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

Optimal. Leaf size=184 \[ \frac {4 f^2 \text {Li}_2\left (-i e^{c+d x}\right )}{a d^3}-\frac {2 i f^2 \cosh (c+d x)}{a d^3}+\frac {4 f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac {2 i f (e+f x) \sinh (c+d x)}{a d^2}-\frac {i (e+f x)^2 \cosh (c+d x)}{a d}-\frac {(e+f x)^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{a d}-\frac {(e+f x)^2}{a d}+\frac {(e+f x)^3}{3 a f} \]

[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 = 0.39, 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 = {5557, 3296, 2638, 32, 3318, 4184, 3716, 2190, 2279, 2391} \[ \frac {4 f^2 \text {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}+\frac {4 f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac {2 i f (e+f x) \sinh (c+d x)}{a d^2}-\frac {2 i f^2 \cosh (c+d x)}{a d^3}-\frac {i (e+f x)^2 \cosh (c+d x)}{a d}-\frac {(e+f x)^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{a d}-\frac {(e+f x)^2}{a d}+\frac {(e+f x)^3}{3 a f} \]

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 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 2638

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

Rule 3296

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 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 ^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 ) \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 {(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 = 3.46, size = 260, normalized size = 1.41 \[ \frac {\frac {6 \left (\frac {d (e+f x) \left (2 \left (e^c-i\right ) f \log \left (1-i e^{-c-d x}\right )-i d (e+f x)\right )}{e^c-i}-2 f^2 \text {Li}_2\left (i e^{-c-d x}\right )\right )}{d^3}-\frac {3 i \cosh (d x) \left (\cosh (c) \left (d^2 (e+f x)^2+2 f^2\right )-2 d f \sinh (c) (e+f x)\right )}{d^3}-\frac {3 i \sinh (d x) \left (\sinh (c) \left (d^2 (e+f x)^2+2 f^2\right )-2 d f \cosh (c) (e+f x)\right )}{d^3}-\frac {6 \sinh \left (\frac {d x}{2}\right ) (e+f x)^2}{d \left (\cosh \left (\frac {c}{2}\right )+i \sinh \left (\frac {c}{2}\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )}+x \left (3 e^2+3 e f x+f^2 x^2\right )}{3 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(3*e^2 + 3*e*f*x + f^2*x^2) + (6*((d*(e + f*x)*((-I)*d*(e + f*x) + 2*(-I + E^c)*f*Log[1 - I*E^(-c - d*x)]))
/(-I + E^c) - 2*f^2*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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fricas [B]  time = 0.51, size = 471, normalized size = 2.56 \[ -\frac {3 \, d^{2} f^{2} x^{2} + 3 \, d^{2} e^{2} + 6 \, d e f + 6 \, f^{2} + 6 \, {\left (d^{2} e f + d f^{2}\right )} x - {\left (24 \, f^{2} e^{\left (2 \, d x + 2 \, c\right )} - 24 i \, f^{2} e^{\left (d x + c\right )}\right )} {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - {\left (-3 i \, d^{2} f^{2} x^{2} - 3 i \, d^{2} e^{2} + 6 i \, d e f - 6 i \, f^{2} + {\left (-6 i \, d^{2} e f + 6 i \, d f^{2}\right )} x\right )} e^{\left (3 \, d x + 3 \, c\right )} - {\left (2 \, d^{3} f^{2} x^{3} - 3 \, d^{2} e^{2} - 6 \, {\left (4 \, c - 1\right )} d e f + 6 \, {\left (2 \, c^{2} - 1\right )} f^{2} + 3 \, {\left (2 \, d^{3} e f - 5 \, d^{2} f^{2}\right )} x^{2} + 6 \, {\left (d^{3} e^{2} - 5 \, d^{2} e f + d f^{2}\right )} x\right )} e^{\left (2 \, d x + 2 \, c\right )} - {\left (-2 i \, d^{3} f^{2} x^{3} - 15 i \, d^{2} e^{2} + {\left (24 i \, c - 6 i\right )} d e f + {\left (-12 i \, c^{2} - 6 i\right )} f^{2} + {\left (-6 i \, d^{3} e f - 3 i \, d^{2} f^{2}\right )} x^{2} + {\left (-6 i \, d^{3} e^{2} - 6 i \, d^{2} e f - 6 i \, d f^{2}\right )} x\right )} e^{\left (d x + c\right )} - {\left (24 \, {\left (d e f - c f^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )} + {\left (-24 i \, d e f + 24 i \, c f^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) - {\left (24 \, {\left (d f^{2} x + c f^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )} + {\left (-24 i \, d f^{2} x - 24 i \, c f^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (i \, e^{\left (d x + c\right )} + 1\right )}{6 \, a d^{3} e^{\left (2 \, d x + 2 \, c\right )} - 6 i \, a d^{3} e^{\left (d x + c\right )}} \]

