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

Optimal. Leaf size=147 \[ \frac{2 (35 A-14 B+18 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{105 d}+\frac{2 a (35 A+49 B+27 C) \sin (c+d x)}{105 d \sqrt{a \cos (c+d x)+a}}+\frac{2 (7 B+C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{35 a d}+\frac{2 C \sin (c+d x) \cos ^2(c+d x) \sqrt{a \cos (c+d x)+a}}{7 d} \]

[Out]

(2*a*(35*A + 49*B + 27*C)*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*(35*A - 14*B + 18*C)*Sqrt[a + a*
Cos[c + d*x]]*Sin[c + d*x])/(105*d) + (2*C*Cos[c + d*x]^2*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(7*d) + (2*(7
*B + C)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(35*a*d)

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Rubi [A]  time = 0.347413, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {3045, 2968, 3023, 2751, 2646} \[ \frac{2 (35 A-14 B+18 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{105 d}+\frac{2 a (35 A+49 B+27 C) \sin (c+d x)}{105 d \sqrt{a \cos (c+d x)+a}}+\frac{2 (7 B+C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{35 a d}+\frac{2 C \sin (c+d x) \cos ^2(c+d x) \sqrt{a \cos (c+d x)+a}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(35*A + 49*B + 27*C)*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*(35*A - 14*B + 18*C)*Sqrt[a + a*
Cos[c + d*x]]*Sin[c + d*x])/(105*d) + (2*C*Cos[c + d*x]^2*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(7*d) + (2*(7
*B + C)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(35*a*d)

Rule 3045

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] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(b*d*(m + n + 2)), Int[(a + b*Sin[e + f*x
])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + (C*(a*d*m - b*c*(m + 1)) + b*B*
d*(m + n + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-1)] && NeQ[m + n + 2, 0]

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 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 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 \cos (c+d x) \sqrt{a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac{2 C \cos ^2(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{7 d}+\frac{2 \int \cos (c+d x) \sqrt{a+a \cos (c+d x)} \left (\frac{1}{2} a (7 A+4 C)+\frac{1}{2} a (7 B+C) \cos (c+d x)\right ) \, dx}{7 a}\\ &=\frac{2 C \cos ^2(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{7 d}+\frac{2 \int \sqrt{a+a \cos (c+d x)} \left (\frac{1}{2} a (7 A+4 C) \cos (c+d x)+\frac{1}{2} a (7 B+C) \cos ^2(c+d x)\right ) \, dx}{7 a}\\ &=\frac{2 C \cos ^2(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{7 d}+\frac{2 (7 B+C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 a d}+\frac{4 \int \sqrt{a+a \cos (c+d x)} \left (\frac{3}{4} a^2 (7 B+C)+\frac{1}{4} a^2 (35 A-14 B+18 C) \cos (c+d x)\right ) \, dx}{35 a^2}\\ &=\frac{2 (35 A-14 B+18 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{105 d}+\frac{2 C \cos ^2(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{7 d}+\frac{2 (7 B+C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 a d}+\frac{1}{105} (35 A+49 B+27 C) \int \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{2 a (35 A+49 B+27 C) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (35 A-14 B+18 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{105 d}+\frac{2 C \cos ^2(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{7 d}+\frac{2 (7 B+C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 a d}\\ \end{align*}

Mathematica [A]  time = 0.407407, size = 86, normalized size = 0.59 \[ \frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)} ((140 A+112 B+141 C) \cos (c+d x)+280 A+6 (7 B+6 C) \cos (2 (c+d x))+266 B+15 C \cos (3 (c+d x))+228 C)}{210 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a*(1 + Cos[c + d*x])]*(280*A + 266*B + 228*C + (140*A + 112*B + 141*C)*Cos[c + d*x] + 6*(7*B + 6*C)*Cos[
2*(c + d*x)] + 15*C*Cos[3*(c + d*x)])*Tan[(c + d*x)/2])/(210*d)

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Maple [A]  time = 0.069, size = 108, normalized size = 0.7 \begin{align*}{\frac{2\,a\sqrt{2}}{105\,d}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( -120\,C \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+ \left ( 84\,B+252\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+ \left ( -70\,A-140\,B-210\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+105\,A+105\,B+105\,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(cos(d*x+c)*(a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)

[Out]

2/105*cos(1/2*d*x+1/2*c)*a*sin(1/2*d*x+1/2*c)*(-120*C*sin(1/2*d*x+1/2*c)^6+(84*B+252*C)*sin(1/2*d*x+1/2*c)^4+(
-70*A-140*B-210*C)*sin(1/2*d*x+1/2*c)^2+105*A+105*B+105*C)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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Maxima [A]  time = 2.26533, size = 205, normalized size = 1.39 \begin{align*} \frac{140 \,{\left (\sqrt{2} \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 3 \, \sqrt{2} \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} A \sqrt{a} + 14 \,{\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 )} B \sqrt{a} + 3 \,{\left (5 \, \sqrt{2} \sin \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ) + 7 \, \sqrt{2} \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 35 \, \sqrt{2} \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 105 \, \sqrt{2} \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} C \sqrt{a}}{420 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/420*(140*(sqrt(2)*sin(3/2*d*x + 3/2*c) + 3*sqrt(2)*sin(1/2*d*x + 1/2*c))*A*sqrt(a) + 14*(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))*B*sqrt(a) + 3*(5*sqrt(2)*sin(7
/2*d*x + 7/2*c) + 7*sqrt(2)*sin(5/2*d*x + 5/2*c) + 35*sqrt(2)*sin(3/2*d*x + 3/2*c) + 105*sqrt(2)*sin(1/2*d*x +
1/2*c))*C*sqrt(a))/d

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Fricas [A]  time = 1.80892, size = 238, normalized size = 1.62 \begin{align*} \frac{2 \,{\left (15 \, C \cos \left (d x + c\right )^{3} + 3 \,{\left (7 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (35 \, A + 28 \, B + 24 \, C\right )} \cos \left (d x + c\right ) + 70 \, A + 56 \, B + 48 \, C\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{105 \,{\left (d \cos \left (d x + c\right ) + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*C*cos(d*x + c)^3 + 3*(7*B + 6*C)*cos(d*x + c)^2 + (35*A + 28*B + 24*C)*cos(d*x + c) + 70*A + 56*B +
48*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(cos(d*x+c)*(a+a*cos(d*x+c))**(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**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(cos(d*x+c)*(a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

Timed out