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

Optimal. Leaf size=111 \[ \frac {16 \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{f^3}-\frac {8 x \sqrt {a+i a \sinh (e+f x)}}{f^2}+\frac {2 x^2 \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{f} \]

[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

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Rubi [A]  time = 0.14, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3319, 3296, 2638} \[ -\frac {8 x \sqrt {a+i a \sinh (e+f x)}}{f^2}+\frac {16 \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{f^3}+\frac {2 x^2 \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{f} \]

Antiderivative was successfully verified.

[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 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^2 \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^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*}

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Mathematica [A]  time = 0.22, size = 105, normalized size = 0.95 \[ \frac {2 \sqrt {a+i a \sinh (e+f x)} \left (\left (f^2 x^2-4 i f x+8\right ) \sinh \left (\frac {1}{2} (e+f x)\right )+i \left (f^2 x^2+4 i f x+8\right ) \cosh \left (\frac {1}{2} (e+f x)\right )\right )}{f^3 \left (\cosh \left (\frac {1}{2} (e+f x)\right )+i \sinh \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^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]))

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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^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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {i \, a \sinh \left (f x + e\right ) + a} x^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.07, size = 128, normalized size = 1.15 \[ \frac {i \sqrt {2}\, \sqrt {a \left (i {\mathrm e}^{2 f x +2 e}+2 \,{\mathrm e}^{f x +e}-i\right ) {\mathrm e}^{-f x -e}}\, \left (i x^{2} f^{2}+f^{2} x^{2} {\mathrm e}^{f x +e}+4 i x f -4 f x \,{\mathrm e}^{f x +e}+8 i+8 \,{\mathrm e}^{f x +e}\right ) \left ({\mathrm e}^{f x +e}-i\right )}{\left (i {\mathrm e}^{2 f x +2 e}+2 \,{\mathrm e}^{f x +e}-i\right ) f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(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^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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {i \, a \sinh \left (f x + e\right ) + a} x^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.29, size = 92, normalized size = 0.83 \[ \frac {\sqrt {2}\,\sqrt {a\,{\mathrm {e}}^{-e-f\,x}\,{\left ({\mathrm {e}}^{e+f\,x}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}\,\left (8\,{\mathrm {e}}^{e+f\,x}+f\,x\,4{}\mathrm {i}+f^2\,x^2\,1{}\mathrm {i}+f^2\,x^2\,{\mathrm {e}}^{e+f\,x}-4\,f\,x\,{\mathrm {e}}^{e+f\,x}+8{}\mathrm {i}\right )}{f^3\,\left ({\mathrm {e}}^{e+f\,x}-\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \sqrt {i a \left (\sinh {\left (e + f x \right )} - i\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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