3.235 \(\int e^{c (a+b x)} \tanh ^2(a c+b c x)^{3/2} \, dx\)

Optimal. Leaf size=193 \[ \frac {e^{c (a+b x)} \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{b c}+\frac {3 e^{c (a+b x)} \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{b c \left (e^{2 c (a+b x)}+1\right )}-\frac {2 e^{c (a+b x)} \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{b c \left (e^{2 c (a+b x)}+1\right )^2}-\frac {3 \tan ^{-1}\left (e^{c (a+b x)}\right ) \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{b c} \]

[Out]

exp(c*(b*x+a))*coth(b*c*x+a*c)*(tanh(b*c*x+a*c)^2)^(1/2)/b/c-2*exp(c*(b*x+a))*coth(b*c*x+a*c)*(tanh(b*c*x+a*c)
^2)^(1/2)/b/c/(1+exp(2*c*(b*x+a)))^2+3*exp(c*(b*x+a))*coth(b*c*x+a*c)*(tanh(b*c*x+a*c)^2)^(1/2)/b/c/(1+exp(2*c
*(b*x+a)))-3*arctan(exp(c*(b*x+a)))*coth(b*c*x+a*c)*(tanh(b*c*x+a*c)^2)^(1/2)/b/c

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Rubi [A]  time = 0.28, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {6720, 2282, 390, 1158, 12, 288, 203} \[ \frac {e^{c (a+b x)} \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{b c}+\frac {3 e^{c (a+b x)} \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{b c \left (e^{2 c (a+b x)}+1\right )}-\frac {2 e^{c (a+b x)} \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{b c \left (e^{2 c (a+b x)}+1\right )^2}-\frac {3 \tan ^{-1}\left (e^{c (a+b x)}\right ) \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{b c} \]

Antiderivative was successfully verified.

[In]

Int[E^(c*(a + b*x))*(Tanh[a*c + b*c*x]^2)^(3/2),x]

[Out]

(E^(c*(a + b*x))*Coth[a*c + b*c*x]*Sqrt[Tanh[a*c + b*c*x]^2])/(b*c) - (2*E^(c*(a + b*x))*Coth[a*c + b*c*x]*Sqr
t[Tanh[a*c + b*c*x]^2])/(b*c*(1 + E^(2*c*(a + b*x)))^2) + (3*E^(c*(a + b*x))*Coth[a*c + b*c*x]*Sqrt[Tanh[a*c +
b*c*x]^2])/(b*c*(1 + E^(2*c*(a + b*x)))) - (3*ArcTan[E^(c*(a + b*x))]*Coth[a*c + b*c*x]*Sqrt[Tanh[a*c + b*c*x
]^2])/(b*c)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 1158

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + c*
x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + c*x^4)^p, d + e*x^2, x], x, 0]}, -Simp[(R*x*(d + e*x
^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*
(2*q + 3), x], x], x]] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rubi steps

\begin {align*} \int e^{c (a+b x)} \tanh ^2(a c+b c x)^{3/2} \, dx &=\left (\coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}\right ) \int e^{c (a+b x)} \tanh ^3(a c+b c x) \, dx\\ &=\frac {\left (\coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}\right ) \operatorname {Subst}\left (\int \frac {\left (-1+x^2\right )^3}{\left (1+x^2\right )^3} \, dx,x,e^{c (a+b x)}\right )}{b c}\\ &=\frac {\left (\coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}\right ) \operatorname {Subst}\left (\int \left (1-\frac {2 \left (1+3 x^4\right )}{\left (1+x^2\right )^3}\right ) \, dx,x,e^{c (a+b x)}\right )}{b c}\\ &=\frac {e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c}-\frac {\left (2 \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}\right ) \operatorname {Subst}\left (\int \frac {1+3 x^4}{\left (1+x^2\right )^3} \, dx,x,e^{c (a+b x)}\right )}{b c}\\ &=\frac {e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c}-\frac {2 e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c \left (1+e^{2 c (a+b x)}\right )^2}+\frac {\left (\coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}\right ) \operatorname {Subst}\left (\int -\frac {12 x^2}{\left (1+x^2\right )^2} \, dx,x,e^{c (a+b x)}\right )}{2 b c}\\ &=\frac {e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c}-\frac {2 e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c \left (1+e^{2 c (a+b x)}\right )^2}-\frac {\left (6 \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (1+x^2\right )^2} \, dx,x,e^{c (a+b x)}\right )}{b c}\\ &=\frac {e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c}-\frac {2 e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c \left (1+e^{2 c (a+b x)}\right )^2}+\frac {3 e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c \left (1+e^{2 c (a+b x)}\right )}-\frac {\left (3 \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,e^{c (a+b x)}\right )}{b c}\\ &=\frac {e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c}-\frac {2 e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c \left (1+e^{2 c (a+b x)}\right )^2}+\frac {3 e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c \left (1+e^{2 c (a+b x)}\right )}-\frac {3 \tan ^{-1}\left (e^{c (a+b x)}\right ) \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 104, normalized size = 0.54 \[ \frac {\left (e^{c (a+b x)} \left (5 e^{2 c (a+b x)}+e^{4 c (a+b x)}+2\right )-3 \left (e^{2 c (a+b x)}+1\right )^2 \tan ^{-1}\left (e^{c (a+b x)}\right )\right ) \sqrt {\tanh ^2(c (a+b x))} \coth (c (a+b x))}{b c \left (e^{2 c (a+b x)}+1\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(c*(a + b*x))*(Tanh[a*c + b*c*x]^2)^(3/2),x]

