3.3.11 \(\int \frac {(a+a \sin (c+d x))^2}{(e \cos (c+d x))^{7/2}} \, dx\) [211]

3.3.11.1 Optimal result
3.3.11.2 Mathematica [C] (verified)
3.3.11.3 Rubi [A] (verified)
3.3.11.4 Maple [B] (verified)
3.3.11.5 Fricas [C] (verification not implemented)
3.3.11.6 Sympy [F(-1)]
3.3.11.7 Maxima [F]
3.3.11.8 Giac [F]
3.3.11.9 Mupad [F(-1)]

3.3.11.1 Optimal result

Integrand size = 25, antiderivative size = 127 \[ \int \frac {(a+a \sin (c+d x))^2}{(e \cos (c+d x))^{7/2}} \, dx=-\frac {2 a^2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^4 \sqrt {\cos (c+d x)}}+\frac {2 a^4 (e \cos (c+d x))^{3/2}}{5 d e^5 (a-a \sin (c+d x))^2}+\frac {2 a^4 (e \cos (c+d x))^{3/2}}{5 d e^5 \left (a^2-a^2 \sin (c+d x)\right )} \]

output
2/5*a^4*(e*cos(d*x+c))^(3/2)/d/e^5/(a-a*sin(d*x+c))^2+2/5*a^4*(e*cos(d*x+c 
))^(3/2)/d/e^5/(a^2-a^2*sin(d*x+c))-2/5*a^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/c 
os(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*(e*cos(d*x+c))^(1/ 
2)/d/e^4/cos(d*x+c)^(1/2)
 
3.3.11.2 Mathematica [C] (verified)

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

Time = 0.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.52 \[ \int \frac {(a+a \sin (c+d x))^2}{(e \cos (c+d x))^{7/2}} \, dx=\frac {2\ 2^{3/4} a^2 \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{4},-\frac {1}{4},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{5/4}}{5 d e (e \cos (c+d x))^{5/2}} \]

input
Integrate[(a + a*Sin[c + d*x])^2/(e*Cos[c + d*x])^(7/2),x]
 
output
(2*2^(3/4)*a^2*Hypergeometric2F1[-5/4, 1/4, -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))
 
3.3.11.3 Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.98, 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, 3160, 3042, 3162, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3149

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3160

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3162

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

input
Int[(a + a*Sin[c + d*x])^2/(e*Cos[c + d*x])^(7/2),x]
 
output
(a^4*((2*(e*Cos[c + d*x])^(3/2))/(5*d*e*(a - a*Sin[c + d*x])^2) + ((-2*Sqr 
t[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(a*d*Sqrt[Cos[c + d*x]]) + (2 
*(e*Cos[c + d*x])^(3/2))/(d*e*(a - a*Sin[c + d*x])))/(5*a)))/e^4
 

3.3.11.3.1 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 3160
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x 
])^m/(a*f*g*(2*m + p + 1))), x] + Simp[(m + p + 1)/(a*(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] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] & 
& IntegersQ[2*m, 2*p]
 

rule 3162
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[b*((g*Cos[e + f*x])^(p + 1)/(a*f*g*(p - 1)*(a + b*S 
in[e + f*x]))), x] + Simp[p/(a*(p - 1))   Int[(g*Cos[e + f*x])^p, x], x] /; 
 FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] &&  !GeQ[p, 1] && Intege 
rQ[2*p]
 
3.3.11.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(305\) vs. \(2(139)=278\).

Time = 6.76 (sec) , antiderivative size = 306, normalized size of antiderivative = 2.41

method result size
default \(\frac {2 \left (8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-4 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{5 \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{3} d}\) \(306\)
parts \(\text {Expression too large to display}\) \(758\)

input
int((a+a*sin(d*x+c))^2/(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*(8*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x 
+1/2*c)-4*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-8*sin(1/2*d*x+1/2 
*c)^4*cos(1/2*d*x+1/2*c)+4*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+ 
6*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(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))+2*sin( 
1/2*d*x+1/2*c))*a^2/d
 
3.3.11.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 313, normalized size of antiderivative = 2.46 \[ \int \frac {(a+a \sin (c+d x))^2}{(e \cos (c+d x))^{7/2}} \, dx=\frac {{\left (-i \, \sqrt {2} a^{2} \cos \left (d x + c\right )^{2} + i \, \sqrt {2} a^{2} \cos \left (d x + c\right ) + 2 i \, \sqrt {2} a^{2} + {\left (-i \, \sqrt {2} a^{2} \cos \left (d x + c\right ) - 2 i \, \sqrt {2} a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + {\left (i \, \sqrt {2} a^{2} \cos \left (d x + c\right )^{2} - i \, \sqrt {2} a^{2} \cos \left (d x + c\right ) - 2 i \, \sqrt {2} a^{2} + {\left (i \, \sqrt {2} a^{2} \cos \left (d x + c\right ) + 2 i \, \sqrt {2} a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \cos \left (d x + c\right ) + a^{2} - {\left (a^{2} \cos \left (d x + c\right ) - a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}}{5 \, {\left (d e^{4} \cos \left (d x + c\right )^{2} - d e^{4} \cos \left (d x + c\right ) - 2 \, d e^{4} + {\left (d e^{4} \cos \left (d x + c\right ) + 2 \, d e^{4}\right )} \sin \left (d x + c\right )\right )}} \]

input
integrate((a+a*sin(d*x+c))^2/(e*cos(d*x+c))^(7/2),x, algorithm="fricas")
 
output
1/5*((-I*sqrt(2)*a^2*cos(d*x + c)^2 + I*sqrt(2)*a^2*cos(d*x + c) + 2*I*sqr 
t(2)*a^2 + (-I*sqrt(2)*a^2*cos(d*x + c) - 2*I*sqrt(2)*a^2)*sin(d*x + c))*s 
qrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I* 
sin(d*x + c))) + (I*sqrt(2)*a^2*cos(d*x + c)^2 - I*sqrt(2)*a^2*cos(d*x + c 
) - 2*I*sqrt(2)*a^2 + (I*sqrt(2)*a^2*cos(d*x + c) + 2*I*sqrt(2)*a^2)*sin(d 
*x + c))*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x 
 + c) - I*sin(d*x + c))) - 2*(a^2*cos(d*x + c)^2 + 2*a^2*cos(d*x + c) + a^ 
2 - (a^2*cos(d*x + c) - a^2)*sin(d*x + c))*sqrt(e*cos(d*x + c)))/(d*e^4*co 
s(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))
 
3.3.11.6 Sympy [F(-1)]

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

input
integrate((a+a*sin(d*x+c))**2/(e*cos(d*x+c))**(7/2),x)
 
output
Timed out
 
3.3.11.7 Maxima [F]

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

input
integrate((a+a*sin(d*x+c))^2/(e*cos(d*x+c))^(7/2),x, algorithm="maxima")
 
output
integrate((a*sin(d*x + c) + a)^2/(e*cos(d*x + c))^(7/2), x)
 
3.3.11.8 Giac [F]

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

input
integrate((a+a*sin(d*x+c))^2/(e*cos(d*x+c))^(7/2),x, algorithm="giac")
 
output
integrate((a*sin(d*x + c) + a)^2/(e*cos(d*x + c))^(7/2), x)
 
3.3.11.9 Mupad [F(-1)]

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

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