3.2.19 \(\int x^3 \sqrt {a+i a \sinh (e+f x)} \, dx\) [119]

Optimal. Leaf size=136 \[ -\frac {96 \sqrt {a+i a \sinh (e+f x)}}{f^4}-\frac {12 x^2 \sqrt {a+i a \sinh (e+f x)}}{f^2}+\frac {48 x \sqrt {a+i a \sinh (e+f x)} \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{f^3}+\frac {2 x^3 \sqrt {a+i a \sinh (e+f x)} \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{f} \]

[Out]

-96*(a+I*a*sinh(f*x+e))^(1/2)/f^4-12*x^2*(a+I*a*sinh(f*x+e))^(1/2)/f^2+48*x*(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^3*(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 = 0.12, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3400, 3377, 2718} \begin {gather*} -\frac {96 \sqrt {a+i a \sinh (e+f x)}}{f^4}+\frac {48 x \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 {12 x^2 \sqrt {a+i a \sinh (e+f x)}}{f^2}+\frac {2 x^3 \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*Sqrt[a + I*a*Sinh[e + f*x]],x]

[Out]

(-96*Sqrt[a + I*a*Sinh[e + f*x]])/f^4 - (12*x^2*Sqrt[a + I*a*Sinh[e + f*x]])/f^2 + (48*x*Sqrt[a + I*a*Sinh[e +
f*x]]*Tanh[e/2 + (I/4)*Pi + (f*x)/2])/f^3 + (2*x^3*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*} \int x^3 \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^3 \sinh \left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right ) \, dx\\ &=\frac {2 x^3 \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 (6 \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}\\ &=-\frac {12 x^2 \sqrt {a+i a \sinh (e+f x)}}{f^2}+\frac {2 x^3 \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 (24 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 \cosh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \, dx}{f^2}\\ &=-\frac {12 x^2 \sqrt {a+i a \sinh (e+f x)}}{f^2}+\frac {48 x \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^3 \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 \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^3}\\ &=-\frac {96 \sqrt {a+i a \sinh (e+f x)}}{f^4}-\frac {12 x^2 \sqrt {a+i a \sinh (e+f x)}}{f^2}+\frac {48 x \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^3 \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 = 0.17, size = 125, normalized size = 0.92 \begin {gather*} \frac {2 \left (i \left (48 i+24 f x+6 i f^2 x^2+f^3 x^3\right ) \cosh \left (\frac {1}{2} (e+f x)\right )+\left (-48 i+24 f x-6 i f^2 x^2+f^3 x^3\right ) \sinh \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a+i a \sinh (e+f x)}}{f^4 \left (\cosh \left (\frac {1}{2} (e+f x)\right )+i \sinh \left (\frac {1}{2} (e+f x)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*Sqrt[a + I*a*Sinh[e + f*x]],x]

[Out]

(2*(I*(48*I + 24*f*x + (6*I)*f^2*x^2 + f^3*x^3)*Cosh[(e + f*x)/2] + (-48*I + 24*f*x - (6*I)*f^2*x^2 + f^3*x^3)
*Sinh[(e + f*x)/2])*Sqrt[a + I*a*Sinh[e + f*x]])/(f^4*(Cosh[(e + f*x)/2] + I*Sinh[(e + f*x)/2]))

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Maple [A]
time = 0.24, size = 151, normalized size = 1.11

method	result	size
risch	\(\frac {i \sqrt {2}\, \sqrt {a \left (i {\mathrm e}^{2 f x +2 e}+2 \,{\mathrm e}^{f x +e}-i\right ) {\mathrm e}^{-f x -e}}\, \left (i x^{3} f^{3}+f^{3} x^{3} {\mathrm e}^{f x +e}+6 i x^{2} f^{2}-6 f^{2} x^{2} {\mathrm e}^{f x +e}+24 i x f +24 f x \,{\mathrm e}^{f x +e}+48 i-48 \,{\mathrm e}^{f x +e}\right ) \left ({\mathrm e}^{f x +e}-i\right )}{\left (i {\mathrm e}^{2 f x +2 e}+2 \,{\mathrm e}^{f x +e}-i\right ) f^{4}}\)	\(151\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(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)+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^3*f^3
+f^3*x^3*exp(f*x+e)+6*I*x^2*f^2-6*f^2*x^2*exp(f*x+e)+24*I*x*f+24*f*x*exp(f*x+e)+48*I-48*exp(f*x+e))*(exp(f*x+e
)-I)/f^4

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+I*a*sinh(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(I*a*sinh(f*x + e) + a)*x^3, x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \sqrt {i a \left (\sinh {\left (e + f x \right )} - i\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+I*a*sinh(f*x+e))**(1/2),x)

[Out]

Integral(x**3*sqrt(I*a*(sinh(e + f*x) - I)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+I*a*sinh(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(I*a*sinh(f*x + e) + a)*x^3, x)

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Mupad [B]
time = 0.36, size = 126, normalized size = 0.93 \begin {gather*} \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 (f^3\,x^3\,{\mathrm {e}}^{e+f\,x}+f\,x\,24{}\mathrm {i}+f^2\,x^2\,6{}\mathrm {i}+f^3\,x^3\,1{}\mathrm {i}-6\,f^2\,x^2\,{\mathrm {e}}^{e+f\,x}-48\,{\mathrm {e}}^{e+f\,x}+24\,f\,x\,{\mathrm {e}}^{e+f\,x}+48{}\mathrm {i}\right )}{f^4\,\left ({\mathrm {e}}^{2\,e+2\,f\,x}+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(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)*(f*x*24i - 48*exp(e + f*x) + f^
2*x^2*6i + f^3*x^3*1i - 6*f^2*x^2*exp(e + f*x) + f^3*x^3*exp(e + f*x) + 24*f*x*exp(e + f*x) + 48i))/(f^4*(exp(
2*e + 2*f*x) + 1))

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