3.3.79 \(\int \frac {(e+f x) \text {sech}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx\) [279]

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

[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 = 0.12, 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 = {5690, 4270, 4269, 3556, 5559, 3853, 3855} \begin {gather*} -\frac {i f \text {ArcTan}(\sinh (c+d x))}{6 a d^2}+\frac {f \text {sech}^2(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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/6*I)*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]*Tan
h[c + d*x])/(a*d^2) + ((e + f*x)*Sech[c + d*x]^2*Tanh[c + d*x])/(3*a*d)

Rule 3556

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

Rule 3853

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 3855

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

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 4270

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] &&
GtQ[n, 1] && NeQ[n, 2]

Rule 5559

Int[((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.)*Tanh[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Sim
p[(-(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 5690

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 = 0.75, size = 194, normalized size = 1.23 \begin {gather*} \frac {2 d (e+f x) (\cosh (2 (c+d x))-2 i \sinh (c+d x))+\cosh (c+d x) \left (-d e-i f+c f-2 f \text {ArcTan}\left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+4 i f \log (\cosh (c+d x))-i \left (d e-c f+2 f \text {ArcTan}\left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )-4 i f \log (\cosh (c+d x))\right ) \sinh (c+d x)\right )}{6 a d^2 \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 ) (-i+\sinh (c+d x))} \end {gather*}

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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Maple [A]
time = 3.48, size = 143, normalized size = 0.91

method	result	size
risch	\(\frac {4 f x}{3 a d}+\frac {4 f c}{3 a \,d^{2}}-\frac {i \left (-8 d f x \,{\mathrm e}^{d x +c}+f \,{\mathrm e}^{3 d x +3 c}-8 d e \,{\mathrm e}^{d x +c}+f \,{\mathrm e}^{d x +c}+4 i d x f +4 i d e \right )}{3 \left ({\mathrm e}^{d x +c}+i\right ) \left ({\mathrm e}^{d x +c}-i\right )^{3} d^{2} a}-\frac {f \ln \left ({\mathrm e}^{d x +c}+i\right )}{2 a \,d^{2}}-\frac {5 f \ln \left ({\mathrm e}^{d x +c}-i\right )}{6 a \,d^{2}}\)	\(143\)

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,method=_RETURNVERBOSE)

[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*x*f+
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 = 0.29, size = 252, normalized size = 1.59 \begin {gather*} \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 )} + \frac {4}{3} \, {\left (\frac {2 \, e^{\left (-d x - c\right )}}{{\left (2 \, a e^{\left (-d x - c\right )} + 2 \, a e^{\left (-3 \, d x - 3 \, c\right )} - i \, a e^{\left (-4 \, d x - 4 \, c\right )} + i \, a\right )} d} + \frac {i}{{\left (2 \, a e^{\left (-d x - c\right )} + 2 \, a e^{\left (-3 \, d x - 3 \, c\right )} - i \, a e^{\left (-4 \, d x - 4 \, c\right )} + i \, a\right )} d}\right )} e \end {gather*}

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

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Fricas [A]
time = 0.37, size = 203, normalized size = 1.28 \begin {gather*} \frac {8 \, d f x e^{\left (4 \, d x + 4 \, c\right )} + 8 \, d e - 2 \, {\left (8 i \, d f x + i \, f\right )} e^{\left (3 \, d x + 3 \, c\right )} - 2 \, {\left (-8 i \, d e + i \, f\right )} e^{\left (d x + c\right )} - 3 \, {\left (f e^{\left (4 \, d x + 4 \, c\right )} - 2 i \, f e^{\left (3 \, d x + 3 \, c\right )} - 2 i \, f e^{\left (d x + c\right )} - f\right )} \log \left (e^{\left (d x + c\right )} + i\right ) - 5 \, {\left (f e^{\left (4 \, d x + 4 \, c\right )} - 2 i \, f e^{\left (3 \, d x + 3 \, c\right )} - 2 i \, f e^{\left (d x + c\right )} - f\right )} \log \left (e^{\left (d x + c\right )} - i\right )}{6 \, {\left (a d^{2} e^{\left (4 \, d x + 4 \, c\right )} - 2 i \, a d^{2} e^{\left (3 \, d x + 3 \, c\right )} - 2 i \, a d^{2} e^{\left (d x + c\right )} - a d^{2}\right )}} \end {gather*}

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]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \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} \end {gather*}

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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Giac [A]
time = 0.43, size = 260, normalized size = 1.65 \begin {gather*} \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 )} + 16 i \, d e e^{\left (d x + 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 )} - 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 \, {\left (a d^{2} e^{\left (4 \, d x + 4 \, c\right )} - 2 i \, a d^{2} e^{\left (3 \, d x + 3 \, c\right )} - 2 i \, a d^{2} e^{\left (d x + c\right )} - a d^{2}\right )}} \end {gather*}

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]

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

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Mupad [B]
time = 2.48, size = 205, normalized size = 1.30 \begin {gather*} \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 )} \end {gather*}

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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