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

Optimal. Leaf size=101 \[ -\frac {4 d (c+d x) \log \left (1+i e^{e+f x}\right )}{a f^2}+\frac {(c+d x)^2 \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{a f}+\frac {(c+d x)^2}{a f}-\frac {4 d^2 \text {Li}_2\left (-i e^{e+f x}\right )}{a f^3} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3318, 4184, 3716, 2190, 2279, 2391} \[ -\frac {4 d^2 \text {PolyLog}\left (2,-i e^{e+f x}\right )}{a f^3}-\frac {4 d (c+d x) \log \left (1+i e^{e+f x}\right )}{a f^2}+\frac {(c+d x)^2 \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{a f}+\frac {(c+d x)^2}{a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rubi steps

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

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Mathematica [A]  time = 2.23, size = 150, normalized size = 1.49 \[ \frac {2 \left (\frac {f^2 (c+d x)^2 \sinh \left (\frac {f x}{2}\right )}{\left (\cosh \left (\frac {e}{2}\right )+i \sinh \left (\frac {e}{2}\right )\right ) \left (\cosh \left (\frac {1}{2} (e+f x)\right )+i \sinh \left (\frac {1}{2} (e+f x)\right )\right )}+\frac {i f (c+d x) \left (f (c+d x)+2 d \left (1+i e^e\right ) \log \left (1-i e^{-e-f x}\right )\right )}{e^e-i}+2 d^2 \text {Li}_2\left (i e^{-e-f x}\right )\right )}{a f^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.56, size = 200, normalized size = 1.98 \[ \frac {2 i \, d^{2} e^{2} - 4 i \, c d e f + 2 i \, c^{2} f^{2} - {\left (4 \, d^{2} e^{\left (f x + e\right )} - 4 i \, d^{2}\right )} {\rm Li}_2\left (-i \, e^{\left (f x + e\right )}\right ) + 2 \, {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x - d^{2} e^{2} + 2 \, c d e f\right )} e^{\left (f x + e\right )} + {\left (-4 i \, d^{2} e + 4 i \, c d f + 4 \, {\left (d^{2} e - c d f\right )} e^{\left (f x + e\right )}\right )} \log \left (e^{\left (f x + e\right )} - i\right ) + {\left (4 i \, d^{2} f x + 4 i \, d^{2} e - 4 \, {\left (d^{2} f x + d^{2} e\right )} e^{\left (f x + e\right )}\right )} \log \left (i \, e^{\left (f x + e\right )} + 1\right )}{a f^{3} e^{\left (f x + e\right )} - i \, a f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.10, size = 227, normalized size = 2.25 \[ \frac {2 i \left (d^{2} x^{2}+2 c d x +c^{2}\right )}{f a \left ({\mathrm e}^{f x +e}-i\right )}-\frac {4 d \ln \left ({\mathrm e}^{f x +e}-i\right ) c}{a \,f^{2}}+\frac {4 d \ln \left ({\mathrm e}^{f x +e}\right ) c}{a \,f^{2}}+\frac {2 d^{2} x^{2}}{a f}+\frac {4 d^{2} e x}{a \,f^{2}}+\frac {2 d^{2} e^{2}}{a \,f^{3}}-\frac {4 d^{2} \ln \left (1+i {\mathrm e}^{f x +e}\right ) x}{a \,f^{2}}-\frac {4 d^{2} \ln \left (1+i {\mathrm e}^{f x +e}\right ) e}{a \,f^{3}}-\frac {4 d^{2} \polylog \left (2, -i {\mathrm e}^{f x +e}\right )}{a \,f^{3}}+\frac {4 d^{2} e \ln \left ({\mathrm e}^{f x +e}-i\right )}{a \,f^{3}}-\frac {4 d^{2} e \ln \left ({\mathrm e}^{f x +e}\right )}{a \,f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I*(d^2*x^2+2*c*d*x+c^2)/f/a/(exp(f*x+e)-I)-4/a/f^2*d*ln(exp(f*x+e)-I)*c+4/a/f^2*d*ln(exp(f*x+e))*c+2/a/f*d^2
*x^2+4/a/f^2*d^2*e*x+2/a/f^3*d^2*e^2-4/a/f^2*d^2*ln(1+I*exp(f*x+e))*x-4/a/f^3*d^2*ln(1+I*exp(f*x+e))*e-4*d^2*p
olylog(2,-I*exp(f*x+e))/a/f^3+4/a/f^3*d^2*e*ln(exp(f*x+e)-I)-4/a/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 \[ d^{2} {\left (\frac {2 i \, x^{2}}{a f e^{\left (f x + e\right )} - i \, a f} - 4 i \, \int \frac {x}{a f e^{\left (f x + e\right )} - i \, a f}\,{d x}\right )} + 4 \, c d {\left (\frac {x e^{\left (f x + e\right )}}{a f e^{\left (f x + e\right )} - i \, a f} - \frac {\log \left ({\left (e^{\left (f x + e\right )} - i\right )} e^{\left (-e\right )}\right )}{a f^{2}}\right )} - \frac {2 \, c^{2}}{{\left (i \, a e^{\left (-f x - e\right )} - a\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^2*(2*I*x^2/(a*f*e^(f*x + e) - I*a*f) - 4*I*integrate(x/(a*f*e^(f*x + e) - I*a*f), x)) + 4*c*d*(x*e^(f*x + e)
/(a*f*e^(f*x + e) - I*a*f) - log((e^(f*x + e) - I)*e^(-e))/(a*f^2)) - 2*c^2/((I*a*e^(-f*x - e) - a)*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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