[Out]

((E^(c*(a + b*x))*(2 + 5*E^(2*c*(a + b*x)) + E^(4*c*(a + b*x))) - 3*(1 + E^(2*c*(a + b*x)))^2*ArcTan[E^(c*(a +
b*x))])*Coth[c*(a + b*x)]*Sqrt[Tanh[c*(a + b*x)]^2])/(b*c*(1 + E^(2*c*(a + b*x)))^2)

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fricas [B]  time = 0.58, size = 458, normalized size = 2.37 \[ \frac {\cosh \left (b c x + a c\right )^{5} + 5 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{4} + \sinh \left (b c x + a c\right )^{5} + 5 \, {\left (2 \, \cosh \left (b c x + a c\right )^{2} + 1\right )} \sinh \left (b c x + a c\right )^{3} + 5 \, \cosh \left (b c x + a c\right )^{3} + 5 \, {\left (2 \, \cosh \left (b c x + a c\right )^{3} + 3 \, \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right )^{2} - 3 \, {\left (\cosh \left (b c x + a c\right )^{4} + 4 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{3} + \sinh \left (b c x + a c\right )^{4} + 2 \, {\left (3 \, \cosh \left (b c x + a c\right )^{2} + 1\right )} \sinh \left (b c x + a c\right )^{2} + 2 \, \cosh \left (b c x + a c\right )^{2} + 4 \, {\left (\cosh \left (b c x + a c\right )^{3} + \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right ) + 1\right )} \arctan \left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right ) + {\left (5 \, \cosh \left (b c x + a c\right )^{4} + 15 \, \cosh \left (b c x + a c\right )^{2} + 2\right )} \sinh \left (b c x + a c\right ) + 2 \, \cosh \left (b c x + a c\right )}{b c \cosh \left (b c x + a c\right )^{4} + 4 \, b c \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{3} + b c \sinh \left (b c x + a c\right )^{4} + 2 \, b c \cosh \left (b c x + a c\right )^{2} + 2 \, {\left (3 \, b c \cosh \left (b c x + a c\right )^{2} + b c\right )} \sinh \left (b c x + a c\right )^{2} + b c + 4 \, {\left (b c \cosh \left (b c x + a c\right )^{3} + b c \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*(tanh(b*c*x+a*c)^2)^(3/2),x, algorithm="fricas")

[Out]

(cosh(b*c*x + a*c)^5 + 5*cosh(b*c*x + a*c)*sinh(b*c*x + a*c)^4 + sinh(b*c*x + a*c)^5 + 5*(2*cosh(b*c*x + a*c)^
2 + 1)*sinh(b*c*x + a*c)^3 + 5*cosh(b*c*x + a*c)^3 + 5*(2*cosh(b*c*x + a*c)^3 + 3*cosh(b*c*x + a*c))*sinh(b*c*
x + a*c)^2 - 3*(cosh(b*c*x + a*c)^4 + 4*cosh(b*c*x + a*c)*sinh(b*c*x + a*c)^3 + sinh(b*c*x + a*c)^4 + 2*(3*cos
h(b*c*x + a*c)^2 + 1)*sinh(b*c*x + a*c)^2 + 2*cosh(b*c*x + a*c)^2 + 4*(cosh(b*c*x + a*c)^3 + cosh(b*c*x + a*c)
)*sinh(b*c*x + a*c) + 1)*arctan(cosh(b*c*x + a*c) + sinh(b*c*x + a*c)) + (5*cosh(b*c*x + a*c)^4 + 15*cosh(b*c*
x + a*c)^2 + 2)*sinh(b*c*x + a*c) + 2*cosh(b*c*x + a*c))/(b*c*cosh(b*c*x + a*c)^4 + 4*b*c*cosh(b*c*x + a*c)*si
nh(b*c*x + a*c)^3 + b*c*sinh(b*c*x + a*c)^4 + 2*b*c*cosh(b*c*x + a*c)^2 + 2*(3*b*c*cosh(b*c*x + a*c)^2 + b*c)*
sinh(b*c*x + a*c)^2 + b*c + 4*(b*c*cosh(b*c*x + a*c)^3 + b*c*cosh(b*c*x + a*c))*sinh(b*c*x + a*c))

