3.3.48 \(\int \frac {\tanh ^4(x)}{(a+b \tanh ^2(x))^{5/2}} \, dx\) [248]

Optimal. Leaf size=90 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {a+b} \tanh (x)}{\sqrt {a+b \tanh ^2(x)}}\right )}{(a+b)^{5/2}}+\frac {a \tanh (x)}{3 b (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}-\frac {(a+4 b) \tanh (x)}{3 b (a+b)^2 \sqrt {a+b \tanh ^2(x)}} \]

[Out]

arctanh((a+b)^(1/2)*tanh(x)/(a+b*tanh(x)^2)^(1/2))/(a+b)^(5/2)-1/3*(a+4*b)*tanh(x)/b/(a+b)^2/(a+b*tanh(x)^2)^(
1/2)+1/3*a*tanh(x)/b/(a+b)/(a+b*tanh(x)^2)^(3/2)

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Rubi [A]
time = 0.09, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {3751, 481, 541, 12, 385, 212} \begin {gather*} -\frac {(a+4 b) \tanh (x)}{3 b (a+b)^2 \sqrt {a+b \tanh ^2(x)}}+\frac {a \tanh (x)}{3 b (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b} \tanh (x)}{\sqrt {a+b \tanh ^2(x)}}\right )}{(a+b)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 212

