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

Optimal result

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

[Out]

-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*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/
sin(1/2*c+1/4*Pi+1/2*d*x)*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/sin(d*x+c)^(
1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2748, 2716, 2721, 2719} \[ \int \frac {a+b \cos (c+d x)}{(e \sin (c+d x))^{3/2}} \, dx=-\frac {2 a E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}-\frac {2 a \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 b}{d e \sqrt {e \sin (c+d x)}} \]

[In]

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

[Out]

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

Rule 2716

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] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2719

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 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Dist[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])

Rubi steps \begin{align*} \text {integral}& = -\frac {2 b}{d e \sqrt {e \sin (c+d x)}}+a \int \frac {1}{(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 {a \int \sqrt {e \sin (c+d x)} \, dx}{e^2} \\ & = -\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 {\left (a \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{e^2 \sqrt {\sin (c+d x)}} \\ & = -\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)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.60 \[ \int \frac {a+b \cos (c+d x)}{(e \sin (c+d x))^{3/2}} \, dx=-\frac {2 \left (b+a \cos (c+d x)-a E\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right ) \sqrt {\sin (c+d x)}\right )}{d e \sqrt {e \sin (c+d x)}} \]

[In]

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

[Out]

(-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]]
)

Maple [A] (verified)

Time = 2.36 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.59

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

[In]

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

[Out]

(2*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(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*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))-2*a
*cos(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 higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.19 \[ \int \frac {a+b \cos (c+d x)}{(e \sin (c+d x))^{3/2}} \, dx=\frac {-i \, \sqrt {2} a \sqrt {-i \, e} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + i \, \sqrt {2} a \sqrt {i \, e} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (a \cos \left (d x + c\right ) + b\right )} \sqrt {e \sin \left (d x + c\right )}}{d e^{2} \sin \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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