Integrand size = 18, antiderivative size = 167 \[ \int e^{c (a+b x)} \tanh ^3(d+e x) \, dx=\frac {e^{c (a+b x)}}{b c}-\frac {6 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {b c}{2 e},1+\frac {b c}{2 e},-e^{2 (d+e x)}\right )}{b c}+\frac {12 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,\frac {b c}{2 e},1+\frac {b c}{2 e},-e^{2 (d+e x)}\right )}{b c}-\frac {8 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (3,\frac {b c}{2 e},1+\frac {b c}{2 e},-e^{2 (d+e x)}\right )}{b c} \]
exp(c*(b*x+a))/b/c-6*exp(c*(b*x+a))*hypergeom([1, 1/2*b*c/e],[1+1/2*b*c/e] ,-exp(2*e*x+2*d))/b/c+12*exp(c*(b*x+a))*hypergeom([2, 1/2*b*c/e],[1+1/2*b* c/e],-exp(2*e*x+2*d))/b/c-8*exp(c*(b*x+a))*hypergeom([3, 1/2*b*c/e],[1+1/2 *b*c/e],-exp(2*e*x+2*d))/b/c
Time = 2.60 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.23 \[ \int e^{c (a+b x)} \tanh ^3(d+e x) \, dx=\frac {1}{2} e^{a c} \left (\frac {2 \left (b^2 c^2+2 e^2\right ) e^{2 d} \left (\frac {e^{(b c+2 e) x} \operatorname {Hypergeometric2F1}\left (1,1+\frac {b c}{2 e},2+\frac {b c}{2 e},-e^{2 (d+e x)}\right )}{b c+2 e}-\frac {e^{b c x} \operatorname {Hypergeometric2F1}\left (1,\frac {b c}{2 e},1+\frac {b c}{2 e},-e^{2 (d+e x)}\right )}{b c}\right )}{e^2 \left (1+e^{2 d}\right )}+\frac {e^{b c x} \text {sech}^2(d+e x)}{e}-\frac {b c e^{b c x} \text {sech}(d) \text {sech}(d+e x) \sinh (e x)}{e^2}+\frac {2 e^{b c x} \tanh (d)}{b c}\right ) \]
(E^(a*c)*((2*(b^2*c^2 + 2*e^2)*E^(2*d)*((E^((b*c + 2*e)*x)*Hypergeometric2 F1[1, 1 + (b*c)/(2*e), 2 + (b*c)/(2*e), -E^(2*(d + e*x))])/(b*c + 2*e) - ( E^(b*c*x)*Hypergeometric2F1[1, (b*c)/(2*e), 1 + (b*c)/(2*e), -E^(2*(d + e* x))])/(b*c)))/(e^2*(1 + E^(2*d))) + (E^(b*c*x)*Sech[d + e*x]^2)/e - (b*c*E ^(b*c*x)*Sech[d]*Sech[d + e*x]*Sinh[e*x])/e^2 + (2*E^(b*c*x)*Tanh[d])/(b*c )))/2
Time = 0.42 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6007, 2009}
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.
\(\displaystyle \int e^{c (a+b x)} \tanh ^3(d+e x) \, dx\) |
\(\Big \downarrow \) 6007 |
\(\displaystyle \int \left (-\frac {6 e^{c (a+b x)}}{e^{2 (d+e x)}+1}+\frac {12 e^{c (a+b x)}}{\left (e^{2 (d+e x)}+1\right )^2}-\frac {8 e^{c (a+b x)}}{\left (e^{2 (d+e x)}+1\right )^3}+e^{c (a+b x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {6 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {b c}{2 e},\frac {b c}{2 e}+1,-e^{2 (d+e x)}\right )}{b c}+\frac {12 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,\frac {b c}{2 e},\frac {b c}{2 e}+1,-e^{2 (d+e x)}\right )}{b c}-\frac {8 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (3,\frac {b c}{2 e},\frac {b c}{2 e}+1,-e^{2 (d+e x)}\right )}{b c}+\frac {e^{c (a+b x)}}{b c}\) |
