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

Optimal. Leaf size=158 \[ \frac {(c+d x) \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{3 a^2 f}+\frac {(c+d x) \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \text {sech}^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{6 a^2 f}+\frac {d \text {sech}^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{6 a^2 f^2}-\frac {2 d \log \left (\cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )\right )}{3 a^2 f^2} \]

[Out]

-2/3*d*ln(cosh(1/2*e+1/4*I*Pi+1/2*f*x))/a^2/f^2+1/6*d*sech(1/2*e+1/4*I*Pi+1/2*f*x)^2/a^2/f^2+1/3*(d*x+c)*tanh(
1/2*e+1/4*I*Pi+1/2*f*x)/a^2/f+1/6*(d*x+c)*sech(1/2*e+1/4*I*Pi+1/2*f*x)^2*tanh(1/2*e+1/4*I*Pi+1/2*f*x)/a^2/f

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Rubi [A]  time = 0.11, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3318, 4185, 4184, 3475} \[ \frac {(c+d x) \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{3 a^2 f}+\frac {(c+d x) \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \text {sech}^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{6 a^2 f}+\frac {d \text {sech}^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{6 a^2 f^2}-\frac {2 d \log \left (\cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )\right )}{3 a^2 f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d*Log[Cosh[e/2 + (I/4)*Pi + (f*x)/2]])/(3*a^2*f^2) + (d*Sech[e/2 + (I/4)*Pi + (f*x)/2]^2)/(6*a^2*f^2) + ((
c + d*x)*Tanh[e/2 + (I/4)*Pi + (f*x)/2])/(3*a^2*f) + ((c + d*x)*Sech[e/2 + (I/4)*Pi + (f*x)/2]^2*Tanh[e/2 + (I
/4)*Pi + (f*x)/2])/(6*a^2*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 3475

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

Rubi steps

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

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Mathematica [A]  time = 1.05, size = 241, normalized size = 1.53 \[ \frac {\left (\sinh \left (\frac {1}{2} (e+f x)\right )-i \cosh \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cosh \left (\frac {3}{2} (e+f x)\right ) \left (2 c f+2 d \tan ^{-1}\left (\tanh \left (\frac {1}{2} (e+f x)\right )\right )-i d \log (\cosh (e+f x))-d e+d f x\right )+2 i \sinh \left (\frac {1}{2} (e+f x)\right ) \left (-3 c f-4 d \tan ^{-1}\left (\tanh \left (\frac {1}{2} (e+f x)\right )\right )+2 i d \log (\cosh (e+f x))+d \cosh (e+f x) \left (-2 \tan ^{-1}\left (\tanh \left (\frac {1}{2} (e+f x)\right )\right )+i \log (\cosh (e+f x))+e+f x\right )+2 d e-d f x-i d\right )+d \cosh \left (\frac {1}{2} (e+f x)\right ) \left (-6 \tan ^{-1}\left (\tanh \left (\frac {1}{2} (e+f x)\right )\right )+3 i \log (\cosh (e+f x))+3 e+3 f x-2 i\right )\right )}{6 a^2 f^2 (\sinh (e+f x)-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-I)*Cosh[(e + f*x)/2] + Sinh[(e + f*x)/2])*(d*Cosh[(e + f*x)/2]*(-2*I + 3*e + 3*f*x - 6*ArcTan[Tanh[(e + f*
x)/2]] + (3*I)*Log[Cosh[e + f*x]]) + Cosh[(3*(e + f*x))/2]*(-(d*e) + 2*c*f + d*f*x + 2*d*ArcTan[Tanh[(e + f*x)
/2]] - I*d*Log[Cosh[e + f*x]]) + (2*I)*((-I)*d + 2*d*e - 3*c*f - d*f*x - 4*d*ArcTan[Tanh[(e + f*x)/2]] + d*Cos
h[e + f*x]*(e + f*x - 2*ArcTan[Tanh[(e + f*x)/2]] + I*Log[Cosh[e + f*x]]) + (2*I)*d*Log[Cosh[e + f*x]])*Sinh[(
e + f*x)/2]))/(6*a^2*f^2*(-I + Sinh[e + f*x])^2)

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fricas [A]  time = 0.44, size = 162, normalized size = 1.03 \[ \frac {2 \, d f x e^{\left (3 \, f x + 3 \, e\right )} - 2 i \, c f + {\left (-6 i \, d f x - 2 i \, d\right )} e^{\left (2 \, f x + 2 \, e\right )} + 2 \, {\left (3 \, c f - d\right )} e^{\left (f x + e\right )} - {\left (2 \, d e^{\left (3 \, f x + 3 \, e\right )} - 6 i \, d e^{\left (2 \, f x + 2 \, e\right )} - 6 \, d e^{\left (f x + e\right )} + 2 i \, d\right )} \log \left (e^{\left (f x + e\right )} - i\right )}{3 \, a^{2} f^{2} e^{\left (3 \, f x + 3 \, e\right )} - 9 i \, a^{2} f^{2} e^{\left (2 \, f x + 2 \, e\right )} - 9 \, a^{2} f^{2} e^{\left (f x + e\right )} + 3 i \, a^{2} f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*d*f*x*e^(3*f*x + 3*e) - 2*I*c*f + (-6*I*d*f*x - 2*I*d)*e^(2*f*x + 2*e) + 2*(3*c*f - d)*e^(f*x + e) - (2*d*e
^(3*f*x + 3*e) - 6*I*d*e^(2*f*x + 2*e) - 6*d*e^(f*x + e) + 2*I*d)*log(e^(f*x + e) - I))/(3*a^2*f^2*e^(3*f*x +
3*e) - 9*I*a^2*f^2*e^(2*f*x + 2*e) - 9*a^2*f^2*e^(f*x + e) + 3*I*a^2*f^2)

