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]

-1/128*d^2*exp(-4*f*x-4*e)/a^2/f^3-1/8*d^2*exp(-2*f*x-2*e)/a^2/f^3-1/32*d*exp(-4*f*x-4*e)*(d*x+c)/a^2/f^2-1/4*
d*exp(-2*f*x-2*e)*(d*x+c)/a^2/f^2-1/16*exp(-4*f*x-4*e)*(d*x+c)^2/a^2/f-1/4*exp(-2*f*x-2*e)*(d*x+c)^2/a^2/f+1/1
2*(d*x+c)^3/a^2/d

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, 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 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]

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]

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 = 0.99, 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)

________________________________________________________________________________________

fricas [B]  time = 0.42, size = 361, normalized size = 2.12 \[ -\frac {96 \, d^{2} f^{2} x^{2} + 96 \, c^{2} f^{2} + 96 \, c d f - {\left (32 \, d^{2} f^{3} x^{3} - 24 \, c^{2} f^{2} - 12 \, c d f + 24 \, {\left (4 \, c d f^{3} - d^{2} f^{2}\right )} x^{2} - 3 \, d^{2} + 12 \, {\left (8 \, c^{2} f^{3} - 4 \, c d f^{2} - d^{2} f\right )} x\right )} \cosh \left (f x + e\right )^{2} - 2 \, {\left (32 \, d^{2} f^{3} x^{3} + 24 \, c^{2} f^{2} + 12 \, c d f + 24 \, {\left (4 \, c d f^{3} + d^{2} f^{2}\right )} x^{2} + 3 \, d^{2} + 12 \, {\left (8 \, c^{2} f^{3} + 4 \, c d f^{2} + d^{2} f\right )} x\right )} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) - {\left (32 \, d^{2} f^{3} x^{3} - 24 \, c^{2} f^{2} - 12 \, c d f + 24 \, {\left (4 \, c d f^{3} - d^{2} f^{2}\right )} x^{2} - 3 \, d^{2} + 12 \, {\left (8 \, c^{2} f^{3} - 4 \, c d f^{2} - d^{2} f\right )} x\right )} \sinh \left (f x + e\right )^{2} + 48 \, d^{2} + 96 \, {\left (2 \, c d f^{2} + d^{2} f\right )} x}{384 \, {\left (a^{2} f^{3} \cosh \left (f x + e\right )^{2} + 2 \, a^{2} f^{3} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + a^{2} f^{3} \sinh \left (f x + e\right )^{2}\right )}} \]

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)

________________________________________________________________________________________

giac [A]  time = 0.15, size = 227, normalized size = 1.34 \[ \frac {{\left (32 \, d^{2} f^{3} x^{3} e^{\left (4 \, f x + 4 \, e\right )} + 96 \, c d f^{3} x^{2} e^{\left (4 \, f x + 4 \, e\right )} + 96 \, c^{2} f^{3} x e^{\left (4 \, f x + 4 \, e\right )} - 96 \, d^{2} f^{2} x^{2} e^{\left (2 \, f x + 2 \, e\right )} - 24 \, d^{2} f^{2} x^{2} - 192 \, c d f^{2} x e^{\left (2 \, f x + 2 \, e\right )} - 48 \, c d f^{2} x - 96 \, c^{2} f^{2} e^{\left (2 \, f x + 2 \, e\right )} - 96 \, d^{2} f x e^{\left (2 \, f x + 2 \, e\right )} - 24 \, c^{2} f^{2} - 12 \, d^{2} f x - 96 \, c d f e^{\left (2 \, f x + 2 \, e\right )} - 12 \, c d f - 48 \, d^{2} e^{\left (2 \, f x + 2 \, e\right )} - 3 \, d^{2}\right )} e^{\left (-4 \, f x - 4 \, e\right )}}{384 \, a^{2} f^{3}} \]

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)

________________________________________________________________________________________

maple [B]  time = 0.24, size = 913, normalized size = 5.37 \[ \text {result too large to display} \]

