3.2.94 \(\int \frac {(e+f x)^2 \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx\) [194]

Optimal. Leaf size=184 \[ -\frac {(e+f x)^2}{a d}+\frac {(e+f x)^3}{3 a f}-\frac {2 i f^2 \cosh (c+d x)}{a d^3}-\frac {i (e+f x)^2 \cosh (c+d x)}{a d}+\frac {4 f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac {4 f^2 \text {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}+\frac {2 i f (e+f x) \sinh (c+d x)}{a d^2}-\frac {(e+f x)^2 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d} \]

[Out]

-(f*x+e)^2/a/d+1/3*(f*x+e)^3/a/f-2*I*f^2*cosh(d*x+c)/a/d^3-I*(f*x+e)^2*cosh(d*x+c)/a/d+4*f*(f*x+e)*ln(1+I*exp(
d*x+c))/a/d^2+4*f^2*polylog(2,-I*exp(d*x+c))/a/d^3+2*I*f*(f*x+e)*sinh(d*x+c)/a/d^2-(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 = 0.28, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {5676, 3377, 2718, 32, 3399, 4269, 3797, 2221, 2317, 2438} \begin {gather*} \frac {4 f^2 \text {Li}_2\left (-i e^{c+d x}\right )}{a d^3}-\frac {2 i f^2 \cosh (c+d x)}{a d^3}+\frac {4 f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac {2 i f (e+f x) \sinh (c+d x)}{a d^2}-\frac {i (e+f x)^2 \cosh (c+d x)}{a d}-\frac {(e+f x)^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{a d}-\frac {(e+f x)^2}{a d}+\frac {(e+f x)^3}{3 a f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((e + f*x)^2/(a*d)) + (e + f*x)^3/(3*a*f) - ((2*I)*f^2*Cosh[c + d*x])/(a*d^3) - (I*(e + f*x)^2*Cosh[c + d*x])
/(a*d) + (4*f*(e + f*x)*Log[1 + I*E^(c + d*x)])/(a*d^2) + (4*f^2*PolyLog[2, (-I)*E^(c + d*x)])/(a*d^3) + ((2*I
)*f*(e + f*x)*Sinh[c + d*x])/(a*d^2) - ((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 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

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

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 ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx &=i \int \frac {(e+f x)^2 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx-\frac {i \int (e+f x)^2 \sinh (c+d x) \, dx}{a}\\ &=-\frac {i (e+f x)^2 \cosh (c+d x)}{a d}+\frac {\int (e+f x)^2 \, dx}{a}+\frac {(2 i f) \int (e+f x) \cosh (c+d x) \, dx}{a d}-\int \frac {(e+f x)^2}{a+i a \sinh (c+d x)} \, dx\\ &=\frac {(e+f x)^3}{3 a f}-\frac {i (e+f x)^2 \cosh (c+d x)}{a d}+\frac {2 i f (e+f x) \sinh (c+d x)}{a d^2}-\frac {\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 {\left (2 i f^2\right ) \int \sinh (c+d x) \, dx}{a d^2}\\ &=\frac {(e+f x)^3}{3 a f}-\frac {2 i f^2 \cosh (c+d x)}{a d^3}-\frac {i (e+f x)^2 \cosh (c+d x)}{a d}+\frac {2 i f (e+f x) \sinh (c+d x)}{a d^2}-\frac {(e+f x)^2 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}+\frac {(2 f) \int (e+f x) \coth \left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \, dx}{a d}\\ &=-\frac {(e+f x)^2}{a d}+\frac {(e+f x)^3}{3 a f}-\frac {2 i f^2 \cosh (c+d x)}{a d^3}-\frac {i (e+f x)^2 \cosh (c+d x)}{a d}+\frac {2 i f (e+f x) \sinh (c+d x)}{a d^2}-\frac {(e+f x)^2 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}+\frac {(4 i 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 {(e+f x)^2}{a d}+\frac {(e+f x)^3}{3 a f}-\frac {2 i f^2 \cosh (c+d x)}{a d^3}-\frac {i (e+f x)^2 \cosh (c+d x)}{a d}+\frac {4 f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac {2 i f (e+f x) \sinh (c+d x)}{a d^2}-\frac {(e+f x)^2 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {\left (4 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 {(e+f x)^2}{a d}+\frac {(e+f x)^3}{3 a f}-\frac {2 i f^2 \cosh (c+d x)}{a d^3}-\frac {i (e+f x)^2 \cosh (c+d x)}{a d}+\frac {4 f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac {2 i f (e+f x) \sinh (c+d x)}{a d^2}-\frac {(e+f x)^2 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {\left (4 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 {(e+f x)^2}{a d}+\frac {(e+f x)^3}{3 a f}-\frac {2 i f^2 \cosh (c+d x)}{a d^3}-\frac {i (e+f x)^2 \cosh (c+d x)}{a d}+\frac {4 f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac {4 f^2 \text {Li}_2\left (-i e^{c+d x}\right )}{a d^3}+\frac {2 i f (e+f x) \sinh (c+d x)}{a d^2}-\frac {(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 = 2.45, size = 249, normalized size = 1.35 \begin {gather*} \frac {x \left (3 e^2+3 e f x+f^2 x^2\right )+\frac {6 f \left (d \left (-\frac {d e^c x (2 e+f x)}{-i+e^c}+2 (e+f x) \log \left (1+i e^{c+d x}\right )\right )+2 f \text {PolyLog}\left (2,-i e^{c+d x}\right )\right )}{d^3}-\frac {3 i \cosh (d x) \left (\left (2 f^2+d^2 (e+f x)^2\right ) \cosh (c)-2 d f (e+f x) \sinh (c)\right )}{d^3}-\frac {3 i \left (-2 d f (e+f x) \cosh (c)+\left (2 f^2+d^2 (e+f x)^2\right ) \sinh (c)\right ) \sinh (d x)}{d^3}-\frac {6 (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]^2)/(a + I*a*Sinh[c + d*x]),x]

