3.1.33 \(\int \frac {(c+d x)^2}{a+a \tanh (e+f x)} \, dx\) [33]

Optimal. Leaf size=122 \[ \frac {d^2 x}{4 a f^2}+\frac {(c+d x)^2}{4 a f}+\frac {(c+d x)^3}{6 a d}-\frac {d^2}{4 f^3 (a+a \tanh (e+f x))}-\frac {d (c+d x)}{2 f^2 (a+a \tanh (e+f x))}-\frac {(c+d x)^2}{2 f (a+a \tanh (e+f x))} \]

[Out]

1/4*d^2*x/a/f^2+1/4*(d*x+c)^2/a/f+1/6*(d*x+c)^3/a/d-1/4*d^2/f^3/(a+a*tanh(f*x+e))-1/2*d*(d*x+c)/f^2/(a+a*tanh(
f*x+e))-1/2*(d*x+c)^2/f/(a+a*tanh(f*x+e))

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Rubi [A]
time = 0.08, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3804, 3560, 8} \begin {gather*} -\frac {d (c+d x)}{2 f^2 (a \tanh (e+f x)+a)}-\frac {(c+d x)^2}{2 f (a \tanh (e+f x)+a)}+\frac {(c+d x)^2}{4 a f}+\frac {(c+d x)^3}{6 a d}-\frac {d^2}{4 f^3 (a \tanh (e+f x)+a)}+\frac {d^2 x}{4 a f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*x)/(4*a*f^2) + (c + d*x)^2/(4*a*f) + (c + d*x)^3/(6*a*d) - d^2/(4*f^3*(a + a*Tanh[e + f*x])) - (d*(c + d*
x))/(2*f^2*(a + a*Tanh[e + f*x])) - (c + d*x)^2/(2*f*(a + a*Tanh[e + f*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3560

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a*((a + b*Tan[c + d*x])^n/(2*b*d*n)), x] +
Dist[1/(2*a), Int[(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0]

Rule 3804

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

Rubi steps

\begin {align*} \int \frac {(c+d x)^2}{a+a \tanh (e+f x)} \, dx &=\frac {(c+d x)^3}{6 a d}-\frac {(c+d x)^2}{2 f (a+a \tanh (e+f x))}+\frac {d \int \frac {c+d x}{a+a \tanh (e+f x)} \, dx}{f}\\ &=\frac {(c+d x)^2}{4 a f}+\frac {(c+d x)^3}{6 a d}-\frac {d (c+d x)}{2 f^2 (a+a \tanh (e+f x))}-\frac {(c+d x)^2}{2 f (a+a \tanh (e+f x))}+\frac {d^2 \int \frac {1}{a+a \tanh (e+f x)} \, dx}{2 f^2}\\ &=\frac {(c+d x)^2}{4 a f}+\frac {(c+d x)^3}{6 a d}-\frac {d^2}{4 f^3 (a+a \tanh (e+f x))}-\frac {d (c+d x)}{2 f^2 (a+a \tanh (e+f x))}-\frac {(c+d x)^2}{2 f (a+a \tanh (e+f x))}+\frac {d^2 \int 1 \, dx}{4 a f^2}\\ &=\frac {d^2 x}{4 a f^2}+\frac {(c+d x)^2}{4 a f}+\frac {(c+d x)^3}{6 a d}-\frac {d^2}{4 f^3 (a+a \tanh (e+f x))}-\frac {d (c+d x)}{2 f^2 (a+a \tanh (e+f x))}-\frac {(c+d x)^2}{2 f (a+a \tanh (e+f x))}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 169, normalized size = 1.39 \begin {gather*} \frac {\text {sech}(e+f x) (\cosh (f x)+\sinh (f x)) \left (\left (2 c^2 f^2+2 c d f (1+2 f x)+d^2 \left (1+2 f x+2 f^2 x^2\right )\right ) \cosh (2 f x) (-\cosh (e)+\sinh (e))+\frac {4}{3} f^3 x \left (3 c^2+3 c d x+d^2 x^2\right ) (\cosh (e)+\sinh (e))+\left (2 c^2 f^2+2 c d f (1+2 f x)+d^2 \left (1+2 f x+2 f^2 x^2\right )\right ) (\cosh (e)-\sinh (e)) \sinh (2 f x)\right )}{8 a f^3 (1+\tanh (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 3.70, size = 103, normalized size = 0.84

