\(\int \frac {\sqrt {a+a \sin (c+d x)}}{(e \cos (c+d x))^{5/2}} \, dx\) [277]

Optimal result

Integrand size = 27, antiderivative size = 74 \[ \int \frac {\sqrt {a+a \sin (c+d x)}}{(e \cos (c+d x))^{5/2}} \, dx=-\frac {2 \sqrt {a+a \sin (c+d x)}}{d e (e \cos (c+d x))^{3/2}}+\frac {4 (a+a \sin (c+d x))^{3/2}}{3 a d e (e \cos (c+d x))^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2751, 2750} \[ \int \frac {\sqrt {a+a \sin (c+d x)}}{(e \cos (c+d x))^{5/2}} \, dx=\frac {4 (a \sin (c+d x)+a)^{3/2}}{3 a d e (e \cos (c+d x))^{3/2}}-\frac {2 \sqrt {a \sin (c+d x)+a}}{d e (e \cos (c+d x))^{3/2}} \]

[In]

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

[Out]

(-2*Sqrt[a + a*Sin[c + d*x]])/(d*e*(e*Cos[c + d*x])^(3/2)) + (4*(a + a*Sin[c + d*x])^(3/2))/(3*a*d*e*(e*Cos[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]

Rule 2751

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*Simplify[2*m + p + 1])), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[2*m + p + 1]), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m
, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \sqrt {a+a \sin (c+d x)}}{d e (e \cos (c+d x))^{3/2}}+\frac {2 \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{5/2}} \, dx}{a} \\ & = -\frac {2 \sqrt {a+a \sin (c+d x)}}{d e (e \cos (c+d x))^{3/2}}+\frac {4 (a+a \sin (c+d x))^{3/2}}{3 a d e (e \cos (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.62 \[ \int \frac {\sqrt {a+a \sin (c+d x)}}{(e \cos (c+d x))^{5/2}} \, dx=\frac {2 \sqrt {a (1+\sin (c+d x))} (-1+2 \sin (c+d x))}{3 d e (e \cos (c+d x))^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.57 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.65

method	result	size
default	\(\frac {2 \sqrt {a \left (1+\sin \left (d x +c \right )\right )}\, \left (2 \tan \left (d x +c \right )-\sec \left (d x +c \right )\right )}{3 d \sqrt {e \cos \left (d x +c \right )}\, e^{2}}\)	\(48\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.65 \[ \int \frac {\sqrt {a+a \sin (c+d x)}}{(e \cos (c+d x))^{5/2}} \, dx=\frac {2 \, \sqrt {e \cos \left (d x + c\right )} \sqrt {a \sin \left (d x + c\right ) + a} {\left (2 \, \sin \left (d x + c\right ) - 1\right )}}{3 \, d e^{3} \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 206 vs. \(2 (64) = 128\).

Time = 0.30 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.78 \[ \int \frac {\sqrt {a+a \sin (c+d x)}}{(e \cos (c+d x))^{5/2}} \, dx=-\frac {2 \, {\left (\sqrt {a} \sqrt {e} - \frac {4 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {4 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {\sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{2}}{3 \, {\left (e^{3} + \frac {2 \, e^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {e^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}\right )} d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {3}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}}} \]

[In]

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

[Out]

-2/3*(sqrt(a)*sqrt(e) - 4*sqrt(a)*sqrt(e)*sin(d*x + c)/(cos(d*x + c) + 1) + 4*sqrt(a)*sqrt(e)*sin(d*x + c)^3/(
cos(d*x + c) + 1)^3 - sqrt(a)*sqrt(e)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^
2 + 1)^2/((e^3 + 2*e^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + e^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)*d*(sin(d
*x + c)/(cos(d*x + c) + 1) + 1)^(3/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2))

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 5.20 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {a+a \sin (c+d x)}}{(e \cos (c+d x))^{5/2}} \, dx=-\frac {4\,\sqrt {a\,\left (\sin \left (c+d\,x\right )+1\right )}\,\left (\cos \left (c+d\,x\right )-\sin \left (2\,c+2\,d\,x\right )\right )}{3\,d\,e^2\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )\,\sqrt {e\,\cos \left (c+d\,x\right )}} \]

[In]

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

[Out]

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