3.253 \(\int \frac {(e+f x)^3 \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx\)

Optimal. Leaf size=139 \[ -\frac {12 i f^3 \text {Li}_4\left (-i e^{c+d x}\right )}{a d^4}+\frac {12 i f^2 (e+f x) \text {Li}_3\left (-i e^{c+d x}\right )}{a d^3}-\frac {6 i f (e+f x)^2 \text {Li}_2\left (-i e^{c+d x}\right )}{a d^2}-\frac {2 i (e+f x)^3 \log \left (1+i e^{c+d x}\right )}{a d}+\frac {i (e+f x)^4}{4 a f} \]

[Out]

1/4*I*(f*x+e)^4/a/f-2*I*(f*x+e)^3*ln(1+I*exp(d*x+c))/a/d-6*I*f*(f*x+e)^2*polylog(2,-I*exp(d*x+c))/a/d^2+12*I*f
^2*(f*x+e)*polylog(3,-I*exp(d*x+c))/a/d^3-12*I*f^3*polylog(4,-I*exp(d*x+c))/a/d^4

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Rubi [A]  time = 0.21, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {5559, 2190, 2531, 6609, 2282, 6589} \[ \frac {12 i f^2 (e+f x) \text {PolyLog}\left (3,-i e^{c+d x}\right )}{a d^3}-\frac {6 i f (e+f x)^2 \text {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^2}-\frac {12 i f^3 \text {PolyLog}\left (4,-i e^{c+d x}\right )}{a d^4}-\frac {2 i (e+f x)^3 \log \left (1+i e^{c+d x}\right )}{a d}+\frac {i (e+f x)^4}{4 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/4)*(e + f*x)^4)/(a*f) - ((2*I)*(e + f*x)^3*Log[1 + I*E^(c + d*x)])/(a*d) - ((6*I)*f*(e + f*x)^2*PolyLog[2,
(-I)*E^(c + d*x)])/(a*d^2) + ((12*I)*f^2*(e + f*x)*PolyLog[3, (-I)*E^(c + d*x)])/(a*d^3) - ((12*I)*f^3*PolyLo
g[4, (-I)*E^(c + d*x)])/(a*d^4)

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 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 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 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 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]

Rule 6609

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

Rubi steps

\begin {align*} \int \frac {(e+f x)^3 \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx &=\frac {i (e+f x)^4}{4 a f}+2 \int \frac {e^{c+d x} (e+f x)^3}{a+i a e^{c+d x}} \, dx\\ &=\frac {i (e+f x)^4}{4 a f}-\frac {2 i (e+f x)^3 \log \left (1+i e^{c+d x}\right )}{a d}+\frac {(6 i f) \int (e+f x)^2 \log \left (1+i e^{c+d x}\right ) \, dx}{a d}\\ &=\frac {i (e+f x)^4}{4 a f}-\frac {2 i (e+f x)^3 \log \left (1+i e^{c+d x}\right )}{a d}-\frac {6 i f (e+f x)^2 \text {Li}_2\left (-i e^{c+d x}\right )}{a d^2}+\frac {\left (12 i f^2\right ) \int (e+f x) \text {Li}_2\left (-i e^{c+d x}\right ) \, dx}{a d^2}\\ &=\frac {i (e+f x)^4}{4 a f}-\frac {2 i (e+f x)^3 \log \left (1+i e^{c+d x}\right )}{a d}-\frac {6 i f (e+f x)^2 \text {Li}_2\left (-i e^{c+d x}\right )}{a d^2}+\frac {12 i f^2 (e+f x) \text {Li}_3\left (-i e^{c+d x}\right )}{a d^3}-\frac {\left (12 i f^3\right ) \int \text {Li}_3\left (-i e^{c+d x}\right ) \, dx}{a d^3}\\ &=\frac {i (e+f x)^4}{4 a f}-\frac {2 i (e+f x)^3 \log \left (1+i e^{c+d x}\right )}{a d}-\frac {6 i f (e+f x)^2 \text {Li}_2\left (-i e^{c+d x}\right )}{a d^2}+\frac {12 i f^2 (e+f x) \text {Li}_3\left (-i e^{c+d x}\right )}{a d^3}-\frac {\left (12 i f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-i x)}{x} \, dx,x,e^{c+d x}\right )}{a d^4}\\ &=\frac {i (e+f x)^4}{4 a f}-\frac {2 i (e+f x)^3 \log \left (1+i e^{c+d x}\right )}{a d}-\frac {6 i f (e+f x)^2 \text {Li}_2\left (-i e^{c+d x}\right )}{a d^2}+\frac {12 i f^2 (e+f x) \text {Li}_3\left (-i e^{c+d x}\right )}{a d^3}-\frac {12 i f^3 \text {Li}_4\left (-i e^{c+d x}\right )}{a d^4}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 118, normalized size = 0.85 \[ \frac {i \left (-\frac {24 f \left (d^2 (e+f x)^2 \text {Li}_2\left (-i e^{c+d x}\right )-2 d f (e+f x) \text {Li}_3\left (-i e^{c+d x}\right )+2 f^2 \text {Li}_4\left (-i e^{c+d x}\right )\right )}{d^4}-\frac {8 (e+f x)^3 \log \left (1+i e^{c+d x}\right )}{d}+\frac {(e+f x)^4}{f}\right )}{4 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/4)*((e + f*x)^4/f - (8*(e + f*x)^3*Log[1 + I*E^(c + d*x)])/d - (24*f*(d^2*(e + f*x)^2*PolyLog[2, (-I)*E^(c
+ d*x)] - 2*d*f*(e + f*x)*PolyLog[3, (-I)*E^(c + d*x)] + 2*f^2*PolyLog[4, (-I)*E^(c + d*x)]))/d^4))/a

