\(\int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{5/2}} \, dx\) [264]

Optimal result

Integrand size = 25, antiderivative size = 37 \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{5/2}} \, dx=\frac {2 (c \sin (a+b x))^{3/2}}{3 b c d (d \cos (a+b x))^{3/2}} \]

[Out]

2/3*(c*sin(b*x+a))^(3/2)/b/c/d/(d*cos(b*x+a))^(3/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2643} \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{5/2}} \, dx=\frac {2 (c \sin (a+b x))^{3/2}}{3 b c d (d \cos (a+b x))^{3/2}} \]

[In]

Int[Sqrt[c*Sin[a + b*x]]/(d*Cos[a + b*x])^(5/2),x]

[Out]

(2*(c*Sin[a + b*x])^(3/2))/(3*b*c*d*(d*Cos[a + b*x])^(3/2))

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 (c \sin (a+b x))^{3/2}}{3 b c d (d \cos (a+b x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{5/2}} \, dx=\frac {2 (c \sin (a+b x))^{3/2}}{3 b c d (d \cos (a+b x))^{3/2}} \]

[In]

Integrate[Sqrt[c*Sin[a + b*x]]/(d*Cos[a + b*x])^(5/2),x]

[Out]

(2*(c*Sin[a + b*x])^(3/2))/(3*b*c*d*(d*Cos[a + b*x])^(3/2))

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.95

method	result	size
default	\(\frac {2 \sqrt {c \sin \left (b x +a \right )}\, \tan \left (b x +a \right )}{3 b \,d^{2} \sqrt {d \cos \left (b x +a \right )}}\)	\(35\)

[In]

int((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/3/b*(c*sin(b*x+a))^(1/2)/d^2/(d*cos(b*x+a))^(1/2)*tan(b*x+a)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.14 \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{5/2}} \, dx=\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {c \sin \left (b x + a\right )} \sin \left (b x + a\right )}{3 \, b d^{3} \cos \left (b x + a\right )^{2}} \]

[In]

integrate((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

2/3*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))*sin(b*x + a)/(b*d^3*cos(b*x + a)^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((c*sin(b*x+a))**(1/2)/(d*cos(b*x+a))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{5/2}} \, dx=\int { \frac {\sqrt {c \sin \left (b x + a\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*sin(b*x + a))/(d*cos(b*x + a))^(5/2), x)

Giac [F]

\[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{5/2}} \, dx=\int { \frac {\sqrt {c \sin \left (b x + a\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(5/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*sin(b*x + a))/(d*cos(b*x + a))^(5/2), x)

Mupad [B] (verification not implemented)

Time = 0.77 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.35 \[ \int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{5/2}} \, dx=\frac {2\,\sin \left (2\,a+2\,b\,x\right )\,\sqrt {c\,\sin \left (a+b\,x\right )}}{3\,b\,d^2\,\left (\cos \left (2\,a+2\,b\,x\right )+1\right )\,\sqrt {d\,\cos \left (a+b\,x\right )}} \]

[In]

int((c*sin(a + b*x))^(1/2)/(d*cos(a + b*x))^(5/2),x)

[Out]

(2*sin(2*a + 2*b*x)*(c*sin(a + b*x))^(1/2))/(3*b*d^2*(cos(2*a + 2*b*x) + 1)*(d*cos(a + b*x))^(1/2))