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

Optimal. Leaf size=140 \[ \frac{2 \left (3 a^2 B+2 a b C+b^2 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 \left (5 a (a C+2 b B)+3 b^2 C\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 b (7 a C+5 b B) \sin (c+d x) \sqrt{\cos (c+d x)}}{15 d}+\frac{2 b C \sin (c+d x) \sqrt{\cos (c+d x)} (a+b \cos (c+d x))}{5 d} \]

[Out]

(2*(3*b^2*C + 5*a*(2*b*B + a*C))*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(3*a^2*B + b^2*B + 2*a*b*C)*EllipticF[(
c + d*x)/2, 2])/(3*d) + (2*b*(5*b*B + 7*a*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(15*d) + (2*b*C*Sqrt[Cos[c + d*x
]]*(a + b*Cos[c + d*x])*Sin[c + d*x])/(5*d)

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Rubi [A]  time = 0.356083, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3029, 2990, 3023, 2748, 2641, 2639} \[ \frac{2 \left (3 a^2 B+2 a b C+b^2 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 \left (5 a (a C+2 b B)+3 b^2 C\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 b (7 a C+5 b B) \sin (c+d x) \sqrt{\cos (c+d x)}}{15 d}+\frac{2 b C \sin (c+d x) \sqrt{\cos (c+d x)} (a+b \cos (c+d x))}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(3*b^2*C + 5*a*(2*b*B + a*C))*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(3*a^2*B + b^2*B + 2*a*b*C)*EllipticF[(
c + d*x)/2, 2])/(3*d) + (2*b*(5*b*B + 7*a*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(15*d) + (2*b*C*Sqrt[Cos[c + d*x
]]*(a + b*Cos[c + d*x])*Sin[c + d*x])/(5*d)

Rule 3029

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

Rule 2990

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x
])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*(m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c -
b*d*(m + n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m,
1] &&  !(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

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 (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx &=\int \frac{(a+b \cos (c+d x))^2 (B+C \cos (c+d x))}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 b C \sqrt{\cos (c+d x)} (a+b \cos (c+d x)) \sin (c+d x)}{5 d}+\frac{2}{5} \int \frac{\frac{1}{2} a (5 a B+b C)+\frac{1}{2} \left (3 b^2 C+5 a (2 b B+a C)\right ) \cos (c+d x)+\frac{1}{2} b (5 b B+7 a C) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 b (5 b B+7 a C) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 b C \sqrt{\cos (c+d x)} (a+b \cos (c+d x)) \sin (c+d x)}{5 d}+\frac{4}{15} \int \frac{\frac{5}{4} \left (3 a^2 B+b^2 B+2 a b C\right )+\frac{3}{4} \left (3 b^2 C+5 a (2 b B+a C)\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 b (5 b B+7 a C) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 b C \sqrt{\cos (c+d x)} (a+b \cos (c+d x)) \sin (c+d x)}{5 d}+\frac{1}{3} \left (3 a^2 B+b^2 B+2 a b C\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\frac{1}{5} \left (3 b^2 C+5 a (2 b B+a C)\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{2 \left (3 b^2 C+5 a (2 b B+a C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 \left (3 a^2 B+b^2 B+2 a b C\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 b (5 b B+7 a C) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 b C \sqrt{\cos (c+d x)} (a+b \cos (c+d x)) \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.569829, size = 106, normalized size = 0.76 \[ \frac{2 \left (5 \left (3 a^2 B+2 a b C+b^2 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+3 \left (5 a^2 C+10 a b B+3 b^2 C\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+b \sin (c+d x) \sqrt{\cos (c+d x)} (10 a C+5 b B+3 b C \cos (c+d x))\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.742, size = 487, normalized size = 3.5 \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*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2),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*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)
^6+(20*B*b^2+40*C*a*b+24*C*b^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-10*B*b^2-20*C*a*b-6*C*b^2)*sin(1/2*d
*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+15*a^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)*Elliptic
F(cos(1/2*d*x+1/2*c),2^(1/2))+5*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)*EllipticF(
cos(1/2*d*x+1/2*c),2^(1/2))-30*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)*EllipticE(cos(1
/2*d*x+1/2*c),2^(1/2))*a*b+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(co
s(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))*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))*b^2)/(-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 (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\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*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

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