\(\int x^2 \sqrt {a+i a \sinh (e+f x)} \, dx\) [120]

Optimal result

Integrand size = 21, antiderivative size = 111 \[ \int x^2 \sqrt {a+i a \sinh (e+f x)} \, dx=-\frac {8 x \sqrt {a+i a \sinh (e+f x)}}{f^2}+\frac {16 \sqrt {a+i a \sinh (e+f x)} \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{f^3}+\frac {2 x^2 \sqrt {a+i a \sinh (e+f x)} \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{f} \]

[Out]

-8*x*(a+I*a*sinh(f*x+e))^(1/2)/f^2+16*(a+I*a*sinh(f*x+e))^(1/2)*tanh(1/2*e+1/4*I*Pi+1/2*f*x)/f^3+2*x^2*(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 = 0.11 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3400, 3377, 2718} \[ \int x^2 \sqrt {a+i a \sinh (e+f x)} \, dx=\frac {16 \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{f^3}-\frac {8 x \sqrt {a+i a \sinh (e+f x)}}{f^2}+\frac {2 x^2 \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{f} \]

[In]

Int[x^2*Sqrt[a + I*a*Sinh[e + f*x]],x]

[Out]

(-8*x*Sqrt[a + I*a*Sinh[e + f*x]])/f^2 + (16*Sqrt[a + I*a*Sinh[e + f*x]]*Tanh[e/2 + (I/4)*Pi + (f*x)/2])/f^3 +
(2*x^2*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^2 \sinh \left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right ) \, dx \\ & = \frac {2 x^2 \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 (4 \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 \cosh \left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right ) \, dx}{f} \\ & = -\frac {8 x \sqrt {a+i a \sinh (e+f x)}}{f^2}+\frac {2 x^2 \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 (8 i \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^2} \\ & = -\frac {8 x \sqrt {a+i a \sinh (e+f x)}}{f^2}+\frac {16 \sqrt {a+i a \sinh (e+f x)} \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{f^3}+\frac {2 x^2 \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 = 0.76 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.95 \[ \int x^2 \sqrt {a+i a \sinh (e+f x)} \, dx=\frac {2 \left (i \left (8+4 i f x+f^2 x^2\right ) \cosh \left (\frac {1}{2} (e+f x)\right )+\left (8-4 i f x+f^2 x^2\right ) \sinh \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a+i a \sinh (e+f x)}}{f^3 \left (\cosh \left (\frac {1}{2} (e+f x)\right )+i \sinh \left (\frac {1}{2} (e+f x)\right )\right )} \]

[In]

Integrate[x^2*Sqrt[a + I*a*Sinh[e + f*x]],x]

[Out]

(2*(I*(8 + (4*I)*f*x + f^2*x^2)*Cosh[(e + f*x)/2] + (8 - (4*I)*f*x + f^2*x^2)*Sinh[(e + f*x)/2])*Sqrt[a + I*a*
Sinh[e + f*x]])/(f^3*(Cosh[(e + f*x)/2] + I*Sinh[(e + f*x)/2]))

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.15

method	result	size
risch	\(\frac {i \sqrt {2}\, \sqrt {a \left (i {\mathrm e}^{2 f x +2 e}-i+2 \,{\mathrm e}^{f x +e}\right ) {\mathrm e}^{-f x -e}}\, \left (i x^{2} f^{2}+f^{2} x^{2} {\mathrm e}^{f x +e}+4 i x f -4 f x \,{\mathrm e}^{f x +e}+8 i+8 \,{\mathrm e}^{f x +e}\right ) \left ({\mathrm e}^{f x +e}-i\right )}{\left (i {\mathrm e}^{2 f x +2 e}-i+2 \,{\mathrm e}^{f x +e}\right ) f^{3}}\)	\(128\)

[In]

int(x^2*(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*x^2*f^2
+f^2*x^2*exp(f*x+e)+4*I*x*f-4*f*x*exp(f*x+e)+8*I+8*exp(f*x+e))*(exp(f*x+e)-I)/f^3

Fricas [F(-2)]

Exception generated. \[ \int x^2 \sqrt {a+i a \sinh (e+f x)} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x^2*(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]

\[ \int x^2 \sqrt {a+i a \sinh (e+f x)} \, dx=\int x^{2} \sqrt {i a \left (\sinh {\left (e + f x \right )} - i\right )}\, dx \]

[In]

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

[Out]

Integral(x**2*sqrt(I*a*(sinh(e + f*x) - I)), x)

Maxima [F]

\[ \int x^2 \sqrt {a+i a \sinh (e+f x)} \, dx=\int { \sqrt {i \, a \sinh \left (f x + e\right ) + a} x^{2} \,d x } \]

[In]

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

[Out]

integrate(sqrt(I*a*sinh(f*x + e) + a)*x^2, x)

Giac [F]

\[ \int x^2 \sqrt {a+i a \sinh (e+f x)} \, dx=\int { \sqrt {i \, a \sinh \left (f x + e\right ) + a} x^{2} \,d x } \]

[In]

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

[Out]

integrate(sqrt(I*a*sinh(f*x + e) + a)*x^2, x)

Mupad [B] (verification not implemented)

Time = 1.07 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.83 \[ \int x^2 \sqrt {a+i a \sinh (e+f x)} \, dx=\frac {\sqrt {2}\,\sqrt {a\,{\mathrm {e}}^{-e-f\,x}\,{\left ({\mathrm {e}}^{e+f\,x}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}\,\left (8\,{\mathrm {e}}^{e+f\,x}+f\,x\,4{}\mathrm {i}+f^2\,x^2\,1{}\mathrm {i}+f^2\,x^2\,{\mathrm {e}}^{e+f\,x}-4\,f\,x\,{\mathrm {e}}^{e+f\,x}+8{}\mathrm {i}\right )}{f^3\,\left ({\mathrm {e}}^{e+f\,x}-\mathrm {i}\right )} \]

[In]

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

[Out]

(2^(1/2)*(a*exp(- e - f*x)*(exp(e + f*x) - 1i)^2*1i)^(1/2)*(8*exp(e + f*x) + f*x*4i + f^2*x^2*1i + f^2*x^2*exp
(e + f*x) - 4*f*x*exp(e + f*x) + 8i))/(f^3*(exp(e + f*x) - 1i))