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]

-(3*d^2*f^2*x^2 + 3*d^2*e^2 + 6*d*e*f + 6*f^2 + 6*(d^2*e*f + d*f^2)*x - (24*f^2*e^(2*d*x + 2*c) - 24*I*f^2*e^(
d*x + c))*dilog(-I*e^(d*x + c)) - (-3*I*d^2*f^2*x^2 - 3*I*d^2*e^2 + 6*I*d*e*f - 6*I*f^2 + (-6*I*d^2*e*f + 6*I*
d*f^2)*x)*e^(3*d*x + 3*c) - (2*d^3*f^2*x^3 - 3*d^2*e^2 - 6*(4*c - 1)*d*e*f + 6*(2*c^2 - 1)*f^2 + 3*(2*d^3*e*f
- 5*d^2*f^2)*x^2 + 6*(d^3*e^2 - 5*d^2*e*f + d*f^2)*x)*e^(2*d*x + 2*c) - (-2*I*d^3*f^2*x^3 - 15*I*d^2*e^2 + (24
*I*c - 6*I)*d*e*f + (-12*I*c^2 - 6*I)*f^2 + (-6*I*d^3*e*f - 3*I*d^2*f^2)*x^2 + (-6*I*d^3*e^2 - 6*I*d^2*e*f - 6
*I*d*f^2)*x)*e^(d*x + c) - (24*(d*e*f - c*f^2)*e^(2*d*x + 2*c) + (-24*I*d*e*f + 24*I*c*f^2)*e^(d*x + c))*log(e
^(d*x + c) - I) - (24*(d*f^2*x + c*f^2)*e^(2*d*x + 2*c) + (-24*I*d*f^2*x - 24*I*c*f^2)*e^(d*x + c))*log(I*e^(d
*x + c) + 1))/(6*a*d^3*e^(2*d*x + 2*c) - 6*I*a*d^3*e^(d*x + c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^2*sinh(d*x+c)^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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maple [B]  time = 0.22, size = 374, normalized size = 2.03 \[ \frac {x^{3} f^{2}}{3 a}+\frac {e f \,x^{2}}{a}+\frac {e^{2} x}{a}-\frac {i \left (d^{2} f^{2} x^{2}+2 d^{2} e f x +d^{2} e^{2}+2 d \,f^{2} x +2 d e f +2 f^{2}\right ) {\mathrm e}^{-d x -c}}{2 a \,d^{3}}-\frac {i \left (d^{2} f^{2} x^{2}+2 d^{2} e f x +d^{2} e^{2}-2 d \,f^{2} x -2 d e f +2 f^{2}\right ) {\mathrm e}^{d x +c}}{2 a \,d^{3}}-\frac {2 i \left (x^{2} f^{2}+2 e f x +e^{2}\right )}{d a \left ({\mathrm e}^{d x +c}-i\right )}+\frac {4 f \ln \left ({\mathrm e}^{d x +c}-i\right ) e}{a \,d^{2}}-\frac {4 f \ln \left ({\mathrm e}^{d x +c}\right ) e}{a \,d^{2}}-\frac {2 f^{2} x^{2}}{a d}-\frac {4 f^{2} c x}{a \,d^{2}}-\frac {2 f^{2} c^{2}}{a \,d^{3}}+\frac {4 f^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right ) x}{a \,d^{2}}+\frac {4 f^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right ) c}{a \,d^{3}}+\frac {4 f^{2} \polylog \left (2, -i {\mathrm e}^{d x +c}\right )}{a \,d^{3}}-\frac {4 f^{2} c \ln \left ({\mathrm e}^{d x +c}-i\right )}{a \,d^{3}}+\frac {4 f^{2} c \ln \left ({\mathrm e}^{d x +c}\right )}{a \,d^{3}} \]