E^(c*(a + b*x))/(b*c) - (6*E^(c*(a + b*x))*Hypergeometric2F1[1, (b*c)/(2*e ), 1 + (b*c)/(2*e), -E^(2*(d + e*x))])/(b*c) + (12*E^(c*(a + b*x))*Hyperge ometric2F1[2, (b*c)/(2*e), 1 + (b*c)/(2*e), -E^(2*(d + e*x))])/(b*c) - (8* E^(c*(a + b*x))*Hypergeometric2F1[3, (b*c)/(2*e), 1 + (b*c)/(2*e), -E^(2*( d + e*x))])/(b*c)
3.3.28.3.1 Defintions of rubi rules used
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Tanh[(d_.) + (e_.)*(x_)]^(n_.), x_Sym bol] :> Int[ExpandIntegrand[F^(c*(a + b*x))*((-1 + E^(2*(d + e*x)))^n/(1 + E^(2*(d + e*x)))^n), x], x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]
\[\int {\mathrm e}^{c \left (b x +a \right )} \tanh \left (e x +d \right )^{3}d x\]
\[ \int e^{c (a+b x)} \tanh ^3(d+e x) \, dx=\int { e^{\left ({\left (b x + a\right )} c\right )} \tanh \left (e x + d\right )^{3} \,d x } \]
\[ \int e^{c (a+b x)} \tanh ^3(d+e x) \, dx=e^{a c} \int e^{b c x} \tanh ^{3}{\left (d + e x \right )}\, dx \]
\[ \int e^{c (a+b x)} \tanh ^3(d+e x) \, dx=\int { e^{\left ({\left (b x + a\right )} c\right )} \tanh \left (e x + d\right )^{3} \,d x } \]
48*(b^2*c^2*e*e^(a*c) + 2*e^3*e^(a*c))*integrate(e^(b*c*x)/(b^3*c^3 - 12*b ^2*c^2*e + 44*b*c*e^2 - 48*e^3 + (b^3*c^3*e^(8*d) - 12*b^2*c^2*e*e^(8*d) + 44*b*c*e^2*e^(8*d) - 48*e^3*e^(8*d))*e^(8*e*x) + 4*(b^3*c^3*e^(6*d) - 12* b^2*c^2*e*e^(6*d) + 44*b*c*e^2*e^(6*d) - 48*e^3*e^(6*d))*e^(6*e*x) + 6*(b^ 3*c^3*e^(4*d) - 12*b^2*c^2*e*e^(4*d) + 44*b*c*e^2*e^(4*d) - 48*e^3*e^(4*d) )*e^(4*e*x) + 4*(b^3*c^3*e^(2*d) - 12*b^2*c^2*e*e^(2*d) + 44*b*c*e^2*e^(2* d) - 48*e^3*e^(2*d))*e^(2*e*x)), x) - (b^3*c^3*e^(a*c) + 36*b^2*c^2*e*e^(a *c) + 44*b*c*e^2*e^(a*c) + 48*e^3*e^(a*c) - (b^3*c^3*e^(a*c + 6*d) - 12*b^ 2*c^2*e*e^(a*c + 6*d) + 44*b*c*e^2*e^(a*c + 6*d) - 48*e^3*e^(a*c + 6*d))*e ^(6*e*x) + 3*(b^3*c^3*e^(a*c + 4*d) - 8*b^2*c^2*e*e^(a*c + 4*d) + 4*b*c*e^ 2*e^(a*c + 4*d) + 48*e^3*e^(a*c + 4*d))*e^(4*e*x) - 3*(b^3*c^3*e^(a*c + 2* d) - 28*b*c*e^2*e^(a*c + 2*d) - 48*e^3*e^(a*c + 2*d))*e^(2*e*x))*e^(b*c*x) /(b^4*c^4 - 12*b^3*c^3*e + 44*b^2*c^2*e^2 - 48*b*c*e^3 + (b^4*c^4*e^(6*d) - 12*b^3*c^3*e*e^(6*d) + 44*b^2*c^2*e^2*e^(6*d) - 48*b*c*e^3*e^(6*d))*e^(6 *e*x) + 3*(b^4*c^4*e^(4*d) - 12*b^3*c^3*e*e^(4*d) + 44*b^2*c^2*e^2*e^(4*d) - 48*b*c*e^3*e^(4*d))*e^(4*e*x) + 3*(b^4*c^4*e^(2*d) - 12*b^3*c^3*e*e^(2* d) + 44*b^2*c^2*e^2*e^(2*d) - 48*b*c*e^3*e^(2*d))*e^(2*e*x))
\[ \int e^{c (a+b x)} \tanh ^3(d+e x) \, dx=\int { e^{\left ({\left (b x + a\right )} c\right )} \tanh \left (e x + d\right )^{3} \,d x } \]
Timed out. \[ \int e^{c (a+b x)} \tanh ^3(d+e x) \, dx=\int {\mathrm {e}}^{c\,\left (a+b\,x\right )}\,{\mathrm {tanh}\left (d+e\,x\right )}^3 \,d x \]