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

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 481

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 541

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

Rule 3751

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

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Mathematica [A]
time = 2.29, size = 132, normalized size = 1.47 \begin {gather*} \frac {\tanh ^3(x) \left (3 \tanh ^{-1}\left (\frac {\sqrt {\frac {(a+b) \tanh ^2(x)}{a}}}{\sqrt {1+\frac {b \tanh ^2(x)}{a}}}\right ) \left (b+a \coth ^2(x)\right )^2 \sqrt {\frac {(a+b) \tanh ^2(x)}{a}}-(a+b) \left (a+4 b+3 a \coth ^2(x)\right ) \sqrt {1+\frac {b \tanh ^2(x)}{a}}\right )}{3 (a+b)^3 \left (a+b \tanh ^2(x)\right )^{3/2} \sqrt {1+\frac {b \tanh ^2(x)}{a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Tanh[x]^3*(3*ArcTanh[Sqrt[((a + b)*Tanh[x]^2)/a]/Sqrt[1 + (b*Tanh[x]^2)/a]]*(b + a*Coth[x]^2)^2*Sqrt[((a + b)
*Tanh[x]^2)/a] - (a + b)*(a + 4*b + 3*a*Coth[x]^2)*Sqrt[1 + (b*Tanh[x]^2)/a]))/(3*(a + b)^3*(a + b*Tanh[x]^2)^
(3/2)*Sqrt[1 + (b*Tanh[x]^2)/a])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(641\) vs. \(2(76)=152\).
time = 0.72, size = 642, normalized size = 7.13 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*tanh(x)/b/(a+b*tanh(x)^2)^(3/2)-1/2*a/b*(1/3*tanh(x)/a/(a+b*tanh(x)^2)^(3/2)+2/3/a^2*tanh(x)/(a+b*tanh(x)^
2)^(1/2))-1/3*tanh(x)/a/(a+b*tanh(x)^2)^(3/2)-2/3/a^2*tanh(x)/(a+b*tanh(x)^2)^(1/2)+1/6/(a+b)/(b*(1+tanh(x))^2
-2*b*(1+tanh(x))+a+b)^(3/2)+1/2*b/(a+b)*(2/3*(2*b*(1+tanh(x))-2*b)/(4*b*(a+b)-4*b^2)/(b*(1+tanh(x))^2-2*b*(1+t
anh(x))+a+b)^(3/2)+16/3*b/(4*b*(a+b)-4*b^2)^2*(2*b*(1+tanh(x))-2*b)/(b*(1+tanh(x))^2-2*b*(1+tanh(x))+a+b)^(1/2
))+1/2/(a+b)*(1/(a+b)/(b*(1+tanh(x))^2-2*b*(1+tanh(x))+a+b)^(1/2)+2*b/(a+b)*(2*b*(1+tanh(x))-2*b)/(4*b*(a+b)-4
*b^2)/(b*(1+tanh(x))^2-2*b*(1+tanh(x))+a+b)^(1/2)-1/(a+b)^(3/2)*ln((2*a+2*b-2*b*(1+tanh(x))+2*(a+b)^(1/2)*(b*(
1+tanh(x))^2-2*b*(1+tanh(x))+a+b)^(1/2))/(1+tanh(x))))-1/6/(a+b)/(b*(tanh(x)-1)^2+2*b*(tanh(x)-1)+a+b)^(3/2)+1
/2*b/(a+b)*(2/3*(2*b*(tanh(x)-1)+2*b)/(4*b*(a+b)-4*b^2)/(b*(tanh(x)-1)^2+2*b*(tanh(x)-1)+a+b)^(3/2)+16/3*b/(4*
b*(a+b)-4*b^2)^2*(2*b*(tanh(x)-1)+2*b)/(b*(tanh(x)-1)^2+2*b*(tanh(x)-1)+a+b)^(1/2))-1/2/(a+b)*(1/(a+b)/(b*(tan
h(x)-1)^2+2*b*(tanh(x)-1)+a+b)^(1/2)-2*b/(a+b)*(2*b*(tanh(x)-1)+2*b)/(4*b*(a+b)-4*b^2)/(b*(tanh(x)-1)^2+2*b*(t
anh(x)-1)+a+b)^(1/2)-1/(a+b)^(3/2)*ln((2*a+2*b+2*b*(tanh(x)-1)+2*(a+b)^(1/2)*(b*(tanh(x)-1)^2+2*b*(tanh(x)-1)+