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fricas [C]  time = 0.58, size = 295, normalized size = 2.12 \[ \frac {i \, d^{4} f^{3} x^{4} + 4 i \, d^{4} e f^{2} x^{3} + 6 i \, d^{4} e^{2} f x^{2} + 4 i \, d^{4} e^{3} x + 8 i \, c d^{3} e^{3} - 12 i \, c^{2} d^{2} e^{2} f + 8 i \, c^{3} d e f^{2} - 2 i \, c^{4} f^{3} - 48 i \, f^{3} {\rm polylog}\left (4, -i \, e^{\left (d x + c\right )}\right ) + {\left (-24 i \, d^{2} f^{3} x^{2} - 48 i \, d^{2} e f^{2} x - 24 i \, d^{2} e^{2} f\right )} {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) + {\left (-8 i \, d^{3} e^{3} + 24 i \, c d^{2} e^{2} f - 24 i \, c^{2} d e f^{2} + 8 i \, c^{3} f^{3}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + {\left (-8 i \, d^{3} f^{3} x^{3} - 24 i \, d^{3} e f^{2} x^{2} - 24 i \, d^{3} e^{2} f x - 24 i \, c d^{2} e^{2} f + 24 i \, c^{2} d e f^{2} - 8 i \, c^{3} f^{3}\right )} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + {\left (48 i \, d f^{3} x + 48 i \, d e f^{2}\right )} {\rm polylog}\left (3, -i \, e^{\left (d x + c\right )}\right )}{4 \, a d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*cosh(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")

[Out]

1/4*(I*d^4*f^3*x^4 + 4*I*d^4*e*f^2*x^3 + 6*I*d^4*e^2*f*x^2 + 4*I*d^4*e^3*x + 8*I*c*d^3*e^3 - 12*I*c^2*d^2*e^2*
f + 8*I*c^3*d*e*f^2 - 2*I*c^4*f^3 - 48*I*f^3*polylog(4, -I*e^(d*x + c)) + (-24*I*d^2*f^3*x^2 - 48*I*d^2*e*f^2*
x - 24*I*d^2*e^2*f)*dilog(-I*e^(d*x + c)) + (-8*I*d^3*e^3 + 24*I*c*d^2*e^2*f - 24*I*c^2*d*e*f^2 + 8*I*c^3*f^3)
*log(e^(d*x + c) - I) + (-8*I*d^3*f^3*x^3 - 24*I*d^3*e*f^2*x^2 - 24*I*d^3*e^2*f*x - 24*I*c*d^2*e^2*f + 24*I*c^
2*d*e*f^2 - 8*I*c^3*f^3)*log(I*e^(d*x + c) + 1) + (48*I*d*f^3*x + 48*I*d*e*f^2)*polylog(3, -I*e^(d*x + c)))/(a
*d^4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )}^{3} \cosh \left (d x + c\right )}{i \, a \sinh \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((f*x + e)^3*cosh(d*x + c)/(I*a*sinh(d*x + c) + a), x)

