\(\int (\frac {x^2}{\sin ^{\frac {3}{2}}(e+f x)}+x^2 \sqrt {\sin (e+f x)}) \, dx\) [68]

Optimal result

Integrand size = 29, antiderivative size = 62 \[ \int \left (\frac {x^2}{\sin ^{\frac {3}{2}}(e+f x)}+x^2 \sqrt {\sin (e+f x)}\right ) \, dx=-\frac {16 E\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right )}{f^3}-\frac {2 x^2 \cos (e+f x)}{f \sqrt {\sin (e+f x)}}+\frac {8 x \sqrt {\sin (e+f x)}}{f^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {3397, 2719} \[ \int \left (\frac {x^2}{\sin ^{\frac {3}{2}}(e+f x)}+x^2 \sqrt {\sin (e+f x)}\right ) \, dx=-\frac {16 E\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right )}{f^3}+\frac {8 x \sqrt {\sin (e+f x)}}{f^2}-\frac {2 x^2 \cos (e+f x)}{f \sqrt {\sin (e+f x)}} \]

[In]

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

[Out]

(-16*EllipticE[(e - Pi/2 + f*x)/2, 2])/f^3 - (2*x^2*Cos[e + f*x])/(f*Sqrt[Sin[e + f*x]]) + (8*x*Sqrt[Sin[e + f
*x]])/f^2

Rule 2719

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

Rule 3397

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2}{\sin ^{\frac {3}{2}}(e+f x)} \, dx+\int x^2 \sqrt {\sin (e+f x)} \, dx \\ & = -\frac {2 x^2 \cos (e+f x)}{f \sqrt {\sin (e+f x)}}+\frac {8 x \sqrt {\sin (e+f x)}}{f^2}-\frac {8 \int \sqrt {\sin (e+f x)} \, dx}{f^2} \\ & = -\frac {16 E\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right )}{f^3}-\frac {2 x^2 \cos (e+f x)}{f \sqrt {\sin (e+f x)}}+\frac {8 x \sqrt {\sin (e+f x)}}{f^2} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 2.44 (sec) , antiderivative size = 163, normalized size of antiderivative = 2.63 \[ \int \left (\frac {x^2}{\sin ^{\frac {3}{2}}(e+f x)}+x^2 \sqrt {\sin (e+f x)}\right ) \, dx=\frac {-\left (\left (8+f^2 x^2\right ) \cos (f x) \sec (e)\right )-\left (-8+f^2 x^2\right ) \cos (2 e+f x) \sec (e)+8 f x \sin (e+f x)+8 \sqrt {\csc ^2(e)} \csc (f x-\arctan (\cot (e))) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(f x-\arctan (\cot (e)))\right ) \sin (e) \sqrt {\sin ^2(f x-\arctan (\cot (e)))}+\frac {4 \csc (e) \sec (e) (\sin (e+f x-\arctan (\cot (e)))+3 \sin (e-f x+\arctan (\cot (e))))}{\sqrt {\csc ^2(e)}}}{f^3 \sqrt {\sin (e+f x)}} \]

[In]

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

[Out]

(-((8 + f^2*x^2)*Cos[f*x]*Sec[e]) - (-8 + f^2*x^2)*Cos[2*e + f*x]*Sec[e] + 8*f*x*Sin[e + f*x] + 8*Sqrt[Csc[e]^
2]*Csc[f*x - ArcTan[Cot[e]]]*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[f*x - ArcTan[Cot[e]]]^2]*Sin[e]*Sqrt[S
in[f*x - ArcTan[Cot[e]]]^2] + (4*Csc[e]*Sec[e]*(Sin[e + f*x - ArcTan[Cot[e]]] + 3*Sin[e - f*x + ArcTan[Cot[e]]
]))/Sqrt[Csc[e]^2])/(f^3*Sqrt[Sin[e + f*x]])

Maple [F]

\[\int \left (\frac {x^{2}}{\sin \left (f x +e \right )^{\frac {3}{2}}}+x^{2} \left (\sqrt {\sin }\left (f x +e \right )\right )\right )d x\]

[In]

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

[Out]

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

Fricas [F(-2)]

Exception generated. \[ \int \left (\frac {x^2}{\sin ^{\frac {3}{2}}(e+f x)}+x^2 \sqrt {\sin (e+f x)}\right ) \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x^2/sin(f*x+e)^(3/2)+x^2*sin(f*x+e)^(1/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

Sympy [F]

\[ \int \left (\frac {x^2}{\sin ^{\frac {3}{2}}(e+f x)}+x^2 \sqrt {\sin (e+f x)}\right ) \, dx=\int \frac {x^{2} \left (\sin ^{2}{\left (e + f x \right )} + 1\right )}{\sin ^{\frac {3}{2}}{\left (e + f x \right )}}\, dx \]

[In]

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

[Out]

Integral(x**2*(sin(e + f*x)**2 + 1)/sin(e + f*x)**(3/2), x)

Maxima [F]

\[ \int \left (\frac {x^2}{\sin ^{\frac {3}{2}}(e+f x)}+x^2 \sqrt {\sin (e+f x)}\right ) \, dx=\int { x^{2} \sqrt {\sin \left (f x + e\right )} + \frac {x^{2}}{\sin \left (f x + e\right )^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

integrate(x^2*sqrt(sin(f*x + e)) + x^2/sin(f*x + e)^(3/2), x)

Giac [F]

\[ \int \left (\frac {x^2}{\sin ^{\frac {3}{2}}(e+f x)}+x^2 \sqrt {\sin (e+f x)}\right ) \, dx=\int { x^{2} \sqrt {\sin \left (f x + e\right )} + \frac {x^{2}}{\sin \left (f x + e\right )^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

integrate(x^2*sqrt(sin(f*x + e)) + x^2/sin(f*x + e)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \left (\frac {x^2}{\sin ^{\frac {3}{2}}(e+f x)}+x^2 \sqrt {\sin (e+f x)}\right ) \, dx=\int x^2\,\sqrt {\sin \left (e+f\,x\right )}+\frac {x^2}{{\sin \left (e+f\,x\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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