\(\int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{9/2}} \, dx\) [295]

Optimal result

Integrand size = 27, antiderivative size = 76 \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{9/2}} \, dx=\frac {2 (a+a \sin (c+d x))^{5/2}}{3 d e (e \cos (c+d x))^{7/2}}-\frac {4 (a+a \sin (c+d x))^{7/2}}{21 a d e (e \cos (c+d x))^{7/2}} \]

[Out]

2/3*(a+a*sin(d*x+c))^(5/2)/d/e/(e*cos(d*x+c))^(7/2)-4/21*(a+a*sin(d*x+c))^(7/2)/a/d/e/(e*cos(d*x+c))^(7/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 76, 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} \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{9/2}} \, dx=\frac {2 (a \sin (c+d x)+a)^{5/2}}{3 d e (e \cos (c+d x))^{7/2}}-\frac {4 (a \sin (c+d x)+a)^{7/2}}{21 a d e (e \cos (c+d x))^{7/2}} \]

[In]

Int[(a + a*Sin[c + d*x])^(5/2)/(e*Cos[c + d*x])^(9/2),x]

[Out]

(2*(a + a*Sin[c + d*x])^(5/2))/(3*d*e*(e*Cos[c + d*x])^(7/2)) - (4*(a + a*Sin[c + d*x])^(7/2))/(21*a*d*e*(e*Co
s[c + d*x])^(7/2))

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 (a+a \sin (c+d x))^{5/2}}{3 d e (e \cos (c+d x))^{7/2}}-\frac {2 \int \frac {(a+a \sin (c+d x))^{7/2}}{(e \cos (c+d x))^{9/2}} \, dx}{3 a} \\ & = \frac {2 (a+a \sin (c+d x))^{5/2}}{3 d e (e \cos (c+d x))^{7/2}}-\frac {4 (a+a \sin (c+d x))^{7/2}}{21 a d e (e \cos (c+d x))^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.71 \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{9/2}} \, dx=-\frac {2 \sqrt {e \cos (c+d x)} \sec ^4(c+d x) (a (1+\sin (c+d x)))^{5/2} (-5+2 \sin (c+d x))}{21 d e^5} \]

[In]

Integrate[(a + a*Sin[c + d*x])^(5/2)/(e*Cos[c + d*x])^(9/2),x]

[Out]

(-2*Sqrt[e*Cos[c + d*x]]*Sec[c + d*x]^4*(a*(1 + Sin[c + d*x]))^(5/2)*(-5 + 2*Sin[c + d*x]))/(21*d*e^5)

Maple [A] (verified)

Time = 2.66 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.01

method	result	size
default	\(\frac {2 \sqrt {a \left (1+\sin \left (d x +c \right )\right )}\, a^{2} \left (2 \tan \left (d x +c \right )-\sec \left (d x +c \right )+6 \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )+6 \left (\sec ^{3}\left (d x +c \right )\right )\right )}{21 d \sqrt {e \cos \left (d x +c \right )}\, e^{4}}\)	\(77\)

[In]

int((a+a*sin(d*x+c))^(5/2)/(e*cos(d*x+c))^(9/2),x,method=_RETURNVERBOSE)

[Out]

2/21/d*(a*(1+sin(d*x+c)))^(1/2)*a^2/(e*cos(d*x+c))^(1/2)/e^4*(2*tan(d*x+c)-sec(d*x+c)+6*tan(d*x+c)*sec(d*x+c)^
2+6*sec(d*x+c)^3)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.99 \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{9/2}} \, dx=\frac {2 \, {\left (2 \, a^{2} \sin \left (d x + c\right ) - 5 \, a^{2}\right )} \sqrt {e \cos \left (d x + c\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{21 \, {\left (d e^{5} \cos \left (d x + c\right )^{2} + 2 \, d e^{5} \sin \left (d x + c\right ) - 2 \, d e^{5}\right )}} \]

[In]

integrate((a+a*sin(d*x+c))^(5/2)/(e*cos(d*x+c))^(9/2),x, algorithm="fricas")

[Out]

2/21*(2*a^2*sin(d*x + c) - 5*a^2)*sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a)/(d*e^5*cos(d*x + c)^2 + 2*d*e^
5*sin(d*x + c) - 2*d*e^5)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{9/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(d*x+c))**(5/2)/(e*cos(d*x+c))**(9/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 207 vs. \(2 (64) = 128\).

Time = 0.30 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.72 \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{9/2}} \, dx=\frac {2 \, {\left (5 \, a^{\frac {5}{2}} \sqrt {e} - \frac {4 \, a^{\frac {5}{2}} \sqrt {e} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {4 \, a^{\frac {5}{2}} \sqrt {e} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {5 \, a^{\frac {5}{2}} \sqrt {e} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}\right )} \sqrt {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{2}}{21 \, {\left (e^{5} + \frac {2 \, e^{5} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {e^{5} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}\right )} d {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {9}{2}}} \]

[In]

integrate((a+a*sin(d*x+c))^(5/2)/(e*cos(d*x+c))^(9/2),x, algorithm="maxima")

[Out]

2/21*(5*a^(5/2)*sqrt(e) - 4*a^(5/2)*sqrt(e)*sin(d*x + c)/(cos(d*x + c) + 1) + 4*a^(5/2)*sqrt(e)*sin(d*x + c)^3
/(cos(d*x + c) + 1)^3 - 5*a^(5/2)*sqrt(e)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)*sqrt(sin(d*x + c)/(cos(d*x + c)
+ 1) + 1)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^2/((e^5 + 2*e^5*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + e^5
*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)*d*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2))

Giac [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{9/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(d*x+c))^(5/2)/(e*cos(d*x+c))^(9/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 6.61 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.26 \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{9/2}} \, dx=\frac {4\,a^2\,\sqrt {a\,\left (\sin \left (c+d\,x\right )+1\right )}\,\left (\cos \left (3\,c+3\,d\,x\right )-11\,\cos \left (c+d\,x\right )+7\,\sin \left (2\,c+2\,d\,x\right )\right )}{21\,d\,e^4\,\sqrt {e\,\cos \left (c+d\,x\right )}\,\left (15\,\sin \left (c+d\,x\right )+6\,\cos \left (2\,c+2\,d\,x\right )-\sin \left (3\,c+3\,d\,x\right )-10\right )} \]

[In]

int((a + a*sin(c + d*x))^(5/2)/(e*cos(c + d*x))^(9/2),x)

[Out]

(4*a^2*(a*(sin(c + d*x) + 1))^(1/2)*(cos(3*c + 3*d*x) - 11*cos(c + d*x) + 7*sin(2*c + 2*d*x)))/(21*d*e^4*(e*co
s(c + d*x))^(1/2)*(15*sin(c + d*x) + 6*cos(2*c + 2*d*x) - sin(3*c + 3*d*x) - 10))