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

Optimal. Leaf size=158 \[ \frac {f \text {sech}^2(c+d x)}{6 a d^2}-\frac {i f \tan ^{-1}(\sinh (c+d x))}{6 a d^2}-\frac {2 f \log (\cosh (c+d x))}{3 a d^2}-\frac {i f \tanh (c+d x) \text {sech}(c+d x)}{6 a d^2}+\frac {2 (e+f x) \tanh (c+d x)}{3 a d}+\frac {i (e+f x) \text {sech}^3(c+d x)}{3 a d}+\frac {(e+f x) \tanh (c+d x) \text {sech}^2(c+d x)}{3 a d} \]

[Out]

-1/6*I*f*arctan(sinh(d*x+c))/a/d^2-2/3*f*ln(cosh(d*x+c))/a/d^2+1/6*f*sech(d*x+c)^2/a/d^2+1/3*I*(f*x+e)*sech(d*
x+c)^3/a/d+2/3*(f*x+e)*tanh(d*x+c)/a/d-1/6*I*f*sech(d*x+c)*tanh(d*x+c)/a/d^2+1/3*(f*x+e)*sech(d*x+c)^2*tanh(d*
x+c)/a/d

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Rubi [A]  time = 0.16, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {5571, 4185, 4184, 3475, 5451, 3768, 3770} \[ \frac {f \text {sech}^2(c+d x)}{6 a d^2}-\frac {i f \tan ^{-1}(\sinh (c+d x))}{6 a d^2}-\frac {2 f \log (\cosh (c+d x))}{3 a d^2}-\frac {i f \tanh (c+d x) \text {sech}(c+d x)}{6 a d^2}+\frac {2 (e+f x) \tanh (c+d x)}{3 a d}+\frac {i (e+f x) \text {sech}^3(c+d x)}{3 a d}+\frac {(e+f x) \tanh (c+d x) \text {sech}^2(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/6)*f*ArcTan[Sinh[c + d*x]])/(a*d^2) - (2*f*Log[Cosh[c + d*x]])/(3*a*d^2) + (f*Sech[c + d*x]^2)/(6*a*d^2)
+ ((I/3)*(e + f*x)*Sech[c + d*x]^3)/(a*d) + (2*(e + f*x)*Tanh[c + d*x])/(3*a*d) - ((I/6)*f*Sech[c + d*x]*Tanh[
c + d*x])/(a*d^2) + ((e + f*x)*Sech[c + d*x]^2*Tanh[c + d*x])/(3*a*d)

Rule 3475

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

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

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

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 4185

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

Rule 5451

Int[((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.)*Tanh[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> -Si
mp[((c + d*x)^m*Sech[a + b*x]^n)/(b*n), x] + Dist[(d*m)/(b*n), Int[(c + d*x)^(m - 1)*Sech[a + b*x]^n, x], x] /
; FreeQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

Rule 5571

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sech[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Sym
bol] :> Dist[1/a, Int[(e + f*x)^m*Sech[c + d*x]^(n + 2), x], x] + Dist[1/b, Int[(e + f*x)^m*Sech[c + d*x]^(n +
1)*Tanh[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && IGtQ[m, 0] && EqQ[a^2 + b^2, 0]

Rubi steps

\begin {align*} \int \frac {(e+f x) \text {sech}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx &=-\frac {i \int (e+f x) \text {sech}^3(c+d x) \tanh (c+d x) \, dx}{a}+\frac {\int (e+f x) \text {sech}^4(c+d x) \, dx}{a}\\ &=\frac {f \text {sech}^2(c+d x)}{6 a d^2}+\frac {i (e+f x) \text {sech}^3(c+d x)}{3 a d}+\frac {(e+f x) \text {sech}^2(c+d x) \tanh (c+d x)}{3 a d}+\frac {2 \int (e+f x) \text {sech}^2(c+d x) \, dx}{3 a}-\frac {(i f) \int \text {sech}^3(c+d x) \, dx}{3 a d}\\ &=\frac {f \text {sech}^2(c+d x)}{6 a d^2}+\frac {i (e+f x) \text {sech}^3(c+d x)}{3 a d}+\frac {2 (e+f x) \tanh (c+d x)}{3 a d}-\frac {i f \text {sech}(c+d x) \tanh (c+d x)}{6 a d^2}+\frac {(e+f x) \text {sech}^2(c+d x) \tanh (c+d x)}{3 a d}-\frac {(i f) \int \text {sech}(c+d x) \, dx}{6 a d}-\frac {(2 f) \int \tanh (c+d x) \, dx}{3 a d}\\ &=-\frac {i f \tan ^{-1}(\sinh (c+d x))}{6 a d^2}-\frac {2 f \log (\cosh (c+d x))}{3 a d^2}+\frac {f \text {sech}^2(c+d x)}{6 a d^2}+\frac {i (e+f x) \text {sech}^3(c+d x)}{3 a d}+\frac {2 (e+f x) \tanh (c+d x)}{3 a d}-\frac {i f \text {sech}(c+d x) \tanh (c+d x)}{6 a d^2}+\frac {(e+f x) \text {sech}^2(c+d x) \tanh (c+d x)}{3 a d}\\ \end {align*}

