3.4.85 \(\int \frac {\text {sech}^3(e+f x)}{(a+b \sinh ^2(e+f x))^{3/2}} \, dx\) [385]

Optimal. Leaf size=142 \[ \frac {(a-4 b) \text {ArcTan}\left (\frac {\sqrt {a-b} \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}\right )}{2 (a-b)^{5/2} f}+\frac {b (a+2 b) \sinh (e+f x)}{2 a (a-b)^2 f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\text {sech}(e+f x) \tanh (e+f x)}{2 (a-b) f \sqrt {a+b \sinh ^2(e+f x)}} \]

[Out]

1/2*(a-4*b)*arctan(sinh(f*x+e)*(a-b)^(1/2)/(a+b*sinh(f*x+e)^2)^(1/2))/(a-b)^(5/2)/f+1/2*b*(a+2*b)*sinh(f*x+e)/
a/(a-b)^2/f/(a+b*sinh(f*x+e)^2)^(1/2)+1/2*sech(f*x+e)*tanh(f*x+e)/(a-b)/f/(a+b*sinh(f*x+e)^2)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.11, antiderivative size = 142, 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 = {3269, 425, 541, 12, 385, 209} \begin {gather*} \frac {(a-4 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)^{5/2}}+\frac {b (a+2 b) \sinh (e+f x)}{2 a f (a-b)^2 \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\tanh (e+f x) \text {sech}(e+f x)}{2 f (a-b) \sqrt {a+b \sinh ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a - 4*b)*ArcTan[(Sqrt[a - b]*Sinh[e + f*x])/Sqrt[a + b*Sinh[e + f*x]^2]])/(2*(a - b)^(5/2)*f) + (b*(a + 2*b)
*Sinh[e + f*x])/(2*a*(a - b)^2*f*Sqrt[a + b*Sinh[e + f*x]^2]) + (Sech[e + f*x]*Tanh[e + f*x])/(2*(a - b)*f*Sqr
t[a + b*Sinh[e + f*x]^2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 425

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[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1
)*(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomi
alQ[a, b, c, d, n, p, q, x]

Rule 541

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(
-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(b*c - a*d)*(p + 1))), x] + Dist[1/(a*n*(b*c - a
*d)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*
f)*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[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)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx &=\frac {\text {Subst}\left (\int \frac {1}{\left (1+x^2\right )^2 \left (a+b x^2\right )^{3/2}} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=\frac {\text {sech}(e+f x) \tanh (e+f x)}{2 (a-b) f \sqrt {a+b \sinh ^2(e+f x)}}-\frac {\text {Subst}\left (\int \frac {-a+2 b-2 b x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\sinh (e+f x)\right )}{2 (a-b) f}\\ &=\frac {b (a+2 b) \sinh (e+f x)}{2 a (a-b)^2 f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\text {sech}(e+f x) \tanh (e+f x)}{2 (a-b) f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {a (a-4 b)}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{2 a (a-b)^2 f}\\ &=\frac {b (a+2 b) \sinh (e+f x)}{2 a (a-b)^2 f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\text {sech}(e+f x) \tanh (e+f x)}{2 (a-b) f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {(a-4 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)^2 f}\\ &=\frac {b (a+2 b) \sinh (e+f x)}{2 a (a-b)^2 f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\text {sech}(e+f x) \tanh (e+f x)}{2 (a-b) f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {(a-4 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)^2 f}\\ &=\frac {(a-4 b) \tan ^{-1}\left (\frac {\sqrt {a-b} \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}\right )}{2 (a-b)^{5/2} f}+\frac {b (a+2 b) \sinh (e+f x)}{2 a (a-b)^2 f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\text {sech}(e+f x) \tanh (e+f x)}{2 (a-b) f \sqrt {a+b \sinh ^2(e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 4.19, size = 231, normalized size = 1.63 \begin {gather*} \frac {\text {sech}^5(e+f x) \left (16 (a-b) \, _3F_2\left (2,2,3;1,\frac {9}{2};\frac {(a-b) \tanh ^2(e+f x)}{a}\right ) \sinh ^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )^2+16 (a-b) \, _2F_1\left (2,3;\frac {9}{2};\frac {(a-b) \tanh ^2(e+f x)}{a}\right ) \sinh ^2(e+f x) \left (4 a^2+7 a b \sinh ^2(e+f x)+3 b^2 \sinh ^4(e+f x)\right )+7 a \cosh ^2(e+f x) \, _2F_1\left (1,2;\frac {7}{2};\frac {(a-b) \tanh ^2(e+f x)}{a}\right ) \left (15 a^2+20 a b \sinh ^2(e+f x)+8 b^2 \sinh ^4(e+f x)\right )\right ) \tanh (e+f x)}{105 a^4 f \sqrt {a+b \sinh ^2(e+f x)}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sech[e + f*x]^5*(16*(a - b)*HypergeometricPFQ[{2, 2, 3}, {1, 9/2}, ((a - b)*Tanh[e + f*x]^2)/a]*Sinh[e + f*x]
^2*(a + b*Sinh[e + f*x]^2)^2 + 16*(a - b)*Hypergeometric2F1[2, 3, 9/2, ((a - b)*Tanh[e + f*x]^2)/a]*Sinh[e + f
*x]^2*(4*a^2 + 7*a*b*Sinh[e + f*x]^2 + 3*b^2*Sinh[e + f*x]^4) + 7*a*Cosh[e + f*x]^2*Hypergeometric2F1[1, 2, 7/
2, ((a - b)*Tanh[e + f*x]^2)/a]*(15*a^2 + 20*a*b*Sinh[e + f*x]^2 + 8*b^2*Sinh[e + f*x]^4))*Tanh[e + f*x])/(105
*a^4*f*Sqrt[a + b*Sinh[e + f*x]^2])