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maple [B]  time = 0.21, size = 635, normalized size = 4.57 \[ -\frac {i e^{3} x}{a}+\frac {i x^{4} f^{3}}{4 a}+\frac {6 i e^{2} f c x}{d a}+\frac {6 i e^{2} f c \ln \left ({\mathrm e}^{d x +c}-i\right )}{d^{2} a}-\frac {6 i e \,f^{2} c^{2} \ln \left ({\mathrm e}^{d x +c}-i\right )}{d^{3} a}-\frac {6 i e^{2} f c \ln \left ({\mathrm e}^{d x +c}\right )}{d^{2} a}+\frac {6 i e \,f^{2} c^{2} \ln \left ({\mathrm e}^{d x +c}\right )}{d^{3} a}+\frac {6 i e \,f^{2} c^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right )}{d^{3} a}-\frac {6 i e^{2} f \ln \left (1+i {\mathrm e}^{d x +c}\right ) x}{d a}-\frac {6 i e^{2} f \ln \left (1+i {\mathrm e}^{d x +c}\right ) c}{d^{2} a}-\frac {6 i e \,f^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right ) x^{2}}{d a}-\frac {12 i e \,f^{2} \polylog \left (2, -i {\mathrm e}^{d x +c}\right ) x}{d^{2} a}-\frac {6 i e \,f^{2} c^{2} x}{d^{2} a}-\frac {2 i f^{3} c^{3} \ln \left ({\mathrm e}^{d x +c}\right )}{d^{4} a}+\frac {2 i f^{3} c^{3} \ln \left ({\mathrm e}^{d x +c}-i\right )}{d^{4} a}-\frac {2 i f^{3} c^{3} \ln \left (1+i {\mathrm e}^{d x +c}\right )}{d^{4} a}-\frac {6 i e^{2} f \polylog \left (2, -i {\mathrm e}^{d x +c}\right )}{d^{2} a}+\frac {12 i e \,f^{2} \polylog \left (3, -i {\mathrm e}^{d x +c}\right )}{d^{3} a}-\frac {6 i f^{3} \polylog \left (2, -i {\mathrm e}^{d x +c}\right ) x^{2}}{d^{2} a}+\frac {12 i f^{3} \polylog \left (3, -i {\mathrm e}^{d x +c}\right ) x}{d^{3} a}+\frac {3 i f^{3} c^{4}}{2 d^{4} a}+\frac {2 i \ln \left ({\mathrm e}^{d x +c}\right ) e^{3}}{d a}-\frac {2 i \ln \left ({\mathrm e}^{d x +c}-i\right ) e^{3}}{d a}-\frac {12 i f^{3} \polylog \left (4, -i {\mathrm e}^{d x +c}\right )}{a \,d^{4}}+\frac {3 i e^{2} f \,x^{2}}{2 a}+\frac {i e \,f^{2} x^{3}}{a}-\frac {2 i f^{3} \ln \left (1+i {\mathrm e}^{d x +c}\right ) x^{3}}{d a}+\frac {3 i e^{2} f \,c^{2}}{d^{2} a}-\frac {4 i e \,f^{2} c^{3}}{d^{3} a}+\frac {2 i f^{3} c^{3} x}{d^{3} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^3*cosh(d*x+c)/(a+I*a*sinh(d*x+c)),x)

[Out]

