3.92 \(\int (\frac {x^2}{\csc ^{\frac {3}{2}}(e+f x)}-\frac {1}{3} x^2 \sqrt {\csc (e+f x)}) \, dx\)

Optimal. Leaf size=111 \[ \frac {16 \cos (e+f x)}{27 f^3 \sqrt {\csc (e+f x)}}-\frac {16 \sqrt {\sin (e+f x)} \sqrt {\csc (e+f x)} F\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right )}{27 f^3}+\frac {8 x}{9 f^2 \csc ^{\frac {3}{2}}(e+f x)}-\frac {2 x^2 \cos (e+f x)}{3 f \sqrt {\csc (e+f x)}} \]

[Out]

8/9*x/f^2/csc(f*x+e)^(3/2)+16/27*cos(f*x+e)/f^3/csc(f*x+e)^(1/2)-2/3*x^2*cos(f*x+e)/f/csc(f*x+e)^(1/2)+16/27*(
sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f*x)*EllipticF(cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2))*csc(
f*x+e)^(1/2)*sin(f*x+e)^(1/2)/f^3

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Rubi [A]  time = 0.21, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {4188, 4189, 3769, 3771, 2641} \[ \frac {8 x}{9 f^2 \csc ^{\frac {3}{2}}(e+f x)}+\frac {16 \cos (e+f x)}{27 f^3 \sqrt {\csc (e+f x)}}-\frac {16 \sqrt {\sin (e+f x)} \sqrt {\csc (e+f x)} F\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right )}{27 f^3}-\frac {2 x^2 \cos (e+f x)}{3 f \sqrt {\csc (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/Csc[e + f*x]^(3/2) - (x^2*Sqrt[Csc[e + f*x]])/3,x]

[Out]

(8*x)/(9*f^2*Csc[e + f*x]^(3/2)) + (16*Cos[e + f*x])/(27*f^3*Sqrt[Csc[e + f*x]]) - (2*x^2*Cos[e + f*x])/(3*f*S
qrt[Csc[e + f*x]]) - (16*Sqrt[Csc[e + f*x]]*EllipticF[(e - Pi/2 + f*x)/2, 2]*Sqrt[Sin[e + f*x]])/(27*f^3)

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 4188

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

Rule 4189

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dist[(b*Sin[e + f*x])^n*(b*C
sc[e + f*x])^n, Int[(c + d*x)^m/(b*Sin[e + f*x])^n, x], x] /; FreeQ[{b, c, d, e, f, m, n}, x] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int \left (\frac {x^2}{\csc ^{\frac {3}{2}}(e+f x)}-\frac {1}{3} x^2 \sqrt {\csc (e+f x)}\right ) \, dx &=-\left (\frac {1}{3} \int x^2 \sqrt {\csc (e+f x)} \, dx\right )+\int \frac {x^2}{\csc ^{\frac {3}{2}}(e+f x)} \, dx\\ &=\frac {8 x}{9 f^2 \csc ^{\frac {3}{2}}(e+f x)}-\frac {2 x^2 \cos (e+f x)}{3 f \sqrt {\csc (e+f x)}}+\frac {1}{3} \int x^2 \sqrt {\csc (e+f x)} \, dx-\frac {8 \int \frac {1}{\csc ^{\frac {3}{2}}(e+f x)} \, dx}{9 f^2}-\frac {1}{3} \left (\sqrt {\csc (e+f x)} \sqrt {\sin (e+f x)}\right ) \int \frac {x^2}{\sqrt {\sin (e+f x)}} \, dx\\ &=\frac {8 x}{9 f^2 \csc ^{\frac {3}{2}}(e+f x)}+\frac {16 \cos (e+f x)}{27 f^3 \sqrt {\csc (e+f x)}}-\frac {2 x^2 \cos (e+f x)}{3 f \sqrt {\csc (e+f x)}}-\frac {8 \int \sqrt {\csc (e+f x)} \, dx}{27 f^2}\\ &=\frac {8 x}{9 f^2 \csc ^{\frac {3}{2}}(e+f x)}+\frac {16 \cos (e+f x)}{27 f^3 \sqrt {\csc (e+f x)}}-\frac {2 x^2 \cos (e+f x)}{3 f \sqrt {\csc (e+f x)}}-\frac {\left (8 \sqrt {\csc (e+f x)} \sqrt {\sin (e+f x)}\right ) \int \frac {1}{\sqrt {\sin (e+f x)}} \, dx}{27 f^2}\\ &=\frac {8 x}{9 f^2 \csc ^{\frac {3}{2}}(e+f x)}+\frac {16 \cos (e+f x)}{27 f^3 \sqrt {\csc (e+f x)}}-\frac {2 x^2 \cos (e+f x)}{3 f \sqrt {\csc (e+f x)}}-\frac {16 \sqrt {\csc (e+f x)} F\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {\sin (e+f x)}}{27 f^3}\\ \end {align*}

