\(\int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{11/2}} \, dx\) [296]

Optimal result

Integrand size = 27, antiderivative size = 113 \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{11/2}} \, dx=\frac {2 (a+a \sin (c+d x))^{5/2}}{d e (e \cos (c+d x))^{9/2}}-\frac {8 (a+a \sin (c+d x))^{7/2}}{5 a d e (e \cos (c+d x))^{9/2}}+\frac {16 (a+a \sin (c+d x))^{9/2}}{45 a^2 d e (e \cos (c+d x))^{9/2}} \]

[Out]

2*(a+a*sin(d*x+c))^(5/2)/d/e/(e*cos(d*x+c))^(9/2)-8/5*(a+a*sin(d*x+c))^(7/2)/a/d/e/(e*cos(d*x+c))^(9/2)+16/45*
(a+a*sin(d*x+c))^(9/2)/a^2/d/e/(e*cos(d*x+c))^(9/2)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 113, 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} \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{11/2}} \, dx=\frac {16 (a \sin (c+d x)+a)^{9/2}}{45 a^2 d e (e \cos (c+d x))^{9/2}}-\frac {8 (a \sin (c+d x)+a)^{7/2}}{5 a d e (e \cos (c+d x))^{9/2}}+\frac {2 (a \sin (c+d x)+a)^{5/2}}{d e (e \cos (c+d x))^{9/2}} \]

[In]

Int[(a + a*Sin[c + d*x])^(5/2)/(e*Cos[c + d*x])^(11/2),x]

[Out]

(2*(a + a*Sin[c + d*x])^(5/2))/(d*e*(e*Cos[c + d*x])^(9/2)) - (8*(a + a*Sin[c + d*x])^(7/2))/(5*a*d*e*(e*Cos[c
+ d*x])^(9/2)) + (16*(a + a*Sin[c + d*x])^(9/2))/(45*a^2*d*e*(e*Cos[c + d*x])^(9/2))

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}}{d e (e \cos (c+d x))^{9/2}}-\frac {4 \int \frac {(a+a \sin (c+d x))^{7/2}}{(e \cos (c+d x))^{11/2}} \, dx}{a} \\ & = \frac {2 (a+a \sin (c+d x))^{5/2}}{d e (e \cos (c+d x))^{9/2}}-\frac {8 (a+a \sin (c+d x))^{7/2}}{5 a d e (e \cos (c+d x))^{9/2}}+\frac {8 \int \frac {(a+a \sin (c+d x))^{9/2}}{(e \cos (c+d x))^{11/2}} \, dx}{5 a^2} \\ & = \frac {2 (a+a \sin (c+d x))^{5/2}}{d e (e \cos (c+d x))^{9/2}}-\frac {8 (a+a \sin (c+d x))^{7/2}}{5 a d e (e \cos (c+d x))^{9/2}}+\frac {16 (a+a \sin (c+d x))^{9/2}}{45 a^2 d e (e \cos (c+d x))^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.57 \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{11/2}} \, dx=\frac {2 \sqrt {e \cos (c+d x)} \sec ^5(c+d x) (a (1+\sin (c+d x)))^{5/2} \left (17-20 \sin (c+d x)+8 \sin ^2(c+d x)\right )}{45 d e^6} \]

[In]

Integrate[(a + a*Sin[c + d*x])^(5/2)/(e*Cos[c + d*x])^(11/2),x]

[Out]

(2*Sqrt[e*Cos[c + d*x]]*Sec[c + d*x]^5*(a*(1 + Sin[c + d*x]))^(5/2)*(17 - 20*Sin[c + d*x] + 8*Sin[c + d*x]^2))
/(45*d*e^6)

Maple [A] (verified)

Time = 2.73 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.65

method	result	size
default	\(\frac {2 \left (8 \left (\cos ^{2}\left (d x +c \right )\right )+20 \sin \left (d x +c \right )-25\right ) a^{2} \sqrt {a \left (1+\sin \left (d x +c \right )\right )}}{45 d \left (\cos ^{2}\left (d x +c \right )+2 \sin \left (d x +c \right )-2\right ) \sqrt {e \cos \left (d x +c \right )}\, e^{5}}\)	\(74\)

[In]

int((a+a*sin(d*x+c))^(5/2)/(e*cos(d*x+c))^(11/2),x,method=_RETURNVERBOSE)

[Out]

2/45/d*(8*cos(d*x+c)^2+20*sin(d*x+c)-25)*a^2*(a*(1+sin(d*x+c)))^(1/2)/(cos(d*x+c)^2+2*sin(d*x+c)-2)/(e*cos(d*x
+c))^(1/2)/e^5