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giac [A]  time = 0.22, size = 211, normalized size = 1.34 \[ \frac {2 \, d f x e^{\left (3 \, f x + 3 \, e\right )} - 6 i \, d f x e^{\left (2 \, f x + 2 \, e\right )} + 6 \, c f e^{\left (f x + e\right )} - 2 \, d e^{\left (3 \, f x + 3 \, e\right )} \log \left (e^{\left (f x + e\right )} - i\right ) + 6 i \, d e^{\left (2 \, f x + 2 \, e\right )} \log \left (e^{\left (f x + e\right )} - i\right ) + 6 \, d e^{\left (f x + e\right )} \log \left (e^{\left (f x + e\right )} - i\right ) - 2 i \, c f - 2 i \, d e^{\left (2 \, f x + 2 \, e\right )} - 2 \, d e^{\left (f x + e\right )} - 2 i \, d \log \left (e^{\left (f x + e\right )} - i\right )}{3 \, a^{2} f^{2} e^{\left (3 \, f x + 3 \, e\right )} - 9 i \, a^{2} f^{2} e^{\left (2 \, f x + 2 \, e\right )} - 9 \, a^{2} f^{2} e^{\left (f x + e\right )} + 3 i \, a^{2} f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*d*f*x*e^(3*f*x + 3*e) - 6*I*d*f*x*e^(2*f*x + 2*e) + 6*c*f*e^(f*x + e) - 2*d*e^(3*f*x + 3*e)*log(e^(f*x + e)
- I) + 6*I*d*e^(2*f*x + 2*e)*log(e^(f*x + e) - I) + 6*d*e^(f*x + e)*log(e^(f*x + e) - I) - 2*I*c*f - 2*I*d*e^
(2*f*x + 2*e) - 2*d*e^(f*x + e) - 2*I*d*log(e^(f*x + e) - I))/(3*a^2*f^2*e^(3*f*x + 3*e) - 9*I*a^2*f^2*e^(2*f*
x + 2*e) - 9*a^2*f^2*e^(f*x + e) + 3*I*a^2*f^2)

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maple [A]  time = 0.23, size = 113, normalized size = 0.72 \[ \frac {2 d x}{3 f \,a^{2}}+\frac {2 d e}{3 f^{2} a^{2}}-\frac {2 i \left (3 i f d x \,{\mathrm e}^{f x +e}+3 i f c \,{\mathrm e}^{f x +e}-i d \,{\mathrm e}^{f x +e}+d f x +{\mathrm e}^{2 f x +2 e} d +c f \right )}{3 \left ({\mathrm e}^{f x +e}-i\right )^{3} f^{2} a^{2}}-\frac {2 d \ln \left ({\mathrm e}^{f x +e}-i\right )}{3 f^{2} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*d/f/a^2*x+2/3*d/f^2/a^2*e-2/3*I*(3*I*f*d*x*exp(f*x+e)+3*I*f*c*exp(f*x+e)-I*d*exp(f*x+e)+d*f*x+exp(2*f*x+2*
e)*d+c*f)/(exp(f*x+e)-I)^3/f^2/a^2-2/3*d/f^2/a^2*ln(exp(f*x+e)-I)

