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

Optimal. Leaf size=106 \[ -\frac{4 i f (e+f x) \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^2}+\frac{4 i f^2 \text{PolyLog}\left (3,-i e^{c+d x}\right )}{a d^3}-\frac{2 i (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d}+\frac{i (e+f x)^3}{3 a f} \]

[Out]

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

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Rubi [A]  time = 0.184023, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {5559, 2190, 2531, 2282, 6589} \[ -\frac{4 i f (e+f x) \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^2}+\frac{4 i f^2 \text{PolyLog}\left (3,-i e^{c+d x}\right )}{a d^3}-\frac{2 i (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d}+\frac{i (e+f x)^3}{3 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5559

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

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 2531

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

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int \frac{(e+f x)^2 \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx &=\frac{i (e+f x)^3}{3 a f}+2 \int \frac{e^{c+d x} (e+f x)^2}{a+i a e^{c+d x}} \, dx\\ &=\frac{i (e+f x)^3}{3 a f}-\frac{2 i (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d}+\frac{(4 i f) \int (e+f x) \log \left (1+i e^{c+d x}\right ) \, dx}{a d}\\ &=\frac{i (e+f x)^3}{3 a f}-\frac{2 i (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d}-\frac{4 i f (e+f x) \text{Li}_2\left (-i e^{c+d x}\right )}{a d^2}+\frac{\left (4 i f^2\right ) \int \text{Li}_2\left (-i e^{c+d x}\right ) \, dx}{a d^2}\\ &=\frac{i (e+f x)^3}{3 a f}-\frac{2 i (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d}-\frac{4 i f (e+f x) \text{Li}_2\left (-i e^{c+d x}\right )}{a d^2}+\frac{\left (4 i f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{c+d x}\right )}{a d^3}\\ &=\frac{i (e+f x)^3}{3 a f}-\frac{2 i (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d}-\frac{4 i f (e+f x) \text{Li}_2\left (-i e^{c+d x}\right )}{a d^2}+\frac{4 i f^2 \text{Li}_3\left (-i e^{c+d x}\right )}{a d^3}\\ \end{align*}

Mathematica [A]  time = 0.0498098, size = 94, normalized size = 0.89 \[ \frac{i \left (-12 d f^2 (e+f x) \text{PolyLog}\left (2,-i e^{c+d x}\right )+12 f^3 \text{PolyLog}\left (3,-i e^{c+d x}\right )+d^2 (e+f x)^2 \left (d (e+f x)-6 f \log \left (1+i e^{c+d x}\right )\right )\right )}{3 a d^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/3)*(d^2*(e + f*x)^2*(d*(e + f*x) - 6*f*Log[1 + I*E^(c + d*x)]) - 12*d*f^2*(e + f*x)*PolyLog[2, (-I)*E^(c +
d*x)] + 12*f^3*PolyLog[3, (-I)*E^(c + d*x)]))/(a*d^3*f)

