3.238 \(\int \frac {\tanh ^5(x)}{(a+b \tanh ^2(x))^{3/2}} \, dx\)
Optimal. Leaf size=72 \[ -\frac {a^2}{b^2 (a+b) \sqrt {a+b \tanh ^2(x)}}-\frac {\sqrt {a+b \tanh ^2(x)}}{b^2}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2}} \]
[Out]
arctanh((a+b*tanh(x)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(3/2)-a^2/b^2/(a+b)/(a+b*tanh(x)^2)^(1/2)-(a+b*tanh(x)^2)^(1/
2)/b^2
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Rubi [A] time = 0.16, antiderivative size = 72, normalized size of antiderivative = 1.00,
number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used =
{3670, 446, 87, 63, 208} \[ -\frac {a^2}{b^2 (a+b) \sqrt {a+b \tanh ^2(x)}}-\frac {\sqrt {a+b \tanh ^2(x)}}{b^2}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[Tanh[x]^5/(a + b*Tanh[x]^2)^(3/2),x]
[Out]
ArcTanh[Sqrt[a + b*Tanh[x]^2]/Sqrt[a + b]]/(a + b)^(3/2) - a^2/(b^2*(a + b)*Sqrt[a + b*Tanh[x]^2]) - Sqrt[a +
b*Tanh[x]^2]/b^2
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 87
Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], ((c + d*x)^n*(e + f*x)^IntegerPart[p])/(a + b*x), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]
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]
Rule 446
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
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]))
Rubi steps
\begin {align*} \int \frac {\tanh ^5(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx &=\operatorname {Subst}\left (\int \frac {x^5}{\left (1-x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\tanh (x)\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2}{(1-x) (a+b x)^{3/2}} \, dx,x,\tanh ^2(x)\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {a^2}{b (a+b) (a+b x)^{3/2}}-\frac {1}{b \sqrt {a+b x}}-\frac {1}{(a+b) (-1+x) \sqrt {a+b x}}\right ) \, dx,x,\tanh ^2(x)\right )\\ &=-\frac {a^2}{b^2 (a+b) \sqrt {a+b \tanh ^2(x)}}-\frac {\sqrt {a+b \tanh ^2(x)}}{b^2}-\frac {\operatorname {Subst}\left (\int \frac {1}{(-1+x) \sqrt {a+b x}} \, dx,x,\tanh ^2(x)\right )}{2 (a+b)}\\ &=-\frac {a^2}{b^2 (a+b) \sqrt {a+b \tanh ^2(x)}}-\frac {\sqrt {a+b \tanh ^2(x)}}{b^2}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \tanh ^2(x)}\right )}{b (a+b)}\\ &=\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {a^2}{b^2 (a+b) \sqrt {a+b \tanh ^2(x)}}-\frac {\sqrt {a+b \tanh ^2(x)}}{b^2}\\ \end {align*}