method	result	size
risch	\(\frac {d^{2} x^{3}}{6 a}+\frac {d c \,x^{2}}{2 a}+\frac {c^{2} x}{2 a}+\frac {c^{3}}{6 a d}-\frac {\left (2 d^{2} x^{2} f^{2}+4 c d \,f^{2} x +2 c^{2} f^{2}+2 d^{2} f x +2 c d f +d^{2}\right ) {\mathrm e}^{-2 f x -2 e}}{8 a \,f^{3}}\)	\(103\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/a*d^2*x^3+1/2/a*d*c*x^2+1/2/a*c^2*x+1/6/a/d*c^3-1/8*(2*d^2*f^2*x^2+4*c*d*f^2*x+2*c^2*f^2+2*d^2*f*x+2*c*d*f
+d^2)/a/f^3*exp(-2*f*x-2*e)

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Maxima [A]
time = 0.32, size = 132, normalized size = 1.08 \begin {gather*} \frac {1}{4} \, c^{2} {\left (\frac {2 \, {\left (f x + e\right )}}{a f} - \frac {e^{\left (-2 \, f x - 2 \, e\right )}}{a f}\right )} + \frac {{\left (2 \, f^{2} x^{2} e^{\left (2 \, e\right )} - {\left (2 \, f x + 1\right )} e^{\left (-2 \, f x\right )}\right )} c d e^{\left (-2 \, e\right )}}{4 \, a f^{2}} + \frac {{\left (4 \, f^{3} x^{3} e^{\left (2 \, e\right )} - 3 \, {\left (2 \, f^{2} x^{2} + 2 \, f x + 1\right )} e^{\left (-2 \, f x\right )}\right )} d^{2} e^{\left (-2 \, e\right )}}{24 \, a f^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*c^2*(2*(f*x + e)/(a*f) - e^(-2*f*x - 2*e)/(a*f)) + 1/4*(2*f^2*x^2*e^(2*e) - (2*f*x + 1)*e^(-2*f*x))*c*d*e^
(-2*e)/(a*f^2) + 1/24*(4*f^3*x^3*e^(2*e) - 3*(2*f^2*x^2 + 2*f*x + 1)*e^(-2*f*x))*d^2*e^(-2*e)/(a*f^3)