a+b)^(1/2))/(tanh(x)-1)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 2545 vs. \(2 (76) = 152\).
time = 0.69, size = 5719, normalized size = 63.54 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(3*((a^2 + 2*a*b + b^2)*cosh(x)^8 + 8*(a^2 + 2*a*b + b^2)*cosh(x)*sinh(x)^7 + (a^2 + 2*a*b + b^2)*sinh(x
)^8 + 4*(a^2 - b^2)*cosh(x)^6 + 4*(7*(a^2 + 2*a*b + b^2)*cosh(x)^2 + a^2 - b^2)*sinh(x)^6 + 8*(7*(a^2 + 2*a*b
+ b^2)*cosh(x)^3 + 3*(a^2 - b^2)*cosh(x))*sinh(x)^5 + 2*(3*a^2 - 2*a*b + 3*b^2)*cosh(x)^4 + 2*(35*(a^2 + 2*a*b
+ b^2)*cosh(x)^4 + 30*(a^2 - b^2)*cosh(x)^2 + 3*a^2 - 2*a*b + 3*b^2)*sinh(x)^4 + 8*(7*(a^2 + 2*a*b + b^2)*cos
h(x)^5 + 10*(a^2 - b^2)*cosh(x)^3 + (3*a^2 - 2*a*b + 3*b^2)*cosh(x))*sinh(x)^3 + 4*(a^2 - b^2)*cosh(x)^2 + 4*(
7*(a^2 + 2*a*b + b^2)*cosh(x)^6 + 15*(a^2 - b^2)*cosh(x)^4 + 3*(3*a^2 - 2*a*b + 3*b^2)*cosh(x)^2 + a^2 - b^2)*
sinh(x)^2 + a^2 + 2*a*b + b^2 + 8*((a^2 + 2*a*b + b^2)*cosh(x)^7 + 3*(a^2 - b^2)*cosh(x)^5 + (3*a^2 - 2*a*b +
3*b^2)*cosh(x)^3 + (a^2 - b^2)*cosh(x))*sinh(x))*sqrt(a + b)*log(-((a*b^2 + b^3)*cosh(x)^8 + 8*(a*b^2 + b^3)*c
osh(x)*sinh(x)^7 + (a*b^2 + b^3)*sinh(x)^8 - 2*(a*b^2 + 2*b^3)*cosh(x)^6 - 2*(a*b^2 + 2*b^3 - 14*(a*b^2 + b^3)
*cosh(x)^2)*sinh(x)^6 + 4*(14*(a*b^2 + b^3)*cosh(x)^3 - 3*(a*b^2 + 2*b^3)*cosh(x))*sinh(x)^5 + (a^3 - a^2*b +
4*a*b^2 + 6*b^3)*cosh(x)^4 + (70*(a*b^2 + b^3)*cosh(x)^4 + a^3 - a^2*b + 4*a*b^2 + 6*b^3 - 30*(a*b^2 + 2*b^3)*
cosh(x)^2)*sinh(x)^4 + 4*(14*(a*b^2 + b^3)*cosh(x)^5 - 10*(a*b^2 + 2*b^3)*cosh(x)^3 + (a^3 - a^2*b + 4*a*b^2 +
6*b^3)*cosh(x))*sinh(x)^3 + a^3 + 3*a^2*b + 3*a*b^2 + b^3 + 2*(a^3 - 3*a*b^2 - 2*b^3)*cosh(x)^2 + 2*(14*(a*b^
2 + b^3)*cosh(x)^6 - 15*(a*b^2 + 2*b^3)*cosh(x)^4 + a^3 - 3*a*b^2 - 2*b^3 + 3*(a^3 - a^2*b + 4*a*b^2 + 6*b^3)*
cosh(x)^2)*sinh(x)^2 + sqrt(2)*(b^2*cosh(x)^6 + 6*b^2*cosh(x)*sinh(x)^5 + b^2*sinh(x)^6 - 3*b^2*cosh(x)^4 + 3*
(5*b^2*cosh(x)^2 - b^2)*sinh(x)^4 + 4*(5*b^2*cosh(x)^3 - 3*b^2*cosh(x))*sinh(x)^3 - (a^2 - 2*a*b - 3*b^2)*cosh
(x)^2 + (15*b^2*cosh(x)^4 - 18*b^2*cosh(x)^2 - a^2 + 2*a*b + 3*b^2)*sinh(x)^2 - a^2 - 2*a*b - b^2 + 2*(3*b^2*c
osh(x)^5 - 6*b^2*cosh(x)^3 - (a^2 - 2*a*b - 3*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*b^2 + b^3)*cosh(x)^7 - 3*(a*b^2
+ 2*b^3)*cosh(x)^5 + (a^3 - a^2*b + 4*a*b^2 + 6*b^3)*cosh(x)^3 + (a^3 - 3*a*b^2 - 2*b^3)*cosh(x))*sinh(x))/(c
osh(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)) + 3*((a^2 + 2*a*b + b^2)*cosh(x)^8 + 8*(a^2 + 2*a*b + b^2)*cosh(x)*sinh(x)^7 +
