\(\int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^3} \, dx\) [257]

Optimal result

Integrand size = 25, antiderivative size = 107 \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^3} \, dx=-\frac {10 e^3 \sqrt {e \cos (c+d x)}}{3 a^3 d}-\frac {10 e^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^3 d \sqrt {e \cos (c+d x)}}-\frac {4 e (e \cos (c+d x))^{5/2}}{3 a d (a+a \sin (c+d x))^2} \] Output:

-10/3*e^3*(e*cos(d*x+c))^(1/2)/a^3/d-10/3*e^4*cos(d*x+c)^(1/2)*InverseJaco 
biAM(1/2*d*x+1/2*c,2^(1/2))/a^3/d/(e*cos(d*x+c))^(1/2)-4/3*e*(e*cos(d*x+c) 
)^(5/2)/a/d/(a+a*sin(d*x+c))^2
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.10 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.62 \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^3} \, dx=-\frac {\sqrt [4]{2} (e \cos (c+d x))^{9/2} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},\frac {9}{4},\frac {13}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{9 a^3 d e (1+\sin (c+d x))^{9/4}} \] Input:

Integrate[(e*Cos[c + d*x])^(7/2)/(a + a*Sin[c + d*x])^3,x]
 

Output:

-1/9*(2^(1/4)*(e*Cos[c + d*x])^(9/2)*Hypergeometric2F1[7/4, 9/4, 13/4, (1 
- Sin[c + d*x])/2])/(a^3*d*e*(1 + Sin[c + d*x])^(9/4))
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3159, 3042, 3161, 3042, 3121, 3042, 3120}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(e \cos (c+d x))^{7/2}}{(a \sin (c+d x)+a)^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(e \cos (c+d x))^{7/2}}{(a \sin (c+d x)+a)^3}dx\)

\(\Big \downarrow \) 3159

\(\displaystyle -\frac {5 e^2 \int \frac {(e \cos (c+d x))^{3/2}}{\sin (c+d x) a+a}dx}{3 a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{3 a d (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {5 e^2 \int \frac {(e \cos (c+d x))^{3/2}}{\sin (c+d x) a+a}dx}{3 a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{3 a d (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3161

\(\displaystyle -\frac {5 e^2 \left (\frac {e^2 \int \frac {1}{\sqrt {e \cos (c+d x)}}dx}{a}+\frac {2 e \sqrt {e \cos (c+d x)}}{a d}\right )}{3 a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{3 a d (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {5 e^2 \left (\frac {e^2 \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}+\frac {2 e \sqrt {e \cos (c+d x)}}{a d}\right )}{3 a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{3 a d (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3121

\(\displaystyle -\frac {5 e^2 \left (\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a \sqrt {e \cos (c+d x)}}+\frac {2 e \sqrt {e \cos (c+d x)}}{a d}\right )}{3 a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{3 a d (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {5 e^2 \left (\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \sqrt {e \cos (c+d x)}}+\frac {2 e \sqrt {e \cos (c+d x)}}{a d}\right )}{3 a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{3 a d (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3120

\(\displaystyle -\frac {5 e^2 \left (\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d \sqrt {e \cos (c+d x)}}+\frac {2 e \sqrt {e \cos (c+d x)}}{a d}\right )}{3 a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{3 a d (a \sin (c+d x)+a)^2}\)

Input:

Int[(e*Cos[c + d*x])^(7/2)/(a + a*Sin[c + d*x])^3,x]
 

Output:

(-5*e^2*((2*e*Sqrt[e*Cos[c + d*x]])/(a*d) + (2*e^2*Sqrt[Cos[c + d*x]]*Elli 
pticF[(c + d*x)/2, 2])/(a*d*Sqrt[e*Cos[c + d*x]])))/(3*a^2) - (4*e*(e*Cos[ 
c + d*x])^(5/2))/(3*a*d*(a + a*Sin[c + d*x])^2)
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3121

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 

rule 3159

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[2*g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f 
*x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Simp[g^2*((p - 1)/(b^2*(2*m + p + 1 
)))   Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; 
FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] & 
& NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]
 

rule 3161

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[g*((g*Cos[e + f*x])^(p - 1)/(b*f*(p - 1))), x] + Si 
mp[g^2/a   Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g}, x 
] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(218\) vs. \(2(94)=188\).

Time = 147.62 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.05

method	result	size
default	\(-\frac {2 \left (-10 \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-12 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+12 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-7 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e^{4}}{3 \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} e +e}\, d}\)	\(219\)

Input:

int((e*cos(d*x+c))^(7/2)/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

-2/3/(2*sin(1/2*d*x+1/2*c)^2-1)/a^3/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2 
*c)^2*e+e)^(1/2)*(-10*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c) 
^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-12*si 
n(1/2*d*x+1/2*c)^5+8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+5*(sin(1/2*d* 
x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1 
/2*c),2^(1/2))+12*sin(1/2*d*x+1/2*c)^3-7*sin(1/2*d*x+1/2*c))*e^4/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.27 \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^3} \, dx=-\frac {2 \, {\left (5 \, \sqrt {\frac {1}{2}} {\left (-i \, e^{3} \sin \left (d x + c\right ) - i \, e^{3}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {\frac {1}{2}} {\left (i \, e^{3} \sin \left (d x + c\right ) + i \, e^{3}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (3 \, e^{3} \sin \left (d x + c\right ) + 7 \, e^{3}\right )} \sqrt {e \cos \left (d x + c\right )}\right )}}{3 \, {\left (a^{3} d \sin \left (d x + c\right ) + a^{3} d\right )}} \] Input:

integrate((e*cos(d*x+c))^(7/2)/(a+a*sin(d*x+c))^3,x, algorithm="fricas")
 

Output:

-2/3*(5*sqrt(1/2)*(-I*e^3*sin(d*x + c) - I*e^3)*sqrt(e)*weierstrassPInvers 
e(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(1/2)*(I*e^3*sin(d*x + c) 
+ I*e^3)*sqrt(e)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) 
 + (3*e^3*sin(d*x + c) + 7*e^3)*sqrt(e*cos(d*x + c)))/(a^3*d*sin(d*x + c) 
+ a^3*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate((e*cos(d*x+c))**(7/2)/(a+a*sin(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^3} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:

integrate((e*cos(d*x+c))^(7/2)/(a+a*sin(d*x+c))^3,x, algorithm="maxima")
 

Output:

integrate((e*cos(d*x + c))^(7/2)/(a*sin(d*x + c) + a)^3, x)
 

Giac [F]

\[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^3} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:

integrate((e*cos(d*x+c))^(7/2)/(a+a*sin(d*x+c))^3,x, algorithm="giac")
 

Output:

integrate((e*cos(d*x + c))^(7/2)/(a*sin(d*x + c) + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^3} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{7/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^3} \,d x \] Input:

int((e*cos(c + d*x))^(7/2)/(a + a*sin(c + d*x))^3,x)
 

Output:

int((e*cos(c + d*x))^(7/2)/(a + a*sin(c + d*x))^3, x)
 

Reduce [F]

\[ \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^3} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{3}}{\sin \left (d x +c \right )^{3}+3 \sin \left (d x +c \right )^{2}+3 \sin \left (d x +c \right )+1}d x \right ) e^{3}}{a^{3}} \] Input:

int((e*cos(d*x+c))^(7/2)/(a+a*sin(d*x+c))^3,x)
 

Output:

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