3.37 \(\int \frac {a+b \cos (c+d x)}{\sqrt {e \sin (c+d x)}} \, dx\)

Optimal. Leaf size=66 \[ \frac {2 a \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d \sqrt {e \sin (c+d x)}}+\frac {2 b \sqrt {e \sin (c+d x)}}{d e} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2669, 2642, 2641} \[ \frac {2 a \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d \sqrt {e \sin (c+d x)}}+\frac {2 b \sqrt {e \sin (c+d x)}}{d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2669

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] + 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*} \int \frac {a+b \cos (c+d x)}{\sqrt {e \sin (c+d x)}} \, dx &=\frac {2 b \sqrt {e \sin (c+d x)}}{d e}+a \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx\\ &=\frac {2 b \sqrt {e \sin (c+d x)}}{d e}+\frac {\left (a \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{\sqrt {e \sin (c+d x)}}\\ &=\frac {2 a F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{d \sqrt {e \sin (c+d x)}}+\frac {2 b \sqrt {e \sin (c+d x)}}{d e}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 54, normalized size = 0.82 \[ \frac {2 \left (b \sin (c+d x)-a \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{4} (-2 c-2 d x+\pi )\right |2\right )\right )}{d \sqrt {e \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.12, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b \cos \left (d x + c\right ) + a\right )} \sqrt {e \sin \left (d x + c\right )}}{e \sin \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))/(e*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral((b*cos(d*x + c) + a)*sqrt(e*sin(d*x + c))/(e*sin(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \cos \left (d x + c\right ) + a}{\sqrt {e \sin \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))/(e*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.16, size = 92, normalized size = 1.39 \[ -\frac {a \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right ) \cos \left (d x +c \right ) b}{\cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))/(e*sin(d*x+c))^(1/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \cos \left (d x + c\right ) + a}{\sqrt {e \sin \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))/(e*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.72, size = 50, normalized size = 0.76 \[ -\frac {2\,\sqrt {\sin \left (c+d\,x\right )}\,\left (a\,\mathrm {F}\left (\frac {\pi }{4}-\frac {c}{2}-\frac {d\,x}{2}\middle |2\right )-b\,\sqrt {\sin \left (c+d\,x\right )}\right )}{d\,\sqrt {e\,\sin \left (c+d\,x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*cos(c + d*x))/(e*sin(c + d*x))^(1/2),x)

[Out]

-(2*sin(c + d*x)^(1/2)*(a*ellipticF(pi/4 - c/2 - (d*x)/2, 2) - b*sin(c + d*x)^(1/2)))/(d*(e*sin(c + d*x))^(1/2
))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \cos {\left (c + d x \right )}}{\sqrt {e \sin {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))/(e*sin(d*x+c))**(1/2),x)

[Out]

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

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