(a^2 + 2*a*b + b^2)*sinh(x)^8 + 4*(a^2 - b^2)*cosh(x)^6 + 4*(7*(a^2 + 2*a*b + b^2)*cosh(x)^2 + a^2 - b^2)*sin
h(x)^6 + 8*(7*(a^2 + 2*a*b + b^2)*cosh(x)^3 + 3*(a^2 - b^2)*cosh(x))*sinh(x)^5 + 2*(3*a^2 - 2*a*b + 3*b^2)*cos
h(x)^4 + 2*(35*(a^2 + 2*a*b + b^2)*cosh(x)^4 + 30*(a^2 - b^2)*cosh(x)^2 + 3*a^2 - 2*a*b + 3*b^2)*sinh(x)^4 + 8
*(7*(a^2 + 2*a*b + b^2)*cosh(x)^5 + 10*(a^2 - b^2)*cosh(x)^3 + (3*a^2 - 2*a*b + 3*b^2)*cosh(x))*sinh(x)^3 + 4*
(a^2 - b^2)*cosh(x)^2 + 4*(7*(a^2 + 2*a*b + b^2)*cosh(x)^6 + 15*(a^2 - b^2)*cosh(x)^4 + 3*(3*a^2 - 2*a*b + 3*b
^2)*cosh(x)^2 + a^2 - b^2)*sinh(x)^2 + a^2 + 2*a*b + b^2 + 8*((a^2 + 2*a*b + b^2)*cosh(x)^7 + 3*(a^2 - b^2)*co
sh(x)^5 + (3*a^2 - 2*a*b + 3*b^2)*cosh(x)^3 + (a^2 - b^2)*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*a*cosh(x)^2 + 2*(3*(a + b)*cosh(x)^2 + a)*sinh(x)^2 + s
qrt(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)) + 4*((a + b)*cosh(x)^3 + a*cosh(x))*sinh(x) + a + b)/(c
osh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)) - 16*sqrt(2)*((a^2 + 2*a*b + b^2)*cosh(x)^6 + 6*(a^2 + 2*a*b + b^2)
*cosh(x)*sinh(x)^5 + (a^2 + 2*a*b + b^2)*sinh(x)^6 - 3*(a*b + b^2)*cosh(x)^4 + 3*(5*(a^2 + 2*a*b + b^2)*cosh(x
)^2 - a*b - b^2)*sinh(x)^4 + 4*(5*(a^2 + 2*a*b + b^2)*cosh(x)^3 - 3*(a*b + b^2)*cosh(x))*sinh(x)^3 + 3*(a*b +
b^2)*cosh(x)^2 + 3*(5*(a^2 + 2*a*b + b^2)*cosh(x)^4 - 6*(a*b + b^2)*cosh(x)^2 + a*b + b^2)*sinh(x)^2 - a^2 - 2
*a*b - b^2 + 6*((a^2 + 2*a*b + b^2)*cosh(x)^5 - 2*(a*b + b^2)*cosh(x)^3 + (a*b + b^2)*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^5 + 5*a^4*b +
10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*cosh(x)^8 + 8*(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b
^5)*cosh(x)*sinh(x)^7 + (a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*sinh(x)^8 + 4*(a^5 + 3*a^4*b
+ 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*cosh(x)^6 + 4*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5
+ 7*(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*cosh(x)^2)*sinh(x)^6 + 8*(7*(a^5 + 5*a^4*b + 10
*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*cosh(x)^3 + 3*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*c
osh(x))*sinh(x)^5 + a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5 + 2*(3*a^5 + 7*a^4*b + 6*a^3*b^2 +
6*a^2*b^3 + 7*a*b^4 + 3*b^5)*cosh(x)^4 + 2*(3*a^5 + 7*a^4*b + 6*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 + 3*b^5 + 35*(a
^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^...