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giac [A]  time = 0.18, size = 129, normalized size = 0.67 \[ -\frac {3 \, \arctan \left (e^{\left (b c x + a c\right )}\right ) \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - e^{\left (b c x + a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - \frac {3 \, e^{\left (3 \, b c x + 3 \, a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) + e^{\left (b c x + a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}{{\left (e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}^{2}}}{b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*(tanh(b*c*x+a*c)^2)^(3/2),x, algorithm="giac")

[Out]

-(3*arctan(e^(b*c*x + a*c))*sgn(e^(2*b*c*x + 2*a*c) - 1) - e^(b*c*x + a*c)*sgn(e^(2*b*c*x + 2*a*c) - 1) - (3*e
^(3*b*c*x + 3*a*c)*sgn(e^(2*b*c*x + 2*a*c) - 1) + e^(b*c*x + a*c)*sgn(e^(2*b*c*x + 2*a*c) - 1))/(e^(2*b*c*x +
2*a*c) + 1)^2)/(b*c)

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maple [C]  time = 0.70, size = 301, normalized size = 1.56 \[ \frac {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, {\mathrm e}^{c \left (b x +a \right )}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) c b}+\frac {\sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, {\mathrm e}^{c \left (b x +a \right )} \left (3 \,{\mathrm e}^{2 c \left (b x +a \right )}+1\right )}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) c b}+\frac {3 i \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \ln \left ({\mathrm e}^{c \left (b x +a \right )}-i\right )}{2 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) c b}-\frac {3 i \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \ln \left ({\mathrm e}^{c \left (b x +a \right )}+i\right )}{2 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) c b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*(b*x+a))*(tanh(b*c*x+a*c)^2)^(3/2),x)

[Out]

1/(exp(2*c*(b*x+a))-1)*(1+exp(2*c*(b*x+a)))*((exp(2*c*(b*x+a))-1)^2/(1+exp(2*c*(b*x+a)))^2)^(1/2)*exp(c*(b*x+a
))/c/b+1/(exp(2*c*(b*x+a))-1)/(1+exp(2*c*(b*x+a)))*((exp(2*c*(b*x+a))-1)^2/(1+exp(2*c*(b*x+a)))^2)^(1/2)*exp(c
*(b*x+a))*(3*exp(2*c*(b*x+a))+1)/c/b+3/2*I/(exp(2*c*(b*x+a))-1)*(1+exp(2*c*(b*x+a)))*((exp(2*c*(b*x+a))-1)^2/(
1+exp(2*c*(b*x+a)))^2)^(1/2)/c/b*ln(exp(c*(b*x+a))-I)-3/2*I/(exp(2*c*(b*x+a))-1)*(1+exp(2*c*(b*x+a)))*((exp(2*
c*(b*x+a))-1)^2/(1+exp(2*c*(b*x+a)))^2)^(1/2)/c/b*ln(exp(c*(b*x+a))+I)

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maxima [A]  time = 0.42, size = 90, normalized size = 0.47 \[ -\frac {3 \, \arctan \left (e^{\left (b c x + a c\right )}\right )}{b c} + \frac {e^{\left (5 \, b c x + 5 \, a c\right )} + 5 \, e^{\left (3 \, b c x + 3 \, a c\right )} + 2 \, e^{\left (b c x + a c\right )}}{b c {\left (e^{\left (4 \, b c x + 4 \, a c\right )} + 2 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*(tanh(b*c*x+a*c)^2)^(3/2),x, algorithm="maxima")

[Out]

-3*arctan(e^(b*c*x + a*c))/(b*c) + (e^(5*b*c*x + 5*a*c) + 5*e^(3*b*c*x + 3*a*c) + 2*e^(b*c*x + a*c))/(b*c*(e^(
4*b*c*x + 4*a*c) + 2*e^(2*b*c*x + 2*a*c) + 1))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {e}}^{c\,\left (a+b\,x\right )}\,{\left ({\mathrm {tanh}\left (a\,c+b\,c\,x\right )}^2\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*(a + b*x))*(tanh(a*c + b*c*x)^2)^(3/2),x)

[Out]

int(exp(c*(a + b*x))*(tanh(a*c + b*c*x)^2)^(3/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*(tanh(b*c*x+a*c)**2)**(3/2),x)

[Out]

Timed out

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