3.268 \(\int \frac {\cosh ^3(c+d x)}{a+i a \sinh (c+d x)} \, dx\)

Optimal. Leaf size=34 \[ \frac {\sinh (c+d x)}{a d}-\frac {i \sinh ^2(c+d x)}{2 a d} \]

[Out]

sinh(d*x+c)/a/d-1/2*I*sinh(d*x+c)^2/a/d

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Rubi [A]  time = 0.05, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {2667} \[ \frac {\sinh (c+d x)}{a d}-\frac {i \sinh ^2(c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

Rubi steps

\begin {align*} \int \frac {\cosh ^3(c+d x)}{a+i a \sinh (c+d x)} \, dx &=-\frac {i \operatorname {Subst}(\int (a-x) \, dx,x,i a \sinh (c+d x))}{a^3 d}\\ &=\frac {\sinh (c+d x)}{a d}-\frac {i \sinh ^2(c+d x)}{2 a d}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 28, normalized size = 0.82 \[ \frac {(2-i \sinh (c+d x)) \sinh (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2 - I*Sinh[c + d*x])*Sinh[c + d*x])/(2*a*d)

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fricas [A]  time = 0.52, size = 49, normalized size = 1.44 \[ \frac {{\left (-i \, e^{\left (4 \, d x + 4 \, c\right )} + 4 \, e^{\left (3 \, d x + 3 \, c\right )} - 4 \, e^{\left (d x + c\right )} - i\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.23, size = 55, normalized size = 1.62 \[ -\frac {\frac {{\left (4 \, e^{\left (d x + c\right )} + i\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{a} + \frac {i \, a e^{\left (2 \, d x + 2 \, c\right )} - 4 \, a e^{\left (d x + c\right )}}{a^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 29, normalized size = 0.85 \[ -\frac {\frac {i \left (\sinh ^{2}\left (d x +c \right )\right )}{2}-\sinh \left (d x +c \right )}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a/d*(1/2*I*sinh(d*x+c)^2-sinh(d*x+c))

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maxima [A]  time = 0.31, size = 60, normalized size = 1.76 \[ -\frac {i \, {\left (4 i \, e^{\left (-d x - c\right )} + 1\right )} e^{\left (2 \, d x + 2 \, c\right )}}{8 \, a d} - \frac {i \, {\left (-4 i \, e^{\left (-d x - c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )}}{8 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.29, size = 29, normalized size = 0.85 \[ \frac {4\,\mathrm {sinh}\left (c+d\,x\right )-\mathrm {cosh}\left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}}{4\,a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(c + d*x)^3/(a + a*sinh(c + d*x)*1i),x)

[Out]

(4*sinh(c + d*x) - cosh(2*c + 2*d*x)*1i)/(4*a*d)

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sympy [A]  time = 0.35, size = 134, normalized size = 3.94 \[ \begin {cases} \frac {\left (- 32 i a^{3} d^{3} e^{5 c} e^{2 d x} + 128 a^{3} d^{3} e^{4 c} e^{d x} - 128 a^{3} d^{3} e^{2 c} e^{- d x} - 32 i a^{3} d^{3} e^{c} e^{- 2 d x}\right ) e^{- 3 c}}{256 a^{4} d^{4}} & \text {for}\: 256 a^{4} d^{4} e^{3 c} \neq 0 \\\frac {x \left (- i e^{4 c} + 2 e^{3 c} + 2 e^{c} + i\right ) e^{- 2 c}}{4 a} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((-32*I*a**3*d**3*exp(5*c)*exp(2*d*x) + 128*a**3*d**3*exp(4*c)*exp(d*x) - 128*a**3*d**3*exp(2*c)*exp
(-d*x) - 32*I*a**3*d**3*exp(c)*exp(-2*d*x))*exp(-3*c)/(256*a**4*d**4), Ne(256*a**4*d**4*exp(3*c), 0)), (x*(-I*
exp(4*c) + 2*exp(3*c) + 2*exp(c) + I)*exp(-2*c)/(4*a), True))

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