3.4.76 \(\int \frac {\text {sech}^3(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx\) [376]

Optimal. Leaf size=97 \[ \frac {(a-2 b) \text {ArcTan}\left (\frac {\sqrt {a-b} \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}\right )}{2 (a-b)^{3/2} f}+\frac {\text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} \tanh (e+f x)}{2 (a-b) f} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3269, 390, 385, 209} \begin {gather*} \frac {(a-2 b) \text {ArcTan}\left (\frac {\sqrt {a-b} \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}\right )}{2 f (a-b)^{3/2}}+\frac {\tanh (e+f x) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{2 f (a-b)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sech[e + f*x]^3/Sqrt[a + b*Sinh[e + f*x]^2],x]

[Out]

((a - 2*b)*ArcTan[(Sqrt[a - b]*Sinh[e + f*x])/Sqrt[a + b*Sinh[e + f*x]^2]])/(2*(a - b)^(3/2)*f) + (Sech[e + f*
x]*Sqrt[a + b*Sinh[e + f*x]^2]*Tanh[e + f*x])/(2*(a - b)*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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a
*d)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && Eq
Q[n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rule 3269

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

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 8.69, size = 443, normalized size = 4.57 \begin {gather*} \frac {\text {sech}^3(e+f x) \left (1+\frac {b \sinh ^2(e+f x)}{a}\right ) \tanh (e+f x) \left (45 a \text {ArcSin}\left (\sqrt {\frac {(a-b) \tanh ^2(e+f x)}{a}}\right )+30 b \text {ArcSin}\left (\sqrt {\frac {(a-b) \tanh ^2(e+f x)}{a}}\right ) \sinh ^2(e+f x)+16 a \, _2F_1\left (2,3;\frac {7}{2};\frac {(a-b) \tanh ^2(e+f x)}{a}\right ) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}} \left (\frac {(a-b) \tanh ^2(e+f x)}{a}\right )^{5/2}+16 b \, _2F_1\left (2,3;\frac {7}{2};\frac {(a-b) \tanh ^2(e+f x)}{a}\right ) \sinh ^2(e+f x) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}} \left (\frac {(a-b) \tanh ^2(e+f x)}{a}\right )^{5/2}-45 a \sqrt {\frac {\text {sech}^2(e+f x) \left (a^2-b^2 \sinh ^2(e+f x)+a b \left (-1+\sinh ^2(e+f x)\right )\right ) \tanh ^2(e+f x)}{a^2}}-30 b \sinh ^2(e+f x) \sqrt {\frac {\text {sech}^2(e+f x) \left (a^2-b^2 \sinh ^2(e+f x)+a b \left (-1+\sinh ^2(e+f x)\right )\right ) \tanh ^2(e+f x)}{a^2}}\right )}{30 a f \sqrt {a+b \sinh ^2(e+f x)} \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}} \left (\frac {(a-b) \tanh ^2(e+f x)}{a}\right )^{3/2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Sech[e + f*x]^3/Sqrt[a + b*Sinh[e + f*x]^2],x]

[Out]

(Sech[e + f*x]^3*(1 + (b*Sinh[e + f*x]^2)/a)*Tanh[e + f*x]*(45*a*ArcSin[Sqrt[((a - b)*Tanh[e + f*x]^2)/a]] + 3
0*b*ArcSin[Sqrt[((a - b)*Tanh[e + f*x]^2)/a]]*Sinh[e + f*x]^2 + 16*a*Hypergeometric2F1[2, 3, 7/2, ((a - b)*Tan
h[e + f*x]^2)/a]*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]*(((a - b)*Tanh[e + f*x]^2)/a)^(5/2) + 16*b*
Hypergeometric2F1[2, 3, 7/2, ((a - b)*Tanh[e + f*x]^2)/a]*Sinh[e + f*x]^2*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e
+ f*x]^2))/a]*(((a - b)*Tanh[e + f*x]^2)/a)^(5/2) - 45*a*Sqrt[(Sech[e + f*x]^2*(a^2 - b^2*Sinh[e + f*x]^2 + a*
b*(-1 + Sinh[e + f*x]^2))*Tanh[e + f*x]^2)/a^2] - 30*b*Sinh[e + f*x]^2*Sqrt[(Sech[e + f*x]^2*(a^2 - b^2*Sinh[e
+ f*x]^2 + a*b*(-1 + Sinh[e + f*x]^2))*Tanh[e + f*x]^2)/a^2]))/(30*a*f*Sqrt[a + b*Sinh[e + f*x]^2]*Sqrt[(Sech
[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]*(((a - b)*Tanh[e + f*x]^2)/a)^(3/2))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 1.95, size = 35, normalized size = 0.36

method	result	size
default	\(\frac {\mathit {`\,int/indef0`\,}\left (\frac {1}{\cosh \left (f x +e \right )^{4} \sqrt {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}}, \sinh \left (f x +e \right )\right )}{f}\)	\(35\)
risch	\(\text {Expression too large to display}\)	\(450266\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

