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

Optimal. Leaf size=303 \[ \frac {32 a \sinh ^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{27 f^3}+\frac {224 a \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{9 f^3}-\frac {32 a x \sqrt {a+i a \sinh (e+f x)}}{3 f^2}-\frac {16 a x \cosh ^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{9 f^2}+\frac {8 a x^2 \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{3 f}+\frac {4 a x^2 \sinh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{3 f} \]

[Out]

-32/3*a*x*(a+I*a*sinh(f*x+e))^(1/2)/f^2-16/9*a*x*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^2*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+224/9*a*(a+I*a
*sinh(f*x+e))^(1/2)*tanh(1/2*e+1/4*I*Pi+1/2*f*x)/f^3+8/3*a*x^2*(a+I*a*sinh(f*x+e))^(1/2)*tanh(1/2*e+1/4*I*Pi+1
/2*f*x)/f+32/27*a*sinh(1/2*e+1/4*I*Pi+1/2*f*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

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Rubi [A]  time = 0.25, antiderivative size = 303, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3319, 3311, 3296, 2638, 2633} \[ -\frac {32 a x \sqrt {a+i a \sinh (e+f x)}}{3 f^2}+\frac {32 a \sinh ^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{27 f^3}+\frac {224 a \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{9 f^3}-\frac {16 a x \cosh ^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{9 f^2}+\frac {8 a x^2 \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{3 f}+\frac {4 a x^2 \sinh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-32*a*x*Sqrt[a + I*a*Sinh[e + f*x]])/(3*f^2) - (16*a*x*Cosh[e/2 + (I/4)*Pi + (f*x)/2]^2*Sqrt[a + I*a*Sinh[e +
f*x]])/(9*f^2) + (4*a*x^2*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) + (224*a*Sqrt[a + I*a*Sinh[e + f*x]]*Tanh[e/2 + (I/4)*Pi + (f*x)/2])/(9*f^3) + (8*a*x^2*Sqrt[a +
I*a*Sinh[e + f*x]]*Tanh[e/2 + (I/4)*Pi + (f*x)/2])/(3*f) + (32*a*Sinh[e/2 + (I/4)*Pi + (f*x)/2]^2*Sqrt[a + I*
a*Sinh[e + f*x]]*Tanh[e/2 + (I/4)*Pi + (f*x)/2])/(27*f^3)

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 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^2 (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^2 \sinh ^3\left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right ) \, dx\right )\\ &=-\frac {16 a x \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^2 \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^2 \sinh \left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right ) \, dx-\frac {\left (16 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 \sinh ^3\left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right ) \, dx}{9 f^2}\\ &=-\frac {16 a x \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^2 \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^2 \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 (32 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 ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cosh \left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right )\right )}{9 f^3}-\frac {\left (16 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 \cosh \left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right ) \, dx}{3 f}\\ &=-\frac {32 a x \sqrt {a+i a \sinh (e+f x)}}{3 f^2}-\frac {16 a x \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^2 \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 {32 a \sqrt {a+i a \sinh (e+f x)} \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{9 f^3}+\frac {8 a x^2 \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 {32 a \sinh ^2\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \sqrt {a+i a \sinh (e+f x)} \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{27 f^3}-\frac {\left (32 i 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^2}\\ &=-\frac {32 a x \sqrt {a+i a \sinh (e+f x)}}{3 f^2}-\frac {16 a x \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^2 \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 {224 a \sqrt {a+i a \sinh (e+f x)} \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{9 f^3}+\frac {8 a x^2 \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 {32 a \sinh ^2\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \sqrt {a+i a \sinh (e+f x)} \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{27 f^3}\\ \end {align*}

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Mathematica [A]  time = 1.06, size = 173, normalized size = 0.57 \[ -\frac {a (\sinh (e+f x)-i) \sqrt {a+i a \sinh (e+f x)} \left (81 \left (f^2 x^2+4 i f x+8\right ) \cosh \left (\frac {1}{2} (e+f x)\right )+\left (9 f^2 x^2-12 i f x+8\right ) \cosh \left (\frac {3}{2} (e+f x)\right )+2 i \sinh \left (\frac {1}{2} (e+f x)\right ) \left (\left (9 f^2 x^2+12 i f x+8\right ) \cosh (e+f x)-4 \left (9 f^2 x^2-42 i f x+80\right )\right )\right )}{27 f^3 \left (\cosh \left (\frac {1}{2} (e+f x)\right )+i \sinh \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/27*(a*(81*(8 + (4*I)*f*x + f^2*x^2)*Cosh[(e + f*x)/2] + (8 - (12*I)*f*x + 9*f^2*x^2)*Cosh[(3*(e + f*x))/2]
+ (2*I)*(-4*(80 - (42*I)*f*x + 9*f^2*x^2) + (8 + (12*I)*f*x + 9*f^2*x^2)*Cosh[e + f*x])*Sinh[(e + f*x)/2])*(-I
+ Sinh[e + f*x])*Sqrt[a + I*a*Sinh[e + f*x]])/(f^3*(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^2*(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 \[ \int {\left (i \, a \sinh \left (f x + e\right ) + a\right )}^{\frac {3}{2}} x^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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