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

Optimal. Leaf size=166 \[ -\frac{2 \left (5 a^2 (A-C)-b^2 (5 A+3 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{4 a b (3 A+C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{4 a b (3 A-C) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d}+\frac{2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt{\cos (c+d x)}}-\frac{2 b^2 (5 A-C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d} \]

[Out]

(-2*(5*a^2*(A - C) - b^2*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a*b*(3*A + C)*EllipticF[(c + d*x)/
2, 2])/(3*d) - (4*a*b*(3*A - C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d) - (2*b^2*(5*A - C)*Cos[c + d*x]^(3/2)*S
in[c + d*x])/(5*d) + (2*A*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])

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Rubi [A]  time = 0.406133, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {3048, 3033, 3023, 2748, 2641, 2639} \[ -\frac{2 \left (5 a^2 (A-C)-b^2 (5 A+3 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{4 a b (3 A+C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{4 a b (3 A-C) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d}+\frac{2 A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt{\cos (c+d x)}}-\frac{2 b^2 (5 A-C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(5*a^2*(A - C) - b^2*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a*b*(3*A + C)*EllipticF[(c + d*x)/
2, 2])/(3*d) - (4*a*b*(3*A - C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d) - (2*b^2*(5*A - C)*Cos[c + d*x]^(3/2)*S
in[c + d*x])/(5*d) + (2*A*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])

Rule 3048

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m
- 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(m + n + 2) + C*(c^2*(
m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3033

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

Rule 3023

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

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 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))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx &=\frac{2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+2 \int \frac{(a+b \cos (c+d x)) \left (2 A b-\frac{1}{2} a (A-C) \cos (c+d x)-\frac{1}{2} b (5 A-C) \cos ^2(c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{2 b^2 (5 A-C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{4}{5} \int \frac{5 a A b-\frac{1}{4} \left (5 a^2 (A-C)-b^2 (5 A+3 C)\right ) \cos (c+d x)-\frac{5}{2} a b (3 A-C) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{4 a b (3 A-C) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}-\frac{2 b^2 (5 A-C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{8}{15} \int \frac{\frac{5}{4} a b (3 A+C)-\frac{3}{8} \left (5 a^2 (A-C)-b^2 (5 A+3 C)\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{4 a b (3 A-C) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}-\frac{2 b^2 (5 A-C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{1}{3} (2 a b (3 A+C)) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\frac{1}{5} \left (-5 a^2 (A-C)+b^2 (5 A+3 C)\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{2 \left (5 a^2 (A-C)-b^2 (5 A+3 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{4 a b (3 A+C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{4 a b (3 A-C) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}-\frac{2 b^2 (5 A-C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.03038, size = 119, normalized size = 0.72 \[ \frac{-6 \left (5 a^2 (A-C)-b^2 (5 A+3 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\frac{\sin (c+d x) \left (30 a^2 A+20 a b C \cos (c+d x)+3 b^2 C \cos (2 (c+d x))+3 b^2 C\right )}{\sqrt{\cos (c+d x)}}+20 a b (3 A+C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*(5*a^2*(A - C) - b^2*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2] + 20*a*b*(3*A + C)*EllipticF[(c + d*x)/2, 2] +
((30*a^2*A + 3*b^2*C + 20*a*b*C*Cos[c + d*x] + 3*b^2*C*Cos[2*(c + d*x)])*Sin[c + d*x])/Sqrt[Cos[c + d*x]])/(1
5*d)

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Maple [B]  time = 0.457, size = 694, normalized size = 4.2 \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+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2),x)

[Out]

-2/15*(-24*C*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+
8*C*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*b*(5*a+3*b)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-2
*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(15*A*a^2+10*C*a*b+3*C*b^2)*sin(1/2*d*x+1/2*c)^2*cos(1/2
*d*x+1/2*c)+30*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)*EllipticF(cos(1/2*d*x+1/2*c
),2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+15*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2
)*a^2-15*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))
*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*b^2+10*a*b*C*(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*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-
15*C*(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))*(-2*s
in(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2-9*C*(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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(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 (C \cos \left (d x + c\right )^{2} + A\right )}{\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*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(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{C b^{2} \cos \left (d x + c\right )^{4} + 2 \, C a b \cos \left (d x + c\right )^{3} + 2 \, A a b \cos \left (d x + c\right ) + A a^{2} +{\left (C a^{2} + A b^{2}\right )} \cos \left (d x + c\right )^{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*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

integral((C*b^2*cos(d*x + c)^4 + 2*C*a*b*cos(d*x + c)^3 + 2*A*a*b*cos(d*x + c) + A*a^2 + (C*a^2 + A*b^2)*cos(d
*x + c)^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*(A+C*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 (C \cos \left (d x + c\right )^{2} + A\right )}{\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*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2),x, algorithm="giac")

[Out]

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