3.285 \(\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx\)

Optimal. Leaf size=61 \[ \frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 C F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d} \]

[Out]

2*B*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/3*C*(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))/d+2/3*C*sin(d*x+c)*cos(d*x+c)^(1/2)/
d

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Rubi [A]  time = 0.08, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3010, 2748, 2639, 2635, 2641} \[ \frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 C F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/Sqrt[Cos[c + d*x]],x]

[Out]

(2*B*EllipticE[(c + d*x)/2, 2])/d + (2*C*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*C*Sqrt[Cos[c + d*x]]*Sin[c + d*
x])/(3*d)

Rule 2635

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

Rule 2639

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

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 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3010

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x
_Symbol] :> Dist[1/b, Int[(b*Sin[e + f*x])^(m + 1)*(B + C*Sin[e + f*x]), x], x] /; FreeQ[{b, e, f, B, C, m}, x
]

Rubi steps

\begin {align*} \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx &=\int \sqrt {\cos (c+d x)} (B+C \cos (c+d x)) \, dx\\ &=B \int \sqrt {\cos (c+d x)} \, dx+C \int \cos ^{\frac {3}{2}}(c+d x) \, dx\\ &=\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 C \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {1}{3} C \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 C F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 C \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 53, normalized size = 0.87 \[ \frac {2 \left (3 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+C \left (F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\sin (c+d x) \sqrt {\cos (c+d x)}\right )\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/Sqrt[Cos[c + d*x]],x]

[Out]

(2*(3*B*EllipticE[(c + d*x)/2, 2] + C*(EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*Sin[c + d*x])))/(3*d)

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \cos \left (d x + c\right ) + B\right )} \sqrt {\cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c) + B)*sqrt(cos(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))/sqrt(cos(d*x + c)), x)

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maple [B]  time = 1.63, size = 229, normalized size = 3.75 \[ \frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-4 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x)

[Out]

2/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-4*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+3*B*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-C*(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))+2*C*cos(1/2*d*x+1/
2*c)*sin(1/2*d*x+1/2*c)^2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*
d*x+1/2*c)^2-1)^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))/sqrt(cos(d*x + c)), x)

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mupad [B]  time = 1.06, size = 53, normalized size = 0.87 \[ \frac {2\,B\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3\,d}+\frac {2\,C\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*cos(c + d*x) + C*cos(c + d*x)^2)/cos(c + d*x)^(1/2),x)

[Out]

(2*B*ellipticE(c/2 + (d*x)/2, 2))/d + (2*C*ellipticF(c/2 + (d*x)/2, 2))/(3*d) + (2*C*cos(c + d*x)^(1/2)*sin(c
+ d*x))/(3*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(1/2),x)

[Out]

Timed out

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