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maxima [B]  time = 0.35, size = 257, normalized size = 1.63 \[ \frac {1}{3} \, d {\left (\frac {3 \, {\left (2 \, f x e^{\left (3 \, f x + 3 \, e\right )} + {\left (-6 i \, f x e^{\left (2 \, e\right )} - 2 i \, e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )} - 2 \, e^{\left (f x + e\right )}\right )}}{3 \, a^{2} f^{2} e^{\left (3 \, f x + 3 \, e\right )} - 9 i \, a^{2} f^{2} e^{\left (2 \, f x + 2 \, e\right )} - 9 \, a^{2} f^{2} e^{\left (f x + e\right )} + 3 i \, a^{2} f^{2}} - \frac {2 \, \log \left (-i \, {\left (i \, e^{\left (f x + e\right )} + 1\right )} e^{\left (-e\right )}\right )}{a^{2} f^{2}}\right )} + 2 \, c {\left (\frac {3 \, e^{\left (-f x - e\right )}}{{\left (9 \, a^{2} e^{\left (-f x - e\right )} - 9 i \, a^{2} e^{\left (-2 \, f x - 2 \, e\right )} - 3 \, a^{2} e^{\left (-3 \, f x - 3 \, e\right )} + 3 i \, a^{2}\right )} f} + \frac {i}{{\left (9 \, a^{2} e^{\left (-f x - e\right )} - 9 i \, a^{2} e^{\left (-2 \, f x - 2 \, e\right )} - 3 \, a^{2} e^{\left (-3 \, f x - 3 \, e\right )} + 3 i \, a^{2}\right )} f}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*d*(3*(2*f*x*e^(3*f*x + 3*e) + (-6*I*f*x*e^(2*e) - 2*I*e^(2*e))*e^(2*f*x) - 2*e^(f*x + e))/(3*a^2*f^2*e^(3*
f*x + 3*e) - 9*I*a^2*f^2*e^(2*f*x + 2*e) - 9*a^2*f^2*e^(f*x + e) + 3*I*a^2*f^2) - 2*log(-I*(I*e^(f*x + e) + 1)
*e^(-e))/(a^2*f^2)) + 2*c*(3*e^(-f*x - e)/((9*a^2*e^(-f*x - e) - 9*I*a^2*e^(-2*f*x - 2*e) - 3*a^2*e^(-3*f*x -
3*e) + 3*I*a^2)*f) + I/((9*a^2*e^(-f*x - e) - 9*I*a^2*e^(-2*f*x - 2*e) - 3*a^2*e^(-3*f*x - 3*e) + 3*I*a^2)*f))

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mupad [B]  time = 0.54, size = 160, normalized size = 1.01 \[ -\frac {\frac {2\,d\,\ln \left ({\mathrm {e}}^{f\,x}\,{\mathrm {e}}^e-\mathrm {i}\right )}{3}+{\mathrm {e}}^{e+f\,x}\,\left (-\frac {d\,2{}\mathrm {i}}{3}+c\,f\,2{}\mathrm {i}+d\,\ln \left ({\mathrm {e}}^{f\,x}\,{\mathrm {e}}^e-\mathrm {i}\right )\,2{}\mathrm {i}\right )+\frac {2\,d\,{\mathrm {e}}^{2\,e+2\,f\,x}}{3}+f\,\left (\frac {2\,c}{3}+2\,d\,x\,{\mathrm {e}}^{2\,e+2\,f\,x}+\frac {d\,x\,{\mathrm {e}}^{3\,e+3\,f\,x}\,2{}\mathrm {i}}{3}\right )-2\,d\,{\mathrm {e}}^{2\,e+2\,f\,x}\,\ln \left ({\mathrm {e}}^{f\,x}\,{\mathrm {e}}^e-\mathrm {i}\right )-\frac {d\,{\mathrm {e}}^{3\,e+3\,f\,x}\,\ln \left ({\mathrm {e}}^{f\,x}\,{\mathrm {e}}^e-\mathrm {i}\right )\,2{}\mathrm {i}}{3}}{a^2\,f^2\,{\left (1+{\mathrm {e}}^{e+f\,x}\,1{}\mathrm {i}\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((2*d*log(exp(f*x)*exp(e) - 1i))/3 + exp(e + f*x)*(c*f*2i - (d*2i)/3 + d*log(exp(f*x)*exp(e) - 1i)*2i) + (2*d
*exp(2*e + 2*f*x))/3 + f*((2*c)/3 + 2*d*x*exp(2*e + 2*f*x) + (d*x*exp(3*e + 3*f*x)*2i)/3) - 2*d*exp(2*e + 2*f*
x)*log(exp(f*x)*exp(e) - 1i) - (d*exp(3*e + 3*f*x)*log(exp(f*x)*exp(e) - 1i)*2i)/3)/(a^2*f^2*(exp(e + f*x)*1i
+ 1)^3)

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sympy [A]  time = 0.52, size = 184, normalized size = 1.16 \[ \frac {2 c f e^{3 e} + 2 d f x e^{3 e} - 2 d e^{e} e^{- 2 f x} + \left (- 6 i c f e^{2 e} - 6 i d f x e^{2 e} - 2 i d e^{2 e}\right ) e^{- f x}}{3 a^{2} f^{2} e^{3 e} - 9 i a^{2} f^{2} e^{2 e} e^{- f x} - 9 a^{2} f^{2} e^{e} e^{- 2 f x} + 3 i a^{2} f^{2} e^{- 3 f x}} - \frac {2 d x}{3 a^{2} f} - \frac {2 d \log {\left (i e^{e} + e^{- f x} \right )}}{3 a^{2} f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*c*f*exp(3*e) + 2*d*f*x*exp(3*e) - 2*d*exp(e)*exp(-2*f*x) + (-6*I*c*f*exp(2*e) - 6*I*d*f*x*exp(2*e) - 2*I*d*
exp(2*e))*exp(-f*x))/(3*a**2*f**2*exp(3*e) - 9*I*a**2*f**2*exp(2*e)*exp(-f*x) - 9*a**2*f**2*exp(e)*exp(-2*f*x)
+ 3*I*a**2*f**2*exp(-3*f*x)) - 2*d*x/(3*a**2*f) - 2*d*log(I*exp(e) + exp(-f*x))/(3*a**2*f**2)

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