3.127 \(\int x (a+i a \sinh (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=185 \[ -\frac {16 a \sqrt {a+i a \sinh (e+f x)}}{3 f^2}-\frac {8 a \cosh ^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{9 f^2}+\frac {8 a x \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{3 f}+\frac {4 a x \sinh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{3 f} \]

[Out]

-16/3*a*(a+I*a*sinh(f*x+e))^(1/2)/f^2-8/9*a*cosh(1/2*e+1/4*I*Pi+1/2*f*x)^2*(a+I*a*sinh(f*x+e))^(1/2)/f^2+4/3*a
*x*cosh(1/2*e+1/4*I*Pi+1/2*f*x)*sinh(1/2*e+1/4*I*Pi+1/2*f*x)*(a+I*a*sinh(f*x+e))^(1/2)/f+8/3*a*x*(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 = 0.13, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {3319, 3310, 3296, 2638} \[ -\frac {16 a \sqrt {a+i a \sinh (e+f x)}}{3 f^2}-\frac {8 a \cosh ^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{9 f^2}+\frac {8 a x \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{3 f}+\frac {4 a x \sinh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + I*a*Sinh[e + f*x])^(3/2),x]

[Out]

(-16*a*Sqrt[a + I*a*Sinh[e + f*x]])/(3*f^2) - (8*a*Cosh[e/2 + (I/4)*Pi + (f*x)/2]^2*Sqrt[a + I*a*Sinh[e + f*x]
])/(9*f^2) + (4*a*x*Cosh[e/2 + (I/4)*Pi + (f*x)/2]*Sinh[e/2 + (I/4)*Pi + (f*x)/2]*Sqrt[a + I*a*Sinh[e + f*x]])
/(3*f) + (8*a*x*Sqrt[a + I*a*Sinh[e + f*x]]*Tanh[e/2 + (I/4)*Pi + (f*x)/2])/(3*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 3310

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*(b*Sin[e + f*x])^n)/(f^2*n
^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[(b*(c + d*x)*Cos[e + f*
x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

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 (a+i a \sinh (e+f x))^{3/2} \, dx &=-\left (\left (2 a \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 ^3\left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right ) \, dx\right )\\ &=-\frac {8 a \cosh ^2\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \sqrt {a+i a \sinh (e+f x)}}{9 f^2}+\frac {4 a x \cosh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \sinh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \sqrt {a+i a \sinh (e+f x)}}{3 f}+\frac {1}{3} \left (4 a \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 {8 a \cosh ^2\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \sqrt {a+i a \sinh (e+f x)}}{9 f^2}+\frac {4 a x \cosh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \sinh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \sqrt {a+i a \sinh (e+f x)}}{3 f}+\frac {8 a x \sqrt {a+i a \sinh (e+f x)} \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{3 f}-\frac {\left (8 a \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}{3 f}\\ &=-\frac {16 a \sqrt {a+i a \sinh (e+f x)}}{3 f^2}-\frac {8 a \cosh ^2\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \sqrt {a+i a \sinh (e+f x)}}{9 f^2}+\frac {4 a x \cosh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \sinh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \sqrt {a+i a \sinh (e+f x)}}{3 f}+\frac {8 a x \sqrt {a+i a \sinh (e+f x)} \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{3 f}\\ \end {align*}

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Mathematica [A]  time = 0.72, size = 138, normalized size = 0.75 \[ -\frac {a (\sinh (e+f x)-i) \sqrt {a+i a \sinh (e+f x)} \left (27 (f x+2 i) \cosh \left (\frac {1}{2} (e+f x)\right )+(3 f x-2 i) \cosh \left (\frac {3}{2} (e+f x)\right )+2 i \sinh \left (\frac {1}{2} (e+f x)\right ) ((3 f x+2 i) \cosh (e+f x)-12 f x+28 i)\right )}{9 f^2 \left (\cosh \left (\frac {1}{2} (e+f x)\right )+i \sinh \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + I*a*Sinh[e + f*x])^(3/2),x]

[Out]

-1/9*(a*(27*(2*I + f*x)*Cosh[(e + f*x)/2] + (-2*I + 3*f*x)*Cosh[(3*(e + f*x))/2] + (2*I)*(28*I - 12*f*x + (2*I
+ 3*f*x)*Cosh[e + f*x])*Sinh[(e + f*x)/2])*(-I + Sinh[e + f*x])*Sqrt[a + I*a*Sinh[e + f*x]])/(f^2*(Cosh[(e +
f*x)/2] + I*Sinh[(e + f*x)/2])^3)

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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*(a+I*a*sinh(f*x+e))^(3/2),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 \[ \int {\left (i \, a \sinh \left (f x + e\right ) + a\right )}^{\frac {3}{2}} x\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int x \left (a +i a \sinh \left (f x +e \right )\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+I*a*sinh(f*x+e))^(3/2),x)

[Out]

int(x*(a+I*a*sinh(f*x+e))^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \sinh \left (f x + e\right ) + a\right )}^{\frac {3}{2}} x\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,{\left (a+a\,\mathrm {sinh}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + a*sinh(e + f*x)*1i)^(3/2),x)

[Out]

int(x*(a + a*sinh(e + f*x)*1i)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \left (i a \left (\sinh {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*(I*a*(sinh(e + f*x) - I))**(3/2), x)

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