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

Optimal. Leaf size=130 \[ -\frac {4 i f^2 \text {Li}_2\left (-i e^{c+d x}\right )}{a d^3}-\frac {4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac {i (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{a d}+\frac {i (e+f x)^2}{a d}-\frac {i (e+f x)^3}{3 a f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.27, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {5557, 32, 3318, 4184, 3716, 2190, 2279, 2391} \[ -\frac {4 i f^2 \text {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}-\frac {4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac {i (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{a d}+\frac {i (e+f x)^2}{a d}-\frac {i (e+f x)^3}{3 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && 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]

Rule 5557

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 2.40, size = 188, normalized size = 1.45 \[ \frac {\frac {3 \left (4 \left (1+i e^c\right ) f^2 \text {Li}_2\left (i e^{-c-d x}\right )-2 d (e+f x) \left (d (e+f x)+2 \left (1+i e^c\right ) f \log \left (1-i e^{-c-d x}\right )\right )\right )}{\left (e^c-i\right ) d^3}+\frac {6 i \sinh \left (\frac {d x}{2}\right ) (e+f x)^2}{d \left (\cosh \left (\frac {c}{2}\right )+i \sinh \left (\frac {c}{2}\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )}-i x \left (3 e^2+3 e f x+f^2 x^2\right )}{3 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*x*(3*e^2 + 3*e*f*x + f^2*x^2) + (3*(-2*d*(e + f*x)*(d*(e + f*x) + 2*(1 + I*E^c)*f*Log[1 - I*E^(-c - d*x)
]) + 4*(1 + I*E^c)*f^2*PolyLog[2, I*E^(-c - d*x)]))/(d^3*(-I + E^c)) + ((6*I)*(e + f*x)^2*Sinh[(d*x)/2])/(d*(C
osh[c/2] + I*Sinh[c/2])*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])))/(3*a)

________________________________________________________________________________________

fricas [B]  time = 0.53, size = 262, normalized size = 2.02 \[ -\frac {d^{3} f^{2} x^{3} + 3 \, d^{3} e f x^{2} + 3 \, d^{3} e^{2} x + 6 \, d^{2} e^{2} - 12 \, c d e f + 6 \, c^{2} f^{2} + 12 \, {\left (i \, f^{2} e^{\left (d x + c\right )} + f^{2}\right )} {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - {\left (-i \, d^{3} f^{2} x^{3} + 12 i \, c d e f - 6 i \, c^{2} f^{2} + {\left (-3 i \, d^{3} e f + 6 i \, d^{2} f^{2}\right )} x^{2} + {\left (-3 i \, d^{3} e^{2} + 12 i \, d^{2} e f\right )} x\right )} e^{\left (d x + c\right )} + {\left (12 \, d e f - 12 \, c f^{2} - {\left (-12 i \, d e f + 12 i \, c f^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + {\left (12 \, d f^{2} x + 12 \, c f^{2} - {\left (-12 i \, d f^{2} x - 12 i \, c f^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (i \, e^{\left (d x + c\right )} + 1\right )}{3 \, {\left (a d^{3} e^{\left (d x + c\right )} - i \, a d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.13, size = 269, normalized size = 2.07 \[ -\frac {i x^{3} f^{2}}{3 a}-\frac {i e f \,x^{2}}{a}-\frac {i e^{2} x}{a}-\frac {2 \left (x^{2} f^{2}+2 e f x +e^{2}\right )}{d a \left ({\mathrm e}^{d x +c}-i\right )}-\frac {4 i \ln \left ({\mathrm e}^{d x +c}-i\right ) e f}{a \,d^{2}}+\frac {4 i \ln \left ({\mathrm e}^{d x +c}\right ) e f}{a \,d^{2}}+\frac {2 i f^{2} x^{2}}{a d}+\frac {4 i f^{2} c x}{a \,d^{2}}+\frac {2 i f^{2} c^{2}}{a \,d^{3}}-\frac {4 i f^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right ) x}{a \,d^{2}}-\frac {4 i f^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right ) c}{a \,d^{3}}-\frac {4 i f^{2} \polylog \left (2, -i {\mathrm e}^{d x +c}\right )}{a \,d^{3}}+\frac {4 i f^{2} c \ln \left ({\mathrm e}^{d x +c}-i\right )}{a \,d^{3}}-\frac {4 i f^{2} c \ln \left ({\mathrm e}^{d x +c}\right )}{a \,d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{6} \, f^{2} {\left (\frac {2 i \, d x^{3} e^{\left (d x + c\right )} + 2 \, d x^{3} + 12 \, x^{2}}{a d e^{\left (d x + c\right )} - i \, a d} - 24 \, \int \frac {x}{a d e^{\left (d x + c\right )} - i \, a d}\,{d x}\right )} + e f {\left (\frac {-i \, d x^{2} + {\left (d x^{2} e^{c} - 4 \, x e^{c}\right )} e^{\left (d x\right )}}{i \, a d e^{\left (d x + c\right )} + a d} - \frac {4 i \, \log \left ({\left (e^{\left (d x + c\right )} - i\right )} e^{\left (-c\right )}\right )}{a d^{2}}\right )} + \frac {1}{2} \, e^{2} {\left (-\frac {2 i \, {\left (d x + c\right )}}{a d} - \frac {4}{{\left (a e^{\left (-d x - c\right )} + i \, a\right )} d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sinh(c + d*x)*(e + f*x)^2)/(a + a*sinh(c + d*x)*1i),x)

[Out]

int((sinh(c + d*x)*(e + f*x)^2)/(a + a*sinh(c + d*x)*1i), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*e**2*exp(c) + 4*e*f*x*exp(c) + 2*f**2*x**2*exp(c))/(-I*a*d*exp(c) - a*d*exp(-d*x)) - I*(Integral(-I*d*e**2/
(exp(c)*exp(d*x) - I), x) + Integral(-I*d*f**2*x**2/(exp(c)*exp(d*x) - I), x) + Integral(d*e**2*exp(c)*exp(d*x
)/(exp(c)*exp(d*x) - I), x) + Integral(4*e*f*exp(c)*exp(d*x)/(exp(c)*exp(d*x) - I), x) + Integral(4*f**2*x*exp
(c)*exp(d*x)/(exp(c)*exp(d*x) - I), x) + Integral(-2*I*d*e*f*x/(exp(c)*exp(d*x) - I), x) + Integral(d*f**2*x**
2*exp(c)*exp(d*x)/(exp(c)*exp(d*x) - I), x) + Integral(2*d*e*f*x*exp(c)*exp(d*x)/(exp(c)*exp(d*x) - I), x))/(a
*d)

________________________________________________________________________________________