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3.6.3 \(\int \frac {\tanh (e+f x)}{(a+b \sinh ^2(e+f x))^{5/2}} \, dx\) [503]

Optimal. Leaf size=99 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a-b}}\right )}{(a-b)^{5/2} f}+\frac {1}{3 (a-b) f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac {1}{(a-b)^2 f \sqrt {a+b \sinh ^2(e+f x)}} \]

[Out]

-arctanh((a+b*sinh(f*x+e)^2)^(1/2)/(a-b)^(1/2))/(a-b)^(5/2)/f+1/3/(a-b)/f/(a+b*sinh(f*x+e)^2)^(3/2)+1/(a-b)^2/
f/(a+b*sinh(f*x+e)^2)^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3273, 53, 65, 214} \begin {gather*} \frac {1}{f (a-b)^2 \sqrt {a+b \sinh ^2(e+f x)}}+\frac {1}{3 f (a-b) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a-b}}\right )}{f (a-b)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tanh[e + f*x]/(a + b*Sinh[e + f*x]^2)^(5/2),x]

[Out]

-(ArcTanh[Sqrt[a + b*Sinh[e + f*x]^2]/Sqrt[a - b]]/((a - b)^(5/2)*f)) + 1/(3*(a - b)*f*(a + b*Sinh[e + f*x]^2)
^(3/2)) + 1/((a - b)^2*f*Sqrt[a + b*Sinh[e + f*x]^2])

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

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 214

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

Rule 3273

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 \frac {\tanh (e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx &=\frac {\text {Subst}\left (\int \frac {1}{(1+x) (a+b x)^{5/2}} \, dx,x,\sinh ^2(e+f x)\right )}{2 f}\\ &=\frac {1}{3 (a-b) f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac {\text {Subst}\left (\int \frac {1}{(1+x) (a+b x)^{3/2}} \, dx,x,\sinh ^2(e+f x)\right )}{2 (a-b) f}\\ &=\frac {1}{3 (a-b) f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac {1}{(a-b)^2 f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {1}{(1+x) \sqrt {a+b x}} \, dx,x,\sinh ^2(e+f x)\right )}{2 (a-b)^2 f}\\ &=\frac {1}{3 (a-b) f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac {1}{(a-b)^2 f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {1}{1-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^2(e+f x)}\right )}{(a-b)^2 b f}\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a-b}}\right )}{(a-b)^{5/2} f}+\frac {1}{3 (a-b) f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac {1}{(a-b)^2 f \sqrt {a+b \sinh ^2(e+f x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.10, size = 60, normalized size = 0.61 \begin {gather*} \frac {\, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};1+\frac {b \cosh ^2(e+f x)}{a-b}\right )}{3 (a-b) f \left (a-b+b \cosh ^2(e+f x)\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tanh[e + f*x]/(a + b*Sinh[e + f*x]^2)^(5/2),x]

[Out]

Hypergeometric2F1[-3/2, 1, -1/2, 1 + (b*Cosh[e + f*x]^2)/(a - b)]/(3*(a - b)*f*(a - b + b*Cosh[e + f*x]^2)^(3/
2))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 1.54, size = 173, normalized size = 1.75

method result size
default \(\frac {\mathit {`\,int/indef0`\,}\left (-\frac {\sinh \left (f x +e \right ) \left (b^{2} \left (\sinh ^{4}\left (f x +e \right )\right )+2 a b \left (\sinh ^{2}\left (f x +e \right )\right )+a^{2}\right )}{\left (-b^{4} \left (\sinh ^{10}\left (f x +e \right )\right )+\left (-4 a \,b^{3}-b^{4}\right ) \left (\sinh ^{8}\left (f x +e \right )\right )+\left (-6 a^{2} b^{2}-4 a \,b^{3}\right ) \left (\sinh ^{6}\left (f x +e \right )\right )+\left (-4 a^{3} b -6 a^{2} b^{2}\right ) \left (\sinh ^{4}\left (f x +e \right )\right )+\left (-a^{4}-4 a^{3} b \right ) \left (\sinh ^{2}\left (f x +e \right )\right )-a^{4}\right ) \sqrt {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}}, \sinh \left (f x +e \right )\right )}{f}\) \(173\)
risch \(\text {Expression too large to display}\) \(47215\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(f*x+e)/(a+b*sinh(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