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)*cosh(f*x+e)^4+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-3/8*(
f*x+e)*cosh(f*x+e)^2)+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)*sinh(f*x+e)+3/16*
(f*x+e)^2-1/16*cosh(f*x+e)^4-3/16*cosh(f*x+e)^2)-4*e*d^2*(1/4*(f*x+e)*sinh(f*x+e)*cosh(f*x+e)^3+3/8*(f*x+e)*co
sh(f*x+e)*sinh(f*x+e)+3/16*(f*x+e)^2-1/16*cosh(f*x+e)^4-3/16*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*cosh(f*x+e)
^4-1/8*(f*x+e)*sinh(f*x+e)*cosh(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*cosh(f*x+e)^
4+3/32*cosh(f*x+e)^2)-4*c*d*f*(1/4*(f*x+e)*cosh(f*x+e)^4-1/16*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*e*d^2*(1/4*(f*x+e)*cosh(f*x+e)^4-1/16*cosh(f*x+e)^3*sinh(f*x+e)-3/32*cosh(f*x+e)*sin
h(f*x+e)-3/32*f*x-3/32*e)-1/2*cosh(f*x+e)^4*c^2*f^2+cosh(f*x+e)^4*c*d*f*e-1/2*cosh(f*x+e)^4*d^2*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*e*d^2*(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))

________________________________________________________________________________________

maxima [A]  time = 0.63, size = 190, normalized size = 1.12 \[ \frac {1}{16} \, c^{2} {\left (\frac {4 \, {\left (f x + e\right )}}{a^{2} f} - \frac {4 \, e^{\left (-2 \, f x - 2 \, e\right )} + e^{\left (-4 \, f x - 4 \, e\right )}}{a^{2} f}\right )} + \frac {{\left (8 \, f^{2} x^{2} e^{\left (4 \, e\right )} - 8 \, {\left (2 \, f x e^{\left (2 \, e\right )} + e^{\left (2 \, e\right )}\right )} e^{\left (-2 \, f x\right )} - {\left (4 \, f x + 1\right )} e^{\left (-4 \, f x\right )}\right )} c d e^{\left (-4 \, e\right )}}{32 \, a^{2} f^{2}} + \frac {{\left (32 \, f^{3} x^{3} e^{\left (4 \, e\right )} - 48 \, {\left (2 \, f^{2} x^{2} e^{\left (2 \, e\right )} + 2 \, f x e^{\left (2 \, e\right )} + e^{\left (2 \, e\right )}\right )} e^{\left (-2 \, f x\right )} - 3 \, {\left (8 \, f^{2} x^{2} + 4 \, f x + 1\right )} e^{\left (-4 \, f x\right )}\right )} d^{2} e^{\left (-4 \, e\right )}}{384 \, a^{2} f^{3}} \]

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)

________________________________________________________________________________________

mupad [B]  time = 1.18, size = 165, normalized size = 0.97 \[ \frac {c^2\,x}{4\,a^2}-{\mathrm {e}}^{-4\,e-4\,f\,x}\,\left (\frac {8\,c^2\,f^2+4\,c\,d\,f+d^2}{128\,a^2\,f^3}+\frac {d^2\,x^2}{16\,a^2\,f}+\frac {d\,x\,\left (d+4\,c\,f\right )}{32\,a^2\,f^2}\right )-{\mathrm {e}}^{-2\,e-2\,f\,x}\,\left (\frac {2\,c^2\,f^2+2\,c\,d\,f+d^2}{8\,a^2\,f^3}+\frac {d^2\,x^2}{4\,a^2\,f}+\frac {d\,x\,\left (d+2\,c\,f\right )}{4\,a^2\,f^2}\right )+\frac {d^2\,x^3}{12\,a^2}+\frac {c\,d\,x^2}{4\,a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c^2*x)/(4*a^2) - exp(- 4*e - 4*f*x)*((d^2 + 8*c^2*f^2 + 4*c*d*f)/(128*a^2*f^3) + (d^2*x^2)/(16*a^2*f) + (d*x*
(d + 4*c*f))/(32*a^2*f^2)) - exp(- 2*e - 2*f*x)*((d^2 + 2*c^2*f^2 + 2*c*d*f)/(8*a^2*f^3) + (d^2*x^2)/(4*a^2*f)
+ (d*x*(d + 2*c*f))/(4*a^2*f^2)) + (d^2*x^3)/(12*a^2) + (c*d*x^2)/(4*a^2)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {c^{2}}{\tanh ^{2}{\left (e + f x \right )} + 2 \tanh {\left (e + f x \right )} + 1}\, dx + \int \frac {d^{2} x^{2}}{\tanh ^{2}{\left (e + f x \right )} + 2 \tanh {\left (e + f x \right )} + 1}\, dx + \int \frac {2 c d x}{\tanh ^{2}{\left (e + f x \right )} + 2 \tanh {\left (e + f x \right )} + 1}\, dx}{a^{2}} \]

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

________________________________________________________________________________________