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)

[Out]

1/3/a*x^3*f^2+1/a*e*f*x^2+1/a*e^2*x-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 \[ -\frac {1}{2} \, e f {\left (\frac {4 \, x e^{\left (d x + c\right )}}{a d e^{\left (d x + c\right )} - i \, a d} - \frac {-2 i \, d^{2} x^{2} e^{c} - 2 i \, d x e^{c} - {\left (2 i \, d x e^{\left (3 \, c\right )} - 2 i \, e^{\left (3 \, c\right )}\right )} e^{\left (2 \, d x\right )} + 2 \, {\left (d^{2} x^{2} e^{\left (2 \, c\right )} - 3 \, d x e^{\left (2 \, c\right )} + e^{\left (2 \, c\right )}\right )} e^{\left (d x\right )} - 2 \, {\left (d x + 1\right )} e^{\left (-d x\right )} - 2 i \, e^{c}}{a d^{2} e^{\left (d x + 2 \, c\right )} - i \, a d^{2} e^{c}} - \frac {8 \, \log \left ({\left (e^{\left (d x + c\right )} - i\right )} e^{\left (-c\right )}\right )}{a d^{2}}\right )} + \frac {1}{12} \, f^{2} {\left (\frac {-4 i \, d^{3} x^{3} - 30 i \, d^{2} x^{2} - 12 i \, d x - {\left (6 i \, d^{2} x^{2} e^{\left (2 \, c\right )} - 12 i \, d x e^{\left (2 \, c\right )} + 12 i \, e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )} + 2 \, {\left (2 \, d^{3} x^{3} e^{c} - 3 \, d^{2} x^{2} e^{c} + 6 \, d x e^{c} - 6 \, e^{c}\right )} e^{\left (d x\right )} - 12 i}{a d^{3} e^{\left (d x + c\right )} - i \, a d^{3}} + 48 i \, \int \frac {x}{a d e^{\left (d x + c\right )} - i \, a d}\,{d x}\right )} + \frac {1}{2} \, e^{2} {\left (\frac {2 \, {\left (d x + c\right )}}{a d} + \frac {-5 i \, e^{\left (-d x - c\right )} + 1}{{\left (i \, a e^{\left (-d x - c\right )} + a e^{\left (-2 \, d x - 2 \, c\right )}\right )} d} - \frac {i \, e^{\left (-d x - c\right )}}{a d}\right )} \]

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]

-1/2*e*f*(4*x*e^(d*x + c)/(a*d*e^(d*x + c) - I*a*d) - (-2*I*d^2*x^2*e^c - 2*I*d*x*e^c - (2*I*d*x*e^(3*c) - 2*I
*e^(3*c))*e^(2*d*x) + 2*(d^2*x^2*e^(2*c) - 3*d*x*e^(2*c) + e^(2*c))*e^(d*x) - 2*(d*x + 1)*e^(-d*x) - 2*I*e^c)/
(a*d^2*e^(d*x + 2*c) - I*a*d^2*e^c) - 8*log((e^(d*x + c) - I)*e^(-c))/(a*d^2)) + 1/12*f^2*((-4*I*d^3*x^3 - 30*
I*d^2*x^2 - 12*I*d*x - (6*I*d^2*x^2*e^(2*c) - 12*I*d*x*e^(2*c) + 12*I*e^(2*c))*e^(2*d*x) + 2*(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) - 12*I)/(a*d^3*e^(d*x + c) - I*a*d^3) + 48*I*integrate(x/(a*d*e^(d*
x + c) - I*a*d), x)) + 1/2*e^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))

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

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

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*exp(c) + 4*I*e*f*x*exp(c) + 2*I*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(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) + Inte
gral(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(2*I*d*e*f*x/(exp(c)*exp(2*d*x) - I*exp(d*x)), x) + Integral(I*d*e**2*
exp(2*c)*exp(2*d*x)/(exp(c)*exp(2*d*x) - I*exp(d*x)), x) + Integral(8*I*e*f*exp(2*c)*exp(2*d*x)/(exp(c)*exp(2*
d*x) - I*exp(d*x)), x) + Integral(8*I*f**2*x*exp(2*c)*exp(2*d*x)/(exp(c)*exp(2*d*x) - I*exp(d*x)), x) + Integr
al(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*exp(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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