∫ (e cos(c+dx))3∕2 ____________________________

Optimal. Leaf size=36 \[ -\frac {2 (e \cos (c+d x))^{5/2}}{5 d e (a+a \sin (c+d x))^{5/2}} \]

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

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Rubi [A]
time = 0.05, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {2750} \begin {gather*} -\frac {2 (e \cos (c+d x))^{5/2}}{5 d e (a \sin (c+d x)+a)^{5/2}} \end {gather*}

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

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

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]

\begin {align*} \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}}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 49, normalized size = 1.36 \begin {gather*} -\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} \end {gather*}

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

(-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)

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method	result	size

default	\(-\frac {2 \left (e \cos \left (d x +c \right )\right )^{\frac {3}{2}} \cos \left (d x +c \right )}{5 d \left (a \left (1+\sin \left (d x +c \right )\right )\right )^{\frac {5}{2}}}\)	\(34\)

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

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 126 vs. \(2 (27) = 54\).
time = 0.55, size = 126, normalized size = 3.50 \begin {gather*} -\frac {2 \, {\left (\sqrt {a} - \frac {\sqrt {a} \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 )} e^{\frac {3}{2}}}{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}}} \end {gather*}

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

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (27) = 54\).
time = 0.36, size = 70, normalized size = 1.94 \begin {gather*} -\frac {2 \, \sqrt {a \sin \left (d x + c\right ) + a} {\left (e^{\frac {3}{2}} \sin \left (d x + c\right ) - e^{\frac {3}{2}}\right )} \sqrt {\cos \left (d x + c\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 )}} \end {gather*}

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

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

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

Integral((e*cos(c + d*x))**(3/2)/(a*(sin(c + d*x) + 1))**(5/2), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

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Mupad [B]
time = 6.57, size = 102, normalized size = 2.83 \begin {gather*} -\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 )} \end {gather*}

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

-(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))

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