3.118 \(\int x^4 \sqrt {a+i a \sinh (e+f x)} \, dx\)

Optimal. Leaf size=181 \[ \frac {768 \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{f^5}-\frac {384 x \sqrt {a+i a \sinh (e+f x)}}{f^4}+\frac {96 x^2 \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{f^3}-\frac {16 x^3 \sqrt {a+i a \sinh (e+f x)}}{f^2}+\frac {2 x^4 \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{f} \]

[Out]

-384*x*(a+I*a*sinh(f*x+e))^(1/2)/f^4-16*x^3*(a+I*a*sinh(f*x+e))^(1/2)/f^2+768*(a+I*a*sinh(f*x+e))^(1/2)*tanh(1
/2*e+1/4*I*Pi+1/2*f*x)/f^5+96*x^2*(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^4*(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 = 0.21, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3319, 3296, 2638} \[ -\frac {16 x^3 \sqrt {a+i a \sinh (e+f x)}}{f^2}+\frac {96 x^2 \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{f^3}-\frac {384 x \sqrt {a+i a \sinh (e+f x)}}{f^4}+\frac {768 \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{f^5}+\frac {2 x^4 \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

Int[x^4*Sqrt[a + I*a*Sinh[e + f*x]],x]

[Out]

(-384*x*Sqrt[a + I*a*Sinh[e + f*x]])/f^4 - (16*x^3*Sqrt[a + I*a*Sinh[e + f*x]])/f^2 + (768*Sqrt[a + I*a*Sinh[e
+ f*x]]*Tanh[e/2 + (I/4)*Pi + (f*x)/2])/f^5 + (96*x^2*Sqrt[a + I*a*Sinh[e + f*x]]*Tanh[e/2 + (I/4)*Pi + (f*x)
/2])/f^3 + (2*x^4*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^4 \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^4 \sinh \left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right ) \, dx\\ &=\frac {2 x^4 \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 \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^3 \cosh \left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right ) \, dx}{f}\\ &=-\frac {16 x^3 \sqrt {a+i a \sinh (e+f x)}}{f^2}+\frac {2 x^4 \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 (48 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 x^2 \cosh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \, dx}{f^2}\\ &=-\frac {16 x^3 \sqrt {a+i a \sinh (e+f x)}}{f^2}+\frac {96 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^3}+\frac {2 x^4 \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 (192 \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^3}\\ &=-\frac {384 x \sqrt {a+i a \sinh (e+f x)}}{f^4}-\frac {16 x^3 \sqrt {a+i a \sinh (e+f x)}}{f^2}+\frac {96 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^3}+\frac {2 x^4 \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 (384 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^4}\\ &=-\frac {384 x \sqrt {a+i a \sinh (e+f x)}}{f^4}-\frac {16 x^3 \sqrt {a+i a \sinh (e+f x)}}{f^2}+\frac {768 \sqrt {a+i a \sinh (e+f x)} \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{f^5}+\frac {96 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^3}+\frac {2 x^4 \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 = 141, normalized size = 0.78 \[ \frac {2 \sqrt {a+i a \sinh (e+f x)} \left (\left (f^4 x^4-8 i f^3 x^3+48 f^2 x^2-192 i f x+384\right ) \sinh \left (\frac {1}{2} (e+f x)\right )+i \left (f^4 x^4+8 i f^3 x^3+48 f^2 x^2+192 i f x+384\right ) \cosh \left (\frac {1}{2} (e+f x)\right )\right )}{f^5 \left (\cosh \left (\frac {1}{2} (e+f x)\right )+i \sinh \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*Sqrt[a + I*a*Sinh[e + f*x]],x]

[Out]

(2*(I*(384 + (192*I)*f*x + 48*f^2*x^2 + (8*I)*f^3*x^3 + f^4*x^4)*Cosh[(e + f*x)/2] + (384 - (192*I)*f*x + 48*f
^2*x^2 - (8*I)*f^3*x^3 + f^4*x^4)*Sinh[(e + f*x)/2])*Sqrt[a + I*a*Sinh[e + f*x]])/(f^5*(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^4*(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 \[ \int \sqrt {i \, a \sinh \left (f x + e\right ) + a} x^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+I*a*sinh(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(I*a*sinh(f*x + e) + a)*x^4, x)

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maple [A]  time = 0.10, size = 174, normalized size = 0.96 \[ \frac {i \sqrt {2}\, \sqrt {a \left (i {\mathrm e}^{2 f x +2 e}+2 \,{\mathrm e}^{f x +e}-i\right ) {\mathrm e}^{-f x -e}}\, \left (i x^{4} f^{4}+f^{4} x^{4} {\mathrm e}^{f x +e}+8 i x^{3} f^{3}-8 f^{3} x^{3} {\mathrm e}^{f x +e}+48 i x^{2} f^{2}+48 f^{2} x^{2} {\mathrm e}^{f x +e}+192 i x f -192 f x \,{\mathrm e}^{f x +e}+384 i+384 \,{\mathrm e}^{f x +e}\right ) \left ({\mathrm e}^{f x +e}-i\right )}{\left (i {\mathrm e}^{2 f x +2 e}+2 \,{\mathrm e}^{f x +e}-i\right ) f^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(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^4*f^4
+f^4*x^4*exp(f*x+e)+8*I*x^3*f^3-8*f^3*x^3*exp(f*x+e)+48*I*x^2*f^2+48*f^2*x^2*exp(f*x+e)+192*I*x*f-192*f*x*exp(
f*x+e)+384*I+384*exp(f*x+e))*(exp(f*x+e)-I)/f^5

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {i \, a \sinh \left (f x + e\right ) + a} x^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+I*a*sinh(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(I*a*sinh(f*x + e) + a)*x^4, x)

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mupad [B]  time = 0.72, size = 149, normalized size = 0.82 \[ \frac {\sqrt {2}\,\left ({\mathrm {e}}^{e+f\,x}+1{}\mathrm {i}\right )\,\sqrt {a\,{\mathrm {e}}^{-e-f\,x}\,{\left ({\mathrm {e}}^{e+f\,x}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}\,\left (384\,{\mathrm {e}}^{e+f\,x}+f\,x\,192{}\mathrm {i}+f^2\,x^2\,48{}\mathrm {i}+f^3\,x^3\,8{}\mathrm {i}+f^4\,x^4\,1{}\mathrm {i}+48\,f^2\,x^2\,{\mathrm {e}}^{e+f\,x}-8\,f^3\,x^3\,{\mathrm {e}}^{e+f\,x}+f^4\,x^4\,{\mathrm {e}}^{e+f\,x}-192\,f\,x\,{\mathrm {e}}^{e+f\,x}+384{}\mathrm {i}\right )}{f^5\,\left ({\mathrm {e}}^{2\,e+2\,f\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a + a*sinh(e + f*x)*1i)^(1/2),x)

[Out]

(2^(1/2)*(exp(e + f*x) + 1i)*(a*exp(- e - f*x)*(exp(e + f*x) - 1i)^2*1i)^(1/2)*(384*exp(e + f*x) + f*x*192i +
f^2*x^2*48i + f^3*x^3*8i + f^4*x^4*1i + 48*f^2*x^2*exp(e + f*x) - 8*f^3*x^3*exp(e + f*x) + f^4*x^4*exp(e + f*x
) - 192*f*x*exp(e + f*x) + 384i))/(f^5*(exp(2*e + 2*f*x) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{4} \sqrt {i a \left (\sinh {\left (e + f x \right )} - i\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+I*a*sinh(f*x+e))**(1/2),x)

[Out]

Integral(x**4*sqrt(I*a*(sinh(e + f*x) - I)), x)

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