3.1.32 \(\int \frac {(c+d x)^3}{a+a \tanh (e+f x)} \, dx\) [32]

Optimal. Leaf size=169 \[ \frac {3 d^3 x}{8 a f^3}+\frac {3 d (c+d x)^2}{8 a f^2}+\frac {(c+d x)^3}{4 a f}+\frac {(c+d x)^4}{8 a d}-\frac {3 d^3}{8 f^4 (a+a \tanh (e+f x))}-\frac {3 d^2 (c+d x)}{4 f^3 (a+a \tanh (e+f x))}-\frac {3 d (c+d x)^2}{4 f^2 (a+a \tanh (e+f x))}-\frac {(c+d x)^3}{2 f (a+a \tanh (e+f x))} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 5, 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 {3 d^2 (c+d x)}{4 f^3 (a \tanh (e+f x)+a)}-\frac {3 d (c+d x)^2}{4 f^2 (a \tanh (e+f x)+a)}-\frac {(c+d x)^3}{2 f (a \tanh (e+f x)+a)}+\frac {3 d (c+d x)^2}{8 a f^2}+\frac {(c+d x)^3}{4 a f}+\frac {(c+d x)^4}{8 a d}-\frac {3 d^3}{8 f^4 (a \tanh (e+f x)+a)}+\frac {3 d^3 x}{8 a f^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3/(a + a*Tanh[e + f*x]),x]

[Out]

(3*d^3*x)/(8*a*f^3) + (3*d*(c + d*x)^2)/(8*a*f^2) + (c + d*x)^3/(4*a*f) + (c + d*x)^4/(8*a*d) - (3*d^3)/(8*f^4
*(a + a*Tanh[e + f*x])) - (3*d^2*(c + d*x))/(4*f^3*(a + a*Tanh[e + f*x])) - (3*d*(c + d*x)^2)/(4*f^2*(a + a*Ta
nh[e + f*x])) - (c + d*x)^3/(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)^3}{a+a \tanh (e+f x)} \, dx &=\frac {(c+d x)^4}{8 a d}-\frac {(c+d x)^3}{2 f (a+a \tanh (e+f x))}+\frac {(3 d) \int \frac {(c+d x)^2}{a+a \tanh (e+f x)} \, dx}{2 f}\\ &=\frac {(c+d x)^3}{4 a f}+\frac {(c+d x)^4}{8 a d}-\frac {3 d (c+d x)^2}{4 f^2 (a+a \tanh (e+f x))}-\frac {(c+d x)^3}{2 f (a+a \tanh (e+f x))}+\frac {\left (3 d^2\right ) \int \frac {c+d x}{a+a \tanh (e+f x)} \, dx}{2 f^2}\\ &=\frac {3 d (c+d x)^2}{8 a f^2}+\frac {(c+d x)^3}{4 a f}+\frac {(c+d x)^4}{8 a d}-\frac {3 d^2 (c+d x)}{4 f^3 (a+a \tanh (e+f x))}-\frac {3 d (c+d x)^2}{4 f^2 (a+a \tanh (e+f x))}-\frac {(c+d x)^3}{2 f (a+a \tanh (e+f x))}+\frac {\left (3 d^3\right ) \int \frac {1}{a+a \tanh (e+f x)} \, dx}{4 f^3}\\ &=\frac {3 d (c+d x)^2}{8 a f^2}+\frac {(c+d x)^3}{4 a f}+\frac {(c+d x)^4}{8 a d}-\frac {3 d^3}{8 f^4 (a+a \tanh (e+f x))}-\frac {3 d^2 (c+d x)}{4 f^3 (a+a \tanh (e+f x))}-\frac {3 d (c+d x)^2}{4 f^2 (a+a \tanh (e+f x))}-\frac {(c+d x)^3}{2 f (a+a \tanh (e+f x))}+\frac {\left (3 d^3\right ) \int 1 \, dx}{8 a f^3}\\ &=\frac {3 d^3 x}{8 a f^3}+\frac {3 d (c+d x)^2}{8 a f^2}+\frac {(c+d x)^3}{4 a f}+\frac {(c+d x)^4}{8 a d}-\frac {3 d^3}{8 f^4 (a+a \tanh (e+f x))}-\frac {3 d^2 (c+d x)}{4 f^3 (a+a \tanh (e+f x))}-\frac {3 d (c+d x)^2}{4 f^2 (a+a \tanh (e+f x))}-\frac {(c+d x)^3}{2 f (a+a \tanh (e+f x))}\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 244, normalized size = 1.44 \begin {gather*} \frac {\text {sech}(e+f x) (\cosh (f x)+\sinh (f x)) \left (\left (4 c^3 f^3+6 c^2 d f^2 (1+2 f x)+6 c d^2 f \left (1+2 f x+2 f^2 x^2\right )+d^3 \left (3+6 f x+6 f^2 x^2+4 f^3 x^3\right )\right ) \cosh (2 f x) (-\cosh (e)+\sinh (e))+2 f^4 x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right ) (\cosh (e)+\sinh (e))+\left (4 c^3 f^3+6 c^2 d f^2 (1+2 f x)+6 c d^2 f \left (1+2 f x+2 f^2 x^2\right )+d^3 \left (3+6 f x+6 f^2 x^2+4 f^3 x^3\right )\right ) (\cosh (e)-\sinh (e)) \sinh (2 f x)\right )}{16 a f^4 (1+\tanh (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^3/(a + a*Tanh[e + f*x]),x]

