3.39 \(\int \frac{(c+d x)^2}{(a+a \tanh (e+f x))^2} \, dx\)

Optimal. Leaf size=170 \[ -\frac{d (c+d x) e^{-4 e-4 f x}}{32 a^2 f^2}-\frac{d (c+d x) e^{-2 e-2 f x}}{4 a^2 f^2}-\frac{(c+d x)^2 e^{-4 e-4 f x}}{16 a^2 f}-\frac{(c+d x)^2 e^{-2 e-2 f x}}{4 a^2 f}+\frac{(c+d x)^3}{12 a^2 d}-\frac{d^2 e^{-4 e-4 f x}}{128 a^2 f^3}-\frac{d^2 e^{-2 e-2 f x}}{8 a^2 f^3} \]

[Out]

-(d^2*E^(-4*e - 4*f*x))/(128*a^2*f^3) - (d^2*E^(-2*e - 2*f*x))/(8*a^2*f^3) - (d*E^(-4*e - 4*f*x)*(c + d*x))/(3
2*a^2*f^2) - (d*E^(-2*e - 2*f*x)*(c + d*x))/(4*a^2*f^2) - (E^(-4*e - 4*f*x)*(c + d*x)^2)/(16*a^2*f) - (E^(-2*e
- 2*f*x)*(c + d*x)^2)/(4*a^2*f) + (c + d*x)^3/(12*a^2*d)

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Rubi [A]  time = 0.18944, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {3729, 2176, 2194} \[ -\frac{d (c+d x) e^{-4 e-4 f x}}{32 a^2 f^2}-\frac{d (c+d x) e^{-2 e-2 f x}}{4 a^2 f^2}-\frac{(c+d x)^2 e^{-4 e-4 f x}}{16 a^2 f}-\frac{(c+d x)^2 e^{-2 e-2 f x}}{4 a^2 f}+\frac{(c+d x)^3}{12 a^2 d}-\frac{d^2 e^{-4 e-4 f x}}{128 a^2 f^3}-\frac{d^2 e^{-2 e-2 f x}}{8 a^2 f^3} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2/(a + a*Tanh[e + f*x])^2,x]

[Out]

-(d^2*E^(-4*e - 4*f*x))/(128*a^2*f^3) - (d^2*E^(-2*e - 2*f*x))/(8*a^2*f^3) - (d*E^(-4*e - 4*f*x)*(c + d*x))/(3
2*a^2*f^2) - (d*E^(-2*e - 2*f*x)*(c + d*x))/(4*a^2*f^2) - (E^(-4*e - 4*f*x)*(c + d*x)^2)/(16*a^2*f) - (E^(-2*e
- 2*f*x)*(c + d*x)^2)/(4*a^2*f) + (c + d*x)^3/(12*a^2*d)

Rule 3729

Int[((c_.) + (d_.)*(x_))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandIntegrand[(c
+ d*x)^m, (1/(2*a) + E^((2*a*(e + f*x))/b)/(2*a))^(-n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2
+ b^2, 0] && ILtQ[n, 0]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

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]

