\(\int \frac {(e \cos (c+d x))^{15/2}}{(a+a \sin (c+d x))^4} \, dx\) [263]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [C] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 25, antiderivative size = 180 \[ \int \frac {(e \cos (c+d x))^{15/2}}{(a+a \sin (c+d x))^4} \, dx=\frac {78 e^8 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{7 a^4 d \sqrt {e \cos (c+d x)}}+\frac {78 e^7 \sqrt {e \cos (c+d x)} \sin (c+d x)}{7 a^4 d}+\frac {234 e^5 (e \cos (c+d x))^{5/2} \sin (c+d x)}{35 a^4 d}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a+a \sin (c+d x))^3}+\frac {52 e^3 (e \cos (c+d x))^{9/2}}{5 d \left (a^4+a^4 \sin (c+d x)\right )} \] Output:

78/7*e^8*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^4/d/(e* 
cos(d*x+c))^(1/2)+78/7*e^7*(e*cos(d*x+c))^(1/2)*sin(d*x+c)/a^4/d+234/35*e^ 
5*(e*cos(d*x+c))^(5/2)*sin(d*x+c)/a^4/d+4*e*(e*cos(d*x+c))^(13/2)/a/d/(a+a 
*sin(d*x+c))^3+52/5*e^3*(e*cos(d*x+c))^(9/2)/d/(a^4+a^4*sin(d*x+c))
 

Mathematica [C] (verified)

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

Time = 0.43 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.37 \[ \int \frac {(e \cos (c+d x))^{15/2}}{(a+a \sin (c+d x))^4} \, dx=-\frac {2 \sqrt [4]{2} (e \cos (c+d x))^{17/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {17}{4},\frac {21}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{17 a^4 d e (1+\sin (c+d x))^{17/4}} \] Input:

Integrate[(e*Cos[c + d*x])^(15/2)/(a + a*Sin[c + d*x])^4,x]
 

Output:

(-2*2^(1/4)*(e*Cos[c + d*x])^(17/2)*Hypergeometric2F1[3/4, 17/4, 21/4, (1 
- Sin[c + d*x])/2])/(17*a^4*d*e*(1 + Sin[c + d*x])^(17/4))
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3159, 3042, 3159, 3042, 3115, 3042, 3115, 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.

\(\displaystyle \int \frac {(e \cos (c+d x))^{15/2}}{(a \sin (c+d x)+a)^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(e \cos (c+d x))^{15/2}}{(a \sin (c+d x)+a)^4}dx\)

\(\Big \downarrow \) 3159

\(\displaystyle \frac {13 e^2 \int \frac {(e \cos (c+d x))^{11/2}}{(\sin (c+d x) a+a)^2}dx}{a^2}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {13 e^2 \int \frac {(e \cos (c+d x))^{11/2}}{(\sin (c+d x) a+a)^2}dx}{a^2}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3159

\(\displaystyle \frac {13 e^2 \left (\frac {9 e^2 \int (e \cos (c+d x))^{7/2}dx}{5 a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {13 e^2 \left (\frac {9 e^2 \int \left (e \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}dx}{5 a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {13 e^2 \left (\frac {9 e^2 \left (\frac {5}{7} e^2 \int (e \cos (c+d x))^{3/2}dx+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 d}\right )}{5 a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {13 e^2 \left (\frac {9 e^2 \left (\frac {5}{7} e^2 \int \left (e \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 d}\right )}{5 a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {13 e^2 \left (\frac {9 e^2 \left (\frac {5}{7} e^2 \left (\frac {1}{3} e^2 \int \frac {1}{\sqrt {e \cos (c+d x)}}dx+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 d}\right )}{5 a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {13 e^2 \left (\frac {9 e^2 \left (\frac {5}{7} e^2 \left (\frac {1}{3} e^2 \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 d}\right )}{5 a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {13 e^2 \left (\frac {9 e^2 \left (\frac {5}{7} e^2 \left (\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 d}\right )}{5 a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {13 e^2 \left (\frac {9 e^2 \left (\frac {5}{7} e^2 \left (\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 d}\right )}{5 a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {13 e^2 \left (\frac {9 e^2 \left (\frac {5}{7} e^2 \left (\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 d}\right )}{5 a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a \sin (c+d x)+a)^3}\)

Input:

Int[(e*Cos[c + d*x])^(15/2)/(a + a*Sin[c + d*x])^4,x]
 

Output:

