\(\int \frac {(e \cos (c+d x))^{3/2}}{(a+a \sin (c+d x))^{5/2}} \, dx\) [316]

Optimal result

Integrand size = 27, antiderivative size = 36 \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {2 (e \cos (c+d x))^{5/2}}{5 d e (a+a \sin (c+d x))^{5/2}} \]

[Out]

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

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} \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {2 (e \cos (c+d x))^{5/2}}{5 d e (a \sin (c+d x)+a)^{5/2}} \]

[In]

Int[(e*Cos[c + d*x])^(3/2)/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-2*(e*Cos[c + d*x])^(5/2))/(5*d*e*(a + a*Sin[c + d*x])^(5/2))

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))^{5/2}}{5 d e (a+a \sin (c+d x))^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.36 \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {2 (e \cos (c+d x))^{5/2} \sqrt {a (1+\sin (c+d x))}}{5 a^3 d e (1+\sin (c+d x))^3} \]

[In]

Integrate[(e*Cos[c + d*x])^(3/2)/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-2*(e*Cos[c + d*x])^(5/2)*Sqrt[a*(1 + Sin[c + d*x])])/(5*a^3*d*e*(1 + Sin[c + d*x])^3)

Maple [A] (verified)

Time = 2.83 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.39

method	result	size
default	\(\frac {2 \left (\sin \left (d x +c \right )-1\right ) \sqrt {e \cos \left (d x +c \right )}\, e}{5 d \left (1+\sin \left (d x +c \right )\right ) \sqrt {a \left (1+\sin \left (d x +c \right )\right )}\, a^{2}}\)	\(50\)

[In]

int((e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (30) = 60\).

Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.94 \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {2 \, \sqrt {e \cos \left (d x + c\right )} \sqrt {a \sin \left (d x + c\right ) + a} {\left (e \sin \left (d x + c\right ) - e\right )}}{5 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \sin \left (d x + c\right ) - 2 \, a^{3} d\right )}} \]

[In]

integrate((e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+a \sin (c+d x))^{5/2}} \, dx=\int \frac {\left (e \cos {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((e*cos(d*x+c))**(3/2)/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Integral((e*cos(c + d*x))**(3/2)/(a*(sin(c + d*x) + 1))**(5/2), 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 = 0.34 (sec) , antiderivative size = 131, normalized size of antiderivative = 3.64 \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {2 \, {\left (\sqrt {a} e^{\frac {3}{2}} - \frac {\sqrt {a} e^{\frac {3}{2}} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {3}{2}} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}}{5 \, {\left (a^{3} + \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}}} \]

[In]

integrate((e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

-2/5*(sqrt(a)*e^(3/2) - sqrt(a)*e^(3/2)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(-sin(d*x + c)/(cos(d*x + c) + 1)
+ 1)^(3/2)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)/((a^3 + a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*d*(sin(
d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2))

Giac [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+a \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 6.16 (sec) , antiderivative size = 102, normalized size of antiderivative = 2.83 \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {4\,e\,\sqrt {e\,\cos \left (c+d\,x\right )}\,\sqrt {a\,\left (\sin \left (c+d\,x\right )+1\right )}\,\left (\sin \left (c+d\,x\right )+2\,\cos \left (2\,c+2\,d\,x\right )+\sin \left (3\,c+3\,d\,x\right )+2\right )}{5\,a^3\,d\,\left (56\,\sin \left (c+d\,x\right )-28\,\cos \left (2\,c+2\,d\,x\right )+\cos \left (4\,c+4\,d\,x\right )-8\,\sin \left (3\,c+3\,d\,x\right )+35\right )} \]

[In]

int((e*cos(c + d*x))^(3/2)/(a + a*sin(c + d*x))^(5/2),x)

[Out]

-(4*e*(e*cos(c + d*x))^(1/2)*(a*(sin(c + d*x) + 1))^(1/2)*(sin(c + d*x) + 2*cos(2*c + 2*d*x) + sin(3*c + 3*d*x
) + 2))/(5*a^3*d*(56*sin(c + d*x) - 28*cos(2*c + 2*d*x) + cos(4*c + 4*d*x) - 8*sin(3*c + 3*d*x) + 35))