Rubi steps

\begin{align*} \int \frac{(c+d x)^2}{(a+a \tanh (e+f x))^2} \, dx &=\int \left (\frac{(c+d x)^2}{4 a^2}+\frac{e^{-4 e-4 f x} (c+d x)^2}{4 a^2}+\frac{e^{-2 e-2 f x} (c+d x)^2}{2 a^2}\right ) \, dx\\ &=\frac{(c+d x)^3}{12 a^2 d}+\frac{\int e^{-4 e-4 f x} (c+d x)^2 \, dx}{4 a^2}+\frac{\int e^{-2 e-2 f x} (c+d x)^2 \, dx}{2 a^2}\\ &=-\frac{e^{-4 e-4 f x} (c+d x)^2}{16 a^2 f}-\frac{e^{-2 e-2 f x} (c+d x)^2}{4 a^2 f}+\frac{(c+d x)^3}{12 a^2 d}+\frac{d \int e^{-4 e-4 f x} (c+d x) \, dx}{8 a^2 f}+\frac{d \int e^{-2 e-2 f x} (c+d x) \, dx}{2 a^2 f}\\ &=-\frac{d e^{-4 e-4 f x} (c+d x)}{32 a^2 f^2}-\frac{d e^{-2 e-2 f x} (c+d x)}{4 a^2 f^2}-\frac{e^{-4 e-4 f x} (c+d x)^2}{16 a^2 f}-\frac{e^{-2 e-2 f x} (c+d x)^2}{4 a^2 f}+\frac{(c+d x)^3}{12 a^2 d}+\frac{d^2 \int e^{-4 e-4 f x} \, dx}{32 a^2 f^2}+\frac{d^2 \int e^{-2 e-2 f x} \, dx}{4 a^2 f^2}\\ &=-\frac{d^2 e^{-4 e-4 f x}}{128 a^2 f^3}-\frac{d^2 e^{-2 e-2 f x}}{8 a^2 f^3}-\frac{d e^{-4 e-4 f x} (c+d x)}{32 a^2 f^2}-\frac{d e^{-2 e-2 f x} (c+d x)}{4 a^2 f^2}-\frac{e^{-4 e-4 f x} (c+d x)^2}{16 a^2 f}-\frac{e^{-2 e-2 f x} (c+d x)^2}{4 a^2 f}+\frac{(c+d x)^3}{12 a^2 d}\\ \end{align*}

Mathematica [A]  time = 0.872055, size = 207, normalized size = 1.22 \[ \frac{\text{sech}^2(e+f x) \left (\left (24 c^2 f^2 (4 f x+1)+12 c d f \left (8 f^2 x^2+4 f x+1\right )+d^2 \left (32 f^3 x^3+24 f^2 x^2+12 f x+3\right )\right ) \sinh (2 (e+f x))+\left (24 c^2 f^2 (4 f x-1)+12 c d f \left (8 f^2 x^2-4 f x-1\right )+d^2 \left (32 f^3 x^3-24 f^2 x^2-12 f x-3\right )\right ) \cosh (2 (e+f x))-48 \left (2 c^2 f^2+2 c d f (2 f x+1)+d^2 \left (2 f^2 x^2+2 f x+1\right )\right )\right )}{384 a^2 f^3 (\tanh (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^2/(a + a*Tanh[e + f*x])^2,x]

[Out]

(Sech[e + f*x]^2*(-48*(2*c^2*f^2 + 2*c*d*f*(1 + 2*f*x) + d^2*(1 + 2*f*x + 2*f^2*x^2)) + (24*c^2*f^2*(-1 + 4*f*
x) + 12*c*d*f*(-1 - 4*f*x + 8*f^2*x^2) + d^2*(-3 - 12*f*x - 24*f^2*x^2 + 32*f^3*x^3))*Cosh[2*(e + f*x)] + (24*
c^2*f^2*(1 + 4*f*x) + 12*c*d*f*(1 + 4*f*x + 8*f^2*x^2) + d^2*(3 + 12*f*x + 24*f^2*x^2 + 32*f^3*x^3))*Sinh[2*(e
+ f*x)]))/(384*a^2*f^3*(1 + Tanh[e + f*x])^2)

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Maple [B]  time = 0.044, size = 1080, normalized size = 6.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^2/(a+a*tanh(f*x+e))^2,x)

[Out]

