+1\right )}-\frac {\sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}}{2 f \left ({\mathrm e}^{2 f x +2 e}+1\right )}+\frac {\left ({\mathrm e}^{2 f x +2 e}-1\right ) \sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}\, {\mathrm e}^{2 f x +2 e}}{f \left ({\mathrm e}^{2 f x +2 e}+1\right )^{3}}+\frac {3 i \ln \left ({\mathrm e}^{f x}-i {\mathrm e}^{-e}\right ) \sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}\, {\mathrm e}^{f x +e}}{2 f \left ({\mathrm e}^{2 f x +2 e}+1\right )}-\frac {3 i \ln \left ({\mathrm e}^{f x}+i {\mathrm e}^{-e}\right ) \sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}\, {\mathrm e}^{f x +e}}{2 f \left ({\mathrm e}^{2 f x +2 e}+1\right )}\)	\(290\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*(3*arctan(sinh(f*x+e))*cosh(f*x+e)^2-2*cosh(f*x+e)^2*sinh(f*x+e)-sinh(f*x+e))/cosh(f*x+e)/(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. 413 vs. \(2 (86) = 172\).
time = 0.53, size = 413, normalized size = 4.54 \begin {gather*} \frac {15 \, \sqrt {a} \arctan \left (e^{\left (-f x - e\right )}\right )}{8 \, f} + \frac {3 \, \sqrt {a} \arctan \left (e^{\left (-f x - e\right )}\right ) + \frac {5 \, \sqrt {a} e^{\left (-f x - e\right )} + 3 \, \sqrt {a} e^{\left (-3 \, f x - 3 \, e\right )}}{2 \, e^{\left (-2 \, f x - 2 \, e\right )} + e^{\left (-4 \, f x - 4 \, e\right )} + 1}}{4 \, f} + \frac {3 \, \sqrt {a} \arctan \left (e^{\left (-f x - e\right )}\right ) - \frac {3 \, \sqrt {a} e^{\left (-f x - e\right )} + 5 \, \sqrt {a} e^{\left (-3 \, f x - 3 \, e\right )}}{2 \, e^{\left (-2 \, f x - 2 \, e\right )} + e^{\left (-4 \, f x - 4 \, e\right )} + 1}}{4 \, f} - \frac {3 \, {\left (\sqrt {a} \arctan \left (e^{\left (-f x - e\right )}\right ) - \frac {\sqrt {a} e^{\left (-f x - e\right )} - \sqrt {a} e^{\left (-3 \, f x - 3 \, e\right )}}{2 \, e^{\left (-2 \, f x - 2 \, e\right )} + e^{\left (-4 \, f x - 4 \, e\right )} + 1}\right )}}{8 \, f} + \frac {25 \, \sqrt {a} e^{\left (-2 \, f x - 2 \, e\right )} + 15 \, \sqrt {a} e^{\left (-4 \, f x - 4 \, e\right )} + 8 \, \sqrt {a}}{16 \, f {\left (e^{\left (-f x - e\right )} + 2 \, e^{\left (-3 \, f x - 3 \, e\right )} + e^{\left (-5 \, f x - 5 \, e\right )}\right )}} - \frac {15 \, \sqrt {a} e^{\left (-f x - e\right )} + 25 \, \sqrt {a} e^{\left (-3 \, f x - 3 \, e\right )} + 8 \, \sqrt {a} e^{\left (-5 \, f x - 5 \, e\right )}}{16 \, f {\left (2 \, e^{\left (-2 \, f x - 2 \, e\right )} + e^{\left (-4 \, f x - 4 \, e\right )} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/8*sqrt(a)*arctan(e^(-f*x - e))/f + 1/4*(3*sqrt(a)*arctan(e^(-f*x - e)) + (5*sqrt(a)*e^(-f*x - e) + 3*sqrt(a
)*e^(-3*f*x - 3*e))/(2*e^(-2*f*x - 2*e) + e^(-4*f*x - 4*e) + 1))/f + 1/4*(3*sqrt(a)*arctan(e^(-f*x - e)) - (3*
sqrt(a)*e^(-f*x - e) + 5*sqrt(a)*e^(-3*f*x - 3*e))/(2*e^(-2*f*x - 2*e) + e^(-4*f*x - 4*e) + 1))/f - 3/8*(sqrt(
a)*arctan(e^(-f*x - e)) - (sqrt(a)*e^(-f*x - e) - sqrt(a)*e^(-3*f*x - 3*e))/(2*e^(-2*f*x - 2*e) + e^(-4*f*x -
4*e) + 1))/f + 1/16*(25*sqrt(a)*e^(-2*f*x - 2*e) + 15*sqrt(a)*e^(-4*f*x - 4*e) + 8*sqrt(a))/(f*(e^(-f*x - e) +
2*e^(-3*f*x - 3*e) + e^(-5*f*x - 5*e))) - 1/16*(15*sqrt(a)*e^(-f*x - e) + 25*sqrt(a)*e^(-3*f*x - 3*e) + 8*sqr
t(a)*e^(-5*f*x - 5*e))/(f*(2*e^(-2*f*x - 2*e) + e^(-4*f*x - 4*e) + 1))

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 742 vs. \(2 (79) = 158\).