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 6.68, size = 95, normalized size = 0.67

method	result	size
default	\(\frac {\mathit {`\,int/indef0`\,}\left (-\frac {\sqrt {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}\, \left (\cosh ^{2}\left (f x +e \right )\right )}{-b^{2} \left (\cosh ^{10}\left (f x +e \right )\right )+\left (-2 a b +2 b^{2}\right ) \left (\cosh ^{8}\left (f x +e \right )\right )+\left (-a^{2}+2 a b -b^{2}\right ) \left (\cosh ^{6}\left (f x +e \right )\right )}, \sinh \left (f x +e \right )\right )}{f}\)	\(95\)
risch	\(\text {Expression too large to display}\)	\(16593815\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

`int/indef0`(-(a+b*sinh(f*x+e)^2)^(1/2)*cosh(f*x+e)^2/(-b^2*cosh(f*x+e)^10+(-2*a*b+2*b^2)*cosh(f*x+e)^8+(-a^2+
2*a*b-b^2)*cosh(f*x+e)^6),sinh(f*x+e))/f

________________________________________________________________________________________

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)^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 2364 vs. \(2 (126) = 252\).
time = 0.69, size = 4845, normalized size = 34.12 \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)^(3/2),x, algorithm="fricas")

[Out]

[1/4*(((a^2*b - 4*a*b^2)*cosh(f*x + e)^8 + 8*(a^2*b - 4*a*b^2)*cosh(f*x + e)*sinh(f*x + e)^7 + (a^2*b - 4*a*b^
2)*sinh(f*x + e)^8 + 4*(a^3 - 4*a^2*b)*cosh(f*x + e)^6 + 4*(a^3 - 4*a^2*b + 7*(a^2*b - 4*a*b^2)*cosh(f*x + e)^
2)*sinh(f*x + e)^6 + 8*(7*(a^2*b - 4*a*b^2)*cosh(f*x + e)^3 + 3*(a^3 - 4*a^2*b)*cosh(f*x + e))*sinh(f*x + e)^5
+ 2*(4*a^3 - 17*a^2*b + 4*a*b^2)*cosh(f*x + e)^4 + 2*(35*(a^2*b - 4*a*b^2)*cosh(f*x + e)^4 + 4*a^3 - 17*a^2*b
+ 4*a*b^2 + 30*(a^3 - 4*a^2*b)*cosh(f*x + e)^2)*sinh(f*x + e)^4 + 8*(7*(a^2*b - 4*a*b^2)*cosh(f*x + e)^5 + 10
*(a^3 - 4*a^2*b)*cosh(f*x + e)^3 + (4*a^3 - 17*a^2*b + 4*a*b^2)*cosh(f*x + e))*sinh(f*x + e)^3 + a^2*b - 4*a*b
^2 + 4*(a^3 - 4*a^2*b)*cosh(f*x + e)^2 + 4*(7*(a^2*b - 4*a*b^2)*cosh(f*x + e)^6 + 15*(a^3 - 4*a^2*b)*cosh(f*x
+ e)^4 + a^3 - 4*a^2*b + 3*(4*a^3 - 17*a^2*b + 4*a*b^2)*cosh(f*x + e)^2)*sinh(f*x + e)^2 + 8*((a^2*b - 4*a*b^2
)*cosh(f*x + e)^7 + 3*(a^3 - 4*a^2*b)*cosh(f*x + e)^5 + (4*a^3 - 17*a^2*b + 4*a*b^2)*cosh(f*x + e)^3 + (a^3 -
4*a^2*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)
*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^2*b + a*
b^2 - 2*b^3)*cosh(f*x + e)^6 + 6*(a^2*b + a*b^2 - 2*b^3)*cosh(f*x + e)*sinh(f*x + e)^5 + (a^2*b + a*b^2 - 2*b^
3)*sinh(f*x + e)^6 + (4*a^3 - 7*a^2*b + 5*a*b^2 - 2*b^3)*cosh(f*x + e)^4 + (4*a^3 - 7*a^2*b + 5*a*b^2 - 2*b^3
+ 15*(a^2*b + a*b^2 - 2*b^3)*cosh(f*x + e)^2)*sinh(f*x + e)^4 + 4*(5*(a^2*b + a*b^2 - 2*b^3)*cosh(f*x + e)^3 +
(4*a^3 - 7*a^2*b + 5*a*b^2 - 2*b^3)*cosh(f*x + e))*sinh(f*x + e)^3 - a^2*b - a*b^2 + 2*b^3 - (4*a^3 - 7*a^2*b
+ 5*a*b^2 - 2*b^3)*cosh(f*x + e)^2 + (15*(a^2*b + a*b^2 - 2*b^3)*cosh(f*x + e)^4 - 4*a^3 + 7*a^2*b - 5*a*b^2
+ 2*b^3 + 6*(4*a^3 - 7*a^2*b + 5*a*b^2 - 2*b^3)*cosh(f*x + e)^2)*sinh(f*x + e)^2 + 2*(3*(a^2*b + a*b^2 - 2*b^3