[Out]

(x*(3*e^2 + 3*e*f*x + f^2*x^2) + (6*f*(d*(-((d*E^c*x*(2*e + f*x))/(-I + E^c)) + 2*(e + f*x)*Log[1 + I*E^(c + d
*x)]) + 2*f*PolyLog[2, (-I)*E^(c + d*x)]))/d^3 - ((3*I)*Cosh[d*x]*((2*f^2 + d^2*(e + f*x)^2)*Cosh[c] - 2*d*f*(
e + f*x)*Sinh[c]))/d^3 - ((3*I)*(-2*d*f*(e + f*x)*Cosh[c] + (2*f^2 + d^2*(e + f*x)^2)*Sinh[c])*Sinh[d*x])/d^3
- (6*(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. 384 vs. \(2 (168 ) = 336\).
time = 1.76, size = 385, normalized size = 2.09

method	result	size
risch	\(\frac {f^{2} x^{3}}{3 a}+\frac {f e \,x^{2}}{a}+\frac {e^{2} x}{a}+\frac {e^{3}}{3 a f}-\frac {i \left (f^{2} x^{2} d^{2}+2 d^{2} e f x +d^{2} e^{2}-2 d \,f^{2} x -2 d e f +2 f^{2}\right ) {\mathrm e}^{d x +c}}{2 a \,d^{3}}-\frac {i \left (f^{2} x^{2} d^{2}+2 d^{2} e f x +d^{2} e^{2}+2 d \,f^{2} x +2 d e f +2 f^{2}\right ) {\mathrm e}^{-d x -c}}{2 a \,d^{3}}-\frac {2 i \left (x^{2} f^{2}+2 e f x +e^{2}\right )}{d a \left ({\mathrm e}^{d x +c}-i\right )}+\frac {4 f \ln \left ({\mathrm e}^{d x +c}-i\right ) e}{a \,d^{2}}-\frac {4 f \ln \left ({\mathrm e}^{d x +c}\right ) e}{a \,d^{2}}-\frac {2 f^{2} x^{2}}{a d}-\frac {4 f^{2} c x}{a \,d^{2}}-\frac {2 f^{2} c^{2}}{a \,d^{3}}+\frac {4 f^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right ) x}{a \,d^{2}}+\frac {4 f^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right ) c}{a \,d^{3}}+\frac {4 f^{2} \polylog \left (2, -i {\mathrm e}^{d x +c}\right )}{a \,d^{3}}-\frac {4 f^{2} c \ln \left ({\mathrm e}^{d x +c}-i\right )}{a \,d^{3}}+\frac {4 f^{2} c \ln \left ({\mathrm e}^{d x +c}\right )}{a \,d^{3}}\)	\(385\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