(4*e*(e*Cos[c + d*x])^(13/2))/(a*d*(a + a*Sin[c + d*x])^3) + (13*e^2*((4*e 
*(e*Cos[c + d*x])^(9/2))/(5*d*(a^2 + a^2*Sin[c + d*x])) + (9*e^2*((2*e*(e* 
Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*d) + (5*e^2*((2*e^2*Sqrt[Cos[c + d*x] 
]*EllipticF[(c + d*x)/2, 2])/(3*d*Sqrt[e*Cos[c + d*x]]) + (2*e*Sqrt[e*Cos[ 
c + d*x]]*Sin[c + d*x])/(3*d)))/7))/(5*a^2)))/a^2
 

Defintions of rubi rules used

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3115
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

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]
 
Maple [A] (verified)

Time = 2.11 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.25

\[-\frac {2 e^{8} \left (80 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-120 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-224 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-280 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+336 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+160 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+195 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+392 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-252 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 a^{4} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} e +e}\, d}\]

Input:

int((e*cos(d*x+c))^(15/2)/(a+a*sin(d*x+c))^4,x)
 

Output:

-2/35/a^4/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^8*(80*c 
os(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-120*sin(1/2*d*x+1/2*c)^6*cos(1/2*d* 
x+1/2*c)-224*sin(1/2*d*x+1/2*c)^7-280*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2 
*c)+336*sin(1/2*d*x+1/2*c)^5+160*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+1 
95*(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))+392*sin(1/2*d*x+1/2*c)^3-252*sin(1/2*d*x+1/2* 
c))/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.65 \[ \int \frac {(e \cos (c+d x))^{15/2}}{(a+a \sin (c+d x))^4} \, dx=-\frac {2 \, {\left (195 i \, \sqrt {\frac {1}{2}} e^{\frac {15}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 195 i \, \sqrt {\frac {1}{2}} e^{\frac {15}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (28 \, e^{7} \cos \left (d x + c\right )^{2} - 280 \, e^{7} - 5 \, {\left (e^{7} \cos \left (d x + c\right )^{2} - 17 \, e^{7}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}\right )}}{35 \, a^{4} d} \] Input:

integrate((e*cos(d*x+c))^(15/2)/(a+a*sin(d*x+c))^4,x, algorithm="fricas")
 

Output:

-2/35*(195*I*sqrt(1/2)*e^(15/2)*weierstrassPInverse(-4, 0, cos(d*x + c) + 
I*sin(d*x + c)) - 195*I*sqrt(1/2)*e^(15/2)*weierstrassPInverse(-4, 0, cos( 
d*x + c) - I*sin(d*x + c)) + (28*e^7*cos(d*x + c)^2 - 280*e^7 - 5*(e^7*cos 
(d*x + c)^2 - 17*e^7)*sin(d*x + c))*sqrt(e*cos(d*x + c)))/(a^4*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^{15/2}}{(a+a \sin (c+d x))^4} \, dx=\text {Timed out} \] Input:

integrate((e*cos(d*x+c))**(15/2)/(a+a*sin(d*x+c))**4,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(e \cos (c+d x))^{15/2}}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {15}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \] Input:

integrate((e*cos(d*x+c))^(15/2)/(a+a*sin(d*x+c))^4,x, algorithm="maxima")
 

Output:

integrate((e*cos(d*x + c))^(15/2)/(a*sin(d*x + c) + a)^4, x)
 

Giac [F]

\[ \int \frac {(e \cos (c+d x))^{15/2}}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {15}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \] Input:

integrate((e*cos(d*x+c))^(15/2)/(a+a*sin(d*x+c))^4,x, algorithm="giac")
 

Output:

integrate((e*cos(d*x + c))^(15/2)/(a*sin(d*x + c) + a)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^{15/2}}{(a+a \sin (c+d x))^4} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{15/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4} \,d x \] Input:

int((e*cos(c + d*x))^(15/2)/(a + a*sin(c + d*x))^4,x)
 

Output:

int((e*cos(c + d*x))^(15/2)/(a + a*sin(c + d*x))^4, x)
 

Reduce [F]

\[ \int \frac {(e \cos (c+d x))^{15/2}}{(a+a \sin (c+d x))^4} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{7}}{\sin \left (d x +c \right )^{4}+4 \sin \left (d x +c \right )^{3}+6 \sin \left (d x +c \right )^{2}+4 \sin \left (d x +c \right )+1}d x \right ) e^{7}}{a^{4}} \] Input:

int((e*cos(d*x+c))^(15/2)/(a+a*sin(d*x+c))^4,x)
                                                                                    
                                                                                    
 

Output:

(sqrt(e)*int((sqrt(cos(c + d*x))*cos(c + d*x)**7)/(sin(c + d*x)**4 + 4*sin 
(c + d*x)**3 + 6*sin(c + d*x)**2 + 4*sin(c + d*x) + 1),x)*e**7)/a**4