3.2.14 \(\int \frac {(c+d x)^2}{(a+i a \sinh (e+f x))^2} \, dx\) [114]

Optimal. Leaf size=241 \[ \frac {(c+d x)^2}{3 a^2 f}-\frac {4 d (c+d x) \log \left (1+i e^{e+f x}\right )}{3 a^2 f^2}-\frac {4 d^2 \text {PolyLog}\left (2,-i e^{e+f x}\right )}{3 a^2 f^3}+\frac {d (c+d x) \text {sech}^2\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{3 a^2 f^2}-\frac {2 d^2 \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{3 a^2 f^3}+\frac {(c+d x)^2 \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{3 a^2 f}+\frac {(c+d x)^2 \text {sech}^2\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{6 a^2 f} \]

[Out]

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

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Rubi [A]
time = 0.21, antiderivative size = 241, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3399, 4271, 3852, 8, 4269, 3797, 2221, 2317, 2438} \begin {gather*} -\frac {4 d (c+d x) \log \left (1+i e^{e+f x}\right )}{3 a^2 f^2}+\frac {d (c+d x) \text {sech}^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{3 a^2 f^2}+\frac {(c+d x)^2 \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{3 a^2 f}+\frac {(c+d x)^2 \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \text {sech}^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{6 a^2 f}+\frac {(c+d x)^2}{3 a^2 f}-\frac {4 d^2 \text {Li}_2\left (-i e^{e+f x}\right )}{3 a^2 f^3}-\frac {2 d^2 \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{3 a^2 f^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c + d*x)^2/(3*a^2*f) - (4*d*(c + d*x)*Log[1 + I*E^(e + f*x)])/(3*a^2*f^2) - (4*d^2*PolyLog[2, (-I)*E^(e + f*x
)])/(3*a^2*f^3) + (d*(c + d*x)*Sech[e/2 + (I/4)*Pi + (f*x)/2]^2)/(3*a^2*f^2) - (2*d^2*Tanh[e/2 + (I/4)*Pi + (f
*x)/2])/(3*a^2*f^3) + ((c + d*x)^2*Tanh[e/2 + (I/4)*Pi + (f*x)/2])/(3*a^2*f) + ((c + d*x)^2*Sech[e/2 + (I/4)*P
i + (f*x)/2]^2*Tanh[e/2 + (I/4)*Pi + (f*x)/2])/(6*a^2*f)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 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 4271

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-b^2)*(c + d*x)^m*Cot[e
+ f*x]*((b*Csc[e + f*x])^(n - 2)/(f*(n - 1))), x] + (Dist[b^2*d^2*m*((m - 1)/(f^2*(n - 1)*(n - 2))), Int[(c +
d*x)^(m - 2)*(b*Csc[e + f*x])^(n - 2), x], x] + Dist[b^2*((n - 2)/(n - 1)), Int[(c + d*x)^m*(b*Csc[e + f*x])^
(n - 2), x], x] - Simp[b^2*d*m*(c + d*x)^(m - 1)*((b*Csc[e + f*x])^(n - 2)/(f^2*(n - 1)*(n - 2))), x]) /; Free
Q[{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]

Rubi steps

\begin {align*} \int \frac {(c+d x)^2}{(a+i a \sinh (e+f x))^2} \, dx &=\frac {\int (c+d x)^2 \csc ^4\left (\frac {1}{2} \left (i e+\frac {\pi }{2}\right )+\frac {i f x}{2}\right ) \, dx}{4 a^2}\\ &=\frac {d (c+d x) \text {sech}^2\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{3 a^2 f^2}+\frac {(c+d x)^2 \text {sech}^2\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{6 a^2 f}-\frac {\int (c+d x)^2 \text {csch}^2\left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right ) \, dx}{6 a^2}+\frac {d^2 \int \text {csch}^2\left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right ) \, dx}{3 a^2 f^2}\\ &=\frac {d (c+d x) \text {sech}^2\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{3 a^2 f^2}+\frac {(c+d x)^2 \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{3 a^2 f}+\frac {(c+d x)^2 \text {sech}^2\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{6 a^2 f}-\frac {\left (2 i d^2\right ) \text {Subst}\left (\int 1 \, dx,x,-i \coth \left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right )\right )}{3 a^2 f^3}-\frac {(2 d) \int (c+d x) \coth \left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right ) \, dx}{3 a^2 f}\\ &=\frac {(c+d x)^2}{3 a^2 f}+\frac {d (c+d x) \text {sech}^2\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{3 a^2 f^2}-\frac {2 d^2 \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{3 a^2 f^3}+\frac {(c+d x)^2 \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{3 a^2 f}+\frac {(c+d x)^2 \text {sech}^2\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{6 a^2 f}-\frac {(4 i d) \int \frac {e^{2 \left (\frac {e}{2}+\frac {f x}{2}\right )} (c+d x)}{1+i e^{2 \left (\frac {e}{2}+\frac {f x}{2}\right )}} \, dx}{3 a^2 f}\\ &=\frac {(c+d x)^2}{3 a^2 f}-\frac {4 d (c+d x) \log \left (1+i e^{e+f x}\right )}{3 a^2 f^2}+\frac {d (c+d x) \text {sech}^2\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{3 a^2 f^2}-\frac {2 d^2 \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{3 a^2 f^3}+\frac {(c+d x)^2 \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{3 a^2 f}+\frac {(c+d x)^2 \text {sech}^2\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{6 a^2 f}+\frac {\left (4 d^2\right ) \int \log \left (1+i e^{2 \left (\frac {e}{2}+\frac {f x}{2}\right )}\right ) \, dx}{3 a^2 f^2}\\ &=\frac {(c+d x)^2}{3 a^2 f}-\frac {4 d (c+d x) \log \left (1+i e^{e+f x}\right )}{3 a^2 f^2}+\frac {d (c+d x) \text {sech}^2\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{3 a^2 f^2}-\frac {2 d^2 \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{3 a^2 f^3}+\frac {(c+d x)^2 \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{3 a^2 f}+\frac {(c+d x)^2 \text {sech}^2\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{6 a^2 f}+\frac {\left (4 d^2\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{2 \left (\frac {e}{2}+\frac {f x}{2}\right )}\right )}{3 a^2 f^3}\\ &=\frac {(c+d x)^2}{3 a^2 f}-\frac {4 d (c+d x) \log \left (1+i e^{e+f x}\right )}{3 a^2 f^2}-\frac {4 d^2 \text {Li}_2\left (-i e^{e+f x}\right )}{3 a^2 f^3}+\frac {d (c+d x) \text {sech}^2\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{3 a^2 f^2}-\frac {2 d^2 \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{3 a^2 f^3}+\frac {(c+d x)^2 \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{3 a^2 f}+\frac {(c+d x)^2 \text {sech}^2\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{6 a^2 f}\\ \end {align*}

