Integrand size = 15, antiderivative size = 118 \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {a+b \tanh ^2(x)}{\sqrt {a+b} \sqrt {a+b \tanh ^4(x)}}\right )}{2 (a+b)^{5/2}}-\frac {a-b \tanh ^2(x)}{6 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}}-\frac {3 a^2-b (5 a+2 b) \tanh ^2(x)}{6 a^2 (a+b)^2 \sqrt {a+b \tanh ^4(x)}} \] Output:
1/2*arctanh((a+b*tanh(x)^2)/(a+b)^(1/2)/(a+b*tanh(x)^4)^(1/2))/(a+b)^(5/2) -1/6*(a-b*tanh(x)^2)/a/(a+b)/(a+b*tanh(x)^4)^(3/2)-1/6*(3*a^2-b*(5*a+2*b)* tanh(x)^2)/a^2/(a+b)^2/(a+b*tanh(x)^4)^(1/2)
Time = 0.56 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.96 \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{5/2}} \, dx=\frac {1}{6} \left (\frac {3 \text {arctanh}\left (\frac {a+b \tanh ^2(x)}{\sqrt {a+b} \sqrt {a+b \tanh ^4(x)}}\right )}{(a+b)^{5/2}}+\frac {-a^2 (4 a+b)+3 a b (2 a+b) \tanh ^2(x)-3 a^2 b \tanh ^4(x)+b^2 (5 a+2 b) \tanh ^6(x)}{a^2 (a+b)^2 \left (a+b \tanh ^4(x)\right )^{3/2}}\right ) \] Input:
Integrate[Tanh[x]/(a + b*Tanh[x]^4)^(5/2),x]
Output:
((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
Time = 0.41 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.14, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.733, Rules used = {3042, 26, 4153, 26, 1577, 496, 25, 686, 27, 488, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i \tan (i x)}{\left (a+b \tan (i x)^4\right )^{5/2}}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {\tan (i x)}{\left (b \tan (i x)^4+a\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle -i \int \frac {i \tanh (x)}{\left (1-\tanh ^2(x)\right ) \left (b \tanh ^4(x)+a\right )^{5/2}}d\tanh (x)\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \int \frac {\tanh (x)}{\left (1-\tanh ^2(x)\right ) \left (a+b \tanh ^4(x)\right )^{5/2}}d\tanh (x)\) |
\(\Big \downarrow \) 1577 |
\(\displaystyle \frac {1}{2} \int \frac {1}{\left (1-\tanh ^2(x)\right ) \left (b \tanh ^4(x)+a\right )^{5/2}}d\tanh ^2(x)\) |
\(\Big \downarrow \) 496 |
\(\displaystyle \frac {1}{2} \left (-\frac {\int -\frac {-2 b \tanh ^2(x)+3 a+2 b}{\left (1-\tanh ^2(x)\right ) \left (b \tanh ^4(x)+a\right )^{3/2}}d\tanh ^2(x)}{3 a (a+b)}-\frac {a-b \tanh ^2(x)}{3 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {-2 b \tanh ^2(x)+3 a+2 b}{\left (1-\tanh ^2(x)\right ) \left (b \tanh ^4(x)+a\right )^{3/2}}d\tanh ^2(x)}{3 a (a+b)}-\frac {a-b \tanh ^2(x)}{3 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 686 |
\(\displaystyle \frac {1}{2} \left (\frac {-\frac {\int -\frac {3 a^2 b}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^4(x)+a}}d\tanh ^2(x)}{a b (a+b)}-\frac {3 a^2-b (5 a+2 b) \tanh ^2(x)}{a (a+b) \sqrt {a+b \tanh ^4(x)}}}{3 a (a+b)}-\frac {a-b \tanh ^2(x)}{3 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {\frac {3 a \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^4(x)+a}}d\tanh ^2(x)}{a+b}-\frac {3 a^2-b (5 a+2 b) \tanh ^2(x)}{a (a+b) \sqrt {a+b \tanh ^4(x)}}}{3 a (a+b)}-\frac {a-b \tanh ^2(x)}{3 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 488 |
