3.286 \(\int \frac{\text{sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx\)

Optimal. Leaf size=91 \[ \frac{i a}{8 d (a+i a \sinh (c+d x))^2}-\frac{i}{8 d (a-i a \sinh (c+d x))}+\frac{i}{4 d (a+i a \sinh (c+d x))}+\frac{3 \tan ^{-1}(\sinh (c+d x))}{8 a d} \]

[Out]

(3*ArcTan[Sinh[c + d*x]])/(8*a*d) - (I/8)/(d*(a - I*a*Sinh[c + d*x])) + ((I/8)*a)/(d*(a + I*a*Sinh[c + d*x])^2
) + (I/4)/(d*(a + I*a*Sinh[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.0811465, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2667, 44, 206} \[ \frac{i a}{8 d (a+i a \sinh (c+d x))^2}-\frac{i}{8 d (a-i a \sinh (c+d x))}+\frac{i}{4 d (a+i a \sinh (c+d x))}+\frac{3 \tan ^{-1}(\sinh (c+d x))}{8 a d} \]

Antiderivative was successfully verified.

[In]

Int[Sech[c + d*x]^3/(a + I*a*Sinh[c + d*x]),x]

[Out]

(3*ArcTan[Sinh[c + d*x]])/(8*a*d) - (I/8)/(d*(a - I*a*Sinh[c + d*x])) + ((I/8)*a)/(d*(a + I*a*Sinh[c + d*x])^2
) + (I/4)/(d*(a + I*a*Sinh[c + d*x]))

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\text{sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx &=-\frac{\left (i a^3\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x)^2 (a+x)^3} \, dx,x,i a \sinh (c+d x)\right )}{d}\\ &=-\frac{\left (i a^3\right ) \operatorname{Subst}\left (\int \left (\frac{1}{8 a^3 (a-x)^2}+\frac{1}{4 a^2 (a+x)^3}+\frac{1}{4 a^3 (a+x)^2}+\frac{3}{8 a^3 \left (a^2-x^2\right )}\right ) \, dx,x,i a \sinh (c+d x)\right )}{d}\\ &=-\frac{i}{8 d (a-i a \sinh (c+d x))}+\frac{i a}{8 d (a+i a \sinh (c+d x))^2}+\frac{i}{4 d (a+i a \sinh (c+d x))}-\frac{(3 i) \operatorname{Subst}\left (\int \frac{1}{a^2-x^2} \, dx,x,i a \sinh (c+d x)\right )}{8 d}\\ &=\frac{3 \tan ^{-1}(\sinh (c+d x))}{8 a d}-\frac{i}{8 d (a-i a \sinh (c+d x))}+\frac{i a}{8 d (a+i a \sinh (c+d x))^2}+\frac{i}{4 d (a+i a \sinh (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.10488, size = 101, normalized size = 1.11 \[ \frac{\text{sech}^2(c+d x) \left (3 \sinh ^3(c+d x) \tan ^{-1}(\sinh (c+d x))+\sinh ^2(c+d x) \left (3-3 i \tan ^{-1}(\sinh (c+d x))\right )+3 \sinh (c+d x) \left (\tan ^{-1}(\sinh (c+d x))-i\right )-3 i \tan ^{-1}(\sinh (c+d x))+2\right )}{8 a d (\sinh (c+d x)-i)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[c + d*x]^3/(a + I*a*Sinh[c + d*x]),x]

[Out]

(Sech[c + d*x]^2*(2 - (3*I)*ArcTan[Sinh[c + d*x]] + 3*(-I + ArcTan[Sinh[c + d*x]])*Sinh[c + d*x] + (3 - (3*I)*
ArcTan[Sinh[c + d*x]])*Sinh[c + d*x]^2 + 3*ArcTan[Sinh[c + d*x]]*Sinh[c + d*x]^3))/(8*a*d*(-I + Sinh[c + d*x])
)