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Mathematica [C] time = 0.11, size = 67, normalized size = 0.93 \[ \frac {b^2 \left (-\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b \tanh ^2(x)+a}{a+b}\right )\right )-(a+b) \left (2 a+b \tanh ^2(x)-b\right )}{b^2 (a+b) \sqrt {a+b \tanh ^2(x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[Tanh[x]^5/(a + b*Tanh[x]^2)^(3/2),x]
[Out]
(-(b^2*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Tanh[x]^2)/(a + b)]) - (a + b)*(2*a - b + b*Tanh[x]^2))/(b^2*(a
+ b)*Sqrt[a + b*Tanh[x]^2])
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fricas [B] time = 0.86, size = 3991, normalized size = 55.43 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(x)^5/(a+b*tanh(x)^2)^(3/2),x, algorithm="fricas")
[Out]
[1/4*(((a*b^2 + b^3)*cosh(x)^6 + 6*(a*b^2 + b^3)*cosh(x)*sinh(x)^5 + (a*b^2 + b^3)*sinh(x)^6 + (3*a*b^2 - b^3)
*cosh(x)^4 + (3*a*b^2 - b^3 + 15*(a*b^2 + b^3)*cosh(x)^2)*sinh(x)^4 + 4*(5*(a*b^2 + b^3)*cosh(x)^3 + (3*a*b^2
- b^3)*cosh(x))*sinh(x)^3 + a*b^2 + b^3 + (3*a*b^2 - b^3)*cosh(x)^2 + (15*(a*b^2 + b^3)*cosh(x)^4 + 3*a*b^2 -
b^3 + 6*(3*a*b^2 - b^3)*cosh(x)^2)*sinh(x)^2 + 2*(3*(a*b^2 + b^3)*cosh(x)^5 + 2*(3*a*b^2 - b^3)*cosh(x)^3 + (3
*a*b^2 - b^3)*cosh(x))*sinh(x))*sqrt(a + b)*log(((a^3 + a^2*b)*cosh(x)^8 + 8*(a^3 + a^2*b)*cosh(x)*sinh(x)^7 +
(a^3 + a^2*b)*sinh(x)^8 + 2*(2*a^3 + a^2*b)*cosh(x)^6 + 2*(2*a^3 + a^2*b + 14*(a^3 + a^2*b)*cosh(x)^2)*sinh(x
)^6 + 4*(14*(a^3 + a^2*b)*cosh(x)^3 + 3*(2*a^3 + a^2*b)*cosh(x))*sinh(x)^5 + (6*a^3 + 4*a^2*b - a*b^2 + b^3)*c
osh(x)^4 + (70*(a^3 + a^2*b)*cosh(x)^4 + 6*a^3 + 4*a^2*b - a*b^2 + b^3 + 30*(2*a^3 + a^2*b)*cosh(x)^2)*sinh(x)
^4 + 4*(14*(a^3 + a^2*b)*cosh(x)^5 + 10*(2*a^3 + a^2*b)*cosh(x)^3 + (6*a^3 + 4*a^2*b - a*b^2 + b^3)*cosh(x))*s
inh(x)^3 + a^3 + 3*a^2*b + 3*a*b^2 + b^3 + 2*(2*a^3 + 3*a^2*b - b^3)*cosh(x)^2 + 2*(14*(a^3 + a^2*b)*cosh(x)^6
+ 15*(2*a^3 + a^2*b)*cosh(x)^4 + 2*a^3 + 3*a^2*b - b^3 + 3*(6*a^3 + 4*a^2*b - a*b^2 + b^3)*cosh(x)^2)*sinh(x)
^2 + sqrt(2)*(a^2*cosh(x)^6 + 6*a^2*cosh(x)*sinh(x)^5 + a^2*sinh(x)^6 + 3*a^2*cosh(x)^4 + 3*(5*a^2*cosh(x)^2 +
a^2)*sinh(x)^4 + 4*(5*a^2*cosh(x)^3 + 3*a^2*cosh(x))*sinh(x)^3 + (3*a^2 + 2*a*b - b^2)*cosh(x)^2 + (15*a^2*co
sh(x)^4 + 18*a^2*cosh(x)^2 + 3*a^2 + 2*a*b - b^2)*sinh(x)^2 + a^2 + 2*a*b + b^2 + 2*(3*a^2*cosh(x)^5 + 6*a^2*c
osh(x)^3 + (3*a^2 + 2*a*b - b^2)*cosh(x))*sinh(x))*sqrt(a + b)*sqrt(((a + b)*cosh(x)^2 + (a + b)*sinh(x)^2 + a
- b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 4*(2*(a^3 + a^2*b)*cosh(x)^7 + 3*(2*a^3 + a^2*b)*cosh(x)^
