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

Optimal. Leaf size=88 \[ -\frac {4 d (c+d x) \log \left (e^{e+f x}+1\right )}{a f^2}+\frac {(c+d x)^2 \tanh \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}+\frac {(c+d x)^2}{a f}-\frac {4 d^2 \text {Li}_2\left (-e^{e+f x}\right )}{a f^3} \]

[Out]

(d*x+c)^2/a/f-4*d*(d*x+c)*ln(1+exp(f*x+e))/a/f^2-4*d^2*polylog(2,-exp(f*x+e))/a/f^3+(d*x+c)^2*tanh(1/2*e+1/2*f
*x)/a/f

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Rubi [A]  time = 0.20, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3318, 4184, 3718, 2190, 2279, 2391} \[ -\frac {4 d^2 \text {PolyLog}\left (2,-e^{e+f x}\right )}{a f^3}-\frac {4 d (c+d x) \log \left (e^{e+f x}+1\right )}{a f^2}+\frac {(c+d x)^2 \tanh \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}+\frac {(c+d x)^2}{a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c + d*x)^2/(a*f) - (4*d*(c + d*x)*Log[1 + E^(e + f*x)])/(a*f^2) - (4*d^2*PolyLog[2, -E^(e + f*x)])/(a*f^3) +
((c + d*x)^2*Tanh[e/2 + (f*x)/2])/(a*f)

Rule 2190

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

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3318

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1*(e + (Pi*a)/(2*b)))/2 + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (GtQ[n, 0] || IGtQ[m, 0])

Rule 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c + d*x)^(m +
1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(1 + E^(2*(-(I*e) + f*fz*x))), x],
x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]

Rule 4184

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

Rubi steps

\begin {align*} \int \frac {(c+d x)^2}{a+a \cosh (e+f x)} \, dx &=\frac {\int (c+d x)^2 \csc ^2\left (\frac {1}{2} (i e+\pi )+\frac {i f x}{2}\right ) \, dx}{2 a}\\ &=\frac {(c+d x)^2 \tanh \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}-\frac {(2 d) \int (c+d x) \tanh \left (\frac {e}{2}+\frac {f x}{2}\right ) \, dx}{a f}\\ &=\frac {(c+d x)^2}{a f}+\frac {(c+d x)^2 \tanh \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}-\frac {(4 d) \int \frac {e^{2 \left (\frac {e}{2}+\frac {f x}{2}\right )} (c+d x)}{1+e^{2 \left (\frac {e}{2}+\frac {f x}{2}\right )}} \, dx}{a f}\\ &=\frac {(c+d x)^2}{a f}-\frac {4 d (c+d x) \log \left (1+e^{e+f x}\right )}{a f^2}+\frac {(c+d x)^2 \tanh \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}+\frac {\left (4 d^2\right ) \int \log \left (1+e^{2 \left (\frac {e}{2}+\frac {f x}{2}\right )}\right ) \, dx}{a f^2}\\ &=\frac {(c+d x)^2}{a f}-\frac {4 d (c+d x) \log \left (1+e^{e+f x}\right )}{a f^2}+\frac {(c+d x)^2 \tanh \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}+\frac {\left (4 d^2\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 \left (\frac {e}{2}+\frac {f x}{2}\right )}\right )}{a f^3}\\ &=\frac {(c+d x)^2}{a f}-\frac {4 d (c+d x) \log \left (1+e^{e+f x}\right )}{a f^2}-\frac {4 d^2 \text {Li}_2\left (-e^{e+f x}\right )}{a f^3}+\frac {(c+d x)^2 \tanh \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}\\ \end {align*}