1/f^3/a^2*(2*d^2*(1/4*(f*x+e)^2*sinh(f*x+e)*cosh(f*x+e)^3+3/8*(f*x+e)^2*cosh(f*x+e)*sinh(f*x+e)+1/8*(f*x+e)^3-
1/8*(f*x+e)*sinh(f*x+e)^2*cosh(f*x+e)^2-1/2*(f*x+e)*cosh(f*x+e)^2+1/32*cosh(f*x+e)^3*sinh(f*x+e)+15/64*cosh(f*
x+e)*sinh(f*x+e)+15/64*f*x+15/64*e)+4*c*d*f*(1/4*(f*x+e)*sinh(f*x+e)*cosh(f*x+e)^3+3/8*(f*x+e)*cosh(f*x+e)*sin
h(f*x+e)+3/16*(f*x+e)^2-1/16*sinh(f*x+e)^2*cosh(f*x+e)^2-1/4*cosh(f*x+e)^2)-4*d^2*e*(1/4*(f*x+e)*sinh(f*x+e)*c
osh(f*x+e)^3+3/8*(f*x+e)*cosh(f*x+e)*sinh(f*x+e)+3/16*(f*x+e)^2-1/16*sinh(f*x+e)^2*cosh(f*x+e)^2-1/4*cosh(f*x+
e)^2)+2*c^2*f^2*((1/4*cosh(f*x+e)^3+3/8*cosh(f*x+e))*sinh(f*x+e)+3/8*f*x+3/8*e)-4*c*d*f*e*((1/4*cosh(f*x+e)^3+
3/8*cosh(f*x+e))*sinh(f*x+e)+3/8*f*x+3/8*e)+2*d^2*e^2*((1/4*cosh(f*x+e)^3+3/8*cosh(f*x+e))*sinh(f*x+e)+3/8*f*x
+3/8*e)-2*d^2*(1/4*(f*x+e)^2*sinh(f*x+e)^2*cosh(f*x+e)^2+1/4*(f*x+e)^2*cosh(f*x+e)^2-1/8*(f*x+e)*sinh(f*x+e)*c
osh(f*x+e)^3-3/16*(f*x+e)*cosh(f*x+e)*sinh(f*x+e)-3/32*(f*x+e)^2+1/32*sinh(f*x+e)^2*cosh(f*x+e)^2+1/8*cosh(f*x
+e)^2)-4*c*d*f*(1/4*(f*x+e)*sinh(f*x+e)^2*cosh(f*x+e)^2+1/4*(f*x+e)*cosh(f*x+e)^2-1/16*cosh(f*x+e)^3*sinh(f*x+
e)-3/32*cosh(f*x+e)*sinh(f*x+e)-3/32*f*x-3/32*e)+4*d^2*e*(1/4*(f*x+e)*sinh(f*x+e)^2*cosh(f*x+e)^2+1/4*(f*x+e)*
cosh(f*x+e)^2-1/16*cosh(f*x+e)^3*sinh(f*x+e)-3/32*cosh(f*x+e)*sinh(f*x+e)-3/32*f*x-3/32*e)-2*c^2*f^2*(1/4*sinh
(f*x+e)^2*cosh(f*x+e)^2+1/4*cosh(f*x+e)^2)+4*c*d*f*e*(1/4*sinh(f*x+e)^2*cosh(f*x+e)^2+1/4*cosh(f*x+e)^2)-2*d^2
*e^2*(1/4*sinh(f*x+e)^2*cosh(f*x+e)^2+1/4*cosh(f*x+e)^2)-d^2*(1/2*(f*x+e)^2*cosh(f*x+e)*sinh(f*x+e)+1/6*(f*x+e
)^3-1/2*(f*x+e)*cosh(f*x+e)^2+1/4*cosh(f*x+e)*sinh(f*x+e)+1/4*f*x+1/4*e)-2*c*d*f*(1/2*(f*x+e)*cosh(f*x+e)*sinh
(f*x+e)+1/4*(f*x+e)^2-1/4*cosh(f*x+e)^2)+2*d^2*e*(1/2*(f*x+e)*cosh(f*x+e)*sinh(f*x+e)+1/4*(f*x+e)^2-1/4*cosh(f
*x+e)^2)-c^2*f^2*(1/2*cosh(f*x+e)*sinh(f*x+e)+1/2*f*x+1/2*e)+2*c*d*f*e*(1/2*cosh(f*x+e)*sinh(f*x+e)+1/2*f*x+1/
2*e)-d^2*e^2*(1/2*cosh(f*x+e)*sinh(f*x+e)+1/2*f*x+1/2*e))

