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

Optimal. Leaf size=232 \[ \frac{8 a^2+24 a b+3 b^2}{8 f (a-b)^4 \sqrt{a+b \sinh ^2(e+f x)}}+\frac{8 a^2+24 a b+3 b^2}{24 f (a-b)^3 \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{\left (8 a^2+24 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a-b}}\right )}{8 f (a-b)^{9/2}}-\frac{\text{sech}^4(e+f x)}{4 f (a-b) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{(8 a-b) \text{sech}^2(e+f x)}{8 f (a-b)^2 \left (a+b \sinh ^2(e+f x)\right )^{3/2}} \]

[Out]

-((8*a^2 + 24*a*b + 3*b^2)*ArcTanh[Sqrt[a + b*Sinh[e + f*x]^2]/Sqrt[a - b]])/(8*(a - b)^(9/2)*f) + (8*a^2 + 24
*a*b + 3*b^2)/(24*(a - b)^3*f*(a + b*Sinh[e + f*x]^2)^(3/2)) + ((8*a - b)*Sech[e + f*x]^2)/(8*(a - b)^2*f*(a +
b*Sinh[e + f*x]^2)^(3/2)) - Sech[e + f*x]^4/(4*(a - b)*f*(a + b*Sinh[e + f*x]^2)^(3/2)) + (8*a^2 + 24*a*b + 3
*b^2)/(8*(a - b)^4*f*Sqrt[a + b*Sinh[e + f*x]^2])

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Rubi [A]  time = 0.315846, antiderivative size = 232, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3194, 89, 78, 51, 63, 208} \[ \frac{8 a^2+24 a b+3 b^2}{8 f (a-b)^4 \sqrt{a+b \sinh ^2(e+f x)}}+\frac{8 a^2+24 a b+3 b^2}{24 f (a-b)^3 \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{\left (8 a^2+24 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a-b}}\right )}{8 f (a-b)^{9/2}}-\frac{\text{sech}^4(e+f x)}{4 f (a-b) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{(8 a-b) \text{sech}^2(e+f x)}{8 f (a-b)^2 \left (a+b \sinh ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((8*a^2 + 24*a*b + 3*b^2)*ArcTanh[Sqrt[a + b*Sinh[e + f*x]^2]/Sqrt[a - b]])/(8*(a - b)^(9/2)*f) + (8*a^2 + 24
*a*b + 3*b^2)/(24*(a - b)^3*f*(a + b*Sinh[e + f*x]^2)^(3/2)) + ((8*a - b)*Sech[e + f*x]^2)/(8*(a - b)^2*f*(a +
b*Sinh[e + f*x]^2)^(3/2)) - Sech[e + f*x]^4/(4*(a - b)*f*(a + b*Sinh[e + f*x]^2)^(3/2)) + (8*a^2 + 24*a*b + 3
*b^2)/(8*(a - b)^4*f*Sqrt[a + b*Sinh[e + f*x]^2])

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]

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 78

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

Rule 51

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 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]