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Mathematica [C]  time = 6.38, size = 472, normalized size = 5.36 \[ \frac {2 \text {sech}\left (\frac {e}{2}\right ) \cosh \left (\frac {e}{2}+\frac {f x}{2}\right ) \left (c^2 \sinh \left (\frac {f x}{2}\right )+2 c d x \sinh \left (\frac {f x}{2}\right )+d^2 x^2 \sinh \left (\frac {f x}{2}\right )\right )}{f (a \cosh (e+f x)+a)}-\frac {8 c d \text {sech}\left (\frac {e}{2}\right ) \cosh ^2\left (\frac {e}{2}+\frac {f x}{2}\right ) \left (\cosh \left (\frac {e}{2}\right ) \log \left (\sinh \left (\frac {e}{2}\right ) \sinh \left (\frac {f x}{2}\right )+\cosh \left (\frac {e}{2}\right ) \cosh \left (\frac {f x}{2}\right )\right )-\frac {1}{2} f x \sinh \left (\frac {e}{2}\right )\right )}{f^2 \left (\cosh ^2\left (\frac {e}{2}\right )-\sinh ^2\left (\frac {e}{2}\right )\right ) (a \cosh (e+f x)+a)}-\frac {8 d^2 \text {csch}\left (\frac {e}{2}\right ) \text {sech}\left (\frac {e}{2}\right ) \cosh ^2\left (\frac {e}{2}+\frac {f x}{2}\right ) \left (\frac {1}{4} f^2 x^2 e^{-\tanh ^{-1}\left (\coth \left (\frac {e}{2}\right )\right )}-\frac {i \coth \left (\frac {e}{2}\right ) \left (i \text {Li}_2\left (e^{2 i \left (\frac {i f x}{2}+i \tanh ^{-1}\left (\coth \left (\frac {e}{2}\right )\right )\right )}\right )-\frac {1}{2} f x \left (-\pi +2 i \tanh ^{-1}\left (\coth \left (\frac {e}{2}\right )\right )\right )-2 \left (i \tanh ^{-1}\left (\coth \left (\frac {e}{2}\right )\right )+\frac {i f x}{2}\right ) \log \left (1-e^{2 i \left (i \tanh ^{-1}\left (\coth \left (\frac {e}{2}\right )\right )+\frac {i f x}{2}\right )}\right )+2 i \tanh ^{-1}\left (\coth \left (\frac {e}{2}\right )\right ) \log \left (i \sinh \left (\tanh ^{-1}\left (\coth \left (\frac {e}{2}\right )\right )+\frac {f x}{2}\right )\right )-\pi \log \left (e^{f x}+1\right )+\pi \log \left (\cosh \left (\frac {f x}{2}\right )\right )\right )}{\sqrt {1-\coth ^2\left (\frac {e}{2}\right )}}\right )}{f^3 \sqrt {\text {csch}^2\left (\frac {e}{2}\right ) \left (\sinh ^2\left (\frac {e}{2}\right )-\cosh ^2\left (\frac {e}{2}\right )\right )} (a \cosh (e+f x)+a)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-8*c*d*Cosh[e/2 + (f*x)/2]^2*Sech[e/2]*(Cosh[e/2]*Log[Cosh[e/2]*Cosh[(f*x)/2] + Sinh[e/2]*Sinh[(f*x)/2]] - (f
*x*Sinh[e/2])/2))/(f^2*(a + a*Cosh[e + f*x])*(Cosh[e/2]^2 - Sinh[e/2]^2)) - (8*d^2*Cosh[e/2 + (f*x)/2]^2*Csch[
e/2]*((f^2*x^2)/(4*E^ArcTanh[Coth[e/2]]) - (I*Coth[e/2]*(-1/2*(f*x*(-Pi + (2*I)*ArcTanh[Coth[e/2]])) - Pi*Log[
1 + E^(f*x)] - 2*((I/2)*f*x + I*ArcTanh[Coth[e/2]])*Log[1 - E^((2*I)*((I/2)*f*x + I*ArcTanh[Coth[e/2]]))] + Pi
*Log[Cosh[(f*x)/2]] + (2*I)*ArcTanh[Coth[e/2]]*Log[I*Sinh[(f*x)/2 + ArcTanh[Coth[e/2]]]] + I*PolyLog[2, E^((2*
I)*((I/2)*f*x + I*ArcTanh[Coth[e/2]]))]))/Sqrt[1 - Coth[e/2]^2])*Sech[e/2])/(f^3*(a + a*Cosh[e + f*x])*Sqrt[Cs
ch[e/2]^2*(-Cosh[e/2]^2 + Sinh[e/2]^2)]) + (2*Cosh[e/2 + (f*x)/2]*Sech[e/2]*(c^2*Sinh[(f*x)/2] + 2*c*d*x*Sinh[
(f*x)/2] + d^2*x^2*Sinh[(f*x)/2]))/(f*(a + a*Cosh[e + f*x]))