[Out]

(Sech[e + f*x]*(Cosh[f*x] + Sinh[f*x])*((4*c^3*f^3 + 6*c^2*d*f^2*(1 + 2*f*x) + 6*c*d^2*f*(1 + 2*f*x + 2*f^2*x^
2) + d^3*(3 + 6*f*x + 6*f^2*x^2 + 4*f^3*x^3))*Cosh[2*f*x]*(-Cosh[e] + Sinh[e]) + 2*f^4*x*(4*c^3 + 6*c^2*d*x +
4*c*d^2*x^2 + d^3*x^3)*(Cosh[e] + Sinh[e]) + (4*c^3*f^3 + 6*c^2*d*f^2*(1 + 2*f*x) + 6*c*d^2*f*(1 + 2*f*x + 2*f
^2*x^2) + d^3*(3 + 6*f*x + 6*f^2*x^2 + 4*f^3*x^3))*(Cosh[e] - Sinh[e])*Sinh[2*f*x]))/(16*a*f^4*(1 + Tanh[e + f
*x]))

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Maple [A]
time = 4.07, size = 165, normalized size = 0.98

method	result	size
risch	\(\frac {d^{3} x^{4}}{8 a}+\frac {d^{2} c \,x^{3}}{2 a}+\frac {3 c^{2} d \,x^{2}}{4 a}+\frac {c^{3} x}{2 a}+\frac {c^{4}}{8 a d}-\frac {\left (4 d^{3} x^{3} f^{3}+12 c \,d^{2} f^{3} x^{2}+12 c^{2} d \,f^{3} x +6 d^{3} f^{2} x^{2}+4 c^{3} f^{3}+12 c \,d^{2} f^{2} x +6 c^{2} d \,f^{2}+6 d^{3} f x +6 c \,d^{2} f +3 d^{3}\right ) {\mathrm e}^{-2 f x -2 e}}{16 a \,f^{4}}\)	\(165\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^3/(a+a*tanh(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

1/8/a*d^3*x^4+1/2/a*d^2*c*x^3+3/4*c^2*d*x^2/a+1/2/a*c^3*x+1/8/a/d*c^4-1/16*(4*d^3*f^3*x^3+12*c*d^2*f^3*x^2+12*
c^2*d*f^3*x+6*d^3*f^2*x^2+4*c^3*f^3+12*c*d^2*f^2*x+6*c^2*d*f^2+6*d^3*f*x+6*c*d^2*f+3*d^3)/a/f^4*exp(-2*f*x-2*e
)

