\(\int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{7/2}} \, dx\) [229]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [B] (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 = 127 \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{7/2}} \, dx=\frac {42 a^4 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^4 \sqrt {\cos (c+d x)}}+\frac {4 a^7 (e \cos (c+d x))^{7/2}}{5 d e^7 (a-a \sin (c+d x))^3}-\frac {28 a^8 (e \cos (c+d x))^{3/2}}{5 d e^5 \left (a^4-a^4 \sin (c+d x)\right )} \] Output:

42/5*a^4*(e*cos(d*x+c))^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/e^4/ 
cos(d*x+c)^(1/2)+4/5*a^7*(e*cos(d*x+c))^(7/2)/d/e^7/(a-a*sin(d*x+c))^3-28/ 
5*a^8*(e*cos(d*x+c))^(3/2)/d/e^5/(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 = 0.06 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.52 \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{7/2}} \, dx=\frac {8\ 2^{3/4} a^4 \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},-\frac {5}{4},-\frac {1}{4},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{5/4}}{5 d e (e \cos (c+d x))^{5/2}} \] Input:

Integrate[(a + a*Sin[c + d*x])^4/(e*Cos[c + d*x])^(7/2),x]
 

Output:

(8*2^(3/4)*a^4*Hypergeometric2F1[-7/4, -5/4, -1/4, (1 - Sin[c + d*x])/2]*( 
1 + Sin[c + d*x])^(5/4))/(5*d*e*(e*Cos[c + d*x])^(5/2))
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3149, 3042, 3159, 3042, 3159, 3042, 3121, 3042, 3119}

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.

\(\displaystyle \int \frac {(a \sin (c+d x)+a)^4}{(e \cos (c+d x))^{7/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (c+d x)+a)^4}{(e \cos (c+d x))^{7/2}}dx\)

\(\Big \downarrow \) 3149

\(\displaystyle \frac {a^8 \int \frac {(e \cos (c+d x))^{9/2}}{(a-a \sin (c+d x))^4}dx}{e^8}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a^8 \int \frac {(e \cos (c+d x))^{9/2}}{(a-a \sin (c+d x))^4}dx}{e^8}\)

\(\Big \downarrow \) 3159

\(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{7/2}}{5 a d (a-a \sin (c+d x))^3}-\frac {7 e^2 \int \frac {(e \cos (c+d x))^{5/2}}{(a-a \sin (c+d x))^2}dx}{5 a^2}\right )}{e^8}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{7/2}}{5 a d (a-a \sin (c+d x))^3}-\frac {7 e^2 \int \frac {(e \cos (c+d x))^{5/2}}{(a-a \sin (c+d x))^2}dx}{5 a^2}\right )}{e^8}\)

\(\Big \downarrow \) 3159

\(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{7/2}}{5 a d (a-a \sin (c+d x))^3}-\frac {7 e^2 \left (\frac {4 e (e \cos (c+d x))^{3/2}}{d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {3 e^2 \int \sqrt {e \cos (c+d x)}dx}{a^2}\right )}{5 a^2}\right )}{e^8}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{7/2}}{5 a d (a-a \sin (c+d x))^3}-\frac {7 e^2 \left (\frac {4 e (e \cos (c+d x))^{3/2}}{d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {3 e^2 \int \sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}\right )}{5 a^2}\right )}{e^8}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{7/2}}{5 a d (a-a \sin (c+d x))^3}-\frac {7 e^2 \left (\frac {4 e (e \cos (c+d x))^{3/2}}{d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {3 e^2 \sqrt {e \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{a^2 \sqrt {\cos (c+d x)}}\right )}{5 a^2}\right )}{e^8}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{7/2}}{5 a d (a-a \sin (c+d x))^3}-\frac {7 e^2 \left (\frac {4 e (e \cos (c+d x))^{3/2}}{d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {3 e^2 \sqrt {e \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2 \sqrt {\cos (c+d x)}}\right )}{5 a^2}\right )}{e^8}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{7/2}}{5 a d (a-a \sin (c+d x))^3}-\frac {7 e^2 \left (\frac {4 e (e \cos (c+d x))^{3/2}}{d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {6 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{a^2 d \sqrt {\cos (c+d x)}}\right )}{5 a^2}\right )}{e^8}\)

Input:

Int[(a + a*Sin[c + d*x])^4/(e*Cos[c + d*x])^(7/2),x]
 

Output:

(a^8*((4*e*(e*Cos[c + d*x])^(7/2))/(5*a*d*(a - a*Sin[c + d*x])^3) - (7*e^2 
*((-6*e^2*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(a^2*d*Sqrt[Cos[ 
c + d*x]]) + (4*e*(e*Cos[c + d*x])^(3/2))/(d*(a^2 - a^2*Sin[c + d*x]))))/( 
5*a^2)))/e^8
 

Defintions of rubi rules used

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

rule 3119
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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 3149
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[(a/g)^(2*m)   Int[(g*Cos[e + f*x])^(2*m + p)/( 
a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2 
, 0] && IntegerQ[m] && LtQ[p, -1] && GeQ[2*m + p, 0]
 

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(331\) vs. \(2(115)=230\).

