\(\int x \sqrt {a+i a \sinh (e+f x)} \, dx\) [121]

Optimal result

Integrand size = 19, antiderivative size = 66 \[ \int x \sqrt {a+i a \sinh (e+f x)} \, dx=-\frac {4 \sqrt {a+i a \sinh (e+f x)}}{f^2}+\frac {2 x \sqrt {a+i a \sinh (e+f x)} \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{f} \]

[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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3400, 3377, 2718} \[ \int x \sqrt {a+i a \sinh (e+f x)} \, dx=\frac {2 x \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{f}-\frac {4 \sqrt {a+i a \sinh (e+f x)}}{f^2} \]

[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 2718

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

Rule 3377

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 3400

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*} \text {integral}& = \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] (verified)

Time = 0.36 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.32 \[ \int x \sqrt {a+i a \sinh (e+f x)} \, dx=\frac {2 \left ((-2+i f x) \cosh \left (\frac {1}{2} (e+f x)\right )+(-2 i+f x) \sinh \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a+i a \sinh (e+f x)}}{f^2 \left (\cosh \left (\frac {1}{2} (e+f x)\right )+i \sinh \left (\frac {1}{2} (e+f x)\right )\right )} \]

[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]))

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.59

method	result	size
risch	\(\frac {i \sqrt {2}\, \sqrt {a \left (i {\mathrm e}^{2 f x +2 e}-i+2 \,{\mathrm e}^{f x +e}\right ) {\mathrm e}^{-f x -e}}\, \left (i x f +f x \,{\mathrm e}^{f x +e}+2 i-2 \,{\mathrm e}^{f x +e}\right ) \left ({\mathrm e}^{f x +e}-i\right )}{\left (i {\mathrm e}^{2 f x +2 e}-i+2 \,{\mathrm e}^{f x +e}\right ) f^{2}}\)	\(105\)

[In]

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

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

Fricas [F(-2)]

Exception generated. \[ \int x \sqrt {a+i a \sinh (e+f x)} \, dx=\text {Exception raised: TypeError} \]

[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)

Sympy [F]

\[ \int x \sqrt {a+i a \sinh (e+f x)} \, dx=\int x \sqrt {i a \left (\sinh {\left (e + f x \right )} - i\right )}\, dx \]

[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)

Maxima [F]

\[ \int x \sqrt {a+i a \sinh (e+f x)} \, dx=\int { \sqrt {i \, a \sinh \left (f x + e\right ) + a} x \,d x } \]

[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)

Giac [F]

\[ \int x \sqrt {a+i a \sinh (e+f x)} \, dx=\int { \sqrt {i \, a \sinh \left (f x + e\right ) + a} x \,d x } \]

[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)

Mupad [B] (verification not implemented)

Time = 1.06 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.21 \[ \int x \sqrt {a+i a \sinh (e+f x)} \, dx=\frac {\sqrt {2}\,\left ({\mathrm {e}}^{e+f\,x}+1{}\mathrm {i}\right )\,\left (f\,x\,{\mathrm {e}}^{e+f\,x}+f\,x\,1{}\mathrm {i}-2\,{\mathrm {e}}^{e+f\,x}+2{}\mathrm {i}\right )\,\sqrt {a\,{\mathrm {e}}^{-e-f\,x}\,{\left ({\mathrm {e}}^{e+f\,x}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}}{f^2\,\left ({\mathrm {e}}^{2\,e+2\,f\,x}+1\right )} \]

[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))