)*cosh(f*x + e)^5 + 2*(4*a^3 - 7*a^2*b + 5*a*b^2 - 2*b^3)*cosh(f*x + e)^3 - (4*a^3 - 7*a^2*b + 5*a*b^2 - 2*b^3
)*cosh(f*x + e))*sinh(f*x + e))*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*co
sh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2)))/((a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*f*cosh(f*x + e)^8 + 8*
(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*f*cosh(f*x + e)*sinh(f*x + e)^7 + (a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b
^4)*f*sinh(f*x + e)^8 + 4*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*f*cosh(f*x + e)^6 + 4*(7*(a^4*b - 3*a^3*b^2 +
3*a^2*b^3 - a*b^4)*f*cosh(f*x + e)^2 + (a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*f)*sinh(f*x + e)^6 + 2*(4*a^5 - 1
3*a^4*b + 15*a^3*b^2 - 7*a^2*b^3 + a*b^4)*f*cosh(f*x + e)^4 + 8*(7*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*f*c
osh(f*x + e)^3 + 3*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*f*cosh(f*x + e))*sinh(f*x + e)^5 + 2*(35*(a^4*b - 3*a
^3*b^2 + 3*a^2*b^3 - a*b^4)*f*cosh(f*x + e)^4 + 30*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*f*cosh(f*x + e)^2 + (
4*a^5 - 13*a^4*b + 15*a^3*b^2 - 7*a^2*b^3 + a*b^4)*f)*sinh(f*x + e)^4 + 4*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3
)*f*cosh(f*x + e)^2 + 8*(7*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*f*cosh(f*x + e)^5 + 10*(a^5 - 3*a^4*b + 3*a
^3*b^2 - a^2*b^3)*f*cosh(f*x + e)^3 + (4*a^5 - 13*a^4*b + 15*a^3*b^2 - 7*a^2*b^3 + a*b^4)*f*cosh(f*x + e))*sin
h(f*x + e)^3 + 4*(7*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*f*cosh(f*x + e)^6 + 15*(a^5 - 3*a^4*b + 3*a^3*b^2
- a^2*b^3)*f*cosh(f*x + e)^4 + 3*(4*a^5 - 13*a^4*b + 15*a^3*b^2 - 7*a^2*b^3 + a*b^4)*f*cosh(f*x + e)^2 + (a^5
- 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*f)*sinh(f*x + e)^2 + (a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*f + 8*((a^4*b -
3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*f*cosh(f*x + e)^7 + 3*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*f*cosh(f*x + e)^5 +
(4*a^5 - 13*a^4*b + 15*a^3*b^2 - 7*a^2*b^3 + a*b^4)*f*cosh(f*x + e)^3 + (a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)
*f*cosh(f*x + e))*sinh(f*x + e)), 1/2*(((a^2*b - 4*a*b^2)*cosh(f*x + e)^8 + 8*(a^2*b - 4*a*b^2)*cosh(f*x + e)*
sinh(f*x + e)^7 + (a^2*b - 4*a*b^2)*sinh(f*x + e)^8 + 4*(a^3 - 4*a^2*b)*cosh(f*x + e)^6 + 4*(a^3 - 4*a^2*b + 7
*(a^2*b - 4*a*b^2)*cosh(f*x + e)^2)*sinh(f*x + e)^6 + 8*(7*(a^2*b - 4*a*b^2)*cosh(f*x + e)^3 + 3*(a^3 - 4*a^2*
b)*cosh(f*x + e))*sinh(f*x + e)^5 + 2*(4*a^3 - 17*a^2*b + 4*a*b^2)*cosh(f*x + e)^4 + 2*(35*(a^2*b - 4*a*b^2)*c
osh(f*x + e)^4 + 4*a^3 - 17*a^2*b + 4*a*b^2 + 30*(a^3 - 4*a^2*b)*cosh(f*x + e)^2)*sinh(f*x + e)^4 + 8*(7*(a^2*
b - 4*a*b^2)*cosh(f*x + e)^5 + 10*(a^3 - 4*a^2*b)*cosh(f*x + e)^3 + (4*a^3 - 17*a^2*b + 4*a*b^2)*cosh(f*x + e)
)*sinh(f*x + e)^3 + a^2*b - 4*a*b^2 + 4*(a^3 - ...