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fricas [B]  time = 0.41, size = 243, normalized size = 2.76 \[ -\frac {2 \, {\left (d^{2} e^{2} - 2 \, c d e f + c^{2} f^{2} - {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x - d^{2} e^{2} + 2 \, c d e f\right )} \cosh \left (f x + e\right ) + 2 \, {\left (d^{2} \cosh \left (f x + e\right ) + d^{2} \sinh \left (f x + e\right ) + d^{2}\right )} {\rm Li}_2\left (-\cosh \left (f x + e\right ) - \sinh \left (f x + e\right )\right ) + 2 \, {\left (d^{2} f x + c d f + {\left (d^{2} f x + c d f\right )} \cosh \left (f x + e\right ) + {\left (d^{2} f x + c d f\right )} \sinh \left (f x + e\right )\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) + 1\right ) - {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x - d^{2} e^{2} + 2 \, c d e f\right )} \sinh \left (f x + e\right )\right )}}{a f^{3} \cosh \left (f x + e\right ) + a f^{3} \sinh \left (f x + e\right ) + a f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(d^2*e^2 - 2*c*d*e*f + c^2*f^2 - (d^2*f^2*x^2 + 2*c*d*f^2*x - d^2*e^2 + 2*c*d*e*f)*cosh(f*x + e) + 2*(d^2*c
osh(f*x + e) + d^2*sinh(f*x + e) + d^2)*dilog(-cosh(f*x + e) - sinh(f*x + e)) + 2*(d^2*f*x + c*d*f + (d^2*f*x
+ c*d*f)*cosh(f*x + e) + (d^2*f*x + c*d*f)*sinh(f*x + e))*log(cosh(f*x + e) + sinh(f*x + e) + 1) - (d^2*f^2*x^
2 + 2*c*d*f^2*x - d^2*e^2 + 2*c*d*e*f)*sinh(f*x + e))/(a*f^3*cosh(f*x + e) + a*f^3*sinh(f*x + e) + a*f^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.21, size = 174, normalized size = 1.98 \[ -\frac {2 \left (d^{2} x^{2}+2 c d x +c^{2}\right )}{f a \left ({\mathrm e}^{f x +e}+1\right )}-\frac {4 d c \ln \left ({\mathrm e}^{f x +e}+1\right )}{a \,f^{2}}+\frac {4 d \ln \left ({\mathrm e}^{f x +e}\right ) c}{a \,f^{2}}+\frac {2 d^{2} x^{2}}{a f}+\frac {4 d^{2} e x}{a \,f^{2}}+\frac {2 d^{2} e^{2}}{a \,f^{3}}-\frac {4 d^{2} \ln \left ({\mathrm e}^{f x +e}+1\right ) x}{a \,f^{2}}-\frac {4 d^{2} \polylog \left (2, -{\mathrm e}^{f x +e}\right )}{a \,f^{3}}-\frac {4 d^{2} e \ln \left ({\mathrm e}^{f x +e}\right )}{a \,f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/f*(d^2*x^2+2*c*d*x+c^2)/a/(exp(f*x+e)+1)-4/a/f^2*d*c*ln(exp(f*x+e)+1)+4/a/f^2*d*ln(exp(f*x+e))*c+2/a/f*d^2*
x^2+4/a/f^2*d^2*e*x+2/a/f^3*d^2*e^2-4/a/f^2*d^2*ln(exp(f*x+e)+1)*x-4*d^2*polylog(2,-exp(f*x+e))/a/f^3-4/a/f^3*
d^2*e*ln(exp(f*x+e))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -2 \, d^{2} {\left (\frac {x^{2}}{a f e^{\left (f x + e\right )} + a f} - 2 \, \int \frac {x}{a f e^{\left (f x + e\right )} + a f}\,{d x}\right )} + 4 \, c d {\left (\frac {x e^{\left (f x + e\right )}}{a f e^{\left (f x + e\right )} + a f} - \frac {\log \left ({\left (e^{\left (f x + e\right )} + 1\right )} e^{\left (-e\right )}\right )}{a f^{2}}\right )} + \frac {2 \, c^{2}}{{\left (a e^{\left (-f x - e\right )} + a\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*d^2*(x^2/(a*f*e^(f*x + e) + a*f) - 2*integrate(x/(a*f*e^(f*x + e) + a*f), x)) + 4*c*d*(x*e^(f*x + e)/(a*f*e
^(f*x + e) + a*f) - log((e^(f*x + e) + 1)*e^(-e))/(a*f^2)) + 2*c^2/((a*e^(-f*x - e) + a)*f)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c+d\,x\right )}^2}{a+a\,\mathrm {cosh}\left (e+f\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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