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Mathematica [A]  time = 1.12, size = 194, normalized size = 1.23 \[ \frac {2 d (e+f x) (\cosh (2 (c+d x))-2 i \sinh (c+d x))+\cosh (c+d x) \left (-i \sinh (c+d x) \left (2 f \tan ^{-1}\left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )-4 i f \log (\cosh (c+d x))-c f+d e\right )-2 f \tan ^{-1}\left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+4 i f \log (\cosh (c+d x))+c f-d e-i f\right )}{6 a d^2 (\sinh (c+d x)-i) \left (\cosh \left (\frac {1}{2} (c+d x)\right )-i \sinh \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d*(e + f*x)*(Cosh[2*(c + d*x)] - (2*I)*Sinh[c + d*x]) + Cosh[c + d*x]*(-(d*e) - I*f + c*f - 2*f*ArcTan[Tanh
[(c + d*x)/2]] + (4*I)*f*Log[Cosh[c + d*x]] - I*(d*e - c*f + 2*f*ArcTan[Tanh[(c + d*x)/2]] - (4*I)*f*Log[Cosh[
c + d*x]])*Sinh[c + d*x]))/(6*a*d^2*(Cosh[(c + d*x)/2] - I*Sinh[(c + d*x)/2])*(Cosh[(c + d*x)/2] + I*Sinh[(c +
d*x)/2])*(-I + Sinh[c + d*x]))

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fricas [A]  time = 0.49, size = 201, normalized size = 1.27 \[ \frac {8 \, d f x e^{\left (4 \, d x + 4 \, c\right )} + 8 \, d e + {\left (-16 i \, d f x - 2 i \, f\right )} e^{\left (3 \, d x + 3 \, c\right )} + {\left (16 i \, d e - 2 i \, f\right )} e^{\left (d x + c\right )} - {\left (3 \, f e^{\left (4 \, d x + 4 \, c\right )} - 6 i \, f e^{\left (3 \, d x + 3 \, c\right )} - 6 i \, f e^{\left (d x + c\right )} - 3 \, f\right )} \log \left (e^{\left (d x + c\right )} + i\right ) - {\left (5 \, f e^{\left (4 \, d x + 4 \, c\right )} - 10 i \, f e^{\left (3 \, d x + 3 \, c\right )} - 10 i \, f e^{\left (d x + c\right )} - 5 \, f\right )} \log \left (e^{\left (d x + c\right )} - i\right )}{6 \, a d^{2} e^{\left (4 \, d x + 4 \, c\right )} - 12 i \, a d^{2} e^{\left (3 \, d x + 3 \, c\right )} - 12 i \, a d^{2} e^{\left (d x + c\right )} - 6 \, a d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8*d*f*x*e^(4*d*x + 4*c) + 8*d*e + (-16*I*d*f*x - 2*I*f)*e^(3*d*x + 3*c) + (16*I*d*e - 2*I*f)*e^(d*x + c) - (3
*f*e^(4*d*x + 4*c) - 6*I*f*e^(3*d*x + 3*c) - 6*I*f*e^(d*x + c) - 3*f)*log(e^(d*x + c) + I) - (5*f*e^(4*d*x + 4
*c) - 10*I*f*e^(3*d*x + 3*c) - 10*I*f*e^(d*x + c) - 5*f)*log(e^(d*x + c) - I))/(6*a*d^2*e^(4*d*x + 4*c) - 12*I
*a*d^2*e^(3*d*x + 3*c) - 12*I*a*d^2*e^(d*x + c) - 6*a*d^2)

