3.572 \(\int \frac{(a+b \cos (c+d x))^2}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=68 \[ -\frac{2 \left (a^2-b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{4 a b F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d} \]

[Out]

(-2*(a^2 - b^2)*EllipticE[(c + d*x)/2, 2])/d + (4*a*b*EllipticF[(c + d*x)/2, 2])/d + (2*a^2*Sin[c + d*x])/(d*S
qrt[Cos[c + d*x]])

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Rubi [A]  time = 0.0856865, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2789, 2641, 3012, 2639} \[ -\frac{2 \left (a^2-b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{4 a b F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^2/Cos[c + d*x]^(3/2),x]

[Out]

(-2*(a^2 - b^2)*EllipticE[(c + d*x)/2, 2])/d + (4*a*b*EllipticF[(c + d*x)/2, 2])/d + (2*a^2*Sin[c + d*x])/(d*S
qrt[Cos[c + d*x]])

Rule 2789

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Dist[(2*c*d)/b
, Int[(b*Sin[e + f*x])^(m + 1), x], x] + Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c,
d, e, f, m}, 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 3012

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*Cos[e
+ f*x]*(b*Sin[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Dist[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)), Int[(b*Sin[e
+ f*x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]

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))^2}{\cos ^{\frac{3}{2}}(c+d x)} \, dx &=(2 a b) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\int \frac{a^2+b^2 \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{4 a b F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}-\left (a^2-b^2\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{2 \left (a^2-b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{4 a b F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.276616, size = 62, normalized size = 0.91 \[ \frac{2 \left (\left (b^2-a^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+a \left (\frac{a \sin (c+d x)}{\sqrt{\cos (c+d x)}}+2 b F\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^2/Cos[c + d*x]^(3/2),x]

[Out]

(2*((-a^2 + b^2)*EllipticE[(c + d*x)/2, 2] + a*(2*b*EllipticF[(c + d*x)/2, 2] + (a*Sin[c + d*x])/Sqrt[Cos[c +
d*x]])))/d

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Maple [A]  time = 3.197, size = 202, normalized size = 3. \begin{align*} -2\,{\frac{2\,ab\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 ) +\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 ){a}^{2}-\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 ){b}^{2}-2\,{a}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\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((a+b*cos(d*x+c))^2/cos(d*x+c)^(3/2),x)

[Out]

-2*(2*a*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+
(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))*a^2-(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))*b^2-2*a^2*cos(1/
2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^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{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2/cos(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c) + a)^2/cos(d*x + c)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}{\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((a+b*cos(d*x+c))^2/cos(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2/cos(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c) + a)^2/cos(d*x + c)^(3/2), x)