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Fricas [A]
time = 0.60, size = 204, normalized size = 1.67 \begin {gather*} \frac {{\left (4 \, d^{2} f^{3} x^{3} - 6 \, c^{2} f^{2} - 6 \, c d f + 6 \, {\left (2 \, c d f^{3} - d^{2} f^{2}\right )} x^{2} - 3 \, d^{2} + 6 \, {\left (2 \, c^{2} f^{3} - 2 \, c d f^{2} - d^{2} f\right )} x\right )} \cosh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + {\left (4 \, d^{2} f^{3} x^{3} + 6 \, c^{2} f^{2} + 6 \, c d f + 6 \, {\left (2 \, c d f^{3} + d^{2} f^{2}\right )} x^{2} + 3 \, d^{2} + 6 \, {\left (2 \, c^{2} f^{3} + 2 \, c d f^{2} + d^{2} f\right )} x\right )} \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right )}{24 \, {\left (a f^{3} \cosh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + a f^{3} \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*((4*d^2*f^3*x^3 - 6*c^2*f^2 - 6*c*d*f + 6*(2*c*d*f^3 - d^2*f^2)*x^2 - 3*d^2 + 6*(2*c^2*f^3 - 2*c*d*f^2 -
d^2*f)*x)*cosh(f*x + cosh(1) + sinh(1)) + (4*d^2*f^3*x^3 + 6*c^2*f^2 + 6*c*d*f + 6*(2*c*d*f^3 + d^2*f^2)*x^2 +
3*d^2 + 6*(2*c^2*f^3 + 2*c*d*f^2 + d^2*f)*x)*sinh(f*x + cosh(1) + sinh(1)))/(a*f^3*cosh(f*x + cosh(1) + sinh(
1)) + a*f^3*sinh(f*x + cosh(1) + sinh(1)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.40, size = 119, normalized size = 0.98 \begin {gather*} \frac {{\left (4 \, d^{2} f^{3} x^{3} e^{\left (2 \, f x + 2 \, e\right )} + 12 \, c d f^{3} x^{2} e^{\left (2 \, f x + 2 \, e\right )} + 12 \, c^{2} f^{3} x e^{\left (2 \, f x + 2 \, e\right )} - 6 \, d^{2} f^{2} x^{2} - 12 \, c d f^{2} x - 6 \, c^{2} f^{2} - 6 \, d^{2} f x - 6 \, c d f - 3 \, d^{2}\right )} e^{\left (-2 \, f x - 2 \, e\right )}}{24 \, a f^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(4*d^2*f^3*x^3*e^(2*f*x + 2*e) + 12*c*d*f^3*x^2*e^(2*f*x + 2*e) + 12*c^2*f^3*x*e^(2*f*x + 2*e) - 6*d^2*f^
2*x^2 - 12*c*d*f^2*x - 6*c^2*f^2 - 6*d^2*f*x - 6*c*d*f - 3*d^2)*e^(-2*f*x - 2*e)/(a*f^3)

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Mupad [B]
time = 1.18, size = 187, normalized size = 1.53 \begin {gather*} \frac {{\mathrm {e}}^{-2\,e-2\,f\,x}\,\left (12\,c^2\,x\,{\mathrm {e}}^{2\,e+2\,f\,x}+4\,d^2\,x^3\,{\mathrm {e}}^{2\,e+2\,f\,x}+12\,c\,d\,x^2\,{\mathrm {e}}^{2\,e+2\,f\,x}\right )}{24\,a}-\frac {\frac {{\mathrm {e}}^{-2\,e-2\,f\,x}\,\left (3\,d^2-3\,d^2\,{\mathrm {e}}^{2\,e+2\,f\,x}\right )}{24}+\frac {f\,{\mathrm {e}}^{-2\,e-2\,f\,x}\,\left (6\,c\,d+6\,d^2\,x-6\,c\,d\,{\mathrm {e}}^{2\,e+2\,f\,x}\right )}{24}+\frac {f^2\,{\mathrm {e}}^{-2\,e-2\,f\,x}\,\left (6\,c^2-6\,c^2\,{\mathrm {e}}^{2\,e+2\,f\,x}+6\,d^2\,x^2+12\,c\,d\,x\right )}{24}}{a\,f^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(- 2*e - 2*f*x)*(12*c^2*x*exp(2*e + 2*f*x) + 4*d^2*x^3*exp(2*e + 2*f*x) + 12*c*d*x^2*exp(2*e + 2*f*x)))/(2
4*a) - ((exp(- 2*e - 2*f*x)*(3*d^2 - 3*d^2*exp(2*e + 2*f*x)))/24 + (f*exp(- 2*e - 2*f*x)*(6*c*d + 6*d^2*x - 6*
c*d*exp(2*e + 2*f*x)))/24 + (f^2*exp(- 2*e - 2*f*x)*(6*c^2 - 6*c^2*exp(2*e + 2*f*x) + 6*d^2*x^2 + 12*c*d*x))/2
4)/(a*f^3)

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