-1/6*f^2*((2*I*d^3*x^3 + 15*I*d^2*x^2 + 6*I*d*x - 3*(-I*d^2*x^2*e^(2*c) + 2*I*d*x*e^(2*c) - 2*I*e^(2*c))*e^(2*
d*x) - (2*d^3*x^3*e^c - 3*d^2*x^2*e^c + 6*d*x*e^c - 6*e^c)*e^(d*x) + 6*I)/(a*d^3*e^(d*x + c) - I*a*d^3) - 24*I
*integrate(x/(a*d*e^(d*x + c) - I*a*d), x)) - f*(2*x*e^(d*x + c)/(a*d*e^(d*x + c) - I*a*d) + (I*d^2*x^2*e^c +
I*d*x*e^c - (-I*d*x*e^(3*c) + I*e^(3*c))*e^(2*d*x) - (d^2*x^2*e^(2*c) - 3*d*x*e^(2*c) + e^(2*c))*e^(d*x) + (d*
x + 1)*e^(-d*x) + I*e^c)/(a*d^2*e^(d*x + 2*c) - I*a*d^2*e^c) - 4*log((e^(d*x + c) - I)*e^(-c))/(a*d^2))*e + 1/
2*(2*(d*x + c)/(a*d) + (-5*I*e^(-d*x - c) + 1)/((I*a*e^(-d*x - c) + a*e^(-2*d*x - 2*c))*d) - I*e^(-d*x - c)/(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. 475 vs. \(2 (167) = 334\).
time = 0.35, size = 475, normalized size = 2.58 \begin {gather*} -\frac {3 \, d^{2} f^{2} x^{2} + 6 \, d f^{2} x + 3 \, d^{2} e^{2} + 6 \, f^{2} - 24 \, {\left (f^{2} e^{\left (2 \, d x + 2 \, c\right )} - i \, f^{2} e^{\left (d x + c\right )}\right )} {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) + 6 \, {\left (d^{2} f x + d f\right )} e + 3 \, {\left (i \, d^{2} f^{2} x^{2} - 2 i \, d f^{2} x + i \, d^{2} e^{2} + 2 i \, f^{2} + 2 \, {\left (i \, d^{2} f x - i \, d f\right )} e\right )} e^{\left (3 \, d x + 3 \, c\right )} - {\left (2 \, d^{3} f^{2} x^{3} - 15 \, d^{2} f^{2} x^{2} + 6 \, d f^{2} x + 6 \, {\left (2 \, c^{2} - 1\right )} f^{2} + 3 \, {\left (2 \, d^{3} x - d^{2}\right )} e^{2} + 6 \, {\left (d^{3} f x^{2} - 5 \, d^{2} f x - {\left (4 \, c - 1\right )} d f\right )} e\right )} e^{\left (2 \, d x + 2 \, c\right )} - {\left (-2 i \, d^{3} f^{2} x^{3} - 3 i \, d^{2} f^{2} x^{2} - 6 i \, d f^{2} x - 6 \, {\left (2 i \, c^{2} + i\right )} f^{2} - 3 \, {\left (2 i \, d^{3} x + 5 i \, d^{2}\right )} e^{2} - 6 \, {\left (i \, d^{3} f x^{2} + i \, d^{2} f x + {\left (-4 i \, c + i\right )} d f\right )} e\right )} e^{\left (d x + c\right )} + 24 \, {\left ({\left (c f^{2} - d f e\right )} e^{\left (2 \, d x + 2 \, c\right )} + {\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 ) - 24 \, {\left ({\left (d f^{2} x + c f^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )} - {\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 )}{6 \, {\left (a d^{3} e^{\left (2 \, d x + 2 \, c\right )} - i \, a d^{3} e^{\left (d x + c\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*d^2*f^2*x^2 + 6*d*f^2*x + 3*d^2*e^2 + 6*f^2 - 24*(f^2*e^(2*d*x + 2*c) - I*f^2*e^(d*x + c))*dilog(-I*e^
(d*x + c)) + 6*(d^2*f*x + d*f)*e + 3*(I*d^2*f^2*x^2 - 2*I*d*f^2*x + I*d^2*e^2 + 2*I*f^2 + 2*(I*d^2*f*x - I*d*f
)*e)*e^(3*d*x + 3*c) - (2*d^3*f^2*x^3 - 15*d^2*f^2*x^2 + 6*d*f^2*x + 6*(2*c^2 - 1)*f^2 + 3*(2*d^3*x - d^2)*e^2
+ 6*(d^3*f*x^2 - 5*d^2*f*x - (4*c - 1)*d*f)*e)*e^(2*d*x + 2*c) - (-2*I*d^3*f^2*x^3 - 3*I*d^2*f^2*x^2 - 6*I*d*
f^2*x - 6*(2*I*c^2 + I)*f^2 - 3*(2*I*d^3*x + 5*I*d^2)*e^2 - 6*(I*d^3*f*x^2 + I*d^2*f*x + (-4*I*c + I)*d*f)*e)*
e^(d*x + c) + 24*((c*f^2 - d*f*e)*e^(2*d*x + 2*c) + (-I*c*f^2 + I*d*f*e)*e^(d*x + c))*log(e^(d*x + c) - I) - 2
