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

Optimal. Leaf size=83 \[ \frac{2 A F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 A \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{2 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 B \sin (c+d x)}{d \sqrt{\cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0593403, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2748, 2636, 2641, 2639} \[ \frac{2 A F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 A \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{2 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 B \sin (c+d x)}{d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.39278, size = 65, normalized size = 0.78 \[ \frac{\frac{2 \sin (c+d x) (A+3 B \cos (c+d x))}{\cos ^{\frac{3}{2}}(c+d x)}+2 A F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-6 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*B*EllipticE[(c + d*x)/2, 2] + 2*A*EllipticF[(c + d*x)/2, 2] + (2*(A + 3*B*Cos[c + d*x])*Sin[c + d*x])/Cos[
c + d*x]^(3/2))/(3*d)

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Maple [B]  time = 6.086, size = 397, normalized size = 4.8 \begin{align*}{\frac{2}{3\,d}\sqrt{- \left ( -2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 2\,A\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 ) \sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+6\,B\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 ) \sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-12\,B\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-A\sqrt{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{2} \right ) +2\,A\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-3\,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 ) +6\,B \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) \right ) \sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 4\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-4\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{-1} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c))/cos(d*x+c)^(5/2),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*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1
)/sin(1/2*d*x+1/2*c)^3*(2*A*(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))*(sin(1/2*d*
x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+6*B*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2-12*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-A*(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*A*cos(1/2*d*x+1/2
*c)*sin(1/2*d*x+1/2*c)^2-3*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))+6*B*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)
^(1/2)/(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{B \cos \left (d x + c\right ) + A}{\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((A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)/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{B \cos \left (d x + c\right ) + A}{\cos \left (d x + c\right )^{\frac{5}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

integral((B*cos(d*x + c) + A)/cos(d*x + c)^(5/2), 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((A+B*cos(d*x+c))/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{B \cos \left (d x + c\right ) + A}{\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((A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x, algorithm="giac")

[Out]

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