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.88 \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{11/2}} \, dx=\frac {2 \, {\left (8 \, a^{2} \cos \left (d x + c\right )^{2} + 20 \, a^{2} \sin \left (d x + c\right ) - 25 \, a^{2}\right )} \sqrt {e \cos \left (d x + c\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{45 \, {\left (d e^{6} \cos \left (d x + c\right )^{3} + 2 \, d e^{6} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, d e^{6} \cos \left (d x + c\right )\right )}} \]

[In]

integrate((a+a*sin(d*x+c))^(5/2)/(e*cos(d*x+c))^(11/2),x, algorithm="fricas")

[Out]

2/45*(8*a^2*cos(d*x + c)^2 + 20*a^2*sin(d*x + c) - 25*a^2)*sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a)/(d*e^
6*cos(d*x + c)^3 + 2*d*e^6*cos(d*x + c)*sin(d*x + c) - 2*d*e^6*cos(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{11/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(d*x+c))**(5/2)/(e*cos(d*x+c))**(11/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 282 vs. \(2 (97) = 194\).

Time = 0.31 (sec) , antiderivative size = 282, normalized size of antiderivative = 2.50 \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{11/2}} \, dx=\frac {2 \, {\left (17 \, a^{\frac {5}{2}} \sqrt {e} - \frac {40 \, a^{\frac {5}{2}} \sqrt {e} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {49 \, a^{\frac {5}{2}} \sqrt {e} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {49 \, a^{\frac {5}{2}} \sqrt {e} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {40 \, a^{\frac {5}{2}} \sqrt {e} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {17 \, a^{\frac {5}{2}} \sqrt {e} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{3}}{45 \, {\left (e^{6} + \frac {3 \, e^{6} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, e^{6} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {e^{6} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )} d \sqrt {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {11}{2}}} \]

[In]

integrate((a+a*sin(d*x+c))^(5/2)/(e*cos(d*x+c))^(11/2),x, algorithm="maxima")

[Out]

2/45*(17*a^(5/2)*sqrt(e) - 40*a^(5/2)*sqrt(e)*sin(d*x + c)/(cos(d*x + c) + 1) + 49*a^(5/2)*sqrt(e)*sin(d*x + c
)^2/(cos(d*x + c) + 1)^2 - 49*a^(5/2)*sqrt(e)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 40*a^(5/2)*sqrt(e)*sin(d*x
+ c)^5/(cos(d*x + c) + 1)^5 - 17*a^(5/2)*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/((e^6 + 3*e^6*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*e^6*sin(d*x + c)^4/(cos(d*x + c) +
1)^4 + e^6*sin(d*x + c)^6/(cos(d*x + c) + 1)^6)*d*sqrt(sin(d*x + c)/(cos(d*x + c) + 1) + 1)*(-sin(d*x + c)/(co
s(d*x + c) + 1) + 1)^(11/2))

Giac [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{11/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(d*x+c))^(5/2)/(e*cos(d*x+c))^(11/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 7.06 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.05 \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{11/2}} \, dx=\frac {8\,a^2\,\sqrt {a\,\left (\sin \left (c+d\,x\right )+1\right )}\,\left (2\,\cos \left (4\,c+4\,d\,x\right )-73\,\cos \left (2\,c+2\,d\,x\right )-162\,\sin \left (c+d\,x\right )+18\,\sin \left (3\,c+3\,d\,x\right )+105\right )}{45\,d\,e^5\,\sqrt {e\,\cos \left (c+d\,x\right )}\,\left (\cos \left (4\,c+4\,d\,x\right )-28\,\cos \left (2\,c+2\,d\,x\right )-56\,\sin \left (c+d\,x\right )+8\,\sin \left (3\,c+3\,d\,x\right )+35\right )} \]

[In]

int((a + a*sin(c + d*x))^(5/2)/(e*cos(c + d*x))^(11/2),x)

[Out]

(8*a^2*(a*(sin(c + d*x) + 1))^(1/2)*(2*cos(4*c + 4*d*x) - 73*cos(2*c + 2*d*x) - 162*sin(c + d*x) + 18*sin(3*c
+ 3*d*x) + 105))/(45*d*e^5*(e*cos(c + d*x))^(1/2)*(cos(4*c + 4*d*x) - 28*cos(2*c + 2*d*x) - 56*sin(c + d*x) +
8*sin(3*c + 3*d*x) + 35))