Time = 12.37 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.61

method result size
default \(-\frac {2 \left (128 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-84 \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-128 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+84 \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+80 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+16 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-21 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-80 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+12 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{4}}{5 \left (4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} e +e}\, e^{3} d}\) \(332\)
parts \(\text {Expression too large to display}\) \(1165\)

Input:

int((a+a*sin(d*x+c))^4/(e*cos(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
 

Output:

-2/5/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)/sin(1/2*d*x+1/2*c)/ 
(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/e^3*(128*sin(1/2*d*x+1/2*c)^6*cos(1/2* 
d*x+1/2*c)-84*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2) 
^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-128*sin(1/2*d 
*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+84*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(s 
in(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/ 
2*c)^2+80*sin(1/2*d*x+1/2*c)^5+16*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)- 
21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE 
(cos(1/2*d*x+1/2*c),2^(1/2))-80*sin(1/2*d*x+1/2*c)^3+12*sin(1/2*d*x+1/2*c) 
)*a^4/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 293, normalized size of antiderivative = 2.31 \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{7/2}} \, dx=-\frac {2 \, {\left (21 \, \sqrt {\frac {1}{2}} {\left (-i \, a^{4} \cos \left (d x + c\right )^{2} + i \, a^{4} \cos \left (d x + c\right ) + 2 i \, a^{4} + {\left (-i \, a^{4} \cos \left (d x + c\right ) - 2 i \, a^{4}\right )} \sin \left (d x + c\right )\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 \, \sqrt {\frac {1}{2}} {\left (i \, a^{4} \cos \left (d x + c\right )^{2} - i \, a^{4} \cos \left (d x + c\right ) - 2 i \, a^{4} + {\left (i \, a^{4} \cos \left (d x + c\right ) + 2 i \, a^{4}\right )} \sin \left (d x + c\right )\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 4 \, {\left (4 \, a^{4} \cos \left (d x + c\right )^{2} + 3 \, a^{4} \cos \left (d x + c\right ) - a^{4} - {\left (4 \, a^{4} \cos \left (d x + c\right ) + a^{4}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}\right )}}{5 \, {\left (d e^{4} \cos \left (d x + c\right )^{2} - d e^{4} \cos \left (d x + c\right ) - 2 \, d e^{4} + {\left (d e^{4} \cos \left (d x + c\right ) + 2 \, d e^{4}\right )} \sin \left (d x + c\right )\right )}} \] Input:

integrate((a+a*sin(d*x+c))^4/(e*cos(d*x+c))^(7/2),x, algorithm="fricas")
 

Output:

-2/5*(21*sqrt(1/2)*(-I*a^4*cos(d*x + c)^2 + I*a^4*cos(d*x + c) + 2*I*a^4 + 
 (-I*a^4*cos(d*x + c) - 2*I*a^4)*sin(d*x + c))*sqrt(e)*weierstrassZeta(-4, 
 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 21*sqrt(1 
/2)*(I*a^4*cos(d*x + c)^2 - I*a^4*cos(d*x + c) - 2*I*a^4 + (I*a^4*cos(d*x 
+ c) + 2*I*a^4)*sin(d*x + c))*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPI 
nverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 4*(4*a^4*cos(d*x + c)^2 + 
3*a^4*cos(d*x + c) - a^4 - (4*a^4*cos(d*x + c) + a^4)*sin(d*x + c))*sqrt(e 
*cos(d*x + c)))/(d*e^4*cos(d*x + c)^2 - d*e^4*cos(d*x + c) - 2*d*e^4 + (d* 
e^4*cos(d*x + c) + 2*d*e^4)*sin(d*x + c))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{7/2}} \, dx=\text {Timed out} \] Input:

integrate((a+a*sin(d*x+c))**4/(e*cos(d*x+c))**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((a+a*sin(d*x+c))^4/(e*cos(d*x+c))^(7/2),x, algorithm="maxima")
 

Output:

integrate((a*sin(d*x + c) + a)^4/(e*cos(d*x + c))^(7/2), x)
 

Giac [F]

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

integrate((a+a*sin(d*x+c))^4/(e*cos(d*x+c))^(7/2),x, algorithm="giac")
 

Output:

integrate((a*sin(d*x + c) + a)^4/(e*cos(d*x + c))^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{7/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{7/2}} \,d x \] Input:

int((a + a*sin(c + d*x))^4/(e*cos(c + d*x))^(7/2),x)
 

Output:

int((a + a*sin(c + d*x))^4/(e*cos(c + d*x))^(7/2), x)
 

Reduce [F]

\[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{7/2}} \, dx=\frac {\sqrt {e}\, a^{4} \left (5 \cos \left (d x +c \right )^{3} \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}}d x \right ) d +5 \cos \left (d x +c \right )^{3} \left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )^{4}}d x \right ) d +20 \cos \left (d x +c \right )^{3} \left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sin \left (d x +c \right )^{3}}{\cos \left (d x +c \right )^{4}}d x \right ) d +30 \cos \left (d x +c \right )^{3} \left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sin \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{4}}d x \right ) d +8 \sqrt {\cos \left (d x +c \right )}\right )}{5 \cos \left (d x +c \right )^{3} d \,e^{4}} \] Input:

int((a+a*sin(d*x+c))^4/(e*cos(d*x+c))^(7/2),x)
 

Output:

(sqrt(e)*a**4*(5*cos(c + d*x)**3*int(sqrt(cos(c + d*x))/cos(c + d*x)**4,x) 
*d + 5*cos(c + d*x)**3*int((sqrt(cos(c + d*x))*sin(c + d*x)**4)/cos(c + d* 
x)**4,x)*d + 20*cos(c + d*x)**3*int((sqrt(cos(c + d*x))*sin(c + d*x)**3)/c 
os(c + d*x)**4,x)*d + 30*cos(c + d*x)**3*int((sqrt(cos(c + d*x))*sin(c + d 
*x)**2)/cos(c + d*x)**4,x)*d + 8*sqrt(cos(c + d*x))))/(5*cos(c + d*x)**3*d 
*e**4)