3.485 \(\int \frac{\tanh ^4(e+f x)}{\sqrt{a+b \sinh ^2(e+f x)}} \, dx\)

Optimal. Leaf size=219 \[ \frac{(3 a-b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \text{EllipticF}\left (\tan ^{-1}(\sinh (e+f x)),1-\frac{b}{a}\right )}{3 f (a-b)^2 \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{\tanh (e+f x) \text{sech}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 f (a-b)}-\frac{2 (2 a-b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 f (a-b)^2 \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

[Out]

(-2*(2*a - b)*EllipticE[ArcTan[Sinh[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(3*(a - b)^
2*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]) + ((3*a - b)*EllipticF[ArcTan[Sinh[e + f*x]], 1 - b/a]*
Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(3*(a - b)^2*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]) +
(Sech[e + f*x]^2*Sqrt[a + b*Sinh[e + f*x]^2]*Tanh[e + f*x])/(3*(a - b)*f)

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Rubi [A]  time = 0.196234, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3196, 470, 525, 418, 411} \[ \frac{\tanh (e+f x) \text{sech}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 f (a-b)}+\frac{(3 a-b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 f (a-b)^2 \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{2 (2 a-b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 f (a-b)^2 \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[e + f*x]^4/Sqrt[a + b*Sinh[e + f*x]^2],x]

[Out]

(-2*(2*a - b)*EllipticE[ArcTan[Sinh[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(3*(a - b)^
2*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]) + ((3*a - b)*EllipticF[ArcTan[Sinh[e + f*x]], 1 - b/a]*
Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(3*(a - b)^2*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]) +
(Sech[e + f*x]^2*Sqrt[a + b*Sinh[e + f*x]^2]*Tanh[e + f*x])/(3*(a - b)*f)

Rule 3196

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

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 525

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)^(3/2)), x_Symbol] :> Dist[(b*e - a*
f)/(b*c - a*d), Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[Sqrt[a + b
*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[b/a] && PosQ[d/c]

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticF[ArcT
an[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rule 411

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticE[ArcTan
[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rubi steps

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

Mathematica [C]  time = 2.10477, size = 206, normalized size = 0.94 \[ \frac{2 i a (a-b) \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} \text{EllipticF}\left (i (e+f x),\frac{b}{a}\right )-\frac{\tanh (e+f x) \text{sech}^2(e+f x) \left (2 \left (4 a^2-3 a b+b^2\right ) \cosh (2 (e+f x))+(2 a-b) (2 a+b \cosh (4 (e+f x))+b)\right )}{\sqrt{2}}-4 i a (2 a-b) \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} E\left (i (e+f x)\left |\frac{b}{a}\right .\right )}{6 f (a-b)^2 \sqrt{2 a+b \cosh (2 (e+f x))-b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[e + f*x]^4/Sqrt[a + b*Sinh[e + f*x]^2],x]

[Out]

((-4*I)*a*(2*a - b)*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticE[I*(e + f*x), b/a] + (2*I)*a*(a - b)*Sqrt
[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticF[I*(e + f*x), b/a] - ((2*(4*a^2 - 3*a*b + b^2)*Cosh[2*(e + f*x)]
+ (2*a - b)*(2*a + b + b*Cosh[4*(e + f*x)]))*Sech[e + f*x]^2*Tanh[e + f*x])/Sqrt[2])/(6*(a - b)^2*f*Sqrt[2*a -
b + b*Cosh[2*(e + f*x)]])

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Maple [A]  time = 0.289, size = 366, normalized size = 1.7 \begin{align*}{\frac{1}{3\, \left ( \cosh \left ( fx+e \right ) \right ) ^{3} \left ( a-b \right ) ^{2}f} \left ( \left ( -4\,\sqrt{-{\frac{b}{a}}}ab+2\,\sqrt{-{\frac{b}{a}}}{b}^{2} \right ) \sinh \left ( fx+e \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{4}+ \left ( -4\,\sqrt{-{\frac{b}{a}}}{a}^{2}+7\,\sqrt{-{\frac{b}{a}}}ab-3\,\sqrt{-{\frac{b}{a}}}{b}^{2} \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{2}\sinh \left ( fx+e \right ) + \left ( \sqrt{-{\frac{b}{a}}}{a}^{2}-2\,\sqrt{-{\frac{b}{a}}}ab+\sqrt{-{\frac{b}{a}}}{b}^{2} \right ) \sinh \left ( fx+e \right ) +\sqrt{ \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{b \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}{a}}+{\frac{a-b}{a}}} \left ( 3\,{\it EllipticF} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ){a}^{2}-5\,{\it EllipticF} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) ab+2\,{\it EllipticF} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ){b}^{2}+4\,{\it EllipticE} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) ab-2\,{\it EllipticE} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ){b}^{2} \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{2} \right ){\frac{1}{\sqrt{-{\frac{b}{a}}}}}{\frac{1}{\sqrt{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((-4*(-1/a*b)^(1/2)*a*b+2*(-1/a*b)^(1/2)*b^2)*sinh(f*x+e)*cosh(f*x+e)^4+(-4*(-1/a*b)^(1/2)*a^2+7*(-1/a*b)^
(1/2)*a*b-3*(-1/a*b)^(1/2)*b^2)*cosh(f*x+e)^2*sinh(f*x+e)+((-1/a*b)^(1/2)*a^2-2*(-1/a*b)^(1/2)*a*b+(-1/a*b)^(1
/2)*b^2)*sinh(f*x+e)+(cosh(f*x+e)^2)^(1/2)*(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(3*EllipticF(sinh(f*x+e)*(-1/a*b)
^(1/2),(a/b)^(1/2))*a^2-5*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*a*b+2*EllipticF(sinh(f*x+e)*(-1/a*
b)^(1/2),(a/b)^(1/2))*b^2+4*EllipticE(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*a*b-2*EllipticE(sinh(f*x+e)*(-1/
a*b)^(1/2),(a/b)^(1/2))*b^2)*cosh(f*x+e)^2)/cosh(f*x+e)^3/(a-b)^2/(-1/a*b)^(1/2)/(a+b*sinh(f*x+e)^2)^(1/2)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh \left (f x + e\right )^{4}}{\sqrt{b \sinh \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(f*x+e)^4/(a+b*sinh(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(tanh(f*x + e)^4/sqrt(b*sinh(f*x + e)^2 + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\tanh \left (f x + e\right )^{4}}{\sqrt{b \sinh \left (f x + e\right )^{2} + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(f*x+e)^4/(a+b*sinh(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

integral(tanh(f*x + e)^4/sqrt(b*sinh(f*x + e)^2 + a), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{4}{\left (e + f x \right )}}{\sqrt{a + b \sinh ^{2}{\left (e + f x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(f*x+e)**4/(a+b*sinh(f*x+e)**2)**(1/2),x)

[Out]

Integral(tanh(e + f*x)**4/sqrt(a + b*sinh(e + f*x)**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh \left (f x + e\right )^{4}}{\sqrt{b \sinh \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(f*x+e)^4/(a+b*sinh(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(tanh(f*x + e)^4/sqrt(b*sinh(f*x + e)^2 + a), x)