\(\int \frac {1}{(e \cos (c+d x))^{5/2} \sqrt {a+a \sin (c+d x)}} \, dx\) [303]

Optimal result

Integrand size = 27, antiderivative size = 115 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} \sqrt {a+a \sin (c+d x)}} \, dx=-\frac {2}{5 d e (e \cos (c+d x))^{3/2} \sqrt {a+a \sin (c+d x)}}-\frac {8 \sqrt {a+a \sin (c+d x)}}{5 a d e (e \cos (c+d x))^{3/2}}+\frac {16 (a+a \sin (c+d x))^{3/2}}{15 a^2 d e (e \cos (c+d x))^{3/2}} \]

[Out]

16/15*(a+a*sin(d*x+c))^(3/2)/a^2/d/e/(e*cos(d*x+c))^(3/2)-2/5/d/e/(e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))^(1/2)-
8/5*(a+a*sin(d*x+c))^(1/2)/a/d/e/(e*cos(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2751, 2750} \[ \int \frac {1}{(e \cos (c+d x))^{5/2} \sqrt {a+a \sin (c+d x)}} \, dx=\frac {16 (a \sin (c+d x)+a)^{3/2}}{15 a^2 d e (e \cos (c+d x))^{3/2}}-\frac {8 \sqrt {a \sin (c+d x)+a}}{5 a d e (e \cos (c+d x))^{3/2}}-\frac {2}{5 d e \sqrt {a \sin (c+d x)+a} (e \cos (c+d x))^{3/2}} \]

[In]

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

[Out]

-2/(5*d*e*(e*Cos[c + d*x])^(3/2)*Sqrt[a + a*Sin[c + d*x]]) - (8*Sqrt[a + a*Sin[c + d*x]])/(5*a*d*e*(e*Cos[c +
d*x])^(3/2)) + (16*(a + a*Sin[c + d*x])^(3/2))/(15*a^2*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}{5 d e (e \cos (c+d x))^{3/2} \sqrt {a+a \sin (c+d x)}}+\frac {4 \int \frac {\sqrt {a+a \sin (c+d x)}}{(e \cos (c+d x))^{5/2}} \, dx}{5 a} \\ & = -\frac {2}{5 d e (e \cos (c+d x))^{3/2} \sqrt {a+a \sin (c+d x)}}-\frac {8 \sqrt {a+a \sin (c+d x)}}{5 a d e (e \cos (c+d x))^{3/2}}+\frac {8 \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{5/2}} \, dx}{5 a^2} \\ & = -\frac {2}{5 d e (e \cos (c+d x))^{3/2} \sqrt {a+a \sin (c+d x)}}-\frac {8 \sqrt {a+a \sin (c+d x)}}{5 a d e (e \cos (c+d x))^{3/2}}+\frac {16 (a+a \sin (c+d x))^{3/2}}{15 a^2 d e (e \cos (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.49 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} \sqrt {a+a \sin (c+d x)}} \, dx=\frac {2 \left (-7+4 \sin (c+d x)+8 \sin ^2(c+d x)\right )}{15 d e (e \cos (c+d x))^{3/2} \sqrt {a (1+\sin (c+d x))}} \]

[In]

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

[Out]

(2*(-7 + 4*Sin[c + d*x] + 8*Sin[c + d*x]^2))/(15*d*e*(e*Cos[c + d*x])^(3/2)*Sqrt[a*(1 + Sin[c + d*x])])

Maple [A] (verified)

Time = 2.82 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.47

method	result	size
default	\(\frac {-\frac {16 \cos \left (d x +c \right )}{15}+\frac {8 \tan \left (d x +c \right )}{15}+\frac {2 \sec \left (d x +c \right )}{15}}{d \,e^{2} \sqrt {a \left (1+\sin \left (d x +c \right )\right )}\, \sqrt {e \cos \left (d x +c \right )}}\)	\(54\)

[In]

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

[Out]

2/15/d/e^2/(a*(1+sin(d*x+c)))^(1/2)/(e*cos(d*x+c))^(1/2)*(-8*cos(d*x+c)+4*tan(d*x+c)+sec(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.70 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} \sqrt {a+a \sin (c+d x)}} \, dx=-\frac {2 \, \sqrt {e \cos \left (d x + c\right )} {\left (8 \, \cos \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right ) - 1\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{15 \, {\left (a d e^{3} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + a d e^{3} \cos \left (d x + c\right )^{2}\right )}} \]

[In]

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

[Out]

-2/15*sqrt(e*cos(d*x + c))*(8*cos(d*x + c)^2 - 4*sin(d*x + c) - 1)*sqrt(a*sin(d*x + c) + a)/(a*d*e^3*cos(d*x +
c)^2*sin(d*x + c) + a*d*e^3*cos(d*x + c)^2)

Sympy [F]

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

[In]

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

[Out]

Integral(1/(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. 287 vs. \(2 (97) = 194\).

Time = 0.31 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.50 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} \sqrt {a+a \sin (c+d x)}} \, dx=-\frac {2 \, {\left (7 \, \sqrt {a} \sqrt {e} - \frac {8 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {25 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {25 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {8 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {7 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{3}}{15 \, {\left (a e^{3} + \frac {3 \, a e^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a e^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a e^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )} d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}}} \]

[In]

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

[Out]

-2/15*(7*sqrt(a)*sqrt(e) - 8*sqrt(a)*sqrt(e)*sin(d*x + c)/(cos(d*x + c) + 1) - 25*sqrt(a)*sqrt(e)*sin(d*x + c)
^2/(cos(d*x + c) + 1)^2 + 25*sqrt(a)*sqrt(e)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 8*sqrt(a)*sqrt(e)*sin(d*x +
c)^5/(cos(d*x + c) + 1)^5 - 7*sqrt(a)*sqrt(e)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6)*(sin(d*x + c)^2/(cos(d*x +
c) + 1)^2 + 1)^3/((a*e^3 + 3*a*e^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a*e^3*sin(d*x + c)^4/(cos(d*x + c)
+ 1)^4 + a*e^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6)*d*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*(-sin(d*x +
c)/(cos(d*x + c) + 1) + 1)^(5/2))

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 6.61 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.04 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} \sqrt {a+a \sin (c+d x)}} \, dx=-\frac {8\,\sqrt {a\,\left (\sin \left (c+d\,x\right )+1\right )}\,\left (8\,\cos \left (c+d\,x\right )+6\,\cos \left (3\,c+3\,d\,x\right )-\sin \left (2\,c+2\,d\,x\right )+2\,\sin \left (4\,c+4\,d\,x\right )\right )}{15\,a\,d\,e^2\,\sqrt {e\,\cos \left (c+d\,x\right )}\,\left (4\,\sin \left (c+d\,x\right )+4\,\cos \left (2\,c+2\,d\,x\right )-\cos \left (4\,c+4\,d\,x\right )+4\,\sin \left (3\,c+3\,d\,x\right )+5\right )} \]

[In]

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

[Out]

-(8*(a*(sin(c + d*x) + 1))^(1/2)*(8*cos(c + d*x) + 6*cos(3*c + 3*d*x) - sin(2*c + 2*d*x) + 2*sin(4*c + 4*d*x))
)/(15*a*d*e^2*(e*cos(c + d*x))^(1/2)*(4*sin(c + d*x) + 4*cos(2*c + 2*d*x) - cos(4*c + 4*d*x) + 4*sin(3*c + 3*d
*x) + 5))