`int/indef0`(1/cosh(f*x+e)^4/(a+b*sinh(f*x+e)^2)^(1/2),sinh(f*x+e))/f

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sech(f*x + e)^3/sqrt(b*sinh(f*x + e)^2 + a), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 693 vs. \(2 (85) = 170\).
time = 0.48, size = 1503, normalized size = 15.49 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[-1/4*(((a - 2*b)*cosh(f*x + e)^4 + 4*(a - 2*b)*cosh(f*x + e)*sinh(f*x + e)^3 + (a - 2*b)*sinh(f*x + e)^4 + 2*
(a - 2*b)*cosh(f*x + e)^2 + 2*(3*(a - 2*b)*cosh(f*x + e)^2 + a - 2*b)*sinh(f*x + e)^2 + 4*((a - 2*b)*cosh(f*x
+ e)^3 + (a - 2*b)*cosh(f*x + e))*sinh(f*x + e) + a - 2*b)*sqrt(-a + b)*log(((a - 2*b)*cosh(f*x + e)^4 + 4*(a
- 2*b)*cosh(f*x + e)*sinh(f*x + e)^3 + (a - 2*b)*sinh(f*x + e)^4 - 2*(3*a - 2*b)*cosh(f*x + e)^2 + 2*(3*(a - 2
*b)*cosh(f*x + e)^2 - 3*a + 2*b)*sinh(f*x + e)^2 - 2*sqrt(2)*(cosh(f*x + e)^2 + 2*cosh(f*x + e)*sinh(f*x + e)
+ sinh(f*x + e)^2 - 1)*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)) + 4*((a - 2*b)*cosh(f*x + e)^3 - (3*a - 2*b)*cosh(f*x + e))*
sinh(f*x + e) + a - 2*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)) - 2*
sqrt(2)*((a - b)*cosh(f*x + e)^2 + 2*(a - b)*cosh(f*x + e)*sinh(f*x + e) + (a - b)*sinh(f*x + e)^2 - a + b)*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)))/((a^2 - 2*a*b + b^2)*f*cosh(f*x + e)^4 + 4*(a^2 - 2*a*b + b^2)*f*cosh(f*x + e)*sinh(f*x + e)^3 +
(a^2 - 2*a*b + b^2)*f*sinh(f*x + e)^4 + 2*(a^2 - 2*a*b + b^2)*f*cosh(f*x + e)^2 + 2*(3*(a^2 - 2*a*b + b^2)*f*c
osh(f*x + e)^2 + (a^2 - 2*a*b + b^2)*f)*sinh(f*x + e)^2 + (a^2 - 2*a*b + b^2)*f + 4*((a^2 - 2*a*b + b^2)*f*cos
h(f*x + e)^3 + (a^2 - 2*a*b + b^2)*f*cosh(f*x + e))*sinh(f*x + e)), 1/2*(((a - 2*b)*cosh(f*x + e)^4 + 4*(a - 2
*b)*cosh(f*x + e)*sinh(f*x + e)^3 + (a - 2*b)*sinh(f*x + e)^4 + 2*(a - 2*b)*cosh(f*x + e)^2 + 2*(3*(a - 2*b)*c
osh(f*x + e)^2 + a - 2*b)*sinh(f*x + e)^2 + 4*((a - 2*b)*cosh(f*x + e)^3 + (a - 2*b)*cosh(f*x + e))*sinh(f*x +
e) + a - 2*b)*sqrt(a - b)*arctan(sqrt(2)*(cosh(f*x + e)^2 + 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2 -
1)*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))/(b*cosh(f*x + e)^4 + 4*b*cosh(f*x + e)*sinh(f*x + e)^3 + b*sinh(f*x + e)^4 + 2*(
2*a - b)*cosh(f*x + e)^2 + 2*(3*b*cosh(f*x + e)^2 + 2*a - b)*sinh(f*x + e)^2 + 4*(b*cosh(f*x + e)^3 + (2*a - b
)*cosh(f*x + e))*sinh(f*x + e) + b)) + sqrt(2)*((a - b)*cosh(f*x + e)^2 + 2*(a - b)*cosh(f*x + e)*sinh(f*x + e
) + (a - b)*sinh(f*x + e)^2 - 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^2 - 2*a*b + b^2)*f*cosh(f*x + e)^4 + 4*(a^2 - 2*a*b +
b^2)*f*cosh(f*x + e)*sinh(f*x + e)^3 + (a^2 - 2*a*b + b^2)*f*sinh(f*x + e)^4 + 2*(a^2 - 2*a*b + b^2)*f*cosh(f*
x + e)^2 + 2*(3*(a^2 - 2*a*b + b^2)*f*cosh(f*x + e)^2 + (a^2 - 2*a*b + b^2)*f)*sinh(f*x + e)^2 + (a^2 - 2*a*b
+ b^2)*f + 4*((a^2 - 2*a*b + b^2)*f*cosh(f*x + e)^3 + (a^2 - 2*a*b + b^2)*f*cosh(f*x + e))*sinh(f*x + e))]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {sech}^{3}{\left (e + f x \right )}}{\sqrt {a + b \sinh ^{2}{\left (e + f x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(f*x+e)**3/(a+b*sinh(f*x+e)**2)**(1/2),x)

