3.263 \(\int \frac{\tanh (x)}{(a+b \tanh ^4(x))^{5/2}} \, dx\)

Optimal. Leaf size=118 \[ -\frac{3 a^2-b (5 a+2 b) \tanh ^2(x)}{6 a^2 (a+b)^2 \sqrt{a+b \tanh ^4(x)}}+\frac{\tanh ^{-1}\left (\frac{a+b \tanh ^2(x)}{\sqrt{a+b} \sqrt{a+b \tanh ^4(x)}}\right )}{2 (a+b)^{5/2}}-\frac{a-b \tanh ^2(x)}{6 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}} \]

[Out]

ArcTanh[(a + b*Tanh[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tanh[x]^4])]/(2*(a + b)^(5/2)) - (a - b*Tanh[x]^2)/(6*a*(a +
b)*(a + b*Tanh[x]^4)^(3/2)) - (3*a^2 - b*(5*a + 2*b)*Tanh[x]^2)/(6*a^2*(a + b)^2*Sqrt[a + b*Tanh[x]^4])

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Rubi [A]  time = 0.198754, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.467, Rules used = {3670, 1248, 741, 823, 12, 725, 206} \[ -\frac{3 a^2-b (5 a+2 b) \tanh ^2(x)}{6 a^2 (a+b)^2 \sqrt{a+b \tanh ^4(x)}}+\frac{\tanh ^{-1}\left (\frac{a+b \tanh ^2(x)}{\sqrt{a+b} \sqrt{a+b \tanh ^4(x)}}\right )}{2 (a+b)^{5/2}}-\frac{a-b \tanh ^2(x)}{6 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]/(a + b*Tanh[x]^4)^(5/2),x]

[Out]

ArcTanh[(a + b*Tanh[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tanh[x]^4])]/(2*(a + b)^(5/2)) - (a - b*Tanh[x]^2)/(6*a*(a +
b)*(a + b*Tanh[x]^4)^(3/2)) - (3*a^2 - b*(5*a + 2*b)*Tanh[x]^2)/(6*a^2*(a + b)^2*Sqrt[a + b*Tanh[x]^4])

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rule 1248

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 12

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

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{5/2}} \, dx &=\operatorname{Subst}\left (\int \frac{x}{\left (1-x^2\right ) \left (a+b x^4\right )^{5/2}} \, dx,x,\tanh (x)\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{(1-x) \left (a+b x^2\right )^{5/2}} \, dx,x,\tanh ^2(x)\right )\\ &=-\frac{a-b \tanh ^2(x)}{6 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{-3 a-2 b+2 b x}{(1-x) \left (a+b x^2\right )^{3/2}} \, dx,x,\tanh ^2(x)\right )}{6 a (a+b)}\\ &=-\frac{a-b \tanh ^2(x)}{6 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}}-\frac{3 a^2-b (5 a+2 b) \tanh ^2(x)}{6 a^2 (a+b)^2 \sqrt{a+b \tanh ^4(x)}}+\frac{\operatorname{Subst}\left (\int \frac{3 a^2 b}{(1-x) \sqrt{a+b x^2}} \, dx,x,\tanh ^2(x)\right )}{6 a^2 b (a+b)^2}\\ &=-\frac{a-b \tanh ^2(x)}{6 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}}-\frac{3 a^2-b (5 a+2 b) \tanh ^2(x)}{6 a^2 (a+b)^2 \sqrt{a+b \tanh ^4(x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{(1-x) \sqrt{a+b x^2}} \, dx,x,\tanh ^2(x)\right )}{2 (a+b)^2}\\ &=-\frac{a-b \tanh ^2(x)}{6 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}}-\frac{3 a^2-b (5 a+2 b) \tanh ^2(x)}{6 a^2 (a+b)^2 \sqrt{a+b \tanh ^4(x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{a+b-x^2} \, dx,x,\frac{-a-b \tanh ^2(x)}{\sqrt{a+b \tanh ^4(x)}}\right )}{2 (a+b)^2}\\ &=\frac{\tanh ^{-1}\left (\frac{a+b \tanh ^2(x)}{\sqrt{a+b} \sqrt{a+b \tanh ^4(x)}}\right )}{2 (a+b)^{5/2}}-\frac{a-b \tanh ^2(x)}{6 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}}-\frac{3 a^2-b (5 a+2 b) \tanh ^2(x)}{6 a^2 (a+b)^2 \sqrt{a+b \tanh ^4(x)}}\\ \end{align*}