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 984 vs. \(2 (126) = 252\).
time = 0.88, size = 984, normalized size = 6.93 \begin {gather*} \frac {{\left (\frac {\frac {{\left (a^{3} b^{2} e^{\left (6 \, e\right )} - 3 \, a^{2} b^{3} e^{\left (6 \, e\right )} + 3 \, a b^{4} e^{\left (6 \, e\right )} - b^{5} e^{\left (6 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a^{6} e^{\left (10 \, e\right )} - 5 \, a^{5} b e^{\left (10 \, e\right )} + 10 \, a^{4} b^{2} e^{\left (10 \, e\right )} - 10 \, a^{3} b^{3} e^{\left (10 \, e\right )} + 5 \, a^{2} b^{4} e^{\left (10 \, e\right )} - a b^{5} e^{\left (10 \, e\right )}} - \frac {a^{3} b^{2} e^{\left (4 \, e\right )} - 3 \, a^{2} b^{3} e^{\left (4 \, e\right )} + 3 \, a b^{4} e^{\left (4 \, e\right )} - b^{5} e^{\left (4 \, e\right )}}{a^{6} e^{\left (10 \, e\right )} - 5 \, a^{5} b e^{\left (10 \, e\right )} + 10 \, a^{4} b^{2} e^{\left (10 \, e\right )} - 10 \, a^{3} b^{3} e^{\left (10 \, e\right )} + 5 \, a^{2} b^{4} e^{\left (10 \, e\right )} - a b^{5} e^{\left (10 \, 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}} + \frac {{\left (a - 4 \, 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^{2} e^{\left (6 \, e\right )} - 2 \, a b e^{\left (6 \, e\right )} + b^{2} e^{\left (6 \, 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 (a^{2} e^{\left (6 \, e\right )} - 2 \, a b e^{\left (6 \, e\right )} + b^{2} e^{\left (6 \, e\right )}\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}}\right )} e^{\left (6 \, 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)^(3/2),x, algorithm="giac")

[Out]

(((a^3*b^2*e^(6*e) - 3*a^2*b^3*e^(6*e) + 3*a*b^4*e^(6*e) - b^5*e^(6*e))*e^(2*f*x)/(a^6*e^(10*e) - 5*a^5*b*e^(1
0*e) + 10*a^4*b^2*e^(10*e) - 10*a^3*b^3*e^(10*e) + 5*a^2*b^4*e^(10*e) - a*b^5*e^(10*e)) - (a^3*b^2*e^(4*e) - 3
*a^2*b^3*e^(4*e) + 3*a*b^4*e^(4*e) - b^5*e^(4*e))/(a^6*e^(10*e) - 5*a^5*b*e^(10*e) + 10*a^4*b^2*e^(10*e) - 10*
a^3*b^3*e^(10*e) + 5*a^2*b^4*e^(10*e) - a*b^5*e^(10*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 - 4*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^2*e^(6*e) - 2*a*b*e^(6*e) + b^2*e^(6*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*s
qrt(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))
^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) - 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*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))/((a^2*e^(6*e) - 2*a*b*e^(6*e) + b^2*e^(6*e))*((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))*e^(6*e)
/f

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________