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

Optimal. Leaf size=213 \[ \frac{2 \left (7 a^2 B+14 a A b+5 b^2 B\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 \left (7 a^2 B+14 a A b+5 b^2 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 \left (5 a^2 A+6 a b B+3 A b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 b (2 a B+A b) \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 b^2 B \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]

[Out]

(2*(5*a^2*A + 3*A*b^2 + 6*a*b*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*(
14*a*A*b + 7*a^2*B + 5*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*b^2
*B*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (2*b*(A*b + 2*a*B)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (2*(14
*a*A*b + 7*a^2*B + 5*b^2*B)*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]])

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Rubi [A]  time = 0.372137, antiderivative size = 213, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used = {2960, 4024, 4047, 3769, 3771, 2641, 4045, 2639} \[ \frac{2 \left (7 a^2 B+14 a A b+5 b^2 B\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 \left (7 a^2 B+14 a A b+5 b^2 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 \left (5 a^2 A+6 a b B+3 A b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 b (2 a B+A b) \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 b^2 B \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(5*a^2*A + 3*A*b^2 + 6*a*b*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*(
14*a*A*b + 7*a^2*B + 5*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*b^2
*B*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (2*b*(A*b + 2*a*B)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (2*(14
*a*A*b + 7*a^2*B + 5*b^2*B)*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]])

Rule 2960

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(
d + c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 4024

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2*(csc[(e_.) + (f_.)*(x_)]*(B_
.) + (A_)), x_Symbol] :> Simp[(a^2*A*Cos[e + f*x]*(d*Csc[e + f*x])^(n + 1))/(d*f*n), x] + Dist[1/(d*n), Int[(d
*Csc[e + f*x])^(n + 1)*(a*(2*A*b + a*B)*n + (2*a*b*B*n + A*(b^2*n + a^2*(n + 1)))*Csc[e + f*x] + b^2*B*n*Csc[e
+ f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, 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 (A+B \cos (c+d x))}{\sqrt{\sec (c+d x)}} \, dx &=\int \frac{(b+a \sec (c+d x))^2 (B+A \sec (c+d x))}{\sec ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 b^2 B \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}-\frac{2}{7} \int \frac{-\frac{7}{2} b (A b+2 a B)+\left (-7 a A b+\left (-\frac{7 a^2}{2}-\frac{5 b^2}{2}\right ) B\right ) \sec (c+d x)-\frac{7}{2} a^2 A \sec ^2(c+d x)}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 b^2 B \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}-\frac{2}{7} \int \frac{-\frac{7}{2} b (A b+2 a B)-\frac{7}{2} a^2 A \sec ^2(c+d x)}{\sec ^{\frac{5}{2}}(c+d x)} \, dx-\frac{1}{7} \left (-14 a A b-7 a^2 B-5 b^2 B\right ) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 b^2 B \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 b (A b+2 a B) \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (14 a A b+7 a^2 B+5 b^2 B\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}-\frac{1}{5} \left (-5 a^2 A-3 A b^2-6 a b B\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx-\frac{1}{21} \left (-14 a A b-7 a^2 B-5 b^2 B\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=\frac{2 b^2 B \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 b (A b+2 a B) \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (14 a A b+7 a^2 B+5 b^2 B\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}-\frac{1}{5} \left (\left (-5 a^2 A-3 A b^2-6 a b B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx-\frac{1}{21} \left (\left (-14 a A b-7 a^2 B-5 b^2 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 \left (5 a^2 A+3 A b^2+6 a b B\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{2 \left (14 a A b+7 a^2 B+5 b^2 B\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{21 d}+\frac{2 b^2 B \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 b (A b+2 a B) \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (14 a A b+7 a^2 B+5 b^2 B\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.33155, size = 161, normalized size = 0.76 \[ \frac{\sqrt{\sec (c+d x)} \left (\sin (2 (c+d x)) \left (5 \left (14 a^2 B+28 a A b+3 b^2 B \cos (2 (c+d x))+13 b^2 B\right )+42 b (2 a B+A b) \cos (c+d x)\right )+20 \left (7 a^2 B+14 a A b+5 b^2 B\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+84 \left (5 a^2 A+6 a b B+3 A b^2\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{210 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sec[c + d*x]]*(84*(5*a^2*A + 3*A*b^2 + 6*a*b*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 20*(14*a*
A*b + 7*a^2*B + 5*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + (42*b*(A*b + 2*a*B)*Cos[c + d*x] + 5*(
28*a*A*b + 14*a^2*B + 13*b^2*B + 3*b^2*B*Cos[2*(c + d*x)]))*Sin[2*(c + d*x)]))/(210*d)

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Maple [B]  time = 3.781, size = 548, normalized size = 2.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c))/sec(d*x+c)^(1/2),x)

[Out]

-2/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*b^2*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^8+(-168*A*b^2-336*B*a*b-360*B*b^2)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(280*A*a*b+168*A*b^2+140*B*a^2+33
6*B*a*b+280*B*b^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-140*A*a*b-42*A*b^2-70*B*a^2-84*B*a*b-80*B*b^2)*si
n(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+70*A*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)*E
llipticF(cos(1/2*d*x+1/2*c),2^(1/2))-105*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)*Ellip
ticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2-63*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)*Ellipt
icE(cos(1/2*d*x+1/2*c),2^(1/2))*b^2+35*B*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)*Ell
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))+25*b^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)*Elli
pticF(cos(1/2*d*x+1/2*c),2^(1/2))-126*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)*Elliptic
E(cos(1/2*d*x+1/2*c),2^(1/2))*a*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)/(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 )}{\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*(A+B*cos(d*x+c))/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*(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 b^{2} \cos \left (d x + c\right )^{3} + A a^{2} +{\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2} +{\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )}{\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*(A+B*cos(d*x+c))/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

integral((B*b^2*cos(d*x + c)^3 + A*a^2 + (2*B*a*b + A*b^2)*cos(d*x + c)^2 + (B*a^2 + 2*A*a*b)*cos(d*x + c))/sq
rt(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 ) \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*(A+B*cos(d*x+c))/sec(d*x+c)**(1/2),x)

[Out]

Integral((A + B*cos(c + d*x))*(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 )}{\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*(A+B*cos(d*x+c))/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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