\(\int \frac {a+b \cos (c+d x)}{(e \sin (c+d x))^{3/2}} \, dx\) [38]

Optimal result

Integrand size = 23, antiderivative size = 96 \[ \int \frac {a+b \cos (c+d x)}{(e \sin (c+d x))^{3/2}} \, dx=-\frac {2 b}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.60 \[ \int \frac {a+b \cos (c+d x)}{(e \sin (c+d x))^{3/2}} \, dx=-\frac {2 \left (b+a \cos (c+d x)-a E\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right ) \sqrt {\sin (c+d x)}\right )}{d e \sqrt {e \sin (c+d x)}} \] Input:

Integrate[(a + b*Cos[c + d*x])/(e*Sin[c + d*x])^(3/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 97, 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, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3148

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3116

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

Input:

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

Output:

(-2*b)/(d*e*Sqrt[e*Sin[c + d*x]]) + a*((-2*Cos[c + d*x])/(d*e*Sqrt[e*Sin[c 
 + d*x]]) - (2*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(d*e 
^2*Sqrt[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 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 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 = 2.22 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.59

method	result	size
default	\(\frac {2 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) a -a \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-2 a \cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right ) b}{e \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\)	\(153\)
parts	\(\frac {a \left (2 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-2 \cos \left (d x +c \right )^{2}\right )}{e \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}-\frac {2 b}{d e \sqrt {e \sin \left (d x +c \right )}}\)	\(162\)

Input:

int((a+cos(d*x+c)*b)/(e*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

(2*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticE( 
(1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a-a*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c) 
)^(1/2)*sin(d*x+c)^(1/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))-2*a*c 
os(d*x+c)^2-2*cos(d*x+c)*b)/e/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 = 108, normalized size of antiderivative = 1.12 \[ \int \frac {a+b \cos (c+d x)}{(e \sin (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left (i \, a \sqrt {-\frac {1}{2} i \, e} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - i \, a \sqrt {\frac {1}{2} i \, e} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + {\left (a \cos \left (d x + c\right ) + b\right )} \sqrt {e \sin \left (d x + c\right )}\right )}}{d e^{2} \sin \left (d x + c\right )} \] Input:

integrate((a+b*cos(d*x+c))/(e*sin(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

-2*(I*a*sqrt(-1/2*I*e)*sin(d*x + c)*weierstrassZeta(4, 0, weierstrassPInve 
rse(4, 0, cos(d*x + c) + I*sin(d*x + c))) - I*a*sqrt(1/2*I*e)*sin(d*x + c) 
*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x 
+ c))) + (a*cos(d*x + c) + b)*sqrt(e*sin(d*x + c)))/(d*e^2*sin(d*x + c))
 

Sympy [F]

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

integrate((a+b*cos(d*x+c))/(e*sin(d*x+c))**(3/2),x)
 

Output:

Integral((a + b*cos(c + d*x))/(e*sin(c + d*x))**(3/2), x)
 

Maxima [F]

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

integrate((a+b*cos(d*x+c))/(e*sin(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

integrate((b*cos(d*x + c) + a)/(e*sin(d*x + c))^(3/2), x)
 

Giac [F]

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

integrate((a+b*cos(d*x+c))/(e*sin(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

integrate((b*cos(d*x + c) + a)/(e*sin(d*x + c))^(3/2), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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