3.427 \(\int \sqrt {a+a \sinh ^2(e+f x)} \tanh ^3(e+f x) \, dx\)

Optimal. Leaf size=38 \[ \frac {a}{f \sqrt {a \cosh ^2(e+f x)}}+\frac {\sqrt {a \cosh ^2(e+f x)}}{f} \]

[Out]

a/f/(a*cosh(f*x+e)^2)^(1/2)+(a*cosh(f*x+e)^2)^(1/2)/f

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Rubi [A]  time = 0.11, antiderivative size = 38, 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 = {3176, 3205, 16, 43} \[ \frac {a}{f \sqrt {a \cosh ^2(e+f x)}}+\frac {\sqrt {a \cosh ^2(e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Sinh[e + f*x]^2]*Tanh[e + f*x]^3,x]

[Out]

a/(f*Sqrt[a*Cosh[e + f*x]^2]) + Sqrt[a*Cosh[e + f*x]^2]/f

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 3205

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFact
ors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(b*ff^(n/2)*x^(n/2))^p)/(1 - ff*x
)^((m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{b, e, f, p}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2
]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 29, normalized size = 0.76 \[ \frac {a \left (\cosh ^2(e+f x)+1\right )}{f \sqrt {a \cosh ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + a*Sinh[e + f*x]^2]*Tanh[e + f*x]^3,x]

[Out]

(a*(1 + Cosh[e + f*x]^2))/(f*Sqrt[a*Cosh[e + f*x]^2])

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fricas [B]  time = 0.70, size = 311, normalized size = 8.18 \[ \frac {{\left (4 \, \cosh \left (f x + e\right ) e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{3} + e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{4} + 6 \, {\left (\cosh \left (f x + e\right )^{2} + 1\right )} e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{2} + 4 \, {\left (\cosh \left (f x + e\right )^{3} + 3 \, \cosh \left (f x + e\right )\right )} e^{\left (f x + e\right )} \sinh \left (f x + e\right ) + {\left (\cosh \left (f x + e\right )^{4} + 6 \, \cosh \left (f x + e\right )^{2} + 1\right )} e^{\left (f x + e\right )}\right )} \sqrt {a e^{\left (4 \, f x + 4 \, e\right )} + 2 \, a e^{\left (2 \, f x + 2 \, e\right )} + a} e^{\left (-f x - e\right )}}{2 \, {\left (f \cosh \left (f x + e\right )^{3} + {\left (f e^{\left (2 \, f x + 2 \, e\right )} + f\right )} \sinh \left (f x + e\right )^{3} + 3 \, {\left (f \cosh \left (f x + e\right ) e^{\left (2 \, f x + 2 \, e\right )} + f \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right )^{2} + f \cosh \left (f x + e\right ) + {\left (f \cosh \left (f x + e\right )^{3} + f \cosh \left (f x + e\right )\right )} e^{\left (2 \, f x + 2 \, e\right )} + {\left (3 \, f \cosh \left (f x + e\right )^{2} + {\left (3 \, f \cosh \left (f x + e\right )^{2} + f\right )} e^{\left (2 \, f x + 2 \, e\right )} + f\right )} \sinh \left (f x + e\right )\right )}} \]

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)^3,x, algorithm="fricas")

[Out]

1/2*(4*cosh(f*x + e)*e^(f*x + e)*sinh(f*x + e)^3 + e^(f*x + e)*sinh(f*x + e)^4 + 6*(cosh(f*x + e)^2 + 1)*e^(f*
x + e)*sinh(f*x + e)^2 + 4*(cosh(f*x + e)^3 + 3*cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e) + (cosh(f*x + e)^4 +
6*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)^3 + (f*e^(2*f*x + 2*e) + f)*sinh(f*x + e)^3 + 3*(f*cosh(f*x + e)*e^(2*f*x + 2*e) + f*cosh(f*x + e))*sinh
(f*x + e)^2 + f*cosh(f*x + e) + (f*cosh(f*x + e)^3 + f*cosh(f*x + e))*e^(2*f*x + 2*e) + (3*f*cosh(f*x + e)^2 +
(3*f*cosh(f*x + e)^2 + f)*e^(2*f*x + 2*e) + f)*sinh(f*x + e))

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giac [A]  time = 0.19, size = 53, normalized size = 1.39 \[ \frac {\sqrt {a} {\left (\frac {{\left (5 \, e^{\left (2 \, f x + 2 \, e\right )} + 1\right )} e^{\left (-e\right )}}{e^{\left (3 \, f x + 2 \, e\right )} + e^{\left (f x\right )}} + e^{\left (f x + e\right )}\right )}}{2 \, f} \]

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)^3,x, algorithm="giac")

[Out]

1/2*sqrt(a)*((5*e^(2*f*x + 2*e) + 1)*e^(-e)/(e^(3*f*x + 2*e) + e^(f*x)) + e^(f*x + e))/f

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maple [C]  time = 0.22, size = 42, normalized size = 1.11 \[ \frac {\mathit {`\,int/indef0`\,}\left (\frac {\left (\sinh ^{3}\left (f x +e \right )\right ) a}{\cosh \left (f x +e \right )^{2} \sqrt {a \left (\cosh ^{2}\left (f x +e \right )\right )}}, \sinh \left (f x +e \right )\right )}{f} \]

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)^3,x)

[Out]

`int/indef0`(sinh(f*x+e)^3*a/cosh(f*x+e)^2/(a*cosh(f*x+e)^2)^(1/2),sinh(f*x+e))/f

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maxima [B]  time = 1.19, size = 106, normalized size = 2.79 \[ \frac {3 \, \sqrt {a} e^{\left (-2 \, f x - 2 \, e\right )}}{f {\left (e^{\left (-f x - e\right )} + e^{\left (-3 \, f x - 3 \, e\right )}\right )}} + \frac {\sqrt {a} e^{\left (-4 \, f x - 4 \, e\right )}}{2 \, f {\left (e^{\left (-f x - e\right )} + e^{\left (-3 \, f x - 3 \, e\right )}\right )}} + \frac {\sqrt {a}}{2 \, f {\left (e^{\left (-f x - e\right )} + e^{\left (-3 \, f x - 3 \, e\right )}\right )}} \]

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)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.91, size = 67, normalized size = 1.76 \[ \frac {\sqrt {a+a\,{\left (\frac {{\mathrm {e}}^{e+f\,x}}{2}-\frac {{\mathrm {e}}^{-e-f\,x}}{2}\right )}^2}\,\left (6\,{\mathrm {e}}^{2\,e+2\,f\,x}+{\mathrm {e}}^{4\,e+4\,f\,x}+1\right )}{f\,{\left ({\mathrm {e}}^{2\,e+2\,f\,x}+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(e + f*x)^3*(a + a*sinh(e + f*x)^2)^(1/2),x)

[Out]

((a + a*(exp(e + f*x)/2 - exp(- e - f*x)/2)^2)^(1/2)*(6*exp(2*e + 2*f*x) + exp(4*e + 4*f*x) + 1))/(f*(exp(2*e
+ 2*f*x) + 1)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \left (\sinh ^{2}{\left (e + f x \right )} + 1\right )} \tanh ^{3}{\left (e + f x \right )}\, dx \]

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)**3,x)

[Out]

Integral(sqrt(a*(sinh(e + f*x)**2 + 1))*tanh(e + f*x)**3, x)

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