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

Optimal. Leaf size=85 \[ \frac{2 A \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt{b \cos (c+d x)}}+\frac{2 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{b^3 d \sqrt{\cos (c+d x)}} \]

[Out]

(2*B*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(b^3*d*Sqrt[Cos[c + d*x]]) + (2*A*Sqrt[Cos[c + d*x]]*Elli
pticF[(c + d*x)/2, 2])/(b^2*d*Sqrt[b*Cos[c + d*x]])

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Rubi [A]  time = 0.0746879, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {16, 2748, 2642, 2641, 2640, 2639} \[ \frac{2 A \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt{b \cos (c+d x)}}+\frac{2 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{b^3 d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*B*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(b^3*d*Sqrt[Cos[c + d*x]]) + (2*A*Sqrt[Cos[c + d*x]]*Elli
pticF[(c + d*x)/2, 2])/(b^2*d*Sqrt[b*Cos[c + d*x]])

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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 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 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 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, 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{\cos ^2(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{5/2}} \, dx &=\frac{\int \frac{A+B \cos (c+d x)}{\sqrt{b \cos (c+d x)}} \, dx}{b^2}\\ &=\frac{A \int \frac{1}{\sqrt{b \cos (c+d x)}} \, dx}{b^2}+\frac{B \int \sqrt{b \cos (c+d x)} \, dx}{b^3}\\ &=\frac{\left (A \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{b^2 \sqrt{b \cos (c+d x)}}+\frac{\left (B \sqrt{b \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{b^3 \sqrt{\cos (c+d x)}}\\ &=\frac{2 B \sqrt{b \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d \sqrt{\cos (c+d x)}}+\frac{2 A \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt{b \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.0455693, size = 57, normalized size = 0.67 \[ \frac{2 \sqrt{\cos (c+d x)} \left (A F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{b^2 d \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 2.763, size = 163, normalized size = 1.9 \begin{align*} -2\,{\frac{\sqrt{b \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1} \left ( A{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -B{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \right ) }{{b}^{2}\sqrt{-b \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}- \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }\sin \left ( 1/2\,dx+c/2 \right ) \sqrt{b \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) }d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(b*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*
c)^2+1)^(1/2)*(A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/b^2/(-b*(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)/(b*(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{{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{\left (b \cos \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*cos(d*x + c) + A)*cos(d*x + c)^2/(b*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{{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt{b \cos \left (d x + c\right )}}{b^{3} \cos \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c))/(b^3*cos(d*x + c)), 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(cos(d*x+c)**2*(A+B*cos(d*x+c))/(b*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{{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{\left (b \cos \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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