`int/indef0`(-sinh(f*x+e)*(b^2*sinh(f*x+e)^4+2*a*b*sinh(f*x+e)^2+a^2)/(-b^4*sinh(f*x+e)^10+(-4*a*b^3-b^4)*sinh
(f*x+e)^8+(-6*a^2*b^2-4*a*b^3)*sinh(f*x+e)^6+(-4*a^3*b-6*a^2*b^2)*sinh(f*x+e)^4+(-a^4-4*a^3*b)*sinh(f*x+e)^2-a
^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(tanh(f*x+e)/(a+b*sinh(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

integrate(tanh(f*x + e)/(b*sinh(f*x + e)^2 + a)^(5/2), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 2002 vs. \(2 (87) = 174\).
time = 0.64, size = 4200, normalized size = 42.42 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(f*x+e)/(a+b*sinh(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(b^2*cosh(f*x + e)^8 + 8*b^2*cosh(f*x + e)*sinh(f*x + e)^7 + b^2*sinh(f*x + e)^8 + 4*(2*a*b - b^2)*cos
h(f*x + e)^6 + 4*(7*b^2*cosh(f*x + e)^2 + 2*a*b - b^2)*sinh(f*x + e)^6 + 8*(7*b^2*cosh(f*x + e)^3 + 3*(2*a*b -
 b^2)*cosh(f*x + e))*sinh(f*x + e)^5 + 2*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^4 + 2*(35*b^2*cosh(f*x + e)^4 +
 30*(2*a*b - b^2)*cosh(f*x + e)^2 + 8*a^2 - 8*a*b + 3*b^2)*sinh(f*x + e)^4 + 8*(7*b^2*cosh(f*x + e)^5 + 10*(2*
a*b - b^2)*cosh(f*x + e)^3 + (8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e))*sinh(f*x + e)^3 + 4*(2*a*b - b^2)*cosh(f*x
 + e)^2 + 4*(7*b^2*cosh(f*x + e)^6 + 15*(2*a*b - b^2)*cosh(f*x + e)^4 + 3*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e
)^2 + 2*a*b - b^2)*sinh(f*x + e)^2 + b^2 + 8*(b^2*cosh(f*x + e)^7 + 3*(2*a*b - b^2)*cosh(f*x + e)^5 + (8*a^2 -
 8*a*b + 3*b^2)*cosh(f*x + e)^3 + (2*a*b - b^2)*cosh(f*x + e))*sinh(f*x + e))*sqrt(a - b)*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*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))*(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)) + 4*sqrt(2)*(3*(a*b - b^2)*cosh(f*x + e)^5 + 15*(a*b - b^2)*cosh(f*x + e)
*sinh(f*x + e)^4 + 3*(a*b - b^2)*sinh(f*x + e)^5 + 2*(8*a^2 - 13*a*b + 5*b^2)*cosh(f*x + e)^3 + 2*(15*(a*b - b
^2)*cosh(f*x + e)^2 + 8*a^2 - 13*a*b + 5*b^2)*sinh(f*x + e)^3 + 6*(5*(a*b - b^2)*cosh(f*x + e)^3 + (8*a^2 - 13
*a*b + 5*b^2)*cosh(f*x + e))*sinh(f*x + e)^2 + 3*(a*b - b^2)*cosh(f*x + e) + 3*(5*(a*b - b^2)*cosh(f*x + e)^4
+ 2*(8*a^2 - 13*a*b + 5*b^2)*cosh(f*x + e)^2 + a*b - b^2)*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*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2)))/((a^3*b^2 - 3*a^2*b^3
+ 3*a*b^4 - b^5)*f*cosh(f*x + e)^8 + 8*(a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*f*cosh(f*x + e)*sinh(f*x + e)^7 +
 (a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*f*sinh(f*x + e)^8 + 4*(2*a^4*b - 7*a^3*b^2 + 9*a^2*b^3 - 5*a*b^4 + b^5)
*f*cosh(f*x + e)^6 + 4*(7*(a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*f*cosh(f*x + e)^2 + (2*a^4*b - 7*a^3*b^2 + 9*a
^2*b^3 - 5*a*b^4 + b^5)*f)*sinh(f*x + e)^6 + 2*(8*a^5 - 32*a^4*b + 51*a^3*b^2 - 41*a^2*b^3 + 17*a*b^4 - 3*b^5)
*f*cosh(f*x + e)^4 + 8*(7*(a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*f*cosh(f*x + e)^3 + 3*(2*a^4*b - 7*a^3*b^2 + 9
*a^2*b^3 - 5*a*b^4 + b^5)*f*cosh(f*x + e))*sinh(f*x + e)^5 + 2*(35*(a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*f*cos
