3.384 \(\int \frac{\text{sech}(e+f x)}{(a+b \sinh ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=85 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \sinh (e+f x)}{\sqrt{a+b \sinh ^2(e+f x)}}\right )}{f (a-b)^{3/2}}-\frac{b \sinh (e+f x)}{a f (a-b) \sqrt{a+b \sinh ^2(e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.0965584, antiderivative size = 85, normalized size of antiderivative = 1., 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 = {3190, 382, 377, 203} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \sinh (e+f x)}{\sqrt{a+b \sinh ^2(e+f x)}}\right )}{f (a-b)^{3/2}}-\frac{b \sinh (e+f x)}{a f (a-b) \sqrt{a+b \sinh ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3190

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]

Rule 382

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] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rule 377

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 203

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

Rubi steps

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

Mathematica [C]  time = 7.55193, size = 315, normalized size = 3.71 \[ \frac{\tanh (e+f x) \text{sech}^7(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \left (4 (a-b)^2 \sinh ^4(e+f x) \left (a+b \sinh ^2(e+f x)\right ) \, _2F_1\left (2,2;\frac{7}{2};\frac{(a-b) \tanh ^2(e+f x)}{a}\right ) \sqrt{\frac{\tanh ^2(e+f x) \text{sech}^2(e+f x) \left (a^2+a b \left (\sinh ^2(e+f x)-1\right )-b^2 \sinh ^2(e+f x)\right )}{a^2}}+15 a \cosh ^2(e+f x) \left (3 a+2 b \sinh ^2(e+f x)\right ) \left (a \cosh ^2(e+f x) \sqrt{\frac{\tanh ^2(e+f x) \text{sech}^2(e+f x) \left (a^2+a b \left (\sinh ^2(e+f x)-1\right )-b^2 \sinh ^2(e+f x)\right )}{a^2}}-\left (a+b \sinh ^2(e+f x)\right ) \sin ^{-1}\left (\sqrt{\frac{(a-b) \tanh ^2(e+f x)}{a}}\right )\right )\right )}{15 a^5 f \left (\frac{(a-b) \tanh ^2(e+f x) \text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a^2}\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sech[e + f*x]^7*Sqrt[a + b*Sinh[e + f*x]^2]*Tanh[e + f*x]*(4*(a - b)^2*Hypergeometric2F1[2, 2, 7/2, ((a - b)*
Tanh[e + f*x]^2)/a]*Sinh[e + f*x]^4*(a + 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] + 15*a*Cosh[e + f*x]^2*(3*a + 2*b*Sinh[e + f*x]^2)*(-(ArcSi
n[Sqrt[((a - b)*Tanh[e + f*x]^2)/a]]*(a + b*Sinh[e + f*x]^2)) + a*Cosh[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])))/(15*a^5*f*(((a - b)*Sech[e + f*x]^
2*(a + b*Sinh[e + f*x]^2)*Tanh[e + f*x]^2)/a^2)^(3/2))

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Maple [C]  time = 0.115, size = 101, normalized size = 1.2 \begin{align*}{\frac{1}{f}\mbox{{\tt ` int/indef0`}} \left ({\frac{-b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}-a}{-{b}^{2} \left ( \sinh \left ( fx+e \right ) \right ) ^{6}+ \left ( -2\,ab-{b}^{2} \right ) \left ( \sinh \left ( fx+e \right ) \right ) ^{4}+ \left ( -{a}^{2}-2\,ab \right ) \left ( \sinh \left ( fx+e \right ) \right ) ^{2}-{a}^{2}}{\frac{1}{\sqrt{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}}}},\sinh \left ( fx+e \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