5 + (6*a^3 + 4*a^2*b - a*b^2 + b^3)*cosh(x)^3 + (2*a^3 + 3*a^2*b - b^3)*cosh(x))*sinh(x))/(cosh(x)^6 + 6*cosh(
x)^5*sinh(x) + 15*cosh(x)^4*sinh(x)^2 + 20*cosh(x)^3*sinh(x)^3 + 15*cosh(x)^2*sinh(x)^4 + 6*cosh(x)*sinh(x)^5
+ sinh(x)^6)) + ((a*b^2 + b^3)*cosh(x)^6 + 6*(a*b^2 + b^3)*cosh(x)*sinh(x)^5 + (a*b^2 + b^3)*sinh(x)^6 + (3*a*
b^2 - b^3)*cosh(x)^4 + (3*a*b^2 - b^3 + 15*(a*b^2 + b^3)*cosh(x)^2)*sinh(x)^4 + 4*(5*(a*b^2 + b^3)*cosh(x)^3 +
(3*a*b^2 - b^3)*cosh(x))*sinh(x)^3 + a*b^2 + b^3 + (3*a*b^2 - b^3)*cosh(x)^2 + (15*(a*b^2 + b^3)*cosh(x)^4 +
3*a*b^2 - b^3 + 6*(3*a*b^2 - b^3)*cosh(x)^2)*sinh(x)^2 + 2*(3*(a*b^2 + b^3)*cosh(x)^5 + 2*(3*a*b^2 - b^3)*cosh
(x)^3 + (3*a*b^2 - b^3)*cosh(x))*sinh(x))*sqrt(a + b)*log(-((a + b)*cosh(x)^4 + 4*(a + b)*cosh(x)*sinh(x)^3 +
(a + b)*sinh(x)^4 - 2*b*cosh(x)^2 + 2*(3*(a + b)*cosh(x)^2 - b)*sinh(x)^2 + sqrt(2)*(cosh(x)^2 + 2*cosh(x)*sin
h(x) + sinh(x)^2 - 1)*sqrt(a + b)*sqrt(((a + b)*cosh(x)^2 + (a + b)*sinh(x)^2 + a - b)/(cosh(x)^2 - 2*cosh(x)*
sinh(x) + sinh(x)^2)) + 4*((a + b)*cosh(x)^3 - b*cosh(x))*sinh(x) + a + b)/(cosh(x)^2 + 2*cosh(x)*sinh(x) + si
nh(x)^2)) - 4*sqrt(2)*((2*a^3 + 4*a^2*b + 3*a*b^2 + b^3)*cosh(x)^4 + 4*(2*a^3 + 4*a^2*b + 3*a*b^2 + b^3)*cosh(
x)*sinh(x)^3 + (2*a^3 + 4*a^2*b + 3*a*b^2 + b^3)*sinh(x)^4 + 2*a^3 + 4*a^2*b + 3*a*b^2 + b^3 + 2*(2*a^3 + 2*a^
2*b - a*b^2 - b^3)*cosh(x)^2 + 2*(2*a^3 + 2*a^2*b - a*b^2 - b^3 + 3*(2*a^3 + 4*a^2*b + 3*a*b^2 + b^3)*cosh(x)^
2)*sinh(x)^2 + 4*((2*a^3 + 4*a^2*b + 3*a*b^2 + b^3)*cosh(x)^3 + (2*a^3 + 2*a^2*b - a*b^2 - b^3)*cosh(x))*sinh(
x))*sqrt(((a + b)*cosh(x)^2 + (a + b)*sinh(x)^2 + a - b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)))/((a^3*b
^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*cosh(x)^6 + 6*(a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*cosh(x)*sinh(x)^5 + (a^3*b
^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*sinh(x)^6 + a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5 + (3*a^3*b^2 + 5*a^2*b^3 + a*b
^4 - b^5)*cosh(x)^4 + (3*a^3*b^2 + 5*a^2*b^3 + a*b^4 - b^5 + 15*(a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*cosh(x)^
2)*sinh(x)^4 + 4*(5*(a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*cosh(x)^3 + (3*a^3*b^2 + 5*a^2*b^3 + a*b^4 - b^5)*co
sh(x))*sinh(x)^3 + (3*a^3*b^2 + 5*a^2*b^3 + a*b^4 - b^5)*cosh(x)^2 + (3*a^3*b^2 + 5*a^2*b^3 + a*b^4 - b^5 + 15