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Maxima [A]
time = 0.33, size = 194, normalized size = 1.15 \begin {gather*} \frac {1}{4} \, c^{3} {\left (\frac {2 \, {\left (f x + e\right )}}{a f} - \frac {e^{\left (-2 \, f x - 2 \, e\right )}}{a f}\right )} + \frac {3 \, {\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^{2} d e^{\left (-2 \, e\right )}}{8 \, 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 )} c d^{2} e^{\left (-2 \, e\right )}}{8 \, a f^{3}} + \frac {{\left (2 \, f^{4} x^{4} e^{\left (2 \, e\right )} - {\left (4 \, f^{3} x^{3} + 6 \, f^{2} x^{2} + 6 \, f x + 3\right )} e^{\left (-2 \, f x\right )}\right )} d^{3} e^{\left (-2 \, e\right )}}{16 \, a f^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3/(a+a*tanh(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 316 vs. \(2 (157) = 314\).
time = 0.38, size = 316, normalized size = 1.87 \begin {gather*} \frac {{\left (2 \, d^{3} f^{4} x^{4} - 4 \, c^{3} f^{3} - 6 \, c^{2} d f^{2} - 6 \, c d^{2} f + 4 \, {\left (2 \, c d^{2} f^{4} - d^{3} f^{3}\right )} x^{3} - 3 \, d^{3} + 6 \, {\left (2 \, c^{2} d f^{4} - 2 \, c d^{2} f^{3} - d^{3} f^{2}\right )} x^{2} + 2 \, {\left (4 \, c^{3} f^{4} - 6 \, c^{2} d f^{3} - 6 \, c d^{2} f^{2} - 3 \, d^{3} f\right )} x\right )} \cosh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + {\left (2 \, d^{3} f^{4} x^{4} + 4 \, c^{3} f^{3} + 6 \, c^{2} d f^{2} + 6 \, c d^{2} f + 4 \, {\left (2 \, c d^{2} f^{4} + d^{3} f^{3}\right )} x^{3} + 3 \, d^{3} + 6 \, {\left (2 \, c^{2} d f^{4} + 2 \, c d^{2} f^{3} + d^{3} f^{2}\right )} x^{2} + 2 \, {\left (4 \, c^{3} f^{4} + 6 \, c^{2} d f^{3} + 6 \, c d^{2} f^{2} + 3 \, d^{3} f\right )} x\right )} \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right )}{16 \, {\left (a f^{4} \cosh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + a f^{4} \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)^3/(a+a*tanh(f*x+e)),x, algorithm="fricas")

[Out]

1/16*((2*d^3*f^4*x^4 - 4*c^3*f^3 - 6*c^2*d*f^2 - 6*c*d^2*f + 4*(2*c*d^2*f^4 - d^3*f^3)*x^3 - 3*d^3 + 6*(2*c^2*
d*f^4 - 2*c*d^2*f^3 - d^3*f^2)*x^2 + 2*(4*c^3*f^4 - 6*c^2*d*f^3 - 6*c*d^2*f^2 - 3*d^3*f)*x)*cosh(f*x + cosh(1)
+ sinh(1)) + (2*d^3*f^4*x^4 + 4*c^3*f^3 + 6*c^2*d*f^2 + 6*c*d^2*f + 4*(2*c*d^2*f^4 + d^3*f^3)*x^3 + 3*d^3 + 6
*(2*c^2*d*f^4 + 2*c*d^2*f^3 + d^3*f^2)*x^2 + 2*(4*c^3*f^4 + 6*c^2*d*f^3 + 6*c*d^2*f^2 + 3*d^3*f)*x)*sinh(f*x +
cosh(1) + sinh(1)))/(a*f^4*cosh(f*x + cosh(1) + sinh(1)) + a*f^4*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^{3}}{\tanh {\left (e + f x \right )} + 1}\, dx + \int \frac {d^{3} x^{3}}{\tanh {\left (e + f x \right )} + 1}\, dx + \int \frac {3 c d^{2} x^{2}}{\tanh {\left (e + f x \right )} + 1}\, dx + \int \frac {3 c^{2} 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)**3/(a+a*tanh(f*x+e)),x)

[Out]

(Integral(c**3/(tanh(e + f*x) + 1), x) + Integral(d**3*x**3/(tanh(e + f*x) + 1), x) + Integral(3*c*d**2*x**2/(
tanh(e + f*x) + 1), x) + Integral(3*c**2*d*x/(tanh(e + f*x) + 1), x))/a

