Integrand size = 15, antiderivative size = 74 \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {a+b \tanh ^2(x)}{\sqrt {a+b} \sqrt {a+b \tanh ^4(x)}}\right )}{2 (a+b)^{3/2}}-\frac {a-b \tanh ^2(x)}{2 a (a+b) \sqrt {a+b \tanh ^4(x)}} \]
1/2*arctanh((a+b*tanh(x)^2)/(a+b)^(1/2)/(a+b*tanh(x)^4)^(1/2))/(a+b)^(3/2) +1/2*(-a+b*tanh(x)^2)/a/(a+b)/(a+b*tanh(x)^4)^(1/2)
Time = 0.54 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.99 \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{3/2}} \, dx=\frac {1}{2} \left (\frac {\text {arctanh}\left (\frac {a+b \tanh ^2(x)}{\sqrt {a+b} \sqrt {a+b \tanh ^4(x)}}\right )}{(a+b)^{3/2}}-\frac {a-b \tanh ^2(x)}{a (a+b) \sqrt {a+b \tanh ^4(x)}}\right ) \]
(ArcTanh[(a + b*Tanh[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tanh[x]^4])]/(a + b)^(3 /2) - (a - b*Tanh[x]^2)/(a*(a + b)*Sqrt[a + b*Tanh[x]^4]))/2
Time = 0.30 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 26, 4153, 26, 1577, 496, 25, 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.
| \(\displaystyle \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {i \tan (i x)}{\left (a+b \tan (i x)^4\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle -i \int \frac {\tan (i x)}{\left (b \tan (i x)^4+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle -i \int \frac {i \tanh (x)}{\left (1-\tanh ^2(x)\right ) \left (b \tanh ^4(x)+a\right )^{3/2}}d\tanh (x)\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \int \frac {\tanh (x)}{\left (1-\tanh ^2(x)\right ) \left (a+b \tanh ^4(x)\right )^{3/2}}d\tanh (x)\) |
| \(\Big \downarrow \) 1577 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{\left (1-\tanh ^2(x)\right ) \left (b \tanh ^4(x)+a\right )^{3/2}}d\tanh ^2(x)\) |
| \(\Big \downarrow \) 496 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\int -\frac {a}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^4(x)+a}}d\tanh ^2(x)}{a (a+b)}-\frac {a-b \tanh ^2(x)}{a (a+b) \sqrt {a+b \tanh ^4(x)}}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int \frac {a}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^4(x)+a}}d\tanh ^2(x)}{a (a+b)}-\frac {a-b \tanh ^2(x)}{a (a+b) \sqrt {a+b \tanh ^4(x)}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^4(x)+a}}d\tanh ^2(x)}{a+b}-\frac {a-b \tanh ^2(x)}{a (a+b) \sqrt {a+b \tanh ^4(x)}}\right )\) |
| \(\Big \downarrow \) 488 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {1}{-\tanh ^4(x)+a+b}d\frac {-b \tanh ^2(x)-a}{\sqrt {b \tanh ^4(x)+a}}}{a+b}-\frac {a-b \tanh ^2(x)}{a (a+b) \sqrt {a+b \tanh ^4(x)}}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\text {arctanh}\left (\frac {-a-b \tanh ^2(x)}{\sqrt {a+b} \sqrt {a+b \tanh ^4(x)}}\right )}{(a+b)^{3/2}}-\frac {a-b \tanh ^2(x)}{a (a+b) \sqrt {a+b \tanh ^4(x)}}\right )\) |
(-(ArcTanh[(-a - b*Tanh[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tanh[x]^4])]/(a + b) ^(3/2)) - (a - b*Tanh[x]^2)/(a*(a + b)*Sqrt[a + b*Tanh[x]^4]))/2
3.3.62.3.1 Defintions of rubi rules used
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[(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 = 0.68 (sec) , antiderivative size = 431, normalized size of antiderivative = 5.82
| method | result | size |
| derivativedivides | \(\frac {b \left (\frac {\tanh \left (x \right )^{3}}{4 a \left (a +b \right )}+\frac {\tanh \left (x \right )^{2}}{4 a \left (a +b \right )}+\frac {\tanh \left (x \right )}{4 a \left (a +b \right )}-\frac {1}{4 \left (a +b \right ) b}\right )}{\sqrt {\left (\tanh \left (x \right )^{4}+\frac {a}{b}\right ) b}}-\frac {-\frac {\operatorname {arctanh}\left (\frac {2 b \tanh \left (x \right )^{2}+2 a}{2 \sqrt {a +b}\, \sqrt {a +b \tanh \left (x \right )^{4}}}\right )}{2 \sqrt {a +b}}-\frac {\sqrt {1-\frac {i \sqrt {b}\, \tanh \left (x \right )^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, \tanh \left (x \right )^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (\tanh \left (x \right ) \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, -\frac {i \sqrt {a}}{\sqrt {b}}, \frac {\sqrt {-\frac {i \sqrt {b}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {a +b \tanh \left (x \right )^{4}}}}{2 \left (a +b \right )}+\frac {b \left (-\frac {\tanh \left (x \right )^{3}}{4 a \left (a +b \right )}+\frac {\tanh \left (x \right )^{2}}{4 a \left (a +b \right )}-\frac {\tanh \left (x \right )}{4 a \left (a +b \right )}-\frac {1}{4 \left (a +b \right ) b}\right )}{\sqrt {\left (\tanh \left (x \right )^{4}+\frac {a}{b}\right ) b}}-\frac {-\frac {\operatorname {arctanh}\left (\frac {2 b \tanh \left (x \right )^{2}+2 a}{2 \sqrt {a +b}\, \sqrt {a +b \tanh \left (x \right )^{4}}}\right )}{2 \sqrt {a +b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, \tanh \left (x \right )^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, \tanh \left (x \right )^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (\tanh \left (x \right ) \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, -\frac {i \sqrt {a}}{\sqrt {b}}, \frac {\sqrt {-\frac {i \sqrt {b}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {a +b \tanh \left (x \right )^{4}}}}{2 \left (a +b \right )}\) | \(431\) |
