\(\int \frac {a+b \cos (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx\) [39]

Optimal result

Integrand size = 23, antiderivative size = 102 \[ \int \frac {a+b \cos (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx=-\frac {2 b}{3 d e (e \sin (c+d x))^{3/2}}-\frac {2 a \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}+\frac {2 a \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{3 d e^2 \sqrt {e \sin (c+d x)}} \] Output:

-2/3*b/d/e/(e*sin(d*x+c))^(3/2)-2/3*a*cos(d*x+c)/d/e/(e*sin(d*x+c))^(3/2)+ 
2/3*a*InverseJacobiAM(1/2*c-1/4*Pi+1/2*d*x,2^(1/2))*sin(d*x+c)^(1/2)/d/e^2 
/(e*sin(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.58 \[ \int \frac {a+b \cos (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx=-\frac {2 \left (b+a \cos (c+d x)+a \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right ) \sin ^{\frac {3}{2}}(c+d x)\right )}{3 d e (e \sin (c+d x))^{3/2}} \] Input:

Integrate[(a + b*Cos[c + d*x])/(e*Sin[c + d*x])^(5/2),x]
 

Output:

(-2*(b + a*Cos[c + d*x] + a*EllipticF[(-2*c + Pi - 2*d*x)/4, 2]*Sin[c + d* 
x]^(3/2)))/(3*d*e*(e*Sin[c + d*x])^(3/2))
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 103, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3148

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3116

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

(-2*b)/(3*d*e*(e*Sin[c + d*x])^(3/2)) + a*((-2*Cos[c + d*x])/(3*d*e*(e*Sin 
[c + d*x])^(3/2)) + (2*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]] 
)/(3*d*e^2*Sqrt[e*Sin[c + d*x]]))
 

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])
 

Maple [A] (verified)

Time = 2.42 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.22

method	result	size
default	\(\frac {-\frac {2 b}{3 e \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {a \left (\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{\frac {5}{2}} \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )^{3}+2 \sin \left (d x +c \right )\right )}{3 e^{2} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(124\)
parts	\(-\frac {a \left (\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{\frac {5}{2}} \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )^{3}+2 \sin \left (d x +c \right )\right )}{3 e^{2} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}-\frac {2 b}{3 d e \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}\)	\(126\)

Input:

int((a+cos(d*x+c)*b)/(e*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

(-2/3*b/e/(e*sin(d*x+c))^(3/2)-1/3*a/e^2*((1-sin(d*x+c))^(1/2)*(2+2*sin(d* 
x+c))^(1/2)*sin(d*x+c)^(5/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))-2 
*sin(d*x+c)^3+2*sin(d*x+c))/sin(d*x+c)^2/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/ 
d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.20 \[ \int \frac {a+b \cos (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx=\frac {2 \, {\left ({\left (a \cos \left (d x + c\right )^{2} - a\right )} \sqrt {-\frac {1}{2} i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (a \cos \left (d x + c\right )^{2} - a\right )} \sqrt {\frac {1}{2} i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (a \cos \left (d x + c\right ) + b\right )} \sqrt {e \sin \left (d x + c\right )}\right )}}{3 \, {\left (d e^{3} \cos \left (d x + c\right )^{2} - d e^{3}\right )}} \] Input:

integrate((a+b*cos(d*x+c))/(e*sin(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

2/3*((a*cos(d*x + c)^2 - a)*sqrt(-1/2*I*e)*weierstrassPInverse(4, 0, cos(d 
*x + c) + I*sin(d*x + c)) + (a*cos(d*x + c)^2 - a)*sqrt(1/2*I*e)*weierstra 
ssPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) + (a*cos(d*x + c) + b)*sqr 
t(e*sin(d*x + c)))/(d*e^3*cos(d*x + c)^2 - d*e^3)
 

Sympy [F]

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

integrate((a+b*cos(d*x+c))/(e*sin(d*x+c))**(5/2),x)
 

Output:

Integral((a + b*cos(c + d*x))/(e*sin(c + d*x))**(5/2), x)
 

Maxima [F]

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

integrate((a+b*cos(d*x+c))/(e*sin(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((a+b*cos(d*x+c))/(e*sin(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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