\(\displaystyle \frac {1}{2} \left (\frac {-\frac {3 a \int \frac {1}{-\tanh ^4(x)+a+b}d\frac {-b \tanh ^2(x)-a}{\sqrt {b \tanh ^4(x)+a}}}{a+b}-\frac {3 a^2-b (5 a+2 b) \tanh ^2(x)}{a (a+b) \sqrt {a+b \tanh ^4(x)}}}{3 a (a+b)}-\frac {a-b \tanh ^2(x)}{3 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {-\frac {3 a^2-b (5 a+2 b) \tanh ^2(x)}{a (a+b) \sqrt {a+b \tanh ^4(x)}}-\frac {3 a \text {arctanh}\left (\frac {-a-b \tanh ^2(x)}{\sqrt {a+b} \sqrt {a+b \tanh ^4(x)}}\right )}{(a+b)^{3/2}}}{3 a (a+b)}-\frac {a-b \tanh ^2(x)}{3 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}}\right )\) |
Input:
Int[Tanh[x]/(a + b*Tanh[x]^4)^(5/2),x]
Output:
(-1/3*(a - b*Tanh[x]^2)/(a*(a + b)*(a + b*Tanh[x]^4)^(3/2)) + ((-3*a*ArcTa nh[(-a - b*Tanh[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tanh[x]^4])])/(a + b)^(3/2) - (3*a^2 - b*(5*a + 2*b)*Tanh[x]^2)/(a*(a + b)*Sqrt[a + b*Tanh[x]^4]))/(3* a*(a + b)))/2
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ [{a, b, c, d}, x]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-(a*d + b*c*x))*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1)*(b*c^2 + a*d^2))), x] + Simp[1/(2*a*(p + 1)*(b*c^2 + a*d^2)) Int[(c + d*x)^n*(a + b*x^2)^(p + 1)*Simp[b*c^2*(2*p + 3) + a*d^2*(n + 2*p + 3) + b*c*d*(n + 2 *p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[p, -1] && IntQuad raticQ[a, 0, b, c, d, n, p, x]
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] + Simp[ 1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)) Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Sim p[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] && LtQ [p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[1/2 Subst[Int[(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; Free Q[{a, c, d, e, p, q}, x]
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]}, Simp[c*(ff/f) Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f f^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] && Ratio nalQ[n]))
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.43 (sec) , antiderivative size = 637, normalized size of antiderivative = 5.40
method | result | size |
derivativedivides | \(-\frac {\left (-\frac {\tanh \left (x \right )^{3}}{6 a \left (a +b \right ) b}-\frac {\tanh \left (x \right )^{2}}{6 a \left (a +b \right ) b}-\frac {\tanh \left (x \right )}{6 a \left (a +b \right ) b}+\frac {1}{6 \left (a +b \right ) b^{2}}\right ) \sqrt {a +b \tanh \left (x \right )^{4}}}{2 \left (\tanh \left (x \right )^{4}+\frac {a}{b}\right )^{2}}+\frac {b \left (\frac {\left (3 a +b \right ) \tanh \left (x \right )^{3}}{8 a^{2} \left (a +b \right )^{2}}+\frac {\left (5 a +2 b \right ) \tanh \left (x \right )^{2}}{12 a^{2} \left (a +b \right )^{2}}+\frac {\left (11 a +5 b \right ) \tanh \left (x \right )}{24 a^{2} \left (a +b \right )^{2}}-\frac {1}{4 \left (a +b \right )^{2} b}\right )}{\sqrt {\left (\tanh \left (x \right )^{4}+\frac {a}{b}\right ) b}}-\frac {-\frac {\operatorname {arctanh}\left (\frac {2 b \tanh \left (x \right )^{2}+2 a}{2 \sqrt {a +b}\, \sqrt {a +b \tanh \left (x \right )^{4}}}\right )}{2 \sqrt {a +b}}-\frac {\sqrt {1-\frac {i \sqrt {b}\, \tanh \left (x \right )^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, \tanh \left (x \right )^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (\tanh \left (x \right ) \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, -\frac {i \sqrt {a}}{\sqrt {b}}, \frac {\sqrt {-\frac {i \sqrt {b}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {a +b \tanh \left (x \right )^{4}}}}{2 \left (a +b \right )^{2}}-\frac {\left (\frac {\tanh \left (x \right )^{3}}{6 a \left (a +b \right ) b}-\frac {\tanh \left (x \right )^{2}}{6 a \left (a +b \right ) b}+\frac {\tanh \left (x \right )}{6 a \left (a +b \right ) b}+\frac {1}{6 \left (a +b \right ) b^{2}}\right ) \sqrt {a +b \tanh \left (x \right )^{4}}}{2 \left (\tanh \left (x \right )^{4}+\frac {a}{b}\right )^{2}}+\frac {b \left (-\frac {\left (3 a +b \right ) \tanh \left (x \right )^{3}}{8 a^{2} \left (a +b \right )^{2}}+\frac {\left (5 a +2 b \right ) \tanh \left (x \right )^{2}}{12 a^{2} \left (a +b \right )^{2}}-\frac {\left (11 a +5 b \right ) \tanh \left (x \right )}{24 a^{2} \left (a +b \right )^{2}}-\frac {1}{4 \left (a +b \right )^{2} b}\right )}{\sqrt {\left (\tanh \left (x \right )^{4}+\frac {a}{b}\right ) b}}-\frac {-\frac {\operatorname {arctanh}\left (\frac {2 b \tanh \left (x \right )^{2}+2 a}{2 \sqrt {a +b}\, \sqrt {a +b \tanh \left (x \right )^{4}}}\right )}{2 \sqrt {a +b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, \tanh \left (x \right )^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, \tanh \left (x \right )^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (\tanh \left (x \right ) \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, -\frac {i \sqrt {a}}{\sqrt {b}}, \frac {\sqrt {-\frac {i \sqrt {b}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {a +b \tanh \left (x \right )^{4}}}}{2 \left (a +b \right )^{2}}\) | \(637\) |
default | \(-\frac {\left (-\frac {\tanh \left (x \right )^{3}}{6 a \left (a +b \right ) b}-\frac {\tanh \left (x \right )^{2}}{6 a \left (a +b \right ) b}-\frac {\tanh \left (x \right )}{6 a \left (a +b \right ) b}+\frac {1}{6 \left (a +b \right ) b^{2}}\right ) \sqrt {a +b \tanh \left (x \right )^{4}}}{2 \left (\tanh \left (x \right )^{4}+\frac {a}{b}\right )^{2}}+\frac {b \left (\frac {\left (3 a +b \right ) \tanh \left (x \right )^{3}}{8 a^{2} \left (a +b \right )^{2}}+\frac {\left (5 a +2 b \right ) \tanh \left (x \right )^{2}}{12 a^{2} \left (a +b \right )^{2}}+\frac {\left (11 a +5 b \right ) \tanh \left (x \right )}{24 a^{2} \left (a +b \right )^{2}}-\frac {1}{4 \left (a +b \right )^{2} b}\right )}{\sqrt {\left (\tanh \left (x \right )^{4}+\frac {a}{b}\right ) b}}-\frac {-\frac {\operatorname {arctanh}\left (\frac {2 b \tanh \left (x \right )^{2}+2 a}{2 \sqrt {a +b}\, \sqrt {a +b \tanh \left (x \right )^{4}}}\right )}{2 \sqrt {a +b}}-\frac {\sqrt {1-\frac {i \sqrt {b}\, \tanh \left (x \right )^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, \tanh \left (x \right )^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (\tanh \left (x \right ) \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, -\frac {i \sqrt {a}}{\sqrt {b}}, \frac {\sqrt {-\frac {i \sqrt {b}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {a +b \tanh \left (x \right )^{4}}}}{2 \left (a +b \right )^{2}}-\frac {\left (\frac {\tanh \left (x \right )^{3}}{6 a \left (a +b \right ) b}-\frac {\tanh \left (x \right )^{2}}{6 a \left (a +b \right ) b}+\frac {\tanh \left (x \right )}{6 a \left (a +b \right ) b}+\frac {1}{6 \left (a +b \right ) b^{2}}\right ) \sqrt {a +b \tanh \left (x \right )^{4}}}{2 \left (\tanh \left (x \right )^{4}+\frac {a}{b}\right )^{2}}+\frac {b \left (-\frac {\left (3 a +b \right ) \tanh \left (x \right )^{3}}{8 a^{2} \left (a +b \right )^{2}}+\frac {\left (5 a +2 b \right ) \tanh \left (x \right )^{2}}{12 a^{2} \left (a +b \right )^{2}}-\frac {\left (11 a +5 b \right ) \tanh \left (x \right )}{24 a^{2} \left (a +b \right )^{2}}-\frac {1}{4 \left (a +b \right )^{2} b}\right )}{\sqrt {\left (\tanh \left (x \right )^{4}+\frac {a}{b}\right ) b}}-\frac {-\frac {\operatorname {arctanh}\left (\frac {2 b \tanh \left (x \right )^{2}+2 a}{2 \sqrt {a +b}\, \sqrt {a +b \tanh \left (x \right )^{4}}}\right )}{2 \sqrt {a +b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, \tanh \left (x \right )^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, \tanh \left (x \right )^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (\tanh \left (x \right ) \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, -\frac {i \sqrt {a}}{\sqrt {b}}, \frac {\sqrt {-\frac {i \sqrt {b}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {a +b \tanh \left (x \right )^{4}}}}{2 \left (a +b \right )^{2}}\) | \(637\) |
Input:
int(tanh(x)/(a+b*tanh(x)^4)^(5/2),x,method=_RETURNVERBOSE)
Output:
-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/a^2*(11*a+5 *b)/(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(t anh(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)/(ta nh(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/a^2*(11*a+5*b)/(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)))
Leaf count of result is larger than twice the leaf count of optimal. 8210 vs. \(2 (102) = 204\).
Time = 1.98 (sec) , antiderivative size = 16463, normalized size of antiderivative = 139.52 \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:
integrate(tanh(x)/(a+b*tanh(x)^4)^(5/2),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{5/2}} \, dx=\int \frac {\tanh {\left (x \right )}}{\left (a + b \tanh ^{4}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(tanh(x)/(a+b*tanh(x)**4)**(5/2),x)
Output:
Integral(tanh(x)/(a + b*tanh(x)**4)**(5/2), x)
\[ \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{5/2}} \, dx=\int { \frac {\tanh \left (x\right )}{{\left (b \tanh \left (x\right )^{4} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(tanh(x)/(a+b*tanh(x)^4)^(5/2),x, algorithm="maxima")
Output:
integrate(tanh(x)/(b*tanh(x)^4 + a)^(5/2), x)
\[ \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{5/2}} \, dx=\int { \frac {\tanh \left (x\right )}{{\left (b \tanh \left (x\right )^{4} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(tanh(x)/(a+b*tanh(x)^4)^(5/2),x, algorithm="giac")
Output:
integrate(tanh(x)/(b*tanh(x)^4 + a)^(5/2), x)
Timed out. \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{5/2}} \, dx=\int \frac {\mathrm {tanh}\left (x\right )}{{\left (b\,{\mathrm {tanh}\left (x\right )}^4+a\right )}^{5/2}} \,d x \] Input:
int(tanh(x)/(a + b*tanh(x)^4)^(5/2),x)
Output:
int(tanh(x)/(a + b*tanh(x)^4)^(5/2), x)
\[ \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\tanh \left (x \right )^{4} b +a}\, \tanh \left (x \right )}{\tanh \left (x \right )^{12} b^{3}+3 \tanh \left (x \right )^{8} a \,b^{2}+3 \tanh \left (x \right )^{4} a^{2} b +a^{3}}d x \] Input:
int(tanh(x)/(a+b*tanh(x)^4)^(5/2),x)
Output:
int((sqrt(tanh(x)**4*b + a)*tanh(x))/(tanh(x)**12*b**3 + 3*tanh(x)**8*a*b* *2 + 3*tanh(x)**4*a**2*b + a**3),x)