Integrand size = 23, antiderivative size = 135 \[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\frac {(b-2 c) \text {arctanh}\left (\frac {b+2 c \tanh ^2(x)}{2 \sqrt {c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{4 c^{3/2}}+\frac {\text {arctanh}\left (\frac {2 a+b+(b+2 c) \tanh ^2(x)}{2 \sqrt {a+b+c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{2 \sqrt {a+b+c}}-\frac {\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}{2 c} \]
1/4*(b-2*c)*arctanh(1/2*(b+2*c*tanh(x)^2)/c^(1/2)/(a+b*tanh(x)^2+c*tanh(x) ^4)^(1/2))/c^(3/2)+1/2*arctanh(1/2*(2*a+b+(b+2*c)*tanh(x)^2)/(a+b+c)^(1/2) /(a+b*tanh(x)^2+c*tanh(x)^4)^(1/2))/(a+b+c)^(1/2)-1/2*(a+b*tanh(x)^2+c*tan h(x)^4)^(1/2)/c
Time = 1.40 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.01 \[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\frac {1}{4} \left (\frac {(-b+2 c) \text {arctanh}\left (\frac {-b-2 c \tanh ^2(x)}{2 \sqrt {c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{c^{3/2}}+\frac {2 \text {arctanh}\left (\frac {2 a+b+(b+2 c) \tanh ^2(x)}{2 \sqrt {a+b+c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{\sqrt {a+b+c}}-\frac {2 \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}{c}\right ) \]
(((-b + 2*c)*ArcTanh[(-b - 2*c*Tanh[x]^2)/(2*Sqrt[c]*Sqrt[a + b*Tanh[x]^2 + c*Tanh[x]^4])])/c^(3/2) + (2*ArcTanh[(2*a + b + (b + 2*c)*Tanh[x]^2)/(2* Sqrt[a + b + c]*Sqrt[a + b*Tanh[x]^2 + c*Tanh[x]^4])])/Sqrt[a + b + c] - ( 2*Sqrt[a + b*Tanh[x]^2 + c*Tanh[x]^4])/c)/4
Result contains complex when optimal does not.
Time = 0.49 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.68, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 26, 4183, 1578, 1267, 27, 1269, 1092, 219, 1154, 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.
\(\displaystyle \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i \tan (i x)^5}{\sqrt {a-b \tan (i x)^2+c \tan (i x)^4}}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {\tan (i x)^5}{\sqrt {c \tan (i x)^4-b \tan (i x)^2+a}}dx\) |
\(\Big \downarrow \) 4183 |
\(\displaystyle -\int \frac {i \tanh ^5(x)}{\left (1-\tanh ^2(x)\right ) \sqrt {c \tanh ^4(x)+b \tanh ^2(x)+a}}d(i \tanh (x))\) |
\(\Big \downarrow \) 1578 |
\(\displaystyle -\frac {1}{2} \int -\frac {\tanh ^2(x)}{\left (1-\tanh ^2(x)\right ) \sqrt {-c \tanh ^2(x)-i b \tanh (x)+a}}d\left (-\tanh ^2(x)\right )\) |
\(\Big \downarrow \) 1267 |
\(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {b-(b-2 c) \tanh ^2(x)}{2 \left (1-\tanh ^2(x)\right ) \sqrt {-c \tanh ^2(x)-i b \tanh (x)+a}}d\left (-\tanh ^2(x)\right )}{c}-\frac {\sqrt {a-i b \tanh (x)-c \tanh ^2(x)}}{c}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {b-(b-2 c) \tanh ^2(x)}{\left (1-\tanh ^2(x)\right ) \sqrt {-c \tanh ^2(x)-i b \tanh (x)+a}}d\left (-\tanh ^2(x)\right )}{2 c}-\frac {\sqrt {a-i b \tanh (x)-c \tanh ^2(x)}}{c}\right )\) |
\(\Big \downarrow \) 1269 |
\(\displaystyle \frac {1}{2} \left (-\frac {(b-2 c) \int \frac {1}{\sqrt {-c \tanh ^2(x)-i b \tanh (x)+a}}d\left (-\tanh ^2(x)\right )+2 c \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {-c \tanh ^2(x)-i b \tanh (x)+a}}d\left (-\tanh ^2(x)\right )}{2 c}-\frac {\sqrt {a-i b \tanh (x)-c \tanh ^2(x)}}{c}\right )\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {1}{2} \left (-\frac {2 (b-2 c) \int \frac {1}{\tanh ^2(x)+4 c}d\left (-\frac {b-2 i c \tanh (x)}{\sqrt {-c \tanh ^2(x)-i b \tanh (x)+a}}\right )+2 c \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {-c \tanh ^2(x)-i b \tanh (x)+a}}d\left (-\tanh ^2(x)\right )}{2 c}-\frac {\sqrt {a-i b \tanh (x)-c \tanh ^2(x)}}{c}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (-\frac {2 c \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {-c \tanh ^2(x)-i b \tanh (x)+a}}d\left (-\tanh ^2(x)\right )+\frac {i (b-2 c) \arctan \left (\frac {\tanh (x)}{2 \sqrt {c}}\right )}{\sqrt {c}}}{2 c}-\frac {\sqrt {a-i b \tanh (x)-c \tanh ^2(x)}}{c}\right )\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \frac {1}{2} \left (-\frac {\frac {i (b-2 c) \arctan \left (\frac {\tanh (x)}{2 \sqrt {c}}\right )}{\sqrt {c}}-4 c \int \frac {1}{\tanh ^2(x)+4 (a+b+c)}d\frac {2 a+b-i (b+2 c) \tanh (x)}{\sqrt {-c \tanh ^2(x)-i b \tanh (x)+a}}}{2 c}-\frac {\sqrt {a-i b \tanh (x)-c \tanh ^2(x)}}{c}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (-\frac {\frac {i (b-2 c) \arctan \left (\frac {\tanh (x)}{2 \sqrt {c}}\right )}{\sqrt {c}}-\frac {2 i c \arctan \left (\frac {\tanh (x)}{2 \sqrt {a+b+c}}\right )}{\sqrt {a+b+c}}}{2 c}-\frac {\sqrt {a-i b \tanh (x)-c \tanh ^2(x)}}{c}\right )\) |
