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*(a+I*a*sinh(f*x+e))^(1/2)/f^2+2*x*(a+I*a*sinh(f*x+e))^(1/2)*tanh(1/2*e+1/4*I*Pi+1/2*f*x)/f

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Rubi [A]  time = 0.08, antiderivative size = 66, normalized size of antiderivative = 1.00, 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 2638

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

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 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])

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*}

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Mathematica [A]  time = 0.16, 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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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {i \, a \sinh \left (f x + e\right ) + a} x\,{d x} \]

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)

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maple [A]  time = 0.06, size = 105, normalized size = 1.59 \[ \frac {i \sqrt {2}\, \sqrt {a \left (i {\mathrm e}^{2 f x +2 e}+2 \,{\mathrm e}^{f x +e}-i\right ) {\mathrm e}^{-f x -e}}\, \left (i x f +f x \,{\mathrm e}^{f x +e}+2 i-2 \,{\mathrm e}^{f x +e}\right ) \left ({\mathrm e}^{f x +e}-i\right )}{\left (i {\mathrm e}^{2 f x +2 e}+2 \,{\mathrm e}^{f x +e}-i\right ) f^{2}} \]

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)+2*exp(f*x+e)-I)*exp(-f*x-e))^(1/2)/(I*exp(2*f*x+2*e)+2*exp(f*x+e)-I)*(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {i \, a \sinh \left (f x + e\right ) + a} x\,{d x} \]

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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mupad [B]  time = 0.27, size = 80, normalized size = 1.21 \[ \frac {\sqrt {2}\,\left ({\mathrm {e}}^{e+f\,x}+1{}\mathrm {i}\right )\,\left (f\,x\,{\mathrm {e}}^{e+f\,x}+f\,x\,1{}\mathrm {i}-2\,{\mathrm {e}}^{e+f\,x}+2{}\mathrm {i}\right )\,\sqrt {a\,{\mathrm {e}}^{-e-f\,x}\,{\left ({\mathrm {e}}^{e+f\,x}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}}{f^2\,\left ({\mathrm {e}}^{2\,e+2\,f\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sqrt {i a \left (\sinh {\left (e + f x \right )} - i\right )}\, dx \]

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(I*a*(sinh(e + f*x) - I)), x)

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