Rubi steps

\begin{align*} \int \frac{\tanh ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{(1+x)^3 (a+b x)^{5/2}} \, dx,x,\sinh ^2(e+f x)\right )}{2 f}\\ &=-\frac{\text{sech}^4(e+f x)}{4 (a-b) f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} (-4 a-3 b)+2 (a-b) x}{(1+x)^2 (a+b x)^{5/2}} \, dx,x,\sinh ^2(e+f x)\right )}{4 (a-b) f}\\ &=\frac{(8 a-b) \text{sech}^2(e+f x)}{8 (a-b)^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{\text{sech}^4(e+f x)}{4 (a-b) f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{\left (8 a^2+24 a b+3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{(1+x) (a+b x)^{5/2}} \, dx,x,\sinh ^2(e+f x)\right )}{16 (a-b)^2 f}\\ &=\frac{8 a^2+24 a b+3 b^2}{24 (a-b)^3 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{(8 a-b) \text{sech}^2(e+f x)}{8 (a-b)^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{\text{sech}^4(e+f x)}{4 (a-b) f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{\left (8 a^2+24 a b+3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{(1+x) (a+b x)^{3/2}} \, dx,x,\sinh ^2(e+f x)\right )}{16 (a-b)^3 f}\\ &=\frac{8 a^2+24 a b+3 b^2}{24 (a-b)^3 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{(8 a-b) \text{sech}^2(e+f x)}{8 (a-b)^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{\text{sech}^4(e+f x)}{4 (a-b) f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{8 a^2+24 a b+3 b^2}{8 (a-b)^4 f \sqrt{a+b \sinh ^2(e+f x)}}+\frac{\left (8 a^2+24 a b+3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{a+b x}} \, dx,x,\sinh ^2(e+f x)\right )}{16 (a-b)^4 f}\\ &=\frac{8 a^2+24 a b+3 b^2}{24 (a-b)^3 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{(8 a-b) \text{sech}^2(e+f x)}{8 (a-b)^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{\text{sech}^4(e+f x)}{4 (a-b) f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{8 a^2+24 a b+3 b^2}{8 (a-b)^4 f \sqrt{a+b \sinh ^2(e+f x)}}+\frac{\left (8 a^2+24 a b+3 b^2\right ) \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 )}{8 (a-b)^4 b f}\\ &=-\frac{\left (8 a^2+24 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a-b}}\right )}{8 (a-b)^{9/2} f}+\frac{8 a^2+24 a b+3 b^2}{24 (a-b)^3 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{(8 a-b) \text{sech}^2(e+f x)}{8 (a-b)^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{\text{sech}^4(e+f x)}{4 (a-b) f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{8 a^2+24 a b+3 b^2}{8 (a-b)^4 f \sqrt{a+b \sinh ^2(e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.58949, size = 114, normalized size = 0.49 \[ \frac{2 \left (8 a^2+24 a b+3 b^2\right ) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{b \sinh ^2(e+f x)+a}{a-b}\right )+3 (a-b) \text{sech}^4(e+f x) ((8 a-b) \cosh (2 (e+f x))+4 a+3 b)}{48 f (a-b)^3 \left (a+b \sinh ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(8*a^2 + 24*a*b + 3*b^2)*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Sinh[e + f*x]^2)/(a - b)] + 3*(a - b)*(4*a
+ 3*b + (8*a - b)*Cosh[2*(e + f*x)])*Sech[e + f*x]^4)/(48*(a - b)^3*f*(a + b*Sinh[e + f*x]^2)^(3/2))

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Maple [C]  time = 0.298, size = 213, normalized size = 0.9 \begin{align*}{\frac{1}{f}\mbox{{\tt ` int/indef0`}} \left ( -{\frac{ \left ( \sinh \left ( fx+e \right ) \right ) ^{5} \left ({b}^{2} \left ( \sinh \left ( fx+e \right ) \right ) ^{4}+2\,ab \left ( \sinh \left ( fx+e \right ) \right ) ^{2}+{a}^{2} \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{4}}{-{b}^{4} \left ( \cosh \left ( fx+e \right ) \right ) ^{18}+ \left ( -4\,a{b}^{3}+4\,{b}^{4} \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{16}+ \left ( -6\,{a}^{2}{b}^{2}+12\,a{b}^{3}-6\,{b}^{4} \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{14}+ \left ( -4\,{a}^{3}b+12\,{a}^{2}{b}^{2}-12\,a{b}^{3}+4\,{b}^{4} \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{12}+ \left ( -{a}^{4}+4\,{a}^{3}b-6\,{a}^{2}{b}^{2}+4\,a{b}^{3}-{b}^{4} \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{10}}{\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(tanh(f*x+e)^5/(a+b*sinh(f*x+e)^2)^(5/2),x)

[Out]

`int/indef0`(-sinh(f*x+e)^5*(b^2*sinh(f*x+e)^4+2*a*b*sinh(f*x+e)^2+a^2)*cosh(f*x+e)^4/(-b^4*cosh(f*x+e)^18+(-4
*a*b^3+4*b^4)*cosh(f*x+e)^16+(-6*a^2*b^2+12*a*b^3-6*b^4)*cosh(f*x+e)^14+(-4*a^3*b+12*a^2*b^2-12*a*b^3+4*b^4)*c
osh(f*x+e)^12+(-a^4+4*a^3*b-6*a^2*b^2+4*a*b^3-b^4)*cosh(f*x+e)^10)/(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{\tanh \left (f x + e\right )^{5}}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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