`int/indef0`((-b*sinh(f*x+e)^2-a)/(-b^2*sinh(f*x+e)^6+(-2*a*b-b^2)*sinh(f*x+e)^4+(-a^2-2*a*b)*sinh(f*x+e)^2-a^
2)/(a+b*sinh(f*x+e)^2)^(1/2),sinh(f*x+e))/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}\left (f x + e\right )}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.48192, size = 4262, normalized size = 50.14 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((a*b*cosh(f*x + e)^4 + 4*a*b*cosh(f*x + e)*sinh(f*x + e)^3 + a*b*sinh(f*x + e)^4 + 2*(2*a^2 - a*b)*cosh(
f*x + e)^2 + 2*(3*a*b*cosh(f*x + e)^2 + 2*a^2 - a*b)*sinh(f*x + e)^2 + a*b + 4*(a*b*cosh(f*x + e)^3 + (2*a^2 -
a*b)*cosh(f*x + e))*sinh(f*x + e))*sqrt(-a + b)*log(((a - 2*b)*cosh(f*x + e)^4 + 4*(a - 2*b)*cosh(f*x + e)*si
nh(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 - b^2)*co
sh(f*x + e)^2 + 2*(a*b - b^2)*cosh(f*x + e)*sinh(f*x + e) + (a*b - b^2)*sinh(f*x + e)^2 - a*b + b^2)*sqrt((b*c
osh(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^3*b - 2*a^2*b^2 + a*b^3)*f*cosh(f*x + e)^4 + 4*(a^3*b - 2*a^2*b^2 + a*b^3)*f*cosh(f*x + e)*sinh(f*x
+ e)^3 + (a^3*b - 2*a^2*b^2 + a*b^3)*f*sinh(f*x + e)^4 + 2*(2*a^4 - 5*a^3*b + 4*a^2*b^2 - a*b^3)*f*cosh(f*x +
e)^2 + 2*(3*(a^3*b - 2*a^2*b^2 + a*b^3)*f*cosh(f*x + e)^2 + (2*a^4 - 5*a^3*b + 4*a^2*b^2 - a*b^3)*f)*sinh(f*x
+ e)^2 + (a^3*b - 2*a^2*b^2 + a*b^3)*f + 4*((a^3*b - 2*a^2*b^2 + a*b^3)*f*cosh(f*x + e)^3 + (2*a^4 - 5*a^3*b +
4*a^2*b^2 - a*b^3)*f*cosh(f*x + e))*sinh(f*x + e)), ((a*b*cosh(f*x + e)^4 + 4*a*b*cosh(f*x + e)*sinh(f*x + e)
^3 + a*b*sinh(f*x + e)^4 + 2*(2*a^2 - a*b)*cosh(f*x + e)^2 + 2*(3*a*b*cosh(f*x + e)^2 + 2*a^2 - a*b)*sinh(f*x
+ e)^2 + a*b + 4*(a*b*cosh(f*x + e)^3 + (2*a^2 - a*b)*cosh(f*x + e))*sinh(f*x + e))*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)) - sq
rt(2)*((a*b - b^2)*cosh(f*x + e)^2 + 2*(a*b - b^2)*cosh(f*x + e)*sinh(f*x + e) + (a*b - b^2)*sinh(f*x + e)^2 -
a*b + b^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)))/((a^3*b - 2*a^2*b^2 + a*b^3)*f*cosh(f*x + e)^4 + 4*(a^3*b - 2*a^2*b^2 + a*b^3)*f*co
sh(f*x + e)*sinh(f*x + e)^3 + (a^3*b - 2*a^2*b^2 + a*b^3)*f*sinh(f*x + e)^4 + 2*(2*a^4 - 5*a^3*b + 4*a^2*b^2 -
a*b^3)*f*cosh(f*x + e)^2 + 2*(3*(a^3*b - 2*a^2*b^2 + a*b^3)*f*cosh(f*x + e)^2 + (2*a^4 - 5*a^3*b + 4*a^2*b^2
- a*b^3)*f)*sinh(f*x + e)^2 + (a^3*b - 2*a^2*b^2 + a*b^3)*f + 4*((a^3*b - 2*a^2*b^2 + a*b^3)*f*cosh(f*x + e)^3
+ (2*a^4 - 5*a^3*b + 4*a^2*b^2 - a*b^3)*f*cosh(f*x + e))*sinh(f*x + e))]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.47401, size = 385, normalized size = 4.53 \begin{align*} -\frac{\frac{{\left (a^{4} b f - 3 \, a^{3} b^{2} f + 3 \, a^{2} b^{3} f - a b^{4} f\right )} e^{\left (2 \, f x + 2 \, e\right )}}{a^{4} b^{3} - 2 \, a^{3} b^{4} + a^{2} b^{5}} - \frac{a^{4} b f - 3 \, a^{3} b^{2} f + 3 \, a^{2} b^{3} f - a b^{4} f}{a^{4} b^{3} - 2 \, a^{3} b^{4} + a^{2} b^{5}}}{8 \, \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}} + \frac{2 \, \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 f - b f\right )} \sqrt{a - b}} - \frac{a \sqrt{b} f - b^{\frac{3}{2}} f}{8 \, a b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((a^4*b*f - 3*a^3*b^2*f + 3*a^2*b^3*f - a*b^4*f)*e^(2*f*x + 2*e)/(a^4*b^3 - 2*a^3*b^4 + a^2*b^5) - (a^4*b
*f - 3*a^3*b^2*f + 3*a^2*b^3*f - a*b^4*f)/(a^4*b^3 - 2*a^3*b^4 + a^2*b^5))/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*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*f - b*f)*sqrt(a - b)) - 1/8*(a*sqrt(b)*f
- b^(3/2)*f)/(a*b^3)