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Mathematica [A]
time = 2.54, size = 259, normalized size = 1.07 \begin {gather*} \frac {\frac {2 d e^e f^2 x (2 c+d x)}{-i+e^e}-4 d f (c+d x) \log \left (1+i e^{e+f x}\right )-4 d^2 \text {PolyLog}\left (2,-i e^{e+f x}\right )+\frac {2 d f (c+d x) \cosh \left (\frac {f x}{2}\right )+2 i d^2 \cosh \left (e+\frac {f x}{2}\right )+i \left (c^2 f^2+2 c d f^2 x+d^2 \left (-2+f^2 x^2\right )\right ) \cosh \left (e+\frac {3 f x}{2}\right )+\left (3 c^2 f^2+6 c d f^2 x+d^2 \left (-4+3 f^2 x^2\right )\right ) \sinh \left (\frac {f x}{2}\right )+2 i d f (c+d x) \sinh \left (e+\frac {f x}{2}\right )}{\left (\cosh \left (\frac {e}{2}\right )+i \sinh \left (\frac {e}{2}\right )\right ) \left (\cosh \left (\frac {1}{2} (e+f x)\right )+i \sinh \left (\frac {1}{2} (e+f x)\right )\right )^3}}{3 a^2 f^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 2.59, size = 374, normalized size = 1.55

