\(\int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^{3/2}} \, dx\) [308]

Optimal result

Integrand size = 27, antiderivative size = 36 \[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {2 (e \cos (c+d x))^{3/2}}{3 d e (a+a \sin (c+d x))^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {2750} \[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {2 (e \cos (c+d x))^{3/2}}{3 d e (a \sin (c+d x)+a)^{3/2}} \]

[In]

Int[Sqrt[e*Cos[c + d*x]]/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

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

Rule 2750

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*m)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && EqQ[Simplify[m + p + 1], 0] &&  !ILtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 (e \cos (c+d x))^{3/2}}{3 d e (a+a \sin (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.36 \[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {2 (e \cos (c+d x))^{3/2} \sqrt {a (1+\sin (c+d x))}}{3 a^2 d e (1+\sin (c+d x))^2} \]

[In]

Integrate[Sqrt[e*Cos[c + d*x]]/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-2*(e*Cos[c + d*x])^(3/2)*Sqrt[a*(1 + Sin[c + d*x])])/(3*a^2*d*e*(1 + Sin[c + d*x])^2)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(62\) vs. \(2(30)=60\).

Time = 2.72 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.75

method	result	size
default	\(\frac {2 \left (-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1\right ) \sqrt {e \cos \left (d x +c \right )}}{3 d \left (1+\cos \left (d x +c \right )+\sin \left (d x +c \right )\right ) a \sqrt {a \left (1+\sin \left (d x +c \right )\right )}}\)	\(63\)

[In]

int((e*cos(d*x+c))^(1/2)/(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (30) = 60\).

Time = 0.29 (sec) , antiderivative size = 100, normalized size of antiderivative = 2.78 \[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {2 \, \sqrt {e \cos \left (d x + c\right )} \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d \cos \left (d x + c\right ) - 2 \, a^{2} d - {\left (a^{2} d \cos \left (d x + c\right ) + 2 \, a^{2} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate((e*cos(d*x+c))^(1/2)/(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

2/3*sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a)*(cos(d*x + c) - sin(d*x + c) + 1)/(a^2*d*cos(d*x + c)^2 - a^
2*d*cos(d*x + c) - 2*a^2*d - (a^2*d*cos(d*x + c) + 2*a^2*d)*sin(d*x + c))

Sympy [F]

\[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^{3/2}} \, dx=\int \frac {\sqrt {e \cos {\left (c + d x \right )}}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate((e*cos(d*x+c))**(1/2)/(a+a*sin(d*x+c))**(3/2),x)

[Out]

Integral(sqrt(e*cos(c + d*x))/(a*(sin(c + d*x) + 1))**(3/2), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 131 vs. \(2 (30) = 60\).

Time = 0.30 (sec) , antiderivative size = 131, normalized size of antiderivative = 3.64 \[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left (\sqrt {a} \sqrt {e} - \frac {\sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} \sqrt {-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}}{3 \, {\left (a^{2} + \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}}} \]

[In]

integrate((e*cos(d*x+c))^(1/2)/(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

-2/3*(sqrt(a)*sqrt(e) - sqrt(a)*sqrt(e)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*sqrt(-sin(d*x + c)/(cos(d*x + c)
+ 1) + 1)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)/((a^2 + a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*d*(sin(d*
x + c)/(cos(d*x + c) + 1) + 1)^(5/2))

Giac [F(-1)]

Timed out. \[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate((e*cos(d*x+c))^(1/2)/(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 5.32 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.28 \[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {4\,\sqrt {e\,\cos \left (c+d\,x\right )}\,\sqrt {a\,\left (\sin \left (c+d\,x\right )+1\right )}\,\left (2\,\cos \left (c+d\,x\right )+\sin \left (2\,c+2\,d\,x\right )\right )}{3\,a^2\,d\,\left (15\,\sin \left (c+d\,x\right )-6\,\cos \left (2\,c+2\,d\,x\right )-\sin \left (3\,c+3\,d\,x\right )+10\right )} \]

[In]

int((e*cos(c + d*x))^(1/2)/(a + a*sin(c + d*x))^(3/2),x)

[Out]

-(4*(e*cos(c + d*x))^(1/2)*(a*(sin(c + d*x) + 1))^(1/2)*(2*cos(c + d*x) + sin(2*c + 2*d*x)))/(3*a^2*d*(15*sin(
c + d*x) - 6*cos(2*c + 2*d*x) - sin(3*c + 3*d*x) + 10))