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

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

[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)

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Rubi [A]  time = 0.218708, antiderivative size = 101, normalized size of antiderivative = 1., 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 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 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 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 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]

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*}

Mathematica [A]  time = 2.21886, size = 150, normalized size = 1.49 \[ \frac{2 \left (2 d^2 \text{PolyLog}\left (2,i e^{-e-f x}\right )+\frac{f^2 (c+d x)^2 \sinh \left (\frac{f x}{2}\right )}{\left (\cosh \left (\frac{e}{2}\right )+i \sinh \left (\frac{e}{2}\right )\right ) \left (\cosh \left (\frac{1}{2} (e+f x)\right )+i \sinh \left (\frac{1}{2} (e+f x)\right )\right )}+\frac{i f (c+d x) \left (f (c+d x)+2 d \left (1+i e^e\right ) \log \left (1-i e^{-e-f x}\right )\right )}{e^e-i}\right )}{a f^3} \]

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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Maple [B]  time = 0.056, size = 227, normalized size = 2.3 \begin{align*}{\frac{2\,i \left ({d}^{2}{x}^{2}+2\,cdx+{c}^{2} \right ) }{fa \left ({{\rm e}^{fx+e}}-i \right ) }}-4\,{\frac{d\ln \left ({{\rm e}^{fx+e}}-i \right ) c}{a{f}^{2}}}+4\,{\frac{d\ln \left ({{\rm e}^{fx+e}} \right ) c}{a{f}^{2}}}+2\,{\frac{{d}^{2}{x}^{2}}{fa}}+4\,{\frac{{d}^{2}ex}{a{f}^{2}}}+2\,{\frac{{d}^{2}{e}^{2}}{{f}^{3}a}}-4\,{\frac{{d}^{2}\ln \left ( 1+i{{\rm e}^{fx+e}} \right ) x}{a{f}^{2}}}-4\,{\frac{{d}^{2}\ln \left ( 1+i{{\rm e}^{fx+e}} \right ) e}{{f}^{3}a}}-4\,{\frac{{d}^{2}{\it polylog} \left ( 2,-i{{\rm e}^{fx+e}} \right ) }{{f}^{3}a}}+4\,{\frac{{d}^{2}e\ln \left ({{\rm e}^{fx+e}}-i \right ) }{{f}^{3}a}}-4\,{\frac{{d}^{2}e\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{3}a}} \end{align*}

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 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} \end{align*}

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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Fricas [B]  time = 2.69656, size = 483, normalized size = 4.78 \begin{align*} \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}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{2}}{i \, a \sinh \left (f x + e\right ) + a}\,{d x} \end{align*}

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)