-12*I*f^3*polylog(4,-I*exp(d*x+c))/a/d^4-I/a*e^3*x+1/4*I/a*x^4*f^3+3/2*I/a*e^2*f*x^2-2*I/d/a*f^3*ln(1+I*exp(d*
x+c))*x^3+3*I/d^2/a*e^2*f*c^2-4*I/d^3/a*e*f^2*c^3+2*I/d^3/a*f^3*c^3*x-2*I/d^4/a*f^3*c^3*ln(exp(d*x+c))-6*I/d^2
/a*e^2*f*polylog(2,-I*exp(d*x+c))+12*I/d^3/a*e*f^2*polylog(3,-I*exp(d*x+c))+2*I/d^4/a*f^3*c^3*ln(exp(d*x+c)-I)
-2*I/d^4/a*f^3*c^3*ln(1+I*exp(d*x+c))-6*I/d^2/a*f^3*polylog(2,-I*exp(d*x+c))*x^2+12*I/d^3/a*f^3*polylog(3,-I*e
xp(d*x+c))*x+3/2*I/d^4/a*f^3*c^4+2*I/d/a*ln(exp(d*x+c))*e^3-2*I/d/a*ln(exp(d*x+c)-I)*e^3+I/a*e*f^2*x^3+6*I/d/a
*e^2*f*c*x+6*I/d^2/a*e^2*f*c*ln(exp(d*x+c)-I)-6*I/d^3/a*e*f^2*c^2*ln(exp(d*x+c)-I)-6*I/d^2/a*e^2*f*c*ln(exp(d*
x+c))+6*I/d^3/a*e*f^2*c^2*ln(exp(d*x+c))+6*I/d^3/a*e*f^2*c^2*ln(1+I*exp(d*x+c))-6*I/d/a*e^2*f*ln(1+I*exp(d*x+c
))*x-6*I/d^2/a*e^2*f*ln(1+I*exp(d*x+c))*c-6*I/d/a*e*f^2*ln(1+I*exp(d*x+c))*x^2-12*I/d^2/a*e*f^2*polylog(2,-I*e
xp(d*x+c))*x-6*I/d^2/a*e*f^2*c^2*x

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maxima [B]  time = 0.54, size = 264, normalized size = 1.90 \[ -\frac {i \, e^{3} \log \left (i \, a \sinh \left (d x + c\right ) + a\right )}{a d} - \frac {6 i \, {\left (d x \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right )\right )} e^{2} f}{a d^{2}} - \frac {i \, {\left (f^{3} x^{4} + 4 \, e f^{2} x^{3} + 6 \, e^{2} f x^{2}\right )}}{4 \, a} - \frac {6 i \, {\left (d^{2} x^{2} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + 2 \, d x {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - 2 \, {\rm Li}_{3}(-i \, e^{\left (d x + c\right )})\right )} e f^{2}}{a d^{3}} - \frac {2 i \, {\left (d^{3} x^{3} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + 3 \, d^{2} x^{2} {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - 6 \, d x {\rm Li}_{3}(-i \, e^{\left (d x + c\right )}) + 6 \, {\rm Li}_{4}(-i \, e^{\left (d x + c\right )})\right )} f^{3}}{a d^{4}} + \frac {i \, d^{4} f^{3} x^{4} + 4 i \, d^{4} e f^{2} x^{3} + 6 i \, d^{4} e^{2} f x^{2}}{2 \, a d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {cosh}\left (c+d\,x\right )\,{\left (e+f\,x\right )}^3}{a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(c + d*x)*(e + f*x)^3)/(a + a*sinh(c + d*x)*1i),x)

[Out]

int((cosh(c + d*x)*(e + f*x)^3)/(a + a*sinh(c + d*x)*1i), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {i \left (\int \frac {e^{3} \cosh {\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {f^{3} x^{3} \cosh {\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {3 e f^{2} x^{2} \cosh {\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {3 e^{2} f x \cosh {\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**3*cosh(d*x+c)/(a+I*a*sinh(d*x+c)),x)

[Out]

-I*(Integral(e**3*cosh(c + d*x)/(sinh(c + d*x) - I), x) + Integral(f**3*x**3*cosh(c + d*x)/(sinh(c + d*x) - I)
, x) + Integral(3*e*f**2*x**2*cosh(c + d*x)/(sinh(c + d*x) - I), x) + Integral(3*e**2*f*x*cosh(c + d*x)/(sinh(
c + d*x) - I), x))/a

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