3.287 \(\int \frac{B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=57 \[ -\frac{2 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 B \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 C F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.0746286, antiderivative size = 57, normalized size of antiderivative = 1., 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, 2636, 2639, 2641} \[ -\frac{2 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 B \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 C F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/Cos[c + d*x]^(5/2),x]

[Out]

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

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 2636

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 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]

Rubi steps

\begin{align*} \int \frac{B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx &=\int \frac{B+C \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=B \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx+C \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 C F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 B \sin (c+d x)}{d \sqrt{\cos (c+d x)}}-B \int \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 )}{d}+\frac{2 B \sin (c+d x)}{d \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.137267, size = 51, normalized size = 0.89 \[ \frac{2 \left (-B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\frac{B \sin (c+d x)}{\sqrt{\cos (c+d x)}}+C F\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/Cos[c + d*x]^(5/2),x]

[Out]

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

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Maple [A]  time = 0.124, size = 148, normalized size = 2.6 \begin{align*} -2\,{\frac{B\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -2\,B\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+C\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) }{\sin \left ( 1/2\,dx+c/2 \right ) \sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}d}} \end{align*}

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)^(5/2),x)

[Out]

-2*(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))-2*B*
cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^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)*Ellipt
icF(cos(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., size = 0, normalized size = 0. \begin{align*} \int \frac{C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

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)^(5/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))/cos(d*x + c)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{C \cos \left (d x + c\right ) + B}{\cos \left (d x + c\right )^{\frac{3}{2}}}, x\right ) \end{align*}

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)^(5/2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c) + B)/cos(d*x + c)^(3/2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

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)^(5/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))/cos(d*x + c)^(5/2), x)