3.77 \(\int \sqrt{a+a \cos (c+d x)} (A+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=95 \[ \frac{2 a (15 A+7 C) \sin (c+d x)}{15 d \sqrt{a \cos (c+d x)+a}}+\frac{2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}-\frac{4 C \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{15 d} \]

[Out]

(2*a*(15*A + 7*C)*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) - (4*C*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/
(15*d) + (2*C*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*a*d)

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Rubi [A]  time = 0.127987, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {3024, 2751, 2646} \[ \frac{2 a (15 A+7 C) \sin (c+d x)}{15 d \sqrt{a \cos (c+d x)+a}}+\frac{2 C \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}-\frac{4 C \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(15*A + 7*C)*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) - (4*C*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/
(15*d) + (2*C*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*a*d)

Rule 3024

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (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) - a*C*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] &&  !LtQ[
m, -1]

Rule 2751

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

Rule 2646

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(-2*b*Cos[c + d*x])/(d*Sqrt[a + b*Sin[c + d*
x]]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \sqrt{a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac{2 C (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 a d}+\frac{2 \int \sqrt{a+a \cos (c+d x)} \left (\frac{1}{2} a (5 A+3 C)-a C \cos (c+d x)\right ) \, dx}{5 a}\\ &=-\frac{4 C \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 C (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 a d}+\frac{1}{15} (15 A+7 C) \int \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{2 a (15 A+7 C) \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}-\frac{4 C \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 C (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 a d}\\ \end{align*}

Mathematica [A]  time = 0.112477, size = 58, normalized size = 0.61 \[ \frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)} (30 A+8 C \cos (c+d x)+3 C \cos (2 (c+d x))+19 C)}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a*(1 + Cos[c + d*x])]*(30*A + 19*C + 8*C*Cos[c + d*x] + 3*C*Cos[2*(c + d*x)])*Tan[(c + d*x)/2])/(15*d)

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Maple [A]  time = 0.042, size = 78, normalized size = 0.8 \begin{align*}{\frac{2\,a\sqrt{2}}{15\,d}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 12\,C \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-4\,C \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+15\,A+7\,C \right ){\frac{1}{\sqrt{a \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*cos(1/2*d*x+1/2*c)*a*sin(1/2*d*x+1/2*c)*(12*C*cos(1/2*d*x+1/2*c)^4-4*C*cos(1/2*d*x+1/2*c)^2+15*A+7*C)*2^(
1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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Maxima [A]  time = 2.0009, size = 97, normalized size = 1.02 \begin{align*} \frac{60 \, \sqrt{2} A \sqrt{a} \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) +{\left (3 \, \sqrt{2} \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 5 \, \sqrt{2} \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 30 \, \sqrt{2} \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} C \sqrt{a}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(60*sqrt(2)*A*sqrt(a)*sin(1/2*d*x + 1/2*c) + (3*sqrt(2)*sin(5/2*d*x + 5/2*c) + 5*sqrt(2)*sin(3/2*d*x + 3/
2*c) + 30*sqrt(2)*sin(1/2*d*x + 1/2*c))*C*sqrt(a))/d

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Fricas [A]  time = 1.40498, size = 159, normalized size = 1.67 \begin{align*} \frac{2 \,{\left (3 \, C \cos \left (d x + c\right )^{2} + 4 \, C \cos \left (d x + c\right ) + 15 \, A + 8 \, C\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{15 \,{\left (d \cos \left (d x + c\right ) + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*C*cos(d*x + c)^2 + 4*C*cos(d*x + c) + 15*A + 8*C)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/(d*cos(d*x + c
) + d)

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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+C*cos(d*x+c)**2)*(a+a*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [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+C*cos(d*x+c)^2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Timed out