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giac [A]  time = 0.27, size = 261, normalized size = 1.65 \[ \frac {8 \, d f x e^{\left (4 \, d x + 4 \, c\right )} - 16 i \, d f x e^{\left (3 \, d x + 3 \, c\right )} - 3 \, f e^{\left (4 \, d x + 4 \, c\right )} \log \left (e^{\left (d x + c\right )} + i\right ) + 6 i \, f e^{\left (3 \, d x + 3 \, c\right )} \log \left (e^{\left (d x + c\right )} + i\right ) + 6 i \, f e^{\left (d x + c\right )} \log \left (e^{\left (d x + c\right )} + i\right ) - 5 \, f e^{\left (4 \, d x + 4 \, c\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + 10 i \, f e^{\left (3 \, d x + 3 \, c\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + 10 i \, f e^{\left (d x + c\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + 8 \, d e - 2 i \, f e^{\left (3 \, d x + 3 \, c\right )} + 16 i \, d e^{\left (d x + c + 1\right )} - 2 i \, f e^{\left (d x + c\right )} + 3 \, f \log \left (e^{\left (d x + c\right )} + i\right ) + 5 \, f \log \left (e^{\left (d x + c\right )} - i\right )}{6 \, a d^{2} e^{\left (4 \, d x + 4 \, c\right )} - 12 i \, a d^{2} e^{\left (3 \, d x + 3 \, c\right )} - 12 i \, a d^{2} e^{\left (d x + c\right )} - 6 \, a d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8*d*f*x*e^(4*d*x + 4*c) - 16*I*d*f*x*e^(3*d*x + 3*c) - 3*f*e^(4*d*x + 4*c)*log(e^(d*x + c) + I) + 6*I*f*e^(3*
d*x + 3*c)*log(e^(d*x + c) + I) + 6*I*f*e^(d*x + c)*log(e^(d*x + c) + I) - 5*f*e^(4*d*x + 4*c)*log(e^(d*x + c)
- I) + 10*I*f*e^(3*d*x + 3*c)*log(e^(d*x + c) - I) + 10*I*f*e^(d*x + c)*log(e^(d*x + c) - I) + 8*d*e - 2*I*f*
e^(3*d*x + 3*c) + 16*I*d*e^(d*x + c + 1) - 2*I*f*e^(d*x + c) + 3*f*log(e^(d*x + c) + I) + 5*f*log(e^(d*x + c)
- I))/(6*a*d^2*e^(4*d*x + 4*c) - 12*I*a*d^2*e^(3*d*x + 3*c) - 12*I*a*d^2*e^(d*x + c) - 6*a*d^2)

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maple [A]  time = 0.27, size = 143, normalized size = 0.91 \[ \frac {4 f x}{3 a d}+\frac {4 f c}{3 a \,d^{2}}-\frac {i \left (-8 d f x \,{\mathrm e}^{d x +c}+f \,{\mathrm e}^{3 d x +3 c}-8 d e \,{\mathrm e}^{d x +c}+f \,{\mathrm e}^{d x +c}+4 i d f x +4 i d e \right )}{3 \left ({\mathrm e}^{d x +c}+i\right ) \left ({\mathrm e}^{d x +c}-i\right )^{3} d^{2} a}-\frac {f \ln \left ({\mathrm e}^{d x +c}+i\right )}{2 a \,d^{2}}-\frac {5 f \ln \left ({\mathrm e}^{d x +c}-i\right )}{6 a \,d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/3*f*x/a/d+4/3*f/a/d^2*c-1/3*I*(-8*d*f*x*exp(d*x+c)+f*exp(3*d*x+3*c)-8*d*e*exp(d*x+c)+f*exp(d*x+c)+4*I*d*f*x+
4*I*d*e)/(exp(d*x+c)+I)/(exp(d*x+c)-I)^3/d^2/a-1/2*f/a/d^2*ln(exp(d*x+c)+I)-5/6*f/a/d^2*ln(exp(d*x+c)-I)