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tanh ^{4}{\left (x \right )}}{\left (a + b \tanh ^{2}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 708 vs. \(2 (76) = 152\).
time = 0.62, size = 708, normalized size = 7.87 \begin {gather*} -\frac {\sqrt {a + b} \log \left ({\left | -{\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}\right )} {\left (a + b\right )} - \sqrt {a + b} {\left (a - b\right )} \right |}\right )}{2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac {\sqrt {a + b} \log \left ({\left | -\sqrt {a + b} e^{\left (2 \, x\right )} + \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b} + \sqrt {a + b} \right |}\right )}{2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} + \frac {\sqrt {a + b} \log \left ({\left | -\sqrt {a + b} e^{\left (2 \, x\right )} + \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b} - \sqrt {a + b} \right |}\right )}{2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac {4 \, {\left ({\left ({\left (\frac {{\left (a^{7} b^{2} + 5 \, a^{6} b^{3} + 10 \, a^{5} b^{4} + 10 \, a^{4} b^{5} + 5 \, a^{3} b^{6} + a^{2} b^{7}\right )} e^{\left (2 \, x\right )}}{a^{8} b^{2} + 6 \, a^{7} b^{3} + 15 \, a^{6} b^{4} + 20 \, a^{5} b^{5} + 15 \, a^{4} b^{6} + 6 \, a^{3} b^{7} + a^{2} b^{8}} - \frac {3 \, {\left (a^{6} b^{3} + 4 \, a^{5} b^{4} + 6 \, a^{4} b^{5} + 4 \, a^{3} b^{6} + a^{2} b^{7}\right )}}{a^{8} b^{2} + 6 \, a^{7} b^{3} + 15 \, a^{6} b^{4} + 20 \, a^{5} b^{5} + 15 \, a^{4} b^{6} + 6 \, a^{3} b^{7} + a^{2} b^{8}}\right )} e^{\left (2 \, x\right )} + \frac {3 \, {\left (a^{6} b^{3} + 4 \, a^{5} b^{4} + 6 \, a^{4} b^{5} + 4 \, a^{3} b^{6} + a^{2} b^{7}\right )}}{a^{8} b^{2} + 6 \, a^{7} b^{3} + 15 \, a^{6} b^{4} + 20 \, a^{5} b^{5} + 15 \, a^{4} b^{6} + 6 \, a^{3} b^{7} + a^{2} b^{8}}\right )} e^{\left (2 \, x\right )} - \frac {a^{7} b^{2} + 5 \, a^{6} b^{3} + 10 \, a^{5} b^{4} + 10 \, a^{4} b^{5} + 5 \, a^{3} b^{6} + a^{2} b^{7}}{a^{8} b^{2} + 6 \, a^{7} b^{3} + 15 \, a^{6} b^{4} + 20 \, a^{5} b^{5} + 15 \, a^{4} b^{6} + 6 \, a^{3} b^{7} + a^{2} b^{8}}\right )}}{3 \, {\left (a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(a + b)*log(abs(-(sqrt(a + b)*e^(2*x) - sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2*x) + a +
b))*(a + b) - sqrt(a + b)*(a - b)))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 1/2*sqrt(a + b)*log(abs(-sqrt(a + b)*e^(
2*x) + sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b) + sqrt(a + b)))/(a^3 + 3*a^2*b + 3*a*b^
2 + b^3) + 1/2*sqrt(a + b)*log(abs(-sqrt(a + b)*e^(2*x) + sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2*
x) + a + b) - sqrt(a + b)))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 4/3*((((a^7*b^2 + 5*a^6*b^3 + 10*a^5*b^4 + 10*a^
4*b^5 + 5*a^3*b^6 + a^2*b^7)*e^(2*x)/(a^8*b^2 + 6*a^7*b^3 + 15*a^6*b^4 + 20*a^5*b^5 + 15*a^4*b^6 + 6*a^3*b^7 +
a^2*b^8) - 3*(a^6*b^3 + 4*a^5*b^4 + 6*a^4*b^5 + 4*a^3*b^6 + a^2*b^7)/(a^8*b^2 + 6*a^7*b^3 + 15*a^6*b^4 + 20*a
^5*b^5 + 15*a^4*b^6 + 6*a^3*b^7 + a^2*b^8))*e^(2*x) + 3*(a^6*b^3 + 4*a^5*b^4 + 6*a^4*b^5 + 4*a^3*b^6 + a^2*b^7
)/(a^8*b^2 + 6*a^7*b^3 + 15*a^6*b^4 + 20*a^5*b^5 + 15*a^4*b^6 + 6*a^3*b^7 + a^2*b^8))*e^(2*x) - (a^7*b^2 + 5*a
^6*b^3 + 10*a^5*b^4 + 10*a^4*b^5 + 5*a^3*b^6 + a^2*b^7)/(a^8*b^2 + 6*a^7*b^3 + 15*a^6*b^4 + 20*a^5*b^5 + 15*a^
4*b^6 + 6*a^3*b^7 + a^2*b^8))/(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b)^(3/2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {tanh}\left (x\right )}^4}{{\left (b\,{\mathrm {tanh}\left (x\right )}^2+a\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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