3.278 \(\int (a+a \cos (c+d x))^{3/2} (B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=138 \[ \frac{8 a^2 (21 B+19 C) \sin (c+d x)}{105 d \sqrt{a \cos (c+d x)+a}}+\frac{2 (7 B-2 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{35 d}+\frac{2 a (21 B+19 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{105 d}+\frac{2 C \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d} \]

[Out]

(8*a^2*(21*B + 19*C)*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(21*B + 19*C)*Sqrt[a + a*Cos[c + d*
x]]*Sin[c + d*x])/(105*d) + (2*(7*B - 2*C)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(35*d) + (2*C*(a + a*Cos[c
+ d*x])^(5/2)*Sin[c + d*x])/(7*a*d)

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Rubi [A]  time = 0.188278, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {3023, 2751, 2647, 2646} \[ \frac{8 a^2 (21 B+19 C) \sin (c+d x)}{105 d \sqrt{a \cos (c+d x)+a}}+\frac{2 (7 B-2 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{35 d}+\frac{2 a (21 B+19 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{105 d}+\frac{2 C \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{7 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^2*(21*B + 19*C)*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(21*B + 19*C)*Sqrt[a + a*Cos[c + d*
x]]*Sin[c + d*x])/(105*d) + (2*(7*B - 2*C)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(35*d) + (2*C*(a + a*Cos[c
+ d*x])^(5/2)*Sin[c + d*x])/(7*a*d)

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 2647

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n -
1))/(d*n), x] + Dist[(a*(2*n - 1))/n, Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && Eq
Q[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

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 (a+a \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac{2 C (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 a d}+\frac{2 \int (a+a \cos (c+d x))^{3/2} \left (\frac{5 a C}{2}+\frac{1}{2} a (7 B-2 C) \cos (c+d x)\right ) \, dx}{7 a}\\ &=\frac{2 (7 B-2 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac{2 C (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 a d}+\frac{1}{35} (21 B+19 C) \int (a+a \cos (c+d x))^{3/2} \, dx\\ &=\frac{2 a (21 B+19 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{105 d}+\frac{2 (7 B-2 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac{2 C (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 a d}+\frac{1}{105} (4 a (21 B+19 C)) \int \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{8 a^2 (21 B+19 C) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a (21 B+19 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{105 d}+\frac{2 (7 B-2 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac{2 C (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{7 a d}\\ \end{align*}

Mathematica [A]  time = 0.393073, size = 81, normalized size = 0.59 \[ \frac{a \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)} ((252 B+253 C) \cos (c+d x)+6 (7 B+13 C) \cos (2 (c+d x))+546 B+15 C \cos (3 (c+d x))+494 C)}{210 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[a*(1 + Cos[c + d*x])]*(546*B + 494*C + (252*B + 253*C)*Cos[c + d*x] + 6*(7*B + 13*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 = 0.065, size = 104, normalized size = 0.8 \begin{align*}{\frac{4\,{a}^{2}\sqrt{2}}{105\,d}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( -60\,C \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+ \left ( 42\,B+168\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+ \left ( -105\,B-175\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+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((a+a*cos(d*x+c))^(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x)

[Out]

4/105*cos(1/2*d*x+1/2*c)*a^2*sin(1/2*d*x+1/2*c)*(-60*C*sin(1/2*d*x+1/2*c)^6+(42*B+168*C)*sin(1/2*d*x+1/2*c)^4+
(-105*B-175*C)*sin(1/2*d*x+1/2*c)^2+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 = 1.86838, size = 166, normalized size = 1.2 \begin{align*} \frac{42 \,{\left (\sqrt{2} a \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 5 \, \sqrt{2} a \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 20 \, \sqrt{2} a \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} B \sqrt{a} +{\left (15 \, \sqrt{2} a \sin \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ) + 63 \, \sqrt{2} a \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 175 \, \sqrt{2} a \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 735 \, \sqrt{2} a \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((a+a*cos(d*x+c))^(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

1/420*(42*(sqrt(2)*a*sin(5/2*d*x + 5/2*c) + 5*sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 20*sqrt(2)*a*sin(1/2*d*x + 1/2*
c))*B*sqrt(a) + (15*sqrt(2)*a*sin(7/2*d*x + 7/2*c) + 63*sqrt(2)*a*sin(5/2*d*x + 5/2*c) + 175*sqrt(2)*a*sin(3/2
*d*x + 3/2*c) + 735*sqrt(2)*a*sin(1/2*d*x + 1/2*c))*C*sqrt(a))/d

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Fricas [A]  time = 1.57489, size = 236, normalized size = 1.71 \begin{align*} \frac{2 \,{\left (15 \, C a \cos \left (d x + c\right )^{3} + 3 \,{\left (7 \, B + 13 \, C\right )} a \cos \left (d x + c\right )^{2} +{\left (63 \, B + 52 \, C\right )} a \cos \left (d x + c\right ) + 2 \,{\left (63 \, B + 52 \, C\right )} a\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((a+a*cos(d*x+c))^(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

2/105*(15*C*a*cos(d*x + c)^3 + 3*(7*B + 13*C)*a*cos(d*x + c)^2 + (63*B + 52*C)*a*cos(d*x + c) + 2*(63*B + 52*C
)*a)*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+a*cos(d*x+c))**(3/2)*(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((a+a*cos(d*x+c))^(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

Timed out