| default | \(\frac {b \left (\frac {\tanh \left (x \right )^{3}}{4 a \left (a +b \right )}+\frac {\tanh \left (x \right )^{2}}{4 a \left (a +b \right )}+\frac {\tanh \left (x \right )}{4 a \left (a +b \right )}-\frac {1}{4 \left (a +b \right ) b}\right )}{\sqrt {\left (\tanh \left (x \right )^{4}+\frac {a}{b}\right ) b}}-\frac {-\frac {\operatorname {arctanh}\left (\frac {2 b \tanh \left (x \right )^{2}+2 a}{2 \sqrt {a +b}\, \sqrt {a +b \tanh \left (x \right )^{4}}}\right )}{2 \sqrt {a +b}}-\frac {\sqrt {1-\frac {i \sqrt {b}\, \tanh \left (x \right )^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, \tanh \left (x \right )^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (\tanh \left (x \right ) \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, -\frac {i \sqrt {a}}{\sqrt {b}}, \frac {\sqrt {-\frac {i \sqrt {b}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {a +b \tanh \left (x \right )^{4}}}}{2 \left (a +b \right )}+\frac {b \left (-\frac {\tanh \left (x \right )^{3}}{4 a \left (a +b \right )}+\frac {\tanh \left (x \right )^{2}}{4 a \left (a +b \right )}-\frac {\tanh \left (x \right )}{4 a \left (a +b \right )}-\frac {1}{4 \left (a +b \right ) b}\right )}{\sqrt {\left (\tanh \left (x \right )^{4}+\frac {a}{b}\right ) b}}-\frac {-\frac {\operatorname {arctanh}\left (\frac {2 b \tanh \left (x \right )^{2}+2 a}{2 \sqrt {a +b}\, \sqrt {a +b \tanh \left (x \right )^{4}}}\right )}{2 \sqrt {a +b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, \tanh \left (x \right )^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, \tanh \left (x \right )^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (\tanh \left (x \right ) \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, -\frac {i \sqrt {a}}{\sqrt {b}}, \frac {\sqrt {-\frac {i \sqrt {b}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {a +b \tanh \left (x \right )^{4}}}}{2 \left (a +b \right )}\) | \(431\) |
b*(1/4/a/(a+b)*tanh(x)^3+1/4/a/(a+b)*tanh(x)^2+1/4/a/(a+b)*tanh(x)-1/4/(a+ b)/b)/((tanh(x)^4+a/b)*b)^(1/2)-1/2/(a+b)*(-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)) )+b*(-1/4/a/(a+b)*tanh(x)^3+1/4/a/(a+b)*tanh(x)^2-1/4/a/(a+b)*tanh(x)-1/4/ (a+b)/b)/((tanh(x)^4+a/b)*b)^(1/2)-1/2/(a+b)*(-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. 1935 vs. \(2 (63) = 126\).
Time = 0.48 (sec) , antiderivative size = 3914, normalized size of antiderivative = 52.89 \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{3/2}} \, dx=\text {Too large to display} \]
[1/4*(((a^2 + a*b)*cosh(x)^8 + 8*(a^2 + a*b)*cosh(x)*sinh(x)^7 + (a^2 + a* b)*sinh(x)^8 + 4*(a^2 - a*b)*cosh(x)^6 + 4*(7*(a^2 + a*b)*cosh(x)^2 + a^2 - a*b)*sinh(x)^6 + 8*(7*(a^2 + a*b)*cosh(x)^3 + 3*(a^2 - a*b)*cosh(x))*sin h(x)^5 + 6*(a^2 + a*b)*cosh(x)^4 + 2*(35*(a^2 + a*b)*cosh(x)^4 + 30*(a^2 - a*b)*cosh(x)^2 + 3*a^2 + 3*a*b)*sinh(x)^4 + 8*(7*(a^2 + a*b)*cosh(x)^5 + 10*(a^2 - a*b)*cosh(x)^3 + 3*(a^2 + a*b)*cosh(x))*sinh(x)^3 + 4*(a^2 - a*b )*cosh(x)^2 + 4*(7*(a^2 + a*b)*cosh(x)^6 + 15*(a^2 - a*b)*cosh(x)^4 + 9*(a ^2 + a*b)*cosh(x)^2 + a^2 - a*b)*sinh(x)^2 + a^2 + a*b + 8*((a^2 + a*b)*co sh(x)^7 + 3*(a^2 - a*b)*cosh(x)^5 + 3*(a^2 + a*b)*cosh(x)^3 + (a^2 - a*b)* cosh(x))*sinh(x))*sqrt(a + b)*log(((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)*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 + sqrt(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)*...
\[ \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{3/2}} \, dx=\int \frac {\tanh {\left (x \right )}}{\left (a + b \tanh ^{4}{\left (x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{3/2}} \, dx=\int { \frac {\tanh \left (x\right )}{{\left (b \tanh \left (x\right )^{4} + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{3/2}} \, dx=\int { \frac {\tanh \left (x\right )}{{\left (b \tanh \left (x\right )^{4} + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{3/2}} \, dx=\int \frac {\mathrm {tanh}\left (x\right )}{{\left (b\,{\mathrm {tanh}\left (x\right )}^4+a\right )}^{3/2}} \,d x \]