3.6.45 \(\int \frac {a+b \sin (c+d x)}{(e \cos (c+d x))^{5/2}} \, dx\) [545]

3.6.45.1 Optimal result
3.6.45.2 Mathematica [A] (verified)
3.6.45.3 Rubi [A] (verified)
3.6.45.4 Maple [A] (verified)
3.6.45.5 Fricas [C] (verification not implemented)
3.6.45.6 Sympy [F(-1)]
3.6.45.7 Maxima [F]
3.6.45.8 Giac [F]
3.6.45.9 Mupad [F(-1)]

3.6.45.1 Optimal result

Integrand size = 23, antiderivative size = 97 \[ \int \frac {a+b \sin (c+d x)}{(e \cos (c+d x))^{5/2}} \, dx=\frac {2 b}{3 d e (e \cos (c+d x))^{3/2}}+\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d e^2 \sqrt {e \cos (c+d x)}}+\frac {2 a \sin (c+d x)}{3 d e (e \cos (c+d x))^{3/2}} \]

output
2/3*b/d/e/(e*cos(d*x+c))^(3/2)+2/3*a*sin(d*x+c)/d/e/(e*cos(d*x+c))^(3/2)+2 
/3*a*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x 
+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/d/e^2/(e*cos(d*x+c))^(1/2)
 
3.6.45.2 Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.57 \[ \int \frac {a+b \sin (c+d x)}{(e \cos (c+d x))^{5/2}} \, dx=\frac {2 \left (b+a \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+a \sin (c+d x)\right )}{3 d e (e \cos (c+d x))^{3/2}} \]

input
Integrate[(a + b*Sin[c + d*x])/(e*Cos[c + d*x])^(5/2),x]
 
output
(2*(b + a*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] + a*Sin[c + d*x]))/ 
(3*d*e*(e*Cos[c + d*x])^(3/2))
 
3.6.45.3 Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3148, 3042, 3116, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3148

\(\displaystyle a \int \frac {1}{(e \cos (c+d x))^{5/2}}dx+\frac {2 b}{3 d e (e \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3116

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

input
Int[(a + b*Sin[c + d*x])/(e*Cos[c + d*x])^(5/2),x]
 
output
(2*b)/(3*d*e*(e*Cos[c + d*x])^(3/2)) + a*((2*Sqrt[Cos[c + d*x]]*EllipticF[ 
(c + d*x)/2, 2])/(3*d*e^2*Sqrt[e*Cos[c + d*x]]) + (2*Sin[c + d*x])/(3*d*e* 
(e*Cos[c + d*x])^(3/2)))
 

3.6.45.3.1 Defintions of rubi rules used

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

rule 3116
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1))   I 
nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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 3148
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + 
 Simp[a   Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && 
 (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
 
3.6.45.4 Maple [A] (verified)

Time = 2.32 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.99

method result size
default \(-\frac {2 \left (2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a +b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{2} d}\) \(193\)
parts \(-\frac {2 a \left (-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{3 e^{2} \sqrt {-e \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d}+\frac {2 b}{3 d e \left (e \cos \left (d x +c \right )\right )^{\frac {3}{2}}}\) \(262\)

input
int((a+b*sin(d*x+c))/(e*cos(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 
output
-2/3/(2*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^2*(2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(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*a+2*cos 
(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a-(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))*a+b*sin(1 
/2*d*x+1/2*c))/d
 
3.6.45.5 Fricas [C] (verification not implemented)

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

Time = 0.09 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.12 \[ \int \frac {a+b \sin (c+d x)}{(e \cos (c+d x))^{5/2}} \, dx=\frac {-i \, \sqrt {2} a \sqrt {e} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} a \sqrt {e} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, \sqrt {e \cos \left (d x + c\right )} {\left (a \sin \left (d x + c\right ) + b\right )}}{3 \, d e^{3} \cos \left (d x + c\right )^{2}} \]

input
integrate((a+b*sin(d*x+c))/(e*cos(d*x+c))^(5/2),x, algorithm="fricas")
 
output
1/3*(-I*sqrt(2)*a*sqrt(e)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d* 
x + c) + I*sin(d*x + c)) + I*sqrt(2)*a*sqrt(e)*cos(d*x + c)^2*weierstrassP 
Inverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*sqrt(e*cos(d*x + c))*(a* 
sin(d*x + c) + b))/(d*e^3*cos(d*x + c)^2)
 
3.6.45.6 Sympy [F(-1)]

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

input
integrate((a+b*sin(d*x+c))/(e*cos(d*x+c))**(5/2),x)
 
output
Timed out
 
3.6.45.7 Maxima [F]

\[ \int \frac {a+b \sin (c+d x)}{(e \cos (c+d x))^{5/2}} \, dx=\int { \frac {b \sin \left (d x + c\right ) + a}{\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]

input
integrate((a+b*sin(d*x+c))/(e*cos(d*x+c))^(5/2),x, algorithm="maxima")
 
output
integrate((b*sin(d*x + c) + a)/(e*cos(d*x + c))^(5/2), x)
 
3.6.45.8 Giac [F]

\[ \int \frac {a+b \sin (c+d x)}{(e \cos (c+d x))^{5/2}} \, dx=\int { \frac {b \sin \left (d x + c\right ) + a}{\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]

input
integrate((a+b*sin(d*x+c))/(e*cos(d*x+c))^(5/2),x, algorithm="giac")
 
output
integrate((b*sin(d*x + c) + a)/(e*cos(d*x + c))^(5/2), x)
 
3.6.45.9 Mupad [F(-1)]

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

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