method	result	size
risch	\(\frac {-\frac {4 i f^{2} c d x}{3}-\frac {4 i f \,d^{2} x \,{\mathrm e}^{2 f x +2 e}}{3}-\frac {4 i f c d \,{\mathrm e}^{2 f x +2 e}}{3}-\frac {4 i d^{2} {\mathrm e}^{2 f x +2 e}}{3}-\frac {4 f \,d^{2} x \,{\mathrm e}^{f x +e}}{3}-\frac {4 f c d \,{\mathrm e}^{f x +e}}{3}-\frac {2 i f^{2} d^{2} x^{2}}{3}-\frac {2 i f^{2} c^{2}}{3}-\frac {8 d^{2} {\mathrm e}^{f x +e}}{3}+\frac {4 i d^{2}}{3}+2 f^{2} d^{2} x^{2} {\mathrm e}^{f x +e}+4 f^{2} c d x \,{\mathrm e}^{f x +e}+2 f^{2} c^{2} {\mathrm e}^{f x +e}}{\left ({\mathrm e}^{f x +e}-i\right )^{3} f^{3} a^{2}}-\frac {4 d \ln \left ({\mathrm e}^{f x +e}-i\right ) c}{3 a^{2} f^{2}}+\frac {4 d \ln \left ({\mathrm e}^{f x +e}\right ) c}{3 a^{2} f^{2}}+\frac {2 d^{2} x^{2}}{3 a^{2} f}+\frac {4 d^{2} e x}{3 a^{2} f^{2}}+\frac {2 d^{2} e^{2}}{3 a^{2} f^{3}}-\frac {4 d^{2} \ln \left (1+i {\mathrm e}^{f x +e}\right ) x}{3 a^{2} f^{2}}-\frac {4 d^{2} \ln \left (1+i {\mathrm e}^{f x +e}\right ) e}{3 a^{2} f^{3}}-\frac {4 d^{2} \polylog \left (2, -i {\mathrm e}^{f x +e}\right )}{3 a^{2} f^{3}}+\frac {4 d^{2} e \ln \left ({\mathrm e}^{f x +e}-i\right )}{3 a^{2} f^{3}}-\frac {4 d^{2} e \ln \left ({\mathrm e}^{f x +e}\right )}{3 a^{2} f^{3}}\)	\(374\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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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. 504 vs. \(2 (187) = 374\).
time = 0.34, size = 504, normalized size = 2.09 \begin {gather*} -\frac {2 \, {\left (i \, c^{2} f^{2} - 2 i \, c d f e + i \, d^{2} e^{2} - 2 i \, d^{2} + 2 \, {\left (d^{2} e^{\left (3 \, f x + 3 \, e\right )} - 3 i \, d^{2} e^{\left (2 \, f x + 2 \, e\right )} - 3 \, d^{2} e^{\left (f x + e\right )} + i \, d^{2}\right )} {\rm Li}_2\left (-i \, e^{\left (f x + e\right )}\right ) - {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + 2 \, c d f e - d^{2} e^{2}\right )} e^{\left (3 \, f x + 3 \, e\right )} + {\left (3 i \, d^{2} f^{2} x^{2} + 6 i \, c d f e + 2 i \, c d f - 3 i \, d^{2} e^{2} + 2 i \, d^{2} + 2 \, {\left (3 i \, c d f^{2} + i \, d^{2} f\right )} x\right )} e^{\left (2 \, f x + 2 \, e\right )} - {\left (3 \, c^{2} f^{2} - 2 \, d^{2} f x - 6 \, c d f e - 2 \, c d f + 3 \, d^{2} e^{2} - 4 \, d^{2}\right )} e^{\left (f x + e\right )} + 2 \, {\left (i \, c d f - i \, d^{2} e + {\left (c d f - d^{2} e\right )} e^{\left (3 \, f x + 3 \, e\right )} + 3 \, {\left (-i \, c d f + i \, d^{2} e\right )} e^{\left (2 \, f x + 2 \, e\right )} - 3 \, {\left (c d f - d^{2} e\right )} e^{\left (f x + e\right )}\right )} \log \left (e^{\left (f x + e\right )} - i\right ) + 2 \, {\left (i \, d^{2} f x + i \, d^{2} e + {\left (d^{2} f x + d^{2} e\right )} e^{\left (3 \, f x + 3 \, e\right )} + 3 \, {\left (-i \, d^{2} f x - i \, d^{2} e\right )} e^{\left (2 \, f x + 2 \, e\right )} - 3 \, {\left (d^{2} f x + d^{2} e\right )} e^{\left (f x + e\right )}\right )} \log \left (i \, e^{\left (f x + e\right )} + 1\right )\right )}}{3 \, {\left (a^{2} f^{3} e^{\left (3 \, f x + 3 \, e\right )} - 3 i \, a^{2} f^{3} e^{\left (2 \, f x + 2 \, e\right )} - 3 \, a^{2} f^{3} e^{\left (f x + e\right )} + i \, a^{2} f^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {- 2 i c^{2} f^{2} - 4 i c d f^{2} x - 2 i d^{2} f^{2} x^{2} + 4 i d^{2} + \left (- 4 i c d f e^{2 e} - 4 i d^{2} f x e^{2 e} - 4 i d^{2} e^{2 e}\right ) e^{2 f x} + \left (6 c^{2} f^{2} e^{e} + 12 c d f^{2} x e^{e} - 4 c d f e^{e} + 6 d^{2} f^{2} x^{2} e^{e} - 4 d^{2} f x e^{e} - 8 d^{2} e^{e}\right ) e^{f x}}{3 a^{2} f^{3} e^{3 e} e^{3 f x} - 9 i a^{2} f^{3} e^{2 e} e^{2 f x} - 9 a^{2} f^{3} e^{e} e^{f x} + 3 i a^{2} f^{3}} - \frac {4 i d \left (\int \frac {c}{e^{e} e^{f x} - i}\, dx + \int \frac {d x}{e^{e} e^{f x} - i}\, dx\right )}{3 a^{2} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*I*c**2*f**2 - 4*I*c*d*f**2*x - 2*I*d**2*f**2*x**2 + 4*I*d**2 + (-4*I*c*d*f*exp(2*e) - 4*I*d**2*f*x*exp(2*e
) - 4*I*d**2*exp(2*e))*exp(2*f*x) + (6*c**2*f**2*exp(e) + 12*c*d*f**2*x*exp(e) - 4*c*d*f*exp(e) + 6*d**2*f**2*
x**2*exp(e) - 4*d**2*f*x*exp(e) - 8*d**2*exp(e))*exp(f*x))/(3*a**2*f**3*exp(3*e)*exp(3*f*x) - 9*I*a**2*f**3*ex
p(2*e)*exp(2*f*x) - 9*a**2*f**3*exp(e)*exp(f*x) + 3*I*a**2*f**3) - 4*I*d*(Integral(c/(exp(e)*exp(f*x) - I), x)
+ Integral(d*x/(exp(e)*exp(f*x) - I), x))/(3*a**2*f)

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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