3.121 \(\int x \sqrt{a+i a \sinh (e+f x)} \, dx\)

Optimal. Leaf size=66 \[ \frac{2 x \tanh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (e+f x)}}{f}-\frac{4 \sqrt{a+i a \sinh (e+f x)}}{f^2} \]

[Out]

(-4*Sqrt[a + I*a*Sinh[e + f*x]])/f^2 + (2*x*Sqrt[a + I*a*Sinh[e + f*x]]*Tanh[e/2 + (I/4)*Pi + (f*x)/2])/f

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Rubi [A]  time = 0.0753615, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {3319, 3296, 2638} \[ \frac{2 x \tanh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right ) \sqrt{a+i a \sinh (e+f x)}}{f}-\frac{4 \sqrt{a+i a \sinh (e+f x)}}{f^2} \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[a + I*a*Sinh[e + f*x]],x]

[Out]

(-4*Sqrt[a + I*a*Sinh[e + f*x]])/f^2 + (2*x*Sqrt[a + I*a*Sinh[e + f*x]]*Tanh[e/2 + (I/4)*Pi + (f*x)/2])/f

Rule 3319

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[((2*a)^IntPart[n
]*(a + b*Sin[e + f*x])^FracPart[n])/Sin[e/2 + (a*Pi)/(4*b) + (f*x)/2]^(2*FracPart[n]), Int[(c + d*x)^m*Sin[e/2
+ (a*Pi)/(4*b) + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[n
+ 1/2] && (GtQ[n, 0] || IGtQ[m, 0])

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2638

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

Rubi steps

\begin{align*} \int x \sqrt{a+i a \sinh (e+f x)} \, dx &=\left (\text{csch}\left (\frac{e}{2}-\frac{i \pi }{4}+\frac{f x}{2}\right ) \sqrt{a+i a \sinh (e+f x)}\right ) \int x \sinh \left (\frac{e}{2}-\frac{i \pi }{4}+\frac{f x}{2}\right ) \, dx\\ &=\frac{2 x \sqrt{a+i a \sinh (e+f x)} \tanh \left (\frac{e}{2}+\frac{i \pi }{4}+\frac{f x}{2}\right )}{f}-\frac{\left (2 \text{csch}\left (\frac{e}{2}-\frac{i \pi }{4}+\frac{f x}{2}\right ) \sqrt{a+i a \sinh (e+f x)}\right ) \int \cosh \left (\frac{e}{2}-\frac{i \pi }{4}+\frac{f x}{2}\right ) \, dx}{f}\\ &=-\frac{4 \sqrt{a+i a \sinh (e+f x)}}{f^2}+\frac{2 x \sqrt{a+i a \sinh (e+f x)} \tanh \left (\frac{e}{2}+\frac{i \pi }{4}+\frac{f x}{2}\right )}{f}\\ \end{align*}

Mathematica [A]  time = 0.16385, size = 87, normalized size = 1.32 \[ \frac{2 \sqrt{a+i a \sinh (e+f x)} \left ((f x-2 i) \sinh \left (\frac{1}{2} (e+f x)\right )+(-2+i f x) \cosh \left (\frac{1}{2} (e+f x)\right )\right )}{f^2 \left (\cosh \left (\frac{1}{2} (e+f x)\right )+i \sinh \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[a + I*a*Sinh[e + f*x]],x]

[Out]

(2*((-2 + I*f*x)*Cosh[(e + f*x)/2] + (-2*I + f*x)*Sinh[(e + f*x)/2])*Sqrt[a + I*a*Sinh[e + f*x]])/(f^2*(Cosh[(
e + f*x)/2] + I*Sinh[(e + f*x)/2]))

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Maple [A]  time = 0.057, size = 105, normalized size = 1.6 \begin{align*}{\frac{i\sqrt{2} \left ( ixf+fx{{\rm e}^{fx+e}}+2\,i-2\,{{\rm e}^{fx+e}} \right ) \left ({{\rm e}^{fx+e}}-i \right ) }{ \left ( i{{\rm e}^{2\,fx+2\,e}}-i+2\,{{\rm e}^{fx+e}} \right ){f}^{2}}\sqrt{a \left ( i{{\rm e}^{2\,fx+2\,e}}-i+2\,{{\rm e}^{fx+e}} \right ){{\rm e}^{-fx-e}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+I*a*sinh(f*x+e))^(1/2),x)

[Out]

I*2^(1/2)*(a*(I*exp(2*f*x+2*e)-I+2*exp(f*x+e))*exp(-f*x-e))^(1/2)/(I*exp(2*f*x+2*e)-I+2*exp(f*x+e))*(I*x*f+f*x
*exp(f*x+e)+2*I-2*exp(f*x+e))*(exp(f*x+e)-I)/f^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{i \, a \sinh \left (f x + e\right ) + a} x\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+I*a*sinh(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(I*a*sinh(f*x + e) + a)*x, x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+I*a*sinh(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \sqrt{a \left (i \sinh{\left (e + f x \right )} + 1\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+I*a*sinh(f*x+e))**(1/2),x)

[Out]

Integral(x*sqrt(a*(I*sinh(e + f*x) + 1)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{i \, a \sinh \left (f x + e\right ) + a} x\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+I*a*sinh(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(I*a*sinh(f*x + e) + a)*x, x)