3.2.88 \(\int \frac {(e+f x)^2 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx\) [188]

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

[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

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Rubi [A]
time = 0.19, 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 = {5676, 32, 3399, 4269, 3797, 2221, 2317, 2438} \begin {gather*} -\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} \end {gather*}

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2438

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 3399

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1/2)*(e + Pi*(a/(2*b))) + 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 3797

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))/(1 + E^(2*((-I)*e + f*
fz*x))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 4269

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 5676

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 ) \text {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*}

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Mathematica [A]
time = 1.70, size = 183, normalized size = 1.41 \begin {gather*} \frac {-i x \left (3 e^2+3 e f x+f^2 x^2\right )+\frac {6 f \left (d \left (d e^c x (2 e+f x)-2 \left (-i+e^c\right ) (e+f x) \log \left (1+i e^{c+d x}\right )\right )-2 \left (-i+e^c\right ) f \text {PolyLog}\left (2,-i e^{c+d x}\right )\right )}{d^3 \left (-1-i e^c\right )}+\frac {6 i (e+f x)^2 \sinh \left (\frac {d x}{2}\right )}{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 )}}{3 a} \end {gather*}

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

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 280 vs. \(2 (112 ) = 224\).
time = 1.28, size = 281, normalized size = 2.16

method	result	size
risch	\(-\frac {i f^{2} x^{3}}{3 a}-\frac {i f e \,x^{2}}{a}-\frac {i e^{2} x}{a}-\frac {i e^{3}}{3 a f}-\frac {2 \left (x^{2} f^{2}+2 e f x +e^{2}\right )}{d a \left ({\mathrm e}^{d x +c}-i\right )}+\frac {4 i \ln \left ({\mathrm e}^{d x +c}\right ) e f}{a \,d^{2}}-\frac {4 i \ln \left ({\mathrm e}^{d x +c}-i\right ) e f}{a \,d^{2}}+\frac {2 i f^{2} x^{2}}{a d}+\frac {4 i f^{2} c x}{a \,d^{2}}+\frac {2 i f^{2} c^{2}}{a \,d^{3}}-\frac {4 i f^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right ) x}{a \,d^{2}}-\frac {4 i f^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right ) c}{a \,d^{3}}-\frac {4 i f^{2} \polylog \left (2, -i {\mathrm e}^{d x +c}\right )}{a \,d^{3}}-\frac {4 i f^{2} c \ln \left ({\mathrm e}^{d x +c}\right )}{a \,d^{3}}+\frac {4 i f^{2} c \ln \left ({\mathrm e}^{d x +c}-i\right )}{a \,d^{3}}\)	\(281\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 263 vs. \(2 (107) = 214\).
time = 0.35, size = 263, normalized size = 2.02 \begin {gather*} -\frac {d^{3} f^{2} x^{3} + 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 ) + 3 \, {\left (d^{3} x + 2 \, d^{2}\right )} e^{2} + 3 \, {\left (d^{3} f x^{2} - 4 \, c d f\right )} e - {\left (-i \, d^{3} f^{2} x^{3} + 6 i \, d^{2} f^{2} x^{2} - 3 i \, d^{3} x e^{2} - 6 i \, c^{2} f^{2} - 3 \, {\left (i \, d^{3} f x^{2} - 4 i \, d^{2} f x - 4 i \, c d f\right )} e\right )} e^{\left (d x + c\right )} - 12 \, {\left (c f^{2} - d f e - {\left (-i \, c f^{2} + i \, d f e\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + 12 \, {\left (d f^{2} x + c f^{2} + {\left (i \, d f^{2} x + 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 )}} \end {gather*}

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 + 6*c^2*f^2 + 12*(I*f^2*e^(d*x + c) + f^2)*dilog(-I*e^(d*x + c)) + 3*(d^3*x + 2*d^2)*e^2 + 3
*(d^3*f*x^2 - 4*c*d*f)*e - (-I*d^3*f^2*x^3 + 6*I*d^2*f^2*x^2 - 3*I*d^3*x*e^2 - 6*I*c^2*f^2 - 3*(I*d^3*f*x^2 -
4*I*d^2*f*x - 4*I*c*d*f)*e)*e^(d*x + c) - 12*(c*f^2 - d*f*e - (-I*c*f^2 + I*d*f*e)*e^(d*x + c))*log(e^(d*x + c
) - I) + 12*(d*f^2*x + c*f^2 + (I*d*f^2*x + 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)

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

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

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)

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