\(\int e^{c (a+b x)} \tanh ^3(d+e x) \, dx\) [228]

Optimal result

Integrand size = 18, antiderivative size = 167 \[ \int e^{c (a+b x)} \tanh ^3(d+e x) \, dx=\frac {e^{c (a+b x)}}{b c}-\frac {6 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {b c}{2 e},1+\frac {b c}{2 e},-e^{2 (d+e x)}\right )}{b c}+\frac {12 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,\frac {b c}{2 e},1+\frac {b c}{2 e},-e^{2 (d+e x)}\right )}{b c}-\frac {8 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (3,\frac {b c}{2 e},1+\frac {b c}{2 e},-e^{2 (d+e x)}\right )}{b c} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5592, 2225, 2283} \[ \int e^{c (a+b x)} \tanh ^3(d+e x) \, dx=-\frac {6 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {b c}{2 e},\frac {b c}{2 e}+1,-e^{2 (d+e x)}\right )}{b c}+\frac {12 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,\frac {b c}{2 e},\frac {b c}{2 e}+1,-e^{2 (d+e x)}\right )}{b c}-\frac {8 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (3,\frac {b c}{2 e},\frac {b c}{2 e}+1,-e^{2 (d+e x)}\right )}{b c}+\frac {e^{c (a+b x)}}{b c} \]

[In]

Int[E^(c*(a + b*x))*Tanh[d + e*x]^3,x]

[Out]

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))*Hypergeometric2F1[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)

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2283

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Simp
[a^p*(G^(h*(f + g*x))/(g*h*Log[G]))*Hypergeometric2F1[-p, g*h*(Log[G]/(d*e*Log[F])), g*h*(Log[G]/(d*e*Log[F]))
+ 1, Simplify[(-b/a)*F^(e*(c + d*x))]], x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x] && (ILtQ[p, 0] || G
tQ[a, 0])

Rule 5592

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Tanh[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> 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] && Integer
Q[n]

Rubi steps \begin{align*} \text {integral}& = \int \left (e^{c (a+b x)}-\frac {8 e^{c (a+b x)}}{\left (1+e^{2 (d+e x)}\right )^3}+\frac {12 e^{c (a+b x)}}{\left (1+e^{2 (d+e x)}\right )^2}-\frac {6 e^{c (a+b x)}}{1+e^{2 (d+e x)}}\right ) \, dx \\ & = -\left (6 \int \frac {e^{c (a+b x)}}{1+e^{2 (d+e x)}} \, dx\right )-8 \int \frac {e^{c (a+b x)}}{\left (1+e^{2 (d+e x)}\right )^3} \, dx+12 \int \frac {e^{c (a+b x)}}{\left (1+e^{2 (d+e x)}\right )^2} \, dx+\int e^{c (a+b 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} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.60 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.23 \[ \int e^{c (a+b x)} \tanh ^3(d+e x) \, dx=\frac {1}{2} e^{a c} \left (\frac {2 \left (b^2 c^2+2 e^2\right ) e^{2 d} \left (\frac {e^{(b c+2 e) x} \operatorname {Hypergeometric2F1}\left (1,1+\frac {b c}{2 e},2+\frac {b c}{2 e},-e^{2 (d+e x)}\right )}{b c+2 e}-\frac {e^{b c x} \operatorname {Hypergeometric2F1}\left (1,\frac {b c}{2 e},1+\frac {b c}{2 e},-e^{2 (d+e x)}\right )}{b c}\right )}{e^2 \left (1+e^{2 d}\right )}+\frac {e^{b c x} \text {sech}^2(d+e x)}{e}-\frac {b c e^{b c x} \text {sech}(d) \text {sech}(d+e x) \sinh (e x)}{e^2}+\frac {2 e^{b c x} \tanh (d)}{b c}\right ) \]

[In]

Integrate[E^(c*(a + b*x))*Tanh[d + e*x]^3,x]

[Out]

(E^(a*c)*((2*(b^2*c^2 + 2*e^2)*E^(2*d)*((E^((b*c + 2*e)*x)*Hypergeometric2F1[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]*Sin
h[e*x])/e^2 + (2*E^(b*c*x)*Tanh[d])/(b*c)))/2

Maple [F]

\[\int {\mathrm e}^{c \left (b x +a \right )} \tanh \left (e x +d \right )^{3}d x\]

[In]

int(exp(c*(b*x+a))*tanh(e*x+d)^3,x)

[Out]

int(exp(c*(b*x+a))*tanh(e*x+d)^3,x)

Fricas [F]

\[ \int e^{c (a+b x)} \tanh ^3(d+e x) \, dx=\int { e^{\left ({\left (b x + a\right )} c\right )} \tanh \left (e x + d\right )^{3} \,d x } \]

[In]

integrate(exp(c*(b*x+a))*tanh(e*x+d)^3,x, algorithm="fricas")

[Out]

integral(e^(b*c*x + a*c)*tanh(e*x + d)^3, x)

Sympy [F]

\[ \int e^{c (a+b x)} \tanh ^3(d+e x) \, dx=e^{a c} \int e^{b c x} \tanh ^{3}{\left (d + e x \right )}\, dx \]

[In]

integrate(exp(c*(b*x+a))*tanh(e*x+d)**3,x)

[Out]

exp(a*c)*Integral(exp(b*c*x)*tanh(d + e*x)**3, x)

Maxima [F]

\[ \int e^{c (a+b x)} \tanh ^3(d+e x) \, dx=\int { e^{\left ({\left (b x + a\right )} c\right )} \tanh \left (e x + d\right )^{3} \,d x } \]

[In]

integrate(exp(c*(b*x+a))*tanh(e*x+d)^3,x, algorithm="maxima")

[Out]

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) - 1
2*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))

Giac [F]

\[ \int e^{c (a+b x)} \tanh ^3(d+e x) \, dx=\int { e^{\left ({\left (b x + a\right )} c\right )} \tanh \left (e x + d\right )^{3} \,d x } \]

[In]

integrate(exp(c*(b*x+a))*tanh(e*x+d)^3,x, algorithm="giac")

[Out]

integrate(e^((b*x + a)*c)*tanh(e*x + d)^3, x)

Mupad [F(-1)]

Timed out. \[ \int e^{c (a+b x)} \tanh ^3(d+e x) \, dx=\int {\mathrm {e}}^{c\,\left (a+b\,x\right )}\,{\mathrm {tanh}\left (d+e\,x\right )}^3 \,d x \]

[In]

int(exp(c*(a + b*x))*tanh(d + e*x)^3,x)

[Out]

int(exp(c*(a + b*x))*tanh(d + e*x)^3, x)