3.259 \(\int \tanh (x) (a+b \tanh ^4(x))^{3/2} \, dx\)

Optimal. Leaf size=124 \[ \frac{1}{2} (a+b)^{3/2} \tanh ^{-1}\left (\frac{a+b \tanh ^2(x)}{\sqrt{a+b} \sqrt{a+b \tanh ^4(x)}}\right )-\frac{1}{6} \left (a+b \tanh ^4(x)\right )^{3/2}-\frac{1}{4} \sqrt{b} (3 a+2 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tanh ^2(x)}{\sqrt{a+b \tanh ^4(x)}}\right )-\frac{1}{4} \left (2 (a+b)+b \tanh ^2(x)\right ) \sqrt{a+b \tanh ^4(x)} \]

[Out]

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

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Rubi [A]  time = 0.241808, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.533, Rules used = {3670, 1248, 735, 815, 844, 217, 206, 725} \[ \frac{1}{2} (a+b)^{3/2} \tanh ^{-1}\left (\frac{a+b \tanh ^2(x)}{\sqrt{a+b} \sqrt{a+b \tanh ^4(x)}}\right )-\frac{1}{6} \left (a+b \tanh ^4(x)\right )^{3/2}-\frac{1}{4} \sqrt{b} (3 a+2 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tanh ^2(x)}{\sqrt{a+b \tanh ^4(x)}}\right )-\frac{1}{4} \left (2 (a+b)+b \tanh ^2(x)\right ) \sqrt{a+b \tanh ^4(x)} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]*(a + b*Tanh[x]^4)^(3/2),x]

[Out]

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

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 735

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + 2*p + 1)), x] + Dist[(2*p)/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[a*e - c*d*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] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !Ration
alQ[m] || LtQ[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 815

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

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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])

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]

Rubi steps

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

Mathematica [A]  time = 4.39531, size = 166, normalized size = 1.34 \[ \frac{1}{12} \left (6 (a+b)^{3/2} \tanh ^{-1}\left (\frac{a+b \tanh ^2(x)}{\sqrt{a+b} \sqrt{a+b \tanh ^4(x)}}\right )-6 \sqrt{b} (a+b) \tanh ^{-1}\left (\frac{\sqrt{b} \tanh ^2(x)}{\sqrt{a+b \tanh ^4(x)}}\right )-\sqrt{a+b \tanh ^4(x)} \left (8 a+2 b \tanh ^4(x)+3 b \tanh ^2(x)+6 b\right )-\frac{3 \sqrt{a} \sqrt{b} \sqrt{a+b \tanh ^4(x)} \sinh ^{-1}\left (\frac{\sqrt{b} \tanh ^2(x)}{\sqrt{a}}\right )}{\sqrt{\frac{b \tanh ^4(x)}{a}+1}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]*(a + b*Tanh[x]^4)^(3/2),x]

[Out]

(-6*Sqrt[b]*(a + b)*ArcTanh[(Sqrt[b]*Tanh[x]^2)/Sqrt[a + b*Tanh[x]^4]] + 6*(a + b)^(3/2)*ArcTanh[(a + b*Tanh[x
]^2)/(Sqrt[a + b]*Sqrt[a + b*Tanh[x]^4])] - Sqrt[a + b*Tanh[x]^4]*(8*a + 6*b + 3*b*Tanh[x]^2 + 2*b*Tanh[x]^4)
- (3*Sqrt[a]*Sqrt[b]*ArcSinh[(Sqrt[b]*Tanh[x]^2)/Sqrt[a]]*Sqrt[a + b*Tanh[x]^4])/Sqrt[1 + (b*Tanh[x]^4)/a])/12

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Maple [C]  time = 0.082, size = 620, normalized size = 5. \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)^(3/2),x)

[Out]

-1/6*b*tanh(x)^4*(a+b*tanh(x)^4)^(1/2)-1/4*b*tanh(x)^2*(a+b*tanh(x)^4)^(1/2)-2/3*(a+b*tanh(x)^4)^(1/2)*a-1/2*b
*(a+b*tanh(x)^4)^(1/2)-1/2*(-5/3*a*b-b^2)/(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)*EllipticF(tanh(x)*(I/a^(1/2)*b^(1/2))^(1/2),I)-3/4*ln(
2*b^(1/2)*tanh(x)^2+2*(a+b*tanh(x)^4)^(1/2))*b^(1/2)*a-1/2*ln(2*b^(1/2)*tanh(x)^2+2*(a+b*tanh(x)^4)^(1/2))*b^(
3/2)-1/2*I*(7/5*a*b+b^2)*a^(1/2)/(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)/b^(1/2)*(EllipticF(tanh(x)*(I/a^(1/2)*b^(1/2))^(1/2),I)-Ellipti
cE(tanh(x)*(I/a^(1/2)*b^(1/2))^(1/2),I))+1/2*a^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))+a*b/(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))+1/2*b^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))-1/2*(5/3*a*b+b^2)/(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)*
EllipticF(tanh(x)*(I/a^(1/2)*b^(1/2))^(1/2),I)-1/2*I*(-7/5*a*b-b^2)*a^(1/2)/(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)/b^(1/2)*(EllipticF(t
anh(x)*(I/a^(1/2)*b^(1/2))^(1/2),I)-EllipticE(tanh(x)*(I/a^(1/2)*b^(1/2))^(1/2),I))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tanh \left (x\right )^{4} + a\right )}^{\frac{3}{2}} \tanh \left (x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)*(a+b*tanh(x)^4)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*tanh(x)^4 + a)^(3/2)*tanh(x), 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)^(3/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tanh ^{4}{\left (x \right )}\right )^{\frac{3}{2}} \tanh{\left (x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)*(a+b*tanh(x)**4)**(3/2),x)

[Out]

Integral((a + b*tanh(x)**4)**(3/2)*tanh(x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tanh \left (x\right )^{4} + a\right )}^{\frac{3}{2}} \tanh \left (x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)*(a+b*tanh(x)^4)^(3/2),x, algorithm="giac")

[Out]

integrate((b*tanh(x)^4 + a)^(3/2)*tanh(x), x)