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Maxima [A]  time = 2.0075, size = 257, normalized size = 1.51 \begin{align*} \frac{1}{16} \, c^{2}{\left (\frac{4 \,{\left (f x + e\right )}}{a^{2} f} - \frac{4 \, e^{\left (-2 \, f x - 2 \, e\right )} + e^{\left (-4 \, f x - 4 \, e\right )}}{a^{2} f}\right )} + \frac{{\left (8 \, f^{2} x^{2} e^{\left (4 \, e\right )} - 8 \,{\left (2 \, f x e^{\left (2 \, e\right )} + e^{\left (2 \, e\right )}\right )} e^{\left (-2 \, f x\right )} -{\left (4 \, f x + 1\right )} e^{\left (-4 \, f x\right )}\right )} c d e^{\left (-4 \, e\right )}}{32 \, a^{2} f^{2}} + \frac{{\left (32 \, f^{3} x^{3} e^{\left (4 \, e\right )} - 48 \,{\left (2 \, f^{2} x^{2} e^{\left (2 \, e\right )} + 2 \, f x e^{\left (2 \, e\right )} + e^{\left (2 \, e\right )}\right )} e^{\left (-2 \, f x\right )} - 3 \,{\left (8 \, f^{2} x^{2} + 4 \, f x + 1\right )} e^{\left (-4 \, f x\right )}\right )} d^{2} e^{\left (-4 \, e\right )}}{384 \, a^{2} f^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*c^2*(4*(f*x + e)/(a^2*f) - (4*e^(-2*f*x - 2*e) + e^(-4*f*x - 4*e))/(a^2*f)) + 1/32*(8*f^2*x^2*e^(4*e) - 8
*(2*f*x*e^(2*e) + e^(2*e))*e^(-2*f*x) - (4*f*x + 1)*e^(-4*f*x))*c*d*e^(-4*e)/(a^2*f^2) + 1/384*(32*f^3*x^3*e^(
4*e) - 48*(2*f^2*x^2*e^(2*e) + 2*f*x*e^(2*e) + e^(2*e))*e^(-2*f*x) - 3*(8*f^2*x^2 + 4*f*x + 1)*e^(-4*f*x))*d^2
*e^(-4*e)/(a^2*f^3)

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Fricas [B]  time = 2.13086, size = 813, normalized size = 4.78 \begin{align*} -\frac{96 \, d^{2} f^{2} x^{2} + 96 \, c^{2} f^{2} + 96 \, c d f -{\left (32 \, d^{2} f^{3} x^{3} - 24 \, c^{2} f^{2} - 12 \, c d f + 24 \,{\left (4 \, c d f^{3} - d^{2} f^{2}\right )} x^{2} - 3 \, d^{2} + 12 \,{\left (8 \, c^{2} f^{3} - 4 \, c d f^{2} - d^{2} f\right )} x\right )} \cosh \left (f x + e\right )^{2} - 2 \,{\left (32 \, d^{2} f^{3} x^{3} + 24 \, c^{2} f^{2} + 12 \, c d f + 24 \,{\left (4 \, c d f^{3} + d^{2} f^{2}\right )} x^{2} + 3 \, d^{2} + 12 \,{\left (8 \, c^{2} f^{3} + 4 \, c d f^{2} + d^{2} f\right )} x\right )} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) -{\left (32 \, d^{2} f^{3} x^{3} - 24 \, c^{2} f^{2} - 12 \, c d f + 24 \,{\left (4 \, c d f^{3} - d^{2} f^{2}\right )} x^{2} - 3 \, d^{2} + 12 \,{\left (8 \, c^{2} f^{3} - 4 \, c d f^{2} - d^{2} f\right )} x\right )} \sinh \left (f x + e\right )^{2} + 48 \, d^{2} + 96 \,{\left (2 \, c d f^{2} + d^{2} f\right )} x}{384 \,{\left (a^{2} f^{3} \cosh \left (f x + e\right )^{2} + 2 \, a^{2} f^{3} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + a^{2} f^{3} \sinh \left (f x + e\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/384*(96*d^2*f^2*x^2 + 96*c^2*f^2 + 96*c*d*f - (32*d^2*f^3*x^3 - 24*c^2*f^2 - 12*c*d*f + 24*(4*c*d*f^3 - d^2
*f^2)*x^2 - 3*d^2 + 12*(8*c^2*f^3 - 4*c*d*f^2 - d^2*f)*x)*cosh(f*x + e)^2 - 2*(32*d^2*f^3*x^3 + 24*c^2*f^2 + 1
2*c*d*f + 24*(4*c*d*f^3 + d^2*f^2)*x^2 + 3*d^2 + 12*(8*c^2*f^3 + 4*c*d*f^2 + d^2*f)*x)*cosh(f*x + e)*sinh(f*x
+ e) - (32*d^2*f^3*x^3 - 24*c^2*f^2 - 12*c*d*f + 24*(4*c*d*f^3 - d^2*f^2)*x^2 - 3*d^2 + 12*(8*c^2*f^3 - 4*c*d*
f^2 - d^2*f)*x)*sinh(f*x + e)^2 + 48*d^2 + 96*(2*c*d*f^2 + d^2*f)*x)/(a^2*f^3*cosh(f*x + e)^2 + 2*a^2*f^3*cosh
(f*x + e)*sinh(f*x + e) + a^2*f^3*sinh(f*x + e)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{c^{2}}{\tanh ^{2}{\left (e + f x \right )} + 2 \tanh{\left (e + f x \right )} + 1}\, dx + \int \frac{d^{2} x^{2}}{\tanh ^{2}{\left (e + f x \right )} + 2 \tanh{\left (e + f x \right )} + 1}\, dx + \int \frac{2 c d x}{\tanh ^{2}{\left (e + f x \right )} + 2 \tanh{\left (e + f x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**2/(a+a*tanh(f*x+e))**2,x)

