\(\int \frac {\sin ^{\frac {3}{2}}(x)}{\cos ^{\frac {7}{2}}(x)} \, dx\) [288]

Optimal result

Integrand size = 13, antiderivative size = 16 \[ \int \frac {\sin ^{\frac {3}{2}}(x)}{\cos ^{\frac {7}{2}}(x)} \, dx=\frac {2 \sin ^{\frac {5}{2}}(x)}{5 \cos ^{\frac {5}{2}}(x)} \]

[Out]

2/5*sin(x)^(5/2)/cos(x)^(5/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2643} \[ \int \frac {\sin ^{\frac {3}{2}}(x)}{\cos ^{\frac {7}{2}}(x)} \, dx=\frac {2 \sin ^{\frac {5}{2}}(x)}{5 \cos ^{\frac {5}{2}}(x)} \]

[In]

Int[Sin[x]^(3/2)/Cos[x]^(7/2),x]

[Out]

(2*Sin[x]^(5/2))/(5*Cos[x]^(5/2))

Rule 2643

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Simp[(a*Sin[e +
f*x])^(m + 1)*((b*Cos[e + f*x])^(n + 1)/(a*b*f*(m + 1))), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n + 2,
0] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \sin ^{\frac {5}{2}}(x)}{5 \cos ^{\frac {5}{2}}(x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {\sin ^{\frac {3}{2}}(x)}{\cos ^{\frac {7}{2}}(x)} \, dx=\frac {2 \sin ^{\frac {5}{2}}(x)}{5 \cos ^{\frac {5}{2}}(x)} \]

[In]

Integrate[Sin[x]^(3/2)/Cos[x]^(7/2),x]

[Out]

(2*Sin[x]^(5/2))/(5*Cos[x]^(5/2))

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69

method	result	size
default	\(\frac {2 \left (\sin ^{\frac {5}{2}}\left (x \right )\right )}{5 \cos \left (x \right )^{\frac {5}{2}}}\)	\(11\)

[In]

int(sin(x)^(3/2)/cos(x)^(7/2),x,method=_RETURNVERBOSE)

[Out]

2/5*sin(x)^(5/2)/cos(x)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {\sin ^{\frac {3}{2}}(x)}{\cos ^{\frac {7}{2}}(x)} \, dx=-\frac {2 \, {\left (\cos \left (x\right )^{2} - 1\right )} \sqrt {\sin \left (x\right )}}{5 \, \cos \left (x\right )^{\frac {5}{2}}} \]

[In]

integrate(sin(x)^(3/2)/cos(x)^(7/2),x, algorithm="fricas")

[Out]

-2/5*(cos(x)^2 - 1)*sqrt(sin(x))/cos(x)^(5/2)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^{\frac {3}{2}}(x)}{\cos ^{\frac {7}{2}}(x)} \, dx=\text {Timed out} \]

[In]

integrate(sin(x)**(3/2)/cos(x)**(7/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sin ^{\frac {3}{2}}(x)}{\cos ^{\frac {7}{2}}(x)} \, dx=\int { \frac {\sin \left (x\right )^{\frac {3}{2}}}{\cos \left (x\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate(sin(x)^(3/2)/cos(x)^(7/2),x, algorithm="maxima")

[Out]

integrate(sin(x)^(3/2)/cos(x)^(7/2), x)

Giac [F]

\[ \int \frac {\sin ^{\frac {3}{2}}(x)}{\cos ^{\frac {7}{2}}(x)} \, dx=\int { \frac {\sin \left (x\right )^{\frac {3}{2}}}{\cos \left (x\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate(sin(x)^(3/2)/cos(x)^(7/2),x, algorithm="giac")

[Out]

integrate(sin(x)^(3/2)/cos(x)^(7/2), x)

Mupad [B] (verification not implemented)

Time = 0.65 (sec) , antiderivative size = 53, normalized size of antiderivative = 3.31 \[ \int \frac {\sin ^{\frac {3}{2}}(x)}{\cos ^{\frac {7}{2}}(x)} \, dx=-\frac {8\,\sqrt {2}\,{\mathrm {tan}\left (\frac {x}{2}\right )}^{5/2}\,\sqrt {1-{\mathrm {tan}\left (\frac {x}{2}\right )}^2}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (5\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2-15\right )+15\right )-5} \]

[In]

int(sin(x)^(3/2)/cos(x)^(7/2),x)

[Out]

-(8*2^(1/2)*tan(x/2)^(5/2)*(1 - tan(x/2)^2)^(1/2))/(tan(x/2)^2*(tan(x/2)^2*(5*tan(x/2)^2 - 15) + 15) - 5)