*(a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*cosh(x)^4 + 6*(3*a^3*b^2 + 5*a^2*b^3 + a*b^4 - b^5)*cosh(x)^2)*sinh(x)^
2 + 2*(3*(a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*cosh(x)^5 + 2*(3*a^3*b^2 + 5*a^2*b^3 + a*b^4 - b^5)*cosh(x)^3 +
(3*a^3*b^2 + 5*a^2*b^3 + a*b^4 - b^5)*cosh(x))*sinh(x)), -1/2*(((a*b^2 + b^3)*cosh(x)^6 + 6*(a*b^2 + b^3)*cos
h(x)*sinh(x)^5 + (a*b^2 + b^3)*sinh(x)^6 + (3*a*b^2 - b^3)*cosh(x)^4 + (3*a*b^2 - b^3 + 15*(a*b^2 + b^3)*cosh(
x)^2)*sinh(x)^4 + 4*(5*(a*b^2 + b^3)*cosh(x)^3 + (3*a*b^2 - b^3)*cosh(x))*sinh(x)^3 + a*b^2 + b^3 + (3*a*b^2 -
b^3)*cosh(x)^2 + (15*(a*b^2 + b^3)*cosh(x)^4 + 3*a*b^2 - b^3 + 6*(3*a*b^2 - b^3)*cosh(x)^2)*sinh(x)^2 + 2*(3*
(a*b^2 + b^3)*cosh(x)^5 + 2*(3*a*b^2 - b^3)*cosh(x)^3 + (3*a*b^2 - b^3)*cosh(x))*sinh(x))*sqrt(-a - b)*arctan(
sqrt(2)*(a*cosh(x)^2 + 2*a*cosh(x)*sinh(x) + a*sinh(x)^2 + a + b)*sqrt(-a - b)*sqrt(((a + b)*cosh(x)^2 + (a +
b)*sinh(x)^2 + a - b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2))/((a^2 + a*b)*cosh(x)^4 + 4*(a^2 + a*b)*cosh
(x)*sinh(x)^3 + (a^2 + a*b)*sinh(x)^4 + (2*a^2 + a*b - b^2)*cosh(x)^2 + (6*(a^2 + a*b)*cosh(x)^2 + 2*a^2 + a*b
- b^2)*sinh(x)^2 + a^2 + 2*a*b + b^2 + 2*(2*(a^2 + a*b)*cosh(x)^3 + (2*a^2 + a*b - b^2)*cosh(x))*sinh(x))) +
((a*b^2 + b^3)*cosh(x)^6 + 6*(a*b^2 + b^3)*cosh(x)*sinh(x)^5 + (a*b^2 + b^3)*sinh(x)^6 + (3*a*b^2 - b^3)*cosh(
x)^4 + (3*a*b^2 - b^3 + 15*(a*b^2 + b^3)*cosh(x)^2)*sinh(x)^4 + 4*(5*(a*b^2 + b^3)*cosh(x)^3 + (3*a*b^2 - b^3)
*cosh(x))*sinh(x)^3 + a*b^2 + b^3 + (3*a*b^2 - b^3)*cosh(x)^2 + (15*(a*b^2 + b^3)*cosh(x)^4 + 3*a*b^2 - b^3 +
6*(3*a*b^2 - b^3)*cosh(x)^2)*sinh(x)^2 + 2*(3*(a*b^2 + b^3)*cosh(x)^5 + 2*(3*a*b^2 - b^3)*cosh(x)^3 + (3*a*b^2
- b^3)*cosh(x))*sinh(x))*sqrt(-a - b)*arctan(sqrt(2)*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1)*sqrt(-a
- b)*sqrt(((a + b)*cosh(x)^2 + (a + b)*sinh(x)^2 + a - b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2))/((a + b
)*cosh(x)^4 + 4*(a + b)*cosh(x)*sinh(x)^3 + (a + b)*sinh(x)^4 + 2*(a - b)*cosh(x)^2 + 2*(3*(a + b)*cosh(x)^2 +
a - b)*sinh(x)^2 + 4*((a + b)*cosh(x)^3 + (a - b)*cosh(x))*sinh(x) + a + b)) + 2*sqrt(2)*((2*a^3 + 4*a^2*b +
3*a*b^2 + b^3)*cosh(x)^4 + 4*(2*a^3 + 4*a^2*b + 3*a*b^2 + b^3)*cosh(x)*sinh(x)^3 + (2*a^3 + 4*a^2*b + 3*a*b^2
+ b^3)*sinh(x)^4 + 2*a^3 + 4*a^2*b + 3*a*b^2 + b^3 + 2*(2*a^3 + 2*a^2*b - a*b^2 - b^3)*cosh(x)^2 + 2*(2*a^3 +
2*a^2*b - a*b^2 - b^3 + 3*(2*a^3 + 4*a^2*b + 3*a*b^2 + b^3)*cosh(x)^2)*sinh(x)^2 + 4*((2*a^3 + 4*a^2*b + 3*a*b