(-1/2*((I*(b - 2*c)*ArcTan[Tanh[x]/(2*Sqrt[c])])/Sqrt[c] - ((2*I)*c*ArcTan [Tanh[x]/(2*Sqrt[a + b + c])])/Sqrt[a + b + c])/c - Sqrt[a - I*b*Tanh[x] - c*Tanh[x]^2]/c)/2
3.2.100.3.1 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/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-2 Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c , d, e}, x]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[g^n*(d + e*x)^(m + n - 1)*((a + b *x + c*x^2)^(p + 1)/(c*e^(n - 1)*(m + n + 2*p + 1))), x] + Simp[1/(c*e^n*(m + n + 2*p + 1)) Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^n*(m + n + 2*p + 1)*(f + g*x)^n - c*g^n*(m + n + 2*p + 1)*(d + e*x)^n - g^n*(d + e*x)^(n - 2)*(b*d*e*(p + 1) + a*e^2*(m + n - 1) - c*d^2*(m + n + 2*p + 1) - e*(2*c*d - b*e)*(m + n + p)*x), x], x], x] /; FreeQ[{a, b, c, d, e, f, g , m, p}, x] && IGtQ[n, 1] && IntegerQ[m] && NeQ[m + n + 2*p + 1, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Simp[(e*f - d*g)/e Int[(d + e*x)^m*(a + b*x + c*x^2)^ p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && !IGtQ[m, 0]
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ )^4)^(p_.), x_Symbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && Int egerQ[(m - 1)/2]
Int[tan[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*((f_.)*tan[(d_.) + (e_.)*( x_)])^(n_.) + (c_.)*((f_.)*tan[(d_.) + (e_.)*(x_)])^(n2_.))^(p_), x_Symbol] :> Simp[f/e Subst[Int[(x/f)^m*((a + b*x^n + c*x^(2*n))^p/(f^2 + x^2)), x ], x, f*Tan[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[n 2, 2*n] && NeQ[b^2 - 4*a*c, 0]
Time = 0.82 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.10
method | result | size |
derivativedivides | \(-\frac {\ln \left (\frac {\frac {b}{2}+c \tanh \left (x \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}\right )}{2 \sqrt {c}}-\frac {\sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}}{2 c}+\frac {b \ln \left (\frac {\frac {b}{2}+c \tanh \left (x \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}\right )}{4 c^{\frac {3}{2}}}+\frac {\operatorname {arctanh}\left (\frac {b \tanh \left (x \right )^{2}+2 c \tanh \left (x \right )^{2}+2 a +b}{2 \sqrt {a +b +c}\, \sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}}\right )}{2 \sqrt {a +b +c}}\) | \(149\) |
default | \(-\frac {\ln \left (\frac {\frac {b}{2}+c \tanh \left (x \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}\right )}{2 \sqrt {c}}-\frac {\sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}}{2 c}+\frac {b \ln \left (\frac {\frac {b}{2}+c \tanh \left (x \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}\right )}{4 c^{\frac {3}{2}}}+\frac {\operatorname {arctanh}\left (\frac {b \tanh \left (x \right )^{2}+2 c \tanh \left (x \right )^{2}+2 a +b}{2 \sqrt {a +b +c}\, \sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}}\right )}{2 \sqrt {a +b +c}}\) | \(149\) |
-1/2*ln((1/2*b+c*tanh(x)^2)/c^(1/2)+(a+b*tanh(x)^2+c*tanh(x)^4)^(1/2))/c^( 1/2)-1/2*(a+b*tanh(x)^2+c*tanh(x)^4)^(1/2)/c+1/4*b/c^(3/2)*ln((1/2*b+c*tan h(x)^2)/c^(1/2)+(a+b*tanh(x)^2+c*tanh(x)^4)^(1/2))+1/2/(a+b+c)^(1/2)*arcta nh(1/2*(b*tanh(x)^2+2*c*tanh(x)^2+2*a+b)/(a+b+c)^(1/2)/(a+b*tanh(x)^2+c*ta nh(x)^4)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 2075 vs. \(2 (111) = 222\).
Time = 1.14 (sec) , antiderivative size = 8891, normalized size of antiderivative = 65.86 \[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\text {Too large to display} \]
\[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\int \frac {\tanh ^{5}{\left (x \right )}}{\sqrt {a + b \tanh ^{2}{\left (x \right )} + c \tanh ^{4}{\left (x \right )}}}\, dx \]
\[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\int { \frac {\tanh \left (x\right )^{5}}{\sqrt {c \tanh \left (x\right )^{4} + b \tanh \left (x\right )^{2} + a}} \,d x } \]
\[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\int { \frac {\tanh \left (x\right )^{5}}{\sqrt {c \tanh \left (x\right )^{4} + b \tanh \left (x\right )^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\int \frac {{\mathrm {tanh}\left (x\right )}^5}{\sqrt {c\,{\mathrm {tanh}\left (x\right )}^4+b\,{\mathrm {tanh}\left (x\right )}^2+a}} \,d x \]