4*((d*f^2*x + c*f^2)*e^(2*d*x + 2*c) - (I*d*f^2*x + I*c*f^2)*e^(d*x + c))*log(I*e^(d*x + c) + 1))/(a*d^3*e^(2*
d*x + 2*c) - I*a*d^3*e^(d*x + c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {- 2 i e^{2} - 4 i e f x - 2 i f^{2} x^{2}}{a d e^{c} e^{d x} - i a d} - \frac {i \left (\int \frac {i d e^{2}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {i d f^{2} x^{2}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {d e^{2} e^{c} e^{d x}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {d e^{2} e^{3 c} e^{3 d x}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \left (- \frac {8 e f e^{c} e^{d x}}{e^{c} e^{2 d x} - i e^{d x}}\right )\, dx + \int \left (- \frac {8 f^{2} x e^{c} e^{d x}}{e^{c} e^{2 d x} - i e^{d x}}\right )\, dx + \int \frac {2 i d e f x}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {i d e^{2} e^{2 c} e^{2 d x}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {d f^{2} x^{2} e^{c} e^{d x}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {d f^{2} x^{2} e^{3 c} e^{3 d x}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {i d f^{2} x^{2} e^{2 c} e^{2 d x}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {2 d e f x e^{c} e^{d x}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {2 d e f x e^{3 c} e^{3 d x}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {2 i d e f x e^{2 c} e^{2 d x}}{e^{c} e^{2 d x} - i e^{d x}}\, dx\right ) e^{- c}}{2 a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*I*e**2 - 4*I*e*f*x - 2*I*f**2*x**2)/(a*d*exp(c)*exp(d*x) - I*a*d) - I*(Integral(I*d*e**2/(exp(c)*exp(2*d*x
) - I*exp(d*x)), x) + Integral(I*d*f**2*x**2/(exp(c)*exp(2*d*x) - I*exp(d*x)), x) + Integral(d*e**2*exp(c)*exp
(d*x)/(exp(c)*exp(2*d*x) - I*exp(d*x)), x) + Integral(d*e**2*exp(3*c)*exp(3*d*x)/(exp(c)*exp(2*d*x) - I*exp(d*
x)), x) + Integral(-8*e*f*exp(c)*exp(d*x)/(exp(c)*exp(2*d*x) - I*exp(d*x)), x) + Integral(-8*f**2*x*exp(c)*exp
(d*x)/(exp(c)*exp(2*d*x) - I*exp(d*x)), x) + Integral(2*I*d*e*f*x/(exp(c)*exp(2*d*x) - I*exp(d*x)), x) + Integ
ral(I*d*e**2*exp(2*c)*exp(2*d*x)/(exp(c)*exp(2*d*x) - I*exp(d*x)), x) + Integral(d*f**2*x**2*exp(c)*exp(d*x)/(
exp(c)*exp(2*d*x) - I*exp(d*x)), x) + Integral(d*f**2*x**2*exp(3*c)*exp(3*d*x)/(exp(c)*exp(2*d*x) - I*exp(d*x)
), x) + Integral(I*d*f**2*x**2*exp(2*c)*exp(2*d*x)/(exp(c)*exp(2*d*x) - I*exp(d*x)), x) + Integral(2*d*e*f*x*e
xp(c)*exp(d*x)/(exp(c)*exp(2*d*x) - I*exp(d*x)), x) + Integral(2*d*e*f*x*exp(3*c)*exp(3*d*x)/(exp(c)*exp(2*d*x
) - I*exp(d*x)), x) + Integral(2*I*d*e*f*x*exp(2*c)*exp(2*d*x)/(exp(c)*exp(2*d*x) - I*exp(d*x)), x))*exp(-c)/(
2*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)^2/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

integrate((f*x + e)^2*sinh(d*x + c)^2/(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 )}^2\,{\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)^2*(e + f*x)^2)/(a + a*sinh(c + d*x)*1i),x)

[Out]

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

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