3.702 \(\int \frac{(a+b \cos (c+d x))^2}{\sqrt{\sec (c+d x)}} \, dx\)

Optimal. Leaf size=141 \[ \frac{2 \left (5 a^2+3 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{4 a b \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{4 a b \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 b^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)} \]

[Out]

(2*(5*a^2 + 3*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a*b*Sqrt[Cos[c
+ d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*b^2*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) +
(4*a*b*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]])

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Rubi [A]  time = 0.14436, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3238, 3788, 3769, 3771, 2641, 4045, 2639} \[ \frac{2 \left (5 a^2+3 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{4 a b \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{4 a b \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 b^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(5*a^2 + 3*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a*b*Sqrt[Cos[c
+ d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*b^2*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) +
(4*a*b*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]])

Rule 3238

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Csc[e + f*x])^(m - n*p)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rule 3788

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[(2*a*b)/
d, Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

Rule 3769

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

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

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 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[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}{\sqrt{\sec (c+d x)}} \, dx &=\int \frac{(b+a \sec (c+d x))^2}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=(2 a b) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx+\int \frac{b^2+a^2 \sec ^2(c+d x)}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 b^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a b \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{1}{3} (2 a b) \int \sqrt{\sec (c+d x)} \, dx-\frac{1}{5} \left (-5 a^2-3 b^2\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{2 b^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a b \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{1}{3} \left (2 a b \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx-\frac{1}{5} \left (\left (-5 a^2-3 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{2 \left (5 a^2+3 b^2\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{5 d}+\frac{4 a b \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 d}+\frac{2 b^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 a b \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.419755, size = 100, normalized size = 0.71 \[ \frac{\sqrt{\sec (c+d x)} \left (6 \left (5 a^2+3 b^2\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+b \sin (2 (c+d x)) (10 a+3 b \cos (c+d x))+20 a b \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sec[c + d*x]]*(6*(5*a^2 + 3*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 20*a*b*Sqrt[Cos[c + d*x]
]*EllipticF[(c + d*x)/2, 2] + b*(10*a + 3*b*Cos[c + d*x])*Sin[2*(c + d*x)]))/(15*d)

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Maple [A]  time = 3.155, size = 321, normalized size = 2.3 \begin{align*} -{\frac{2}{15\,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 ( -24\,{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+ \left ( 40\,ab+24\,{b}^{2} \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( -20\,ab-6\,{b}^{2} \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +10\,\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) ab-15\,\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}-9\,\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} \right ){\frac{1}{\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 ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\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))^2/sec(d*x+c)^(1/2),x)

[Out]

-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-24*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6
+(40*a*b+24*b^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-20*a*b-6*b^2)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*
c)+10*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b-
15*(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^2-9*(
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*si
n(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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

integral((b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)/sqrt(sec(d*x + c)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \cos{\left (c + d x \right )}\right )^{2}}{\sqrt{\sec{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**2/sec(d*x+c)**(1/2),x)

[Out]

Integral((a + b*cos(c + d*x))**2/sqrt(sec(c + d*x)), x)

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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}}{\sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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