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Maple [B]  time = 0.073, size = 393, normalized size = 3.7 \begin{align*}{\frac{-2\,i{f}^{2}\ln \left ( 1+i{{\rm e}^{dx+c}} \right ){x}^{2}}{da}}+{\frac{2\,ief{c}^{2}}{a{d}^{2}}}-{\frac{4\,i{f}^{2}{\it polylog} \left ( 2,-i{{\rm e}^{dx+c}} \right ) x}{a{d}^{2}}}-{\frac{2\,i\ln \left ({{\rm e}^{dx+c}}-i \right ){e}^{2}}{da}}-{\frac{2\,i{f}^{2}{c}^{2}x}{a{d}^{2}}}+{\frac{4\,i{f}^{2}{\it polylog} \left ( 3,-i{{\rm e}^{dx+c}} \right ) }{a{d}^{3}}}-{\frac{2\,i{f}^{2}{c}^{2}\ln \left ({{\rm e}^{dx+c}}-i \right ) }{a{d}^{3}}}+{\frac{2\,i\ln \left ({{\rm e}^{dx+c}} \right ){e}^{2}}{da}}-{\frac{i{e}^{2}x}{a}}-{\frac{4\,iefc\ln \left ({{\rm e}^{dx+c}} \right ) }{a{d}^{2}}}-{\frac{4\,ief\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) x}{da}}+{\frac{{\frac{i}{3}}{x}^{3}{f}^{2}}{a}}-{\frac{4\,ief\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) c}{a{d}^{2}}}-{\frac{4\,ief{\it polylog} \left ( 2,-i{{\rm e}^{dx+c}} \right ) }{a{d}^{2}}}+{\frac{2\,i{f}^{2}{c}^{2}\ln \left ({{\rm e}^{dx+c}} \right ) }{a{d}^{3}}}-{\frac{{\frac{4\,i}{3}}{f}^{2}{c}^{3}}{a{d}^{3}}}+{\frac{4\,iefc\ln \left ({{\rm e}^{dx+c}}-i \right ) }{a{d}^{2}}}+{\frac{ief{x}^{2}}{a}}+{\frac{4\,iefcx}{da}}+{\frac{2\,i{f}^{2}{c}^{2}\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) }{a{d}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.56212, size = 221, normalized size = 2.08 \begin{align*} -\frac{i \, e^{2} \log \left (i \, a \sinh \left (d x + c\right ) + a\right )}{a d} - \frac{i \, f^{2} x^{3} + 3 i \, e f x^{2}}{3 \, a} - \frac{4 i \,{\left (d x \log \left (i \, e^{\left (d x + c\right )} + 1\right ) +{\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right )\right )} e f}{a d^{2}} - \frac{2 i \,{\left (d^{2} x^{2} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + 2 \, d x{\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - 2 \,{\rm Li}_{3}(-i \, e^{\left (d x + c\right )})\right )} f^{2}}{a d^{3}} + \frac{2 i \, d^{3} f^{2} x^{3} + 6 i \, d^{3} e f x^{2}}{3 \, a d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*e^2*log(I*a*sinh(d*x + c) + a)/(a*d) - 1/3*(I*f^2*x^3 + 3*I*e*f*x^2)/a - 4*I*(d*x*log(I*e^(d*x + c) + 1) +
dilog(-I*e^(d*x + c)))*e*f/(a*d^2) - 2*I*(d^2*x^2*log(I*e^(d*x + c) + 1) + 2*d*x*dilog(-I*e^(d*x + c)) - 2*pol
ylog(3, -I*e^(d*x + c)))*f^2/(a*d^3) + 1/3*(2*I*d^3*f^2*x^3 + 6*I*d^3*e*f*x^2)/(a*d^3)

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Fricas [C]  time = 2.22861, size = 482, normalized size = 4.55 \begin{align*} \frac{i \, d^{3} f^{2} x^{3} + 3 i \, d^{3} e f x^{2} + 3 i \, d^{3} e^{2} x + 6 i \, c d^{2} e^{2} - 6 i \, c^{2} d e f + 2 i \, c^{3} f^{2} + 12 i \, f^{2}{\rm polylog}\left (3, -i \, e^{\left (d x + c\right )}\right ) +{\left (-12 i \, d f^{2} x - 12 i \, d e f\right )}{\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) +{\left (-6 i \, d^{2} e^{2} + 12 i \, c d e f - 6 i \, c^{2} f^{2}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) +{\left (-6 i \, d^{2} f^{2} x^{2} - 12 i \, d^{2} e f x - 12 i \, c d e f + 6 i \, c^{2} f^{2}\right )} \log \left (i \, e^{\left (d x + c\right )} + 1\right )}{3 \, a d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(I*d^3*f^2*x^3 + 3*I*d^3*e*f*x^2 + 3*I*d^3*e^2*x + 6*I*c*d^2*e^2 - 6*I*c^2*d*e*f + 2*I*c^3*f^2 + 12*I*f^2*
polylog(3, -I*e^(d*x + c)) + (-12*I*d*f^2*x - 12*I*d*e*f)*dilog(-I*e^(d*x + c)) + (-6*I*d^2*e^2 + 12*I*c*d*e*f
- 6*I*c^2*f^2)*log(e^(d*x + c) - I) + (-6*I*d^2*f^2*x^2 - 12*I*d^2*e*f*x - 12*I*c*d*e*f + 6*I*c^2*f^2)*log(I*
e^(d*x + c) + 1))/(a*d^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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