Mathematica [A]  time = 0.785528, size = 113, normalized size = 0.96 \[ \frac{1}{6} \left (\frac{-3 a^2 b \tanh ^4(x)-a^2 (4 a+b)+b^2 (5 a+2 b) \tanh ^6(x)+3 a b (2 a+b) \tanh ^2(x)}{a^2 (a+b)^2 \left (a+b \tanh ^4(x)\right )^{3/2}}+\frac{3 \tanh ^{-1}\left (\frac{a+b \tanh ^2(x)}{\sqrt{a+b} \sqrt{a+b \tanh ^4(x)}}\right )}{(a+b)^{5/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]/(a + b*Tanh[x]^4)^(5/2),x]

[Out]

((3*ArcTanh[(a + b*Tanh[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tanh[x]^4])])/(a + b)^(5/2) + (-(a^2*(4*a + b)) + 3*a*b*
(2*a + b)*Tanh[x]^2 - 3*a^2*b*Tanh[x]^4 + b^2*(5*a + 2*b)*Tanh[x]^6)/(a^2*(a + b)^2*(a + b*Tanh[x]^4)^(3/2)))/
6

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Maple [C]  time = 0.053, size = 637, normalized size = 5.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)/(a+b*tanh(x)^4)^(5/2),x)

[Out]

-1/2*(1/6/a/(a+b)/b*tanh(x)^3-1/6/a/(a+b)/b*tanh(x)^2+1/6/a/(a+b)/b*tanh(x)+1/6/(a+b)/b^2)*(a+b*tanh(x)^4)^(1/
2)/(tanh(x)^4+a/b)^2+b*(-1/8*(3*a+b)/a^2/(a+b)^2*tanh(x)^3+1/12*(5*a+2*b)/a^2/(a+b)^2*tanh(x)^2-1/24*(11*a+5*b
)/a^2/(a+b)^2*tanh(x)-1/4/(a+b)^2/b)/((tanh(x)^4+a/b)*b)^(1/2)-1/2/(a+b)^2*(-1/2/(a+b)^(1/2)*arctanh(1/2*(2*b*
tanh(x)^2+2*a)/(a+b)^(1/2)/(a+b*tanh(x)^4)^(1/2))+1/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*tanh(x)^2)^
(1/2)*(1+I/a^(1/2)*b^(1/2)*tanh(x)^2)^(1/2)/(a+b*tanh(x)^4)^(1/2)*EllipticPi(tanh(x)*(I/a^(1/2)*b^(1/2))^(1/2)
,-I*a^(1/2)/b^(1/2),(-I/a^(1/2)*b^(1/2))^(1/2)/(I/a^(1/2)*b^(1/2))^(1/2)))-1/2*(-1/6/a/(a+b)/b*tanh(x)^3-1/6/a
/(a+b)/b*tanh(x)^2-1/6/a/(a+b)/b*tanh(x)+1/6/(a+b)/b^2)*(a+b*tanh(x)^4)^(1/2)/(tanh(x)^4+a/b)^2+b*(1/8*(3*a+b)
/a^2/(a+b)^2*tanh(x)^3+1/12*(5*a+2*b)/a^2/(a+b)^2*tanh(x)^2+1/24*(11*a+5*b)/a^2/(a+b)^2*tanh(x)-1/4/(a+b)^2/b)
/((tanh(x)^4+a/b)*b)^(1/2)-1/2/(a+b)^2*(-1/2/(a+b)^(1/2)*arctanh(1/2*(2*b*tanh(x)^2+2*a)/(a+b)^(1/2)/(a+b*tanh
(x)^4)^(1/2))-1/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*tanh(x)^2)^(1/2)*(1+I/a^(1/2)*b^(1/2)*tanh(x)^2
)^(1/2)/(a+b*tanh(x)^4)^(1/2)*EllipticPi(tanh(x)*(I/a^(1/2)*b^(1/2))^(1/2),-I*a^(1/2)/b^(1/2),(-I/a^(1/2)*b^(1
/2))^(1/2)/(I/a^(1/2)*b^(1/2))^(1/2)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh \left (x\right )}{{\left (b \tanh \left (x\right )^{4} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)/(a+b*tanh(x)^4)^(5/2),x, algorithm="maxima")

[Out]

integrate(tanh(x)/(b*tanh(x)^4 + 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(x)/(a+b*tanh(x)^4)^(5/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh{\left (x \right )}}{\left (a + b \tanh ^{4}{\left (x \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)/(a+b*tanh(x)**4)**(5/2),x)

[Out]

Integral(tanh(x)/(a + b*tanh(x)**4)**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh \left (x\right )}{{\left (b \tanh \left (x\right )^{4} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)/(a+b*tanh(x)^4)^(5/2),x, algorithm="giac")

[Out]

integrate(tanh(x)/(b*tanh(x)^4 + a)^(5/2), x)