[Out]

Integral(sech(e + f*x)**3/sqrt(a + b*sinh(e + f*x)**2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 693 vs. \(2 (85) = 170\).
time = 0.55, size = 693, normalized size = 7.14 \begin {gather*} \frac {{\left (\frac {{\left (a - 2 \, b\right )} \arctan \left (-\frac {\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b} + \sqrt {b}}{2 \, \sqrt {a - b}}\right )}{{\left (a e^{\left (4 \, e\right )} - b e^{\left (4 \, e\right )}\right )} \sqrt {a - b}} - \frac {2 \, {\left ({\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )}^{3} a - 2 \, {\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )}^{3} b - 5 \, {\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )}^{2} a \sqrt {b} + 2 \, {\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )}^{2} b^{\frac {3}{2}} - 4 \, {\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )} a^{2} - {\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )} a b + 2 \, {\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )} b^{2} - 4 \, a^{2} \sqrt {b} + 5 \, a b^{\frac {3}{2}} - 2 \, b^{\frac {5}{2}}\right )}}{{\left ({\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )}^{2} + 2 \, {\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )} \sqrt {b} + 4 \, a - 3 \, b\right )}^{2} {\left (a e^{\left (4 \, e\right )} - b e^{\left (4 \, e\right )}\right )}}\right )} e^{\left (4 \, e\right )}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

((a - 2*b)*arctan(-1/2*(sqrt(b)*e^(2*f*x + 2*e) - sqrt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x
+ 2*e) + b) + sqrt(b))/sqrt(a - b))/((a*e^(4*e) - b*e^(4*e))*sqrt(a - b)) - 2*((sqrt(b)*e^(2*f*x + 2*e) - sqrt
(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))^3*a - 2*(sqrt(b)*e^(2*f*x + 2*e) - sqrt(b
*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))^3*b - 5*(sqrt(b)*e^(2*f*x + 2*e) - sqrt(b*e
^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))^2*a*sqrt(b) + 2*(sqrt(b)*e^(2*f*x + 2*e) - sq
rt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))^2*b^(3/2) - 4*(sqrt(b)*e^(2*f*x + 2*e)
- sqrt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))*a^2 - (sqrt(b)*e^(2*f*x + 2*e) - sq
rt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))*a*b + 2*(sqrt(b)*e^(2*f*x + 2*e) - sqrt
(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))*b^2 - 4*a^2*sqrt(b) + 5*a*b^(3/2) - 2*b^(
5/2))/(((sqrt(b)*e^(2*f*x + 2*e) - sqrt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))^2
+ 2*(sqrt(b)*e^(2*f*x + 2*e) - sqrt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))*sqrt(b
) + 4*a - 3*b)^2*(a*e^(4*e) - b*e^(4*e))))*e^(4*e)/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\mathrm {cosh}\left (e+f\,x\right )}^3\,\sqrt {b\,{\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(1/(cosh(e + f*x)^3*(a + b*sinh(e + f*x)^2)^(1/2)),x)

[Out]

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

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