3.137 \(\int \tanh (c+d x) (a+b \tanh ^2(c+d x)) \, dx\)

Optimal. Leaf size=31 \[ \frac{(a+b) \log (\cosh (c+d x))}{d}-\frac{b \tanh ^2(c+d x)}{2 d} \]

[Out]

((a + b)*Log[Cosh[c + d*x]])/d - (b*Tanh[c + d*x]^2)/(2*d)

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Rubi [A]  time = 0.0319005, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3631, 3475} \[ \frac{(a+b) \log (\cosh (c+d x))}{d}-\frac{b \tanh ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[c + d*x]*(a + b*Tanh[c + d*x]^2),x]

[Out]

((a + b)*Log[Cosh[c + d*x]])/d - (b*Tanh[c + d*x]^2)/(2*d)

Rule 3631

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp
[(C*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Dist[A - C, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[
{a, b, e, f, A, C, m}, x] && NeQ[A*b^2 + a^2*C, 0] &&  !LeQ[m, -1]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx &=-\frac{b \tanh ^2(c+d x)}{2 d}-(-a-b) \int \tanh (c+d x) \, dx\\ &=\frac{(a+b) \log (\cosh (c+d x))}{d}-\frac{b \tanh ^2(c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.0234953, size = 41, normalized size = 1.32 \[ \frac{a \log (\cosh (c+d x))}{d}-\frac{b \tanh ^2(c+d x)}{2 d}+\frac{b \log (\cosh (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[c + d*x]*(a + b*Tanh[c + d*x]^2),x]

[Out]

(a*Log[Cosh[c + d*x]])/d + (b*Log[Cosh[c + d*x]])/d - (b*Tanh[c + d*x]^2)/(2*d)

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Maple [B]  time = 0.004, size = 76, normalized size = 2.5 \begin{align*} -{\frac{b \left ( \tanh \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{\ln \left ( \tanh \left ( dx+c \right ) -1 \right ) a}{2\,d}}-{\frac{\ln \left ( \tanh \left ( dx+c \right ) -1 \right ) b}{2\,d}}-{\frac{\ln \left ( \tanh \left ( dx+c \right ) +1 \right ) a}{2\,d}}-{\frac{\ln \left ( \tanh \left ( dx+c \right ) +1 \right ) b}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(d*x+c)*(a+b*tanh(d*x+c)^2),x)

[Out]

-1/2*b*tanh(d*x+c)^2/d-1/2/d*ln(tanh(d*x+c)-1)*a-1/2/d*ln(tanh(d*x+c)-1)*b-1/2/d*ln(tanh(d*x+c)+1)*a-1/2/d*ln(
tanh(d*x+c)+1)*b

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Maxima [B]  time = 1.69827, size = 103, normalized size = 3.32 \begin{align*} b{\left (x + \frac{c}{d} + \frac{\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} + \frac{2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d{\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} + \frac{a \log \left (\cosh \left (d x + c\right )\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*(x + c/d + log(e^(-2*d*x - 2*c) + 1)/d + 2*e^(-2*d*x - 2*c)/(d*(2*e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c) + 1)))
+ a*log(cosh(d*x + c))/d

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Fricas [B]  time = 1.89958, size = 1114, normalized size = 35.94 \begin{align*} -\frac{{\left (a + b\right )} d x \cosh \left (d x + c\right )^{4} + 4 \,{\left (a + b\right )} d x \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} +{\left (a + b\right )} d x \sinh \left (d x + c\right )^{4} +{\left (a + b\right )} d x + 2 \,{\left ({\left (a + b\right )} d x - b\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (3 \,{\left (a + b\right )} d x \cosh \left (d x + c\right )^{2} +{\left (a + b\right )} d x - b\right )} \sinh \left (d x + c\right )^{2} -{\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{4} + 4 \,{\left (a + b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} +{\left (a + b\right )} \sinh \left (d x + c\right )^{4} + 2 \,{\left (a + b\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (3 \,{\left (a + b\right )} \cosh \left (d x + c\right )^{2} + a + b\right )} \sinh \left (d x + c\right )^{2} + 4 \,{\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{3} +{\left (a + b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + a + b\right )} \log \left (\frac{2 \, \cosh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + 4 \,{\left ({\left (a + b\right )} d x \cosh \left (d x + c\right )^{3} +{\left ({\left (a + b\right )} d x - b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{d \cosh \left (d x + c\right )^{4} + 4 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + d \sinh \left (d x + c\right )^{4} + 2 \, d \cosh \left (d x + c\right )^{2} + 2 \,{\left (3 \, d \cosh \left (d x + c\right )^{2} + d\right )} \sinh \left (d x + c\right )^{2} + 4 \,{\left (d \cosh \left (d x + c\right )^{3} + d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a + b)*d*x*cosh(d*x + c)^4 + 4*(a + b)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + (a + b)*d*x*sinh(d*x + c)^4 + (a
+ b)*d*x + 2*((a + b)*d*x - b)*cosh(d*x + c)^2 + 2*(3*(a + b)*d*x*cosh(d*x + c)^2 + (a + b)*d*x - b)*sinh(d*x
+ c)^2 - ((a + b)*cosh(d*x + c)^4 + 4*(a + b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a + b)*sinh(d*x + c)^4 + 2*(a
+ b)*cosh(d*x + c)^2 + 2*(3*(a + b)*cosh(d*x + c)^2 + a + b)*sinh(d*x + c)^2 + 4*((a + b)*cosh(d*x + c)^3 + (a
+ b)*cosh(d*x + c))*sinh(d*x + c) + a + b)*log(2*cosh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) + 4*((a + b)*
d*x*cosh(d*x + c)^3 + ((a + b)*d*x - b)*cosh(d*x + c))*sinh(d*x + c))/(d*cosh(d*x + c)^4 + 4*d*cosh(d*x + c)*s
inh(d*x + c)^3 + d*sinh(d*x + c)^4 + 2*d*cosh(d*x + c)^2 + 2*(3*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^2 + 4*(d*
cosh(d*x + c)^3 + d*cosh(d*x + c))*sinh(d*x + c) + d)

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Sympy [A]  time = 0.319899, size = 60, normalized size = 1.94 \begin{align*} \begin{cases} a x - \frac{a \log{\left (\tanh{\left (c + d x \right )} + 1 \right )}}{d} + b x - \frac{b \log{\left (\tanh{\left (c + d x \right )} + 1 \right )}}{d} - \frac{b \tanh ^{2}{\left (c + d x \right )}}{2 d} & \text{for}\: d \neq 0 \\x \left (a + b \tanh ^{2}{\left (c \right )}\right ) \tanh{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)*(a+b*tanh(d*x+c)**2),x)

[Out]

Piecewise((a*x - a*log(tanh(c + d*x) + 1)/d + b*x - b*log(tanh(c + d*x) + 1)/d - b*tanh(c + d*x)**2/(2*d), Ne(
d, 0)), (x*(a + b*tanh(c)**2)*tanh(c), True))

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Giac [B]  time = 1.16461, size = 82, normalized size = 2.65 \begin{align*} -\frac{{\left (d x + c\right )}{\left (a + b\right )}}{d} + \frac{{\left (a + b\right )} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}{d} + \frac{2 \, b e^{\left (2 \, d x + 2 \, c\right )}}{d{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(d*x + c)*(a + b)/d + (a + b)*log(e^(2*d*x + 2*c) + 1)/d + 2*b*e^(2*d*x + 2*c)/(d*(e^(2*d*x + 2*c) + 1)^2)