h(f*x + e)^4 + 30*(2*a^4*b - 7*a^3*b^2 + 9*a^2*b^3 - 5*a*b^4 + b^5)*f*cosh(f*x + e)^2 + (8*a^5 - 32*a^4*b + 51
*a^3*b^2 - 41*a^2*b^3 + 17*a*b^4 - 3*b^5)*f)*sinh(f*x + e)^4 + 4*(2*a^4*b - 7*a^3*b^2 + 9*a^2*b^3 - 5*a*b^4 +
b^5)*f*cosh(f*x + e)^2 + 8*(7*(a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*f*cosh(f*x + e)^5 + 10*(2*a^4*b - 7*a^3*b^
2 + 9*a^2*b^3 - 5*a*b^4 + b^5)*f*cosh(f*x + e)^3 + (8*a^5 - 32*a^4*b + 51*a^3*b^2 - 41*a^2*b^3 + 17*a*b^4 - 3*
b^5)*f*cosh(f*x + e))*sinh(f*x + e)^3 + 4*(7*(a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*f*cosh(f*x + e)^6 + 15*(2*a
^4*b - 7*a^3*b^2 + 9*a^2*b^3 - 5*a*b^4 + b^5)*f*cosh(f*x + e)^4 + 3*(8*a^5 - 32*a^4*b + 51*a^3*b^2 - 41*a^2*b^
3 + 17*a*b^4 - 3*b^5)*f*cosh(f*x + e)^2 + (2*a^4*b - 7*a^3*b^2 + 9*a^2*b^3 - 5*a*b^4 + b^5)*f)*sinh(f*x + e)^2
 + (a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*f + 8*((a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*f*cosh(f*x + e)^7 + 3*(2
*a^4*b - 7*a^3*b^2 + 9*a^2*b^3 - 5*a*b^4 + b^5)*f*cosh(f*x + e)^5 + (8*a^5 - 32*a^4*b + 51*a^3*b^2 - 41*a^2*b^
3 + 17*a*b^4 - 3*b^5)*f*cosh(f*x + e)^3 + (2*a^4*b - 7*a^3*b^2 + 9*a^2*b^3 - 5*a*b^4 + b^5)*f*cosh(f*x + e))*s
inh(f*x + e)), -1/3*(3*(b^2*cosh(f*x + e)^8 + 8*b^2*cosh(f*x + e)*sinh(f*x + e)^7 + b^2*sinh(f*x + e)^8 + 4*(2
*a*b - b^2)*cosh(f*x + e)^6 + 4*(7*b^2*cosh(f*x + e)^2 + 2*a*b - b^2)*sinh(f*x + e)^6 + 8*(7*b^2*cosh(f*x + e)
^3 + 3*(2*a*b - b^2)*cosh(f*x + e))*sinh(f*x + e)^5 + 2*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^4 + 2*(35*b^2*co
sh(f*x + e)^4 + 30*(2*a*b - b^2)*cosh(f*x + e)^2 + 8*a^2 - 8*a*b + 3*b^2)*sinh(f*x + e)^4 + 8*(7*b^2*cosh(f*x
+ e)^5 + 10*(2*a*b - b^2)*cosh(f*x + e)^3 + (8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e))*sinh(f*x + e)^3 + 4*(2*a*b
- b^2)*cosh(f*x + e)^2 + 4*(7*b^2*cosh(f*x + e)^6 + 15*(2*a*b - b^2)*cosh(f*x + e)^4 + 3*(8*a^2 - 8*a*b + 3*b^
2)*cosh(f*x + e)^2 + 2*a*b - b^2)*sinh(f*x + e)^2 + b^2 + 8*(b^2*cosh(f*x + e)^7 + 3*(2*a*b - b^2)*cosh(f*x +
e)^5 + (8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^3 + (2*a*b - b^2)*cosh(f*x + e))*sinh(f*x + e))*sqrt(-a + b)*arct
an(-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))) - 2*sqrt(2)*(3*(a*
b - b^2)*cosh(f*x + e)^5 + 15*(a*b - b^2)*cosh(...

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tanh {\left (e + f x \right )}}{\left (a + b \sinh ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(f*x+e)/(a+b*sinh(f*x+e)**2)**(5/2),x)

[Out]

Integral(tanh(e + f*x)/(a + b*sinh(e + f*x)**2)**(5/2), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(f*x+e)/(a+b*sinh(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Evaluation time:
1.19Error: Bad Argument Type

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {tanh}\left (e+f\,x\right )}{{\left (b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(e + f*x)/(a + b*sinh(e + f*x)^2)^(5/2),x)

[Out]

int(tanh(e + f*x)/(a + b*sinh(e + f*x)^2)^(5/2), x)

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