[Out]

(Integral(c**2/(tanh(e + f*x)**2 + 2*tanh(e + f*x) + 1), x) + Integral(d**2*x**2/(tanh(e + f*x)**2 + 2*tanh(e
+ f*x) + 1), x) + Integral(2*c*d*x/(tanh(e + f*x)**2 + 2*tanh(e + f*x) + 1), x))/a**2

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Giac [A]  time = 1.2111, size = 306, normalized size = 1.8 \begin{align*} \frac{{\left (32 \, d^{2} f^{3} x^{3} e^{\left (4 \, f x + 4 \, e\right )} + 96 \, c d f^{3} x^{2} e^{\left (4 \, f x + 4 \, e\right )} + 96 \, c^{2} f^{3} x e^{\left (4 \, f x + 4 \, e\right )} - 96 \, d^{2} f^{2} x^{2} e^{\left (2 \, f x + 2 \, e\right )} - 24 \, d^{2} f^{2} x^{2} - 192 \, c d f^{2} x e^{\left (2 \, f x + 2 \, e\right )} - 48 \, c d f^{2} x - 96 \, c^{2} f^{2} e^{\left (2 \, f x + 2 \, e\right )} - 96 \, d^{2} f x e^{\left (2 \, f x + 2 \, e\right )} - 24 \, c^{2} f^{2} - 12 \, d^{2} f x - 96 \, c d f e^{\left (2 \, f x + 2 \, e\right )} - 12 \, c d f - 48 \, d^{2} e^{\left (2 \, f x + 2 \, e\right )} - 3 \, d^{2}\right )} e^{\left (-4 \, f x - 4 \, e\right )}}{384 \, a^{2} f^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(32*d^2*f^3*x^3*e^(4*f*x + 4*e) + 96*c*d*f^3*x^2*e^(4*f*x + 4*e) + 96*c^2*f^3*x*e^(4*f*x + 4*e) - 96*d^2
*f^2*x^2*e^(2*f*x + 2*e) - 24*d^2*f^2*x^2 - 192*c*d*f^2*x*e^(2*f*x + 2*e) - 48*c*d*f^2*x - 96*c^2*f^2*e^(2*f*x
+ 2*e) - 96*d^2*f*x*e^(2*f*x + 2*e) - 24*c^2*f^2 - 12*d^2*f*x - 96*c*d*f*e^(2*f*x + 2*e) - 12*c*d*f - 48*d^2*
e^(2*f*x + 2*e) - 3*d^2)*e^(-4*f*x - 4*e)/(a^2*f^3)