^2 + b^3)*cosh(x)^3 + (2*a^3 + 2*a^2*b - a*b^2 - b^3)*cosh(x))*sinh(x))*sqrt(((a + b)*cosh(x)^2 + (a + b)*sinh
(x)^2 + a - b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)))/((a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*cosh(x)^6
+ 6*(a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*cosh(x)*sinh(x)^5 + (a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*sinh(x)^6
+ a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5 + (3*a^3*b^2 + 5*a^2*b^3 + a*b^4 - b^5)*cosh(x)^4 + (3*a^3*b^2 + 5*a^2*b
^3 + a*b^4 - b^5 + 15*(a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*cosh(x)^2)*sinh(x)^4 + 4*(5*(a^3*b^2 + 3*a^2*b^3 +
3*a*b^4 + b^5)*cosh(x)^3 + (3*a^3*b^2 + 5*a^2*b^3 + a*b^4 - b^5)*cosh(x))*sinh(x)^3 + (3*a^3*b^2 + 5*a^2*b^3
+ a*b^4 - b^5)*cosh(x)^2 + (3*a^3*b^2 + 5*a^2*b^3 + a*b^4 - b^5 + 15*(a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*cos
h(x)^4 + 6*(3*a^3*b^2 + 5*a^2*b^3 + a*b^4 - b^5)*cosh(x)^2)*sinh(x)^2 + 2*(3*(a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 +
b^5)*cosh(x)^5 + 2*(3*a^3*b^2 + 5*a^2*b^3 + a*b^4 - b^5)*cosh(x)^3 + (3*a^3*b^2 + 5*a^2*b^3 + a*b^4 - b^5)*cos
h(x))*sinh(x))]
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(x)^5/(a+b*tanh(x)^2)^(3/2),x, algorithm="giac")
[Out]
Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep+1)]Evaluation time: 1.35Error: Bad Argument Type
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maple [B] time = 0.08, size = 322, normalized size = 4.47 \[ -\frac {\tanh ^{2}\relax (x )}{b \sqrt {a +b \left (\tanh ^{2}\relax (x )\right )}}-\frac {2 a}{b^{2} \sqrt {a +b \left (\tanh ^{2}\relax (x )\right )}}+\frac {1}{b \sqrt {a +b \left (\tanh ^{2}\relax (x )\right )}}-\frac {1}{2 \left (a +b \right ) \sqrt {\left (\tanh \relax (x )-1\right )^{2} b +2 \left (\tanh \relax (x )-1\right ) b +a +b}}+\frac {b \left (2 \left (\tanh \relax (x )-1\right ) b +2 b \right )}{\left (a +b \right ) \left (4 b \left (a +b \right )-4 b^{2}\right ) \sqrt {\left (\tanh \relax (x )-1\right )^{2} b +2 \left (\tanh \relax (x )-1\right ) b +a +b}}+\frac {\ln \left (\frac {2 a +2 b +2 \left (\tanh \relax (x )-1\right ) b +2 \sqrt {a +b}\, \sqrt {\left (\tanh \relax (x )-1\right )^{2} b +2 \left (\tanh \relax (x )-1\right ) b +a +b}}{\tanh \relax (x )-1}\right )}{2 \left (a +b \right )^{\frac {3}{2}}}-\frac {1}{2 \left (a +b \right ) \sqrt {\left (1+\tanh \relax (x )\right )^{2} b -2 \left (1+\tanh \relax (x )\right ) b +a +b}}-\frac {b \left (2 \left (1+\tanh \relax (x )\right ) b -2 b \right )}{\left (a +b \right ) \left (4 b \left (a +b \right )-4 b^{2}\right ) \sqrt {\left (1+\tanh \relax (x )\right )^{2} b -2 \left (1+\tanh \relax (x )\right ) b +a +b}}+\frac {\ln \left (\frac {2 a +2 b -2 \left (1+\tanh \relax (x )\right ) b +2 \sqrt {a +b}\, \sqrt {\left (1+\tanh \relax (x )\right )^{2} b -2 \left (1+\tanh \relax (x )\right ) b +a +b}}{1+\tanh \relax (x )}\right )}{2 \left (a +b \right )^{\frac {3}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tanh(x)^5/(a+b*tanh(x)^2)^(3/2),x)
[Out]
-tanh(x)^2/b/(a+b*tanh(x)^2)^(1/2)-2*a/b^2/(a+b*tanh(x)^2)^(1/2)+1/b/(a+b*tanh(x)^2)^(1/2)-1/2/(a+b)/((tanh(x)
-1)^2*b+2*(tanh(x)-1)*b+a+b)^(1/2)+b/(a+b)*(2*(tanh(x)-1)*b+2*b)/(4*b*(a+b)-4*b^2)/((tanh(x)-1)^2*b+2*(tanh(x)
-1)*b+a+b)^(1/2)+1/2/(a+b)^(3/2)*ln((2*a+2*b+2*(tanh(x)-1)*b+2*(a+b)^(1/2)*((tanh(x)-1)^2*b+2*(tanh(x)-1)*b+a+
b)^(1/2))/(tanh(x)-1))-1/2/(a+b)/((1+tanh(x))^2*b-2*(1+tanh(x))*b+a+b)^(1/2)-b/(a+b)*(2*(1+tanh(x))*b-2*b)/(4*
b*(a+b)-4*b^2)/((1+tanh(x))^2*b-2*(1+tanh(x))*b+a+b)^(1/2)+1/2/(a+b)^(3/2)*ln((2*a+2*b-2*(1+tanh(x))*b+2*(a+b)
^(1/2)*((1+tanh(x))^2*b-2*(1+tanh(x))*b+a+b)^(1/2))/(1+tanh(x)))
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh \relax (x)^{5}}{{\left (b \tanh \relax (x)^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(x)^5/(a+b*tanh(x)^2)^(3/2),x, algorithm="maxima")
[Out]
integrate(tanh(x)^5/(b*tanh(x)^2 + a)^(3/2), x)
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mupad [B] time = 2.52, size = 70, normalized size = 0.97 \[ \frac {\mathrm {atanh}\left (\frac {\sqrt {b\,{\mathrm {tanh}\relax (x)}^2+a}\,\left (2\,a+2\,b\right )}{2\,{\left (a+b\right )}^{3/2}}\right )}{{\left (a+b\right )}^{3/2}}-\frac {\sqrt {b\,{\mathrm {tanh}\relax (x)}^2+a}}{b^2}-\frac {a^2}{b^2\,\left (a+b\right )\,\sqrt {b\,{\mathrm {tanh}\relax (x)}^2+a}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tanh(x)^5/(a + b*tanh(x)^2)^(3/2),x)
[Out]
atanh(((a + b*tanh(x)^2)^(1/2)*(2*a + 2*b))/(2*(a + b)^(3/2)))/(a + b)^(3/2) - (a + b*tanh(x)^2)^(1/2)/b^2 - a
^2/(b^2*(a + b)*(a + b*tanh(x)^2)^(1/2))
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh ^{5}{\relax (x )}}{\left (a + b \tanh ^{2}{\relax (x )}\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(x)**5/(a+b*tanh(x)**2)**(3/2),x)
[Out]
Integral(tanh(x)**5/(a + b*tanh(x)**2)**(3/2), x)
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