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Giac [A]
time = 0.40, size = 188, normalized size = 1.11 \begin {gather*} \frac {{\left (2 \, d^{3} f^{4} x^{4} e^{\left (2 \, f x + 2 \, e\right )} + 8 \, c d^{2} f^{4} x^{3} e^{\left (2 \, f x + 2 \, e\right )} + 12 \, c^{2} d f^{4} x^{2} e^{\left (2 \, f x + 2 \, e\right )} - 4 \, d^{3} f^{3} x^{3} + 8 \, c^{3} f^{4} x e^{\left (2 \, f x + 2 \, e\right )} - 12 \, c d^{2} f^{3} x^{2} - 12 \, c^{2} d f^{3} x - 6 \, d^{3} f^{2} x^{2} - 4 \, c^{3} f^{3} - 12 \, c d^{2} f^{2} x - 6 \, c^{2} d f^{2} - 6 \, d^{3} f x - 6 \, c d^{2} f - 3 \, d^{3}\right )} e^{\left (-2 \, f x - 2 \, e\right )}}{16 \, a f^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3/(a+a*tanh(f*x+e)),x, algorithm="giac")

[Out]

1/16*(2*d^3*f^4*x^4*e^(2*f*x + 2*e) + 8*c*d^2*f^4*x^3*e^(2*f*x + 2*e) + 12*c^2*d*f^4*x^2*e^(2*f*x + 2*e) - 4*d
^3*f^3*x^3 + 8*c^3*f^4*x*e^(2*f*x + 2*e) - 12*c*d^2*f^3*x^2 - 12*c^2*d*f^3*x - 6*d^3*f^2*x^2 - 4*c^3*f^3 - 12*
c*d^2*f^2*x - 6*c^2*d*f^2 - 6*d^3*f*x - 6*c*d^2*f - 3*d^3)*e^(-2*f*x - 2*e)/(a*f^4)

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Mupad [B]
time = 1.31, size = 273, normalized size = 1.62 \begin {gather*} \frac {{\mathrm {e}}^{-2\,e-2\,f\,x}\,\left (8\,c^3\,x\,{\mathrm {e}}^{2\,e+2\,f\,x}+2\,d^3\,x^4\,{\mathrm {e}}^{2\,e+2\,f\,x}+12\,c^2\,d\,x^2\,{\mathrm {e}}^{2\,e+2\,f\,x}+8\,c\,d^2\,x^3\,{\mathrm {e}}^{2\,e+2\,f\,x}\right )}{16\,a}-\frac {\frac {{\mathrm {e}}^{-2\,e-2\,f\,x}\,\left (3\,d^3-3\,d^3\,{\mathrm {e}}^{2\,e+2\,f\,x}\right )}{16}+\frac {f^2\,{\mathrm {e}}^{-2\,e-2\,f\,x}\,\left (6\,c^2\,d+6\,d^3\,x^2-6\,c^2\,d\,{\mathrm {e}}^{2\,e+2\,f\,x}+12\,c\,d^2\,x\right )}{16}+\frac {f\,{\mathrm {e}}^{-2\,e-2\,f\,x}\,\left (6\,c\,d^2+6\,d^3\,x-6\,c\,d^2\,{\mathrm {e}}^{2\,e+2\,f\,x}\right )}{16}+\frac {f^3\,{\mathrm {e}}^{-2\,e-2\,f\,x}\,\left (4\,c^3-4\,c^3\,{\mathrm {e}}^{2\,e+2\,f\,x}+4\,d^3\,x^3+12\,c\,d^2\,x^2+12\,c^2\,d\,x\right )}{16}}{a\,f^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^3/(a + a*tanh(e + f*x)),x)

[Out]

(exp(- 2*e - 2*f*x)*(8*c^3*x*exp(2*e + 2*f*x) + 2*d^3*x^4*exp(2*e + 2*f*x) + 12*c^2*d*x^2*exp(2*e + 2*f*x) + 8
*c*d^2*x^3*exp(2*e + 2*f*x)))/(16*a) - ((exp(- 2*e - 2*f*x)*(3*d^3 - 3*d^3*exp(2*e + 2*f*x)))/16 + (f^2*exp(-
2*e - 2*f*x)*(6*c^2*d + 6*d^3*x^2 - 6*c^2*d*exp(2*e + 2*f*x) + 12*c*d^2*x))/16 + (f*exp(- 2*e - 2*f*x)*(6*c*d^
2 + 6*d^3*x - 6*c*d^2*exp(2*e + 2*f*x)))/16 + (f^3*exp(- 2*e - 2*f*x)*(4*c^3 - 4*c^3*exp(2*e + 2*f*x) + 4*d^3*
x^3 + 12*c*d^2*x^2 + 12*c^2*d*x))/16)/(a*f^4)

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