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maxima [A]  time = 0.39, size = 251, normalized size = 1.59 \[ \frac {1}{6} \, f {\left (\frac {24 \, {\left (4 i \, d x e^{\left (4 \, d x + 4 \, c\right )} + {\left (8 \, d x e^{\left (3 \, c\right )} + e^{\left (3 \, c\right )}\right )} e^{\left (3 \, d x\right )} + e^{\left (d x + c\right )}\right )}}{12 i \, a d^{2} e^{\left (4 \, d x + 4 \, c\right )} + 24 \, a d^{2} e^{\left (3 \, d x + 3 \, c\right )} + 24 \, a d^{2} e^{\left (d x + c\right )} - 12 i \, a d^{2}} - \frac {3 \, \log \left ({\left (e^{\left (d x + c\right )} + i\right )} e^{\left (-c\right )}\right )}{a d^{2}} - \frac {5 \, \log \left (-i \, {\left (i \, e^{\left (d x + c\right )} + 1\right )} e^{\left (-c\right )}\right )}{a d^{2}}\right )} + 4 \, e {\left (\frac {2 \, e^{\left (-d x - c\right )}}{{\left (6 \, a e^{\left (-d x - c\right )} + 6 \, a e^{\left (-3 \, d x - 3 \, c\right )} - 3 i \, a e^{\left (-4 \, d x - 4 \, c\right )} + 3 i \, a\right )} d} + \frac {i}{{\left (6 \, a e^{\left (-d x - c\right )} + 6 \, a e^{\left (-3 \, d x - 3 \, c\right )} - 3 i \, a e^{\left (-4 \, d x - 4 \, c\right )} + 3 i \, a\right )} d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*f*(24*(4*I*d*x*e^(4*d*x + 4*c) + (8*d*x*e^(3*c) + e^(3*c))*e^(3*d*x) + e^(d*x + c))/(12*I*a*d^2*e^(4*d*x +
4*c) + 24*a*d^2*e^(3*d*x + 3*c) + 24*a*d^2*e^(d*x + c) - 12*I*a*d^2) - 3*log((e^(d*x + c) + I)*e^(-c))/(a*d^2
) - 5*log(-I*(I*e^(d*x + c) + 1)*e^(-c))/(a*d^2)) + 4*e*(2*e^(-d*x - c)/((6*a*e^(-d*x - c) + 6*a*e^(-3*d*x - 3
*c) - 3*I*a*e^(-4*d*x - 4*c) + 3*I*a)*d) + I/((6*a*e^(-d*x - c) + 6*a*e^(-3*d*x - 3*c) - 3*I*a*e^(-4*d*x - 4*c
) + 3*I*a)*d))

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mupad [B]  time = 2.48, size = 205, normalized size = 1.30 \[ \frac {4\,f\,x}{3\,a\,d}-\frac {f+3\,d\,e+3\,d\,f\,x}{3\,a\,d^2\,\left (1-{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{c+d\,x}\,2{}\mathrm {i}\right )}-\frac {5\,f\,\ln \left (f+f\,{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )}{6\,a\,d^2}-\frac {\left (e+f\,x\right )\,2{}\mathrm {i}}{3\,a\,d\,\left (3\,{\mathrm {e}}^{c+d\,x}+{\mathrm {e}}^{2\,c+2\,d\,x}\,3{}\mathrm {i}-{\mathrm {e}}^{3\,c+3\,d\,x}-\mathrm {i}\right )}-\frac {\left (e+f\,x\right )\,1{}\mathrm {i}}{2\,a\,d\,\left ({\mathrm {e}}^{c+d\,x}+1{}\mathrm {i}\right )}-\frac {f\,\ln \left (-1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )}{2\,a\,d^2}+\frac {\left (3\,d\,e-2\,f+3\,d\,f\,x\right )\,1{}\mathrm {i}}{6\,a\,d^2\,\left ({\mathrm {e}}^{c+d\,x}-\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*f*x)/(3*a*d) - (f + 3*d*e + 3*d*f*x)/(3*a*d^2*(exp(c + d*x)*2i - exp(2*c + 2*d*x) + 1)) - (5*f*log(f + f*ex
p(c + d*x)*1i))/(6*a*d^2) - ((e + f*x)*2i)/(3*a*d*(3*exp(c + d*x) + exp(2*c + 2*d*x)*3i - exp(3*c + 3*d*x) - 1
i)) - ((e + f*x)*1i)/(2*a*d*(exp(c + d*x) + 1i)) - (f*log(exp(c + d*x)*1i - 1))/(2*a*d^2) + ((3*d*e - 2*f + 3*
d*f*x)*1i)/(6*a*d^2*(exp(c + d*x) - 1i))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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