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

Optimal. Leaf size=62 \[ \frac {\sqrt {a+b \sinh ^2(e+f x)}}{f}-\frac {\sqrt {a-b} \tanh ^{-1}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a-b}}\right )}{f} \]

[Out]

-arctanh((a+b*sinh(f*x+e)^2)^(1/2)/(a-b)^(1/2))*(a-b)^(1/2)/f+(a+b*sinh(f*x+e)^2)^(1/2)/f

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Rubi [A]  time = 0.06, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3194, 50, 63, 208} \[ \frac {\sqrt {a+b \sinh ^2(e+f x)}}{f}-\frac {\sqrt {a-b} \tanh ^{-1}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a-b}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[a - b]*ArcTanh[Sqrt[a + b*Sinh[e + f*x]^2]/Sqrt[a - b]])/f) + Sqrt[a + b*Sinh[e + f*x]^2]/f

Rule 50

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 63

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 208

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

Rule 3194

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

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 65, normalized size = 1.05 \[ \frac {\sqrt {a+b \cosh ^2(e+f x)-b}-\sqrt {a-b} \tanh ^{-1}\left (\frac {\sqrt {a+b \cosh ^2(e+f x)-b}}{\sqrt {a-b}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.29, size = 624, normalized size = 10.06 \[ \left [\frac {\sqrt {a - b} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )} \log \left (\frac {b \cosh \left (f x + e\right )^{4} + 4 \, b \cosh \left (f x + e\right ) \sinh \left (f x + e\right )^{3} + b \sinh \left (f x + e\right )^{4} + 2 \, {\left (4 \, a - 3 \, b\right )} \cosh \left (f x + e\right )^{2} + 2 \, {\left (3 \, b \cosh \left (f x + e\right )^{2} + 4 \, a - 3 \, b\right )} \sinh \left (f x + e\right )^{2} - 4 \, \sqrt {2} \sqrt {a - b} \sqrt {\frac {b \cosh \left (f x + e\right )^{2} + b \sinh \left (f x + e\right )^{2} + 2 \, a - b}{\cosh \left (f x + e\right )^{2} - 2 \, \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + \sinh \left (f x + e\right )^{2}}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )} + 4 \, {\left (b \cosh \left (f x + e\right )^{3} + {\left (4 \, a - 3 \, b\right )} \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right ) + b}{\cosh \left (f x + e\right )^{4} + 4 \, \cosh \left (f x + e\right ) \sinh \left (f x + e\right )^{3} + \sinh \left (f x + e\right )^{4} + 2 \, {\left (3 \, \cosh \left (f x + e\right )^{2} + 1\right )} \sinh \left (f x + e\right )^{2} + 2 \, \cosh \left (f x + e\right )^{2} + 4 \, {\left (\cosh \left (f x + e\right )^{3} + \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right ) + 1}\right ) + \sqrt {2} \sqrt {\frac {b \cosh \left (f x + e\right )^{2} + b \sinh \left (f x + e\right )^{2} + 2 \, a - b}{\cosh \left (f x + e\right )^{2} - 2 \, \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + \sinh \left (f x + e\right )^{2}}}}{2 \, {\left (f \cosh \left (f x + e\right ) + f \sinh \left (f x + e\right )\right )}}, -\frac {2 \, \sqrt {-a + b} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )} \arctan \left (-\frac {\sqrt {2} \sqrt {-a + b} \sqrt {\frac {b \cosh \left (f x + e\right )^{2} + b \sinh \left (f x + e\right )^{2} + 2 \, a - b}{\cosh \left (f x + e\right )^{2} - 2 \, \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + \sinh \left (f x + e\right )^{2}}}}{2 \, {\left ({\left (a - b\right )} \cosh \left (f x + e\right ) + {\left (a - b\right )} \sinh \left (f x + e\right )\right )}}\right ) - \sqrt {2} \sqrt {\frac {b \cosh \left (f x + e\right )^{2} + b \sinh \left (f x + e\right )^{2} + 2 \, a - b}{\cosh \left (f x + e\right )^{2} - 2 \, \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + \sinh \left (f x + e\right )^{2}}}}{2 \, {\left (f \cosh \left (f x + e\right ) + f \sinh \left (f x + e\right )\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(a - b)*(cosh(f*x + e) + sinh(f*x + e))*log((b*cosh(f*x + e)^4 + 4*b*cosh(f*x + e)*sinh(f*x + e)^3 +
b*sinh(f*x + e)^4 + 2*(4*a - 3*b)*cosh(f*x + e)^2 + 2*(3*b*cosh(f*x + e)^2 + 4*a - 3*b)*sinh(f*x + e)^2 - 4*s
qrt(2)*sqrt(a - b)*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*s
inh(f*x + e) + sinh(f*x + e)^2))*(cosh(f*x + e) + sinh(f*x + e)) + 4*(b*cosh(f*x + e)^3 + (4*a - 3*b)*cosh(f*x
+ e))*sinh(f*x + e) + b)/(cosh(f*x + e)^4 + 4*cosh(f*x + e)*sinh(f*x + e)^3 + sinh(f*x + e)^4 + 2*(3*cosh(f*x
+ e)^2 + 1)*sinh(f*x + e)^2 + 2*cosh(f*x + e)^2 + 4*(cosh(f*x + e)^3 + cosh(f*x + e))*sinh(f*x + e) + 1)) + s
qrt(2)*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e)
+ sinh(f*x + e)^2)))/(f*cosh(f*x + e) + f*sinh(f*x + e)), -1/2*(2*sqrt(-a + b)*(cosh(f*x + e) + sinh(f*x + e)
)*arctan(-1/2*sqrt(2)*sqrt(-a + b)*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2
*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2))/((a - b)*cosh(f*x + e) + (a - b)*sinh(f*x + e))) - sqrt(2)*sq
rt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f
*x + e)^2)))/(f*cosh(f*x + e) + f*sinh(f*x + e))]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, need to choose a branch for the root of a polynomial with parameters. This might be wrong.Non regular val
ue [0] was discarded and replaced randomly by 0=[-27]Warning, need to choose a branch for the root of a polyno
mial with parameters. This might be wrong.Non regular value [0] was discarded and replaced randomly by 0=[14]W
arning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.Non regular
value [0] was discarded and replaced randomly by 0=[-85]Warning, need to choose a branch for the root of a pol
ynomial with parameters. This might be wrong.Non regular value [0] was discarded and replaced randomly by 0=[-
68]Precision problem choosing root in common_EXT, current precision 14Warning, need to choose a branch for the
root of a polynomial with parameters. This might be wrong.Non regular value [0] was discarded and replaced ra
ndomly by 0=[-96]Warning, need to choose a branch for the root of a polynomial with parameters. This might be
wrong.Non regular value [0] was discarded and replaced randomly by 0=[86]Warning, need to choose a branch for
the root of a polynomial with parameters. This might be wrong.Non regular value [0] was discarded and replaced
randomly by 0=[22]Warning, need to choose a branch for the root of a polynomial with parameters. This might b
e wrong.Non regular value [0] was discarded and replaced randomly by 0=[98]Evaluation time: 0.57index.cc index
_m operator + Error: Bad Argument Value

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maple [C]  time = 0.17, size = 41, normalized size = 0.66 \[ \frac {\mathit {`\,int/indef0`\,}\left (\frac {\sqrt {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}\, \sinh \left (f x +e \right )}{\cosh \left (f x +e \right )^{2}}, \sinh \left (f x +e \right )\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*sinh(f*x + e)^2 + a)*tanh(f*x + e), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \mathrm {tanh}\left (e+f\,x\right )\,\sqrt {b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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