________________________________________________________________________________________

Maple [B]  time = 0.056, size = 180, normalized size = 2. \begin{align*}{\frac{{\frac{i}{4}}}{da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +i \right ) ^{-2}}+{\frac{{\frac{3\,i}{8}}}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +i \right ) }-{\frac{1}{4\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +i \right ) ^{-1}}+{\frac{{\frac{i}{2}}}{da} \left ( -i+\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-4}}-{\frac{{\frac{3\,i}{8}}}{da}\ln \left ( -i+\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-{\frac{{\frac{3\,i}{2}}}{da} \left ( -i+\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}+{\frac{1}{da} \left ( -i+\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}-{\frac{1}{da} \left ( -i+\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(d*x+c)^3/(a+I*a*sinh(d*x+c)),x)

[Out]

1/4*I/d/a/(tanh(1/2*d*x+1/2*c)+I)^2+3/8*I/d/a*ln(tanh(1/2*d*x+1/2*c)+I)-1/4/d/a/(tanh(1/2*d*x+1/2*c)+I)+1/2*I/
d/a/(-I+tanh(1/2*d*x+1/2*c))^4-3/8*I/d/a*ln(-I+tanh(1/2*d*x+1/2*c))-3/2*I/d/a/(-I+tanh(1/2*d*x+1/2*c))^2+1/d/a
/(-I+tanh(1/2*d*x+1/2*c))^3-1/d/a/(-I+tanh(1/2*d*x+1/2*c))

________________________________________________________________________________________

Maxima [B]  time = 1.14857, size = 243, normalized size = 2.67 \begin{align*} -\frac{8 \,{\left (3 \, e^{\left (-d x - c\right )} - 6 i \, e^{\left (-2 \, d x - 2 \, c\right )} + 2 \, e^{\left (-3 \, d x - 3 \, c\right )} + 6 i \, e^{\left (-4 \, d x - 4 \, c\right )} + 3 \, e^{\left (-5 \, d x - 5 \, c\right )}\right )}}{{\left (64 i \, a e^{\left (-d x - c\right )} - 32 \, a e^{\left (-2 \, d x - 2 \, c\right )} + 128 i \, a e^{\left (-3 \, d x - 3 \, c\right )} + 32 \, a e^{\left (-4 \, d x - 4 \, c\right )} + 64 i \, a e^{\left (-5 \, d x - 5 \, c\right )} + 32 \, a e^{\left (-6 \, d x - 6 \, c\right )} - 32 \, a\right )} d} - \frac{3 i \, \log \left (e^{\left (-d x - c\right )} + i\right )}{8 \, a d} + \frac{3 i \, \log \left (e^{\left (-d x - c\right )} - i\right )}{8 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^3/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")

[Out]

-8*(3*e^(-d*x - c) - 6*I*e^(-2*d*x - 2*c) + 2*e^(-3*d*x - 3*c) + 6*I*e^(-4*d*x - 4*c) + 3*e^(-5*d*x - 5*c))/((
64*I*a*e^(-d*x - c) - 32*a*e^(-2*d*x - 2*c) + 128*I*a*e^(-3*d*x - 3*c) + 32*a*e^(-4*d*x - 4*c) + 64*I*a*e^(-5*
d*x - 5*c) + 32*a*e^(-6*d*x - 6*c) - 32*a)*d) - 3/8*I*log(e^(-d*x - c) + I)/(a*d) + 3/8*I*log(e^(-d*x - c) - I
)/(a*d)