time = 0.49, size = 742, normalized size = 8.15 \begin {gather*} \frac {{\left (6 \, \cosh \left (f x + e\right ) e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{5} + e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{6} + 3 \, {\left (5 \, \cosh \left (f x + e\right )^{2} + 1\right )} e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{4} + 4 \, {\left (5 \, \cosh \left (f x + e\right )^{3} + 3 \, \cosh \left (f x + e\right )\right )} e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{3} + 3 \, {\left (5 \, \cosh \left (f x + e\right )^{4} + 6 \, \cosh \left (f x + e\right )^{2} - 1\right )} e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{2} + 6 \, {\left (\cosh \left (f x + e\right )^{5} + 2 \, \cosh \left (f x + e\right )^{3} - \cosh \left (f x + e\right )\right )} e^{\left (f x + e\right )} \sinh \left (f x + e\right ) - 6 \, {\left (5 \, \cosh \left (f x + e\right ) e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{4} + e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{5} + 2 \, {\left (5 \, \cosh \left (f x + e\right )^{2} + 1\right )} e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{3} + 2 \, {\left (5 \, \cosh \left (f x + e\right )^{3} + 3 \, \cosh \left (f x + e\right )\right )} e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{2} + {\left (5 \, \cosh \left (f x + e\right )^{4} + 6 \, \cosh \left (f x + e\right )^{2} + 1\right )} e^{\left (f x + e\right )} \sinh \left (f x + e\right ) + {\left (\cosh \left (f x + e\right )^{5} + 2 \, \cosh \left (f x + e\right )^{3} + \cosh \left (f x + e\right )\right )} e^{\left (f x + e\right )}\right )} \arctan \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right ) + {\left (\cosh \left (f x + e\right )^{6} + 3 \, \cosh \left (f x + e\right )^{4} - 3 \, \cosh \left (f x + e\right )^{2} - 1\right )} e^{\left (f x + e\right )}\right )} \sqrt {a e^{\left (4 \, f x + 4 \, e\right )} + 2 \, a e^{\left (2 \, f x + 2 \, e\right )} + a} e^{\left (-f x - e\right )}}{2 \, {\left (f \cosh \left (f x + e\right )^{5} + {\left (f e^{\left (2 \, f x + 2 \, e\right )} + f\right )} \sinh \left (f x + e\right )^{5} + 5 \, {\left (f \cosh \left (f x + e\right ) e^{\left (2 \, f x + 2 \, e\right )} + f \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right )^{4} + 2 \, f \cosh \left (f x + e\right )^{3} + 2 \, {\left (5 \, f \cosh \left (f x + e\right )^{2} + {\left (5 \, f \cosh \left (f x + e\right )^{2} + f\right )} e^{\left (2 \, f x + 2 \, e\right )} + f\right )} \sinh \left (f x + e\right )^{3} + 2 \, {\left (5 \, f \cosh \left (f x + e\right )^{3} + 3 \, f \cosh \left (f x + e\right ) + {\left (5 \, f \cosh \left (f x + e\right )^{3} + 3 \, f \cosh \left (f x + e\right )\right )} e^{\left (2 \, f x + 2 \, e\right )}\right )} \sinh \left (f x + e\right )^{2} + f \cosh \left (f x + e\right ) + {\left (f \cosh \left (f x + e\right )^{5} + 2 \, f \cosh \left (f x + e\right )^{3} + f \cosh \left (f x + e\right )\right )} e^{\left (2 \, f x + 2 \, e\right )} + {\left (5 \, f \cosh \left (f x + e\right )^{4} + 6 \, f \cosh \left (f x + e\right )^{2} + {\left (5 \, f \cosh \left (f x + e\right )^{4} + 6 \, f \cosh \left (f x + e\right )^{2} + f\right )} e^{\left (2 \, f x + 2 \, e\right )} + f\right )} \sinh \left (f x + e\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(6*cosh(f*x + e)*e^(f*x + e)*sinh(f*x + e)^5 + e^(f*x + e)*sinh(f*x + e)^6 + 3*(5*cosh(f*x + e)^2 + 1)*e^(
f*x + e)*sinh(f*x + e)^4 + 4*(5*cosh(f*x + e)^3 + 3*cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e)^3 + 3*(5*cosh(f*x
+ e)^4 + 6*cosh(f*x + e)^2 - 1)*e^(f*x + e)*sinh(f*x + e)^2 + 6*(cosh(f*x + e)^5 + 2*cosh(f*x + e)^3 - cosh(f
*x + e))*e^(f*x + e)*sinh(f*x + e) - 6*(5*cosh(f*x + e)*e^(f*x + e)*sinh(f*x + e)^4 + e^(f*x + e)*sinh(f*x + e
)^5 + 2*(5*cosh(f*x + e)^2 + 1)*e^(f*x + e)*sinh(f*x + e)^3 + 2*(5*cosh(f*x + e)^3 + 3*cosh(f*x + e))*e^(f*x +
e)*sinh(f*x + e)^2 + (5*cosh(f*x + e)^4 + 6*cosh(f*x + e)^2 + 1)*e^(f*x + e)*sinh(f*x + e) + (cosh(f*x + e)^5
+ 2*cosh(f*x + e)^3 + cosh(f*x + e))*e^(f*x + e))*arctan(cosh(f*x + e) + sinh(f*x + e)) + (cosh(f*x + e)^6 +
3*cosh(f*x + e)^4 - 3*cosh(f*x + e)^2 - 1)*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)/(f*cosh(f*x + e)^5 + (f*e^(2*f*x + 2*e) + f)*sinh(f*x + e)^5 + 5*(f*cosh(f*x + e)*e^(2*f*x + 2*e) + f
*cosh(f*x + e))*sinh(f*x + e)^4 + 2*f*cosh(f*x + e)^3 + 2*(5*f*cosh(f*x + e)^2 + (5*f*cosh(f*x + e)^2 + f)*e^(
2*f*x + 2*e) + f)*sinh(f*x + e)^3 + 2*(5*f*cosh(f*x + e)^3 + 3*f*cosh(f*x + e) + (5*f*cosh(f*x + e)^3 + 3*f*co
sh(f*x + e))*e^(2*f*x + 2*e))*sinh(f*x + e)^2 + f*cosh(f*x + e) + (f*cosh(f*x + e)^5 + 2*f*cosh(f*x + e)^3 + f
*cosh(f*x + e))*e^(2*f*x + 2*e) + (5*f*cosh(f*x + e)^4 + 6*f*cosh(f*x + e)^2 + (5*f*cosh(f*x + e)^4 + 6*f*cosh
(f*x + e)^2 + f)*e^(2*f*x + 2*e) + f)*sinh(f*x + e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*(sinh(e + f*x)**2 + 1))*tanh(e + f*x)**4, x)

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Giac [A]
time = 0.46, size = 100, normalized size = 1.10 \begin {gather*} -\frac {{\left (3 \, \pi - \frac {4 \, {\left (e^{\left (f x + e\right )} - e^{\left (-f x - e\right )}\right )}}{{\left (e^{\left (f x + e\right )} - e^{\left (-f x - e\right )}\right )}^{2} + 4} + 6 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, f x + 2 \, e\right )} - 1\right )} e^{\left (-f x - e\right )}\right ) - 2 \, e^{\left (f x + e\right )} + 2 \, e^{\left (-f x - e\right )}\right )} \sqrt {a}}{4 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(3*pi - 4*(e^(f*x + e) - e^(-f*x - e))/((e^(f*x + e) - e^(-f*x - e))^2 + 4) + 6*arctan(1/2*(e^(2*f*x + 2*
e) - 1)*e^(-f*x - e)) - 2*e^(f*x + e) + 2*e^(-f*x - e))*sqrt(a)/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\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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