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Mathematica [A]  time = 0.59, size = 87, normalized size = 0.78 \[ -\frac {\sqrt {\csc (e+f x)} \left (9 f^2 x^2 \sin (2 (e+f x))-8 \sin (2 (e+f x))+12 f x \cos (2 (e+f x))-16 \sqrt {\sin (e+f x)} F\left (\left .\frac {1}{4} (-2 e-2 f x+\pi )\right |2\right )-12 f x\right )}{27 f^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/Csc[e + f*x]^(3/2) - (x^2*Sqrt[Csc[e + f*x]])/3,x]

[Out]

-1/27*(Sqrt[Csc[e + f*x]]*(-12*f*x + 12*f*x*Cos[2*(e + f*x)] - 16*EllipticF[(-2*e + Pi - 2*f*x)/4, 2]*Sqrt[Sin
[e + f*x]] - 8*Sin[2*(e + f*x)] + 9*f^2*x^2*Sin[2*(e + f*x)]))/f^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/csc(f*x+e)^(3/2)-1/3*x^2*csc(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 \[ \int -\frac {1}{3} \, x^{2} \sqrt {\csc \left (f x + e\right )} + \frac {x^{2}}{\csc \left (f x + e\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/csc(f*x+e)^(3/2)-1/3*x^2*csc(f*x+e)^(1/2),x, algorithm="giac")

[Out]

integrate(-1/3*x^2*sqrt(csc(f*x + e)) + x^2/csc(f*x + e)^(3/2), x)

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maple [F]  time = 0.22, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\csc \left (f x +e \right )^{\frac {3}{2}}}-\frac {x^{2} \left (\sqrt {\csc }\left (f x +e \right )\right )}{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/csc(f*x+e)^(3/2)-1/3*x^2*csc(f*x+e)^(1/2),x)

[Out]

int(x^2/csc(f*x+e)^(3/2)-1/3*x^2*csc(f*x+e)^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {1}{3} \, x^{2} \sqrt {\csc \left (f x + e\right )} + \frac {x^{2}}{\csc \left (f x + e\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/csc(f*x+e)^(3/2)-1/3*x^2*csc(f*x+e)^(1/2),x, algorithm="maxima")

[Out]

integrate(-1/3*x^2*sqrt(csc(f*x + e)) + x^2/csc(f*x + e)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(1/sin(e + f*x))^(3/2) - (x^2*(1/sin(e + f*x))^(1/2))/3,x)

[Out]

int(x^2/(1/sin(e + f*x))^(3/2) - (x^2*(1/sin(e + f*x))^(1/2))/3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/csc(f*x+e)**(3/2)-1/3*x**2*csc(f*x+e)**(1/2),x)

[Out]

-(Integral(-3*x**2/csc(e + f*x)**(3/2), x) + Integral(x**2*sqrt(csc(e + f*x)), x))/3

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