________________________________________________________________________________________

Fricas [B]  time = 2.10379, size = 763, normalized size = 8.38 \begin{align*} \frac{{\left (3 i \, e^{\left (6 \, d x + 6 \, c\right )} + 6 \, e^{\left (5 \, d x + 5 \, c\right )} + 3 i \, e^{\left (4 \, d x + 4 \, c\right )} + 12 \, e^{\left (3 \, d x + 3 \, c\right )} - 3 i \, e^{\left (2 \, d x + 2 \, c\right )} + 6 \, e^{\left (d x + c\right )} - 3 i\right )} \log \left (e^{\left (d x + c\right )} + i\right ) +{\left (-3 i \, e^{\left (6 \, d x + 6 \, c\right )} - 6 \, e^{\left (5 \, d x + 5 \, c\right )} - 3 i \, e^{\left (4 \, d x + 4 \, c\right )} - 12 \, e^{\left (3 \, d x + 3 \, c\right )} + 3 i \, e^{\left (2 \, d x + 2 \, c\right )} - 6 \, e^{\left (d x + c\right )} + 3 i\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + 6 \, e^{\left (5 \, d x + 5 \, c\right )} - 12 i \, e^{\left (4 \, d x + 4 \, c\right )} + 4 \, e^{\left (3 \, d x + 3 \, c\right )} + 12 i \, e^{\left (2 \, d x + 2 \, c\right )} + 6 \, e^{\left (d x + c\right )}}{8 \, a d e^{\left (6 \, d x + 6 \, c\right )} - 16 i \, a d e^{\left (5 \, d x + 5 \, c\right )} + 8 \, a d e^{\left (4 \, d x + 4 \, c\right )} - 32 i \, a d e^{\left (3 \, d x + 3 \, c\right )} - 8 \, a d e^{\left (2 \, d x + 2 \, c\right )} - 16 i \, a d e^{\left (d x + c\right )} - 8 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^3/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")

[Out]

((3*I*e^(6*d*x + 6*c) + 6*e^(5*d*x + 5*c) + 3*I*e^(4*d*x + 4*c) + 12*e^(3*d*x + 3*c) - 3*I*e^(2*d*x + 2*c) + 6
*e^(d*x + c) - 3*I)*log(e^(d*x + c) + I) + (-3*I*e^(6*d*x + 6*c) - 6*e^(5*d*x + 5*c) - 3*I*e^(4*d*x + 4*c) - 1
2*e^(3*d*x + 3*c) + 3*I*e^(2*d*x + 2*c) - 6*e^(d*x + c) + 3*I)*log(e^(d*x + c) - I) + 6*e^(5*d*x + 5*c) - 12*I
*e^(4*d*x + 4*c) + 4*e^(3*d*x + 3*c) + 12*I*e^(2*d*x + 2*c) + 6*e^(d*x + c))/(8*a*d*e^(6*d*x + 6*c) - 16*I*a*d
*e^(5*d*x + 5*c) + 8*a*d*e^(4*d*x + 4*c) - 32*I*a*d*e^(3*d*x + 3*c) - 8*a*d*e^(2*d*x + 2*c) - 16*I*a*d*e^(d*x
+ c) - 8*a*d)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\operatorname{sech}^{3}{\left (c + d x \right )}}{i \sinh{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)**3/(a+I*a*sinh(d*x+c)),x)

[Out]

Integral(sech(c + d*x)**3/(I*sinh(c + d*x) + 1), x)/a

________________________________________________________________________________________

Giac [B]  time = 1.19123, size = 250, normalized size = 2.75 \begin{align*} \frac{3 i \, \log \left (-i \, e^{\left (d x + c\right )} + i \, e^{\left (-d x - c\right )} + 2\right )}{16 \, a d} - \frac{3 i \, \log \left (-i \, e^{\left (d x + c\right )} + i \, e^{\left (-d x - c\right )} - 2\right )}{16 \, a d} + \frac{3 \, e^{\left (d x + c\right )} - 3 \, e^{\left (-d x - c\right )} + 10 i}{16 \, a d{\left (i \, e^{\left (d x + c\right )} - i \, e^{\left (-d x - c\right )} - 2\right )}} - \frac{-9 i \,{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} - 52 \, e^{\left (d x + c\right )} + 52 \, e^{\left (-d x - c\right )} + 84 i}{32 \, a d{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )} - 2 i\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^3/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

3/16*I*log(-I*e^(d*x + c) + I*e^(-d*x - c) + 2)/(a*d) - 3/16*I*log(-I*e^(d*x + c) + I*e^(-d*x - c) - 2)/(a*d)
+ 1/16*(3*e^(d*x + c) - 3*e^(-d*x - c) + 10*I)/(a*d*(I*e^(d*x + c) - I*e^(-d*x - c) - 2)) - 1/32*(-9*I*(e^(d*x
+ c) - e^(-d*x - c))^2 - 52*e^(d*x + c) + 52*e^(-d*x - c) + 84*I)/(a*d*(e^(d*x + c) - e^(-d*x - c) - 2*I)^2)