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

Optimal. Leaf size=187 \[ \frac{8 a^2 (63 A+57 B+47 C) \sin (c+d x)}{315 d \sqrt{a \cos (c+d x)+a}}+\frac{2 (63 A-18 B+22 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{315 d}+\frac{2 a (63 A+57 B+47 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{315 d}+\frac{2 (3 B+C) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{21 a d}+\frac{2 C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^{3/2}}{9 d} \]

[Out]

(8*a^2*(63*A + 57*B + 47*C)*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(63*A + 57*B + 47*C)*Sqrt[a
+ a*Cos[c + d*x]]*Sin[c + d*x])/(315*d) + (2*(63*A - 18*B + 22*C)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(31
5*d) + (2*C*Cos[c + d*x]^2*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(9*d) + (2*(3*B + C)*(a + a*Cos[c + d*x])^
(5/2)*Sin[c + d*x])/(21*a*d)

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Rubi [A]  time = 0.415421, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {3045, 2968, 3023, 2751, 2647, 2646} \[ \frac{8 a^2 (63 A+57 B+47 C) \sin (c+d x)}{315 d \sqrt{a \cos (c+d x)+a}}+\frac{2 (63 A-18 B+22 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{315 d}+\frac{2 a (63 A+57 B+47 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{315 d}+\frac{2 (3 B+C) \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{21 a d}+\frac{2 C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^{3/2}}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^2*(63*A + 57*B + 47*C)*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(63*A + 57*B + 47*C)*Sqrt[a
+ a*Cos[c + d*x]]*Sin[c + d*x])/(315*d) + (2*(63*A - 18*B + 22*C)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(31
5*d) + (2*C*Cos[c + d*x]^2*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(9*d) + (2*(3*B + C)*(a + a*Cos[c + d*x])^
(5/2)*Sin[c + d*x])/(21*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 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 \cos (c+d x) (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac{2 C \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{9 d}+\frac{2 \int \cos (c+d x) (a+a \cos (c+d x))^{3/2} \left (\frac{1}{2} a (9 A+4 C)+\frac{3}{2} a (3 B+C) \cos (c+d x)\right ) \, dx}{9 a}\\ &=\frac{2 C \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{9 d}+\frac{2 \int (a+a \cos (c+d x))^{3/2} \left (\frac{1}{2} a (9 A+4 C) \cos (c+d x)+\frac{3}{2} a (3 B+C) \cos ^2(c+d x)\right ) \, dx}{9 a}\\ &=\frac{2 C \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{9 d}+\frac{2 (3 B+C) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{21 a d}+\frac{4 \int (a+a \cos (c+d x))^{3/2} \left (\frac{15}{4} a^2 (3 B+C)+\frac{1}{4} a^2 (63 A-18 B+22 C) \cos (c+d x)\right ) \, dx}{63 a^2}\\ &=\frac{2 (63 A-18 B+22 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{315 d}+\frac{2 C \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{9 d}+\frac{2 (3 B+C) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{21 a d}+\frac{1}{105} (63 A+57 B+47 C) \int (a+a \cos (c+d x))^{3/2} \, dx\\ &=\frac{2 a (63 A+57 B+47 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac{2 (63 A-18 B+22 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{315 d}+\frac{2 C \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{9 d}+\frac{2 (3 B+C) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{21 a d}+\frac{1}{315} (4 a (63 A+57 B+47 C)) \int \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{8 a^2 (63 A+57 B+47 C) \sin (c+d x)}{315 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a (63 A+57 B+47 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac{2 (63 A-18 B+22 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{315 d}+\frac{2 C \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{9 d}+\frac{2 (3 B+C) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{21 a d}\\ \end{align*}

Mathematica [A]  time = 0.755938, size = 113, normalized size = 0.6 \[ \frac{a \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)} (2 (756 A+759 B+799 C) \cos (c+d x)+4 (63 A+117 B+137 C) \cos (2 (c+d x))+3276 A+90 B \cos (3 (c+d x))+2964 B+170 C \cos (3 (c+d x))+35 C \cos (4 (c+d x))+2689 C)}{1260 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[a*(1 + Cos[c + d*x])]*(3276*A + 2964*B + 2689*C + 2*(756*A + 759*B + 799*C)*Cos[c + d*x] + 4*(63*A + 1
17*B + 137*C)*Cos[2*(c + d*x)] + 90*B*Cos[3*(c + d*x)] + 170*C*Cos[3*(c + d*x)] + 35*C*Cos[4*(c + d*x)])*Tan[(
c + d*x)/2])/(1260*d)

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Maple [A]  time = 0.066, size = 132, normalized size = 0.7 \begin{align*}{\frac{4\,{a}^{2}\sqrt{2}}{315\,d}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 280\,C \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}+ \left ( -180\,B-900\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{6}+ \left ( 126\,A+504\,B+1134\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+ \left ( -315\,A-525\,B-735\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+315\,A+315\,B+315\,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))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)

[Out]

4/315*cos(1/2*d*x+1/2*c)*a^2*sin(1/2*d*x+1/2*c)*(280*C*sin(1/2*d*x+1/2*c)^8+(-180*B-900*C)*sin(1/2*d*x+1/2*c)^
6+(126*A+504*B+1134*C)*sin(1/2*d*x+1/2*c)^4+(-315*A-525*B-735*C)*sin(1/2*d*x+1/2*c)^2+315*A+315*B+315*C)*2^(1/
2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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Maxima [A]  time = 2.21449, size = 277, normalized size = 1.48 \begin{align*} \frac{252 \,{\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 )} A \sqrt{a} + 6 \,{\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 )} B \sqrt{a} +{\left (35 \, \sqrt{2} a \sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ) + 135 \, \sqrt{2} a \sin \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ) + 378 \, \sqrt{2} a \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 1050 \, \sqrt{2} a \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 3780 \, \sqrt{2} a \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} C \sqrt{a}}{2520 \, 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))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

1/2520*(252*(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))*A*sqrt(a) + 6*(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))*B*sqrt(a) + (35*sqrt(2)*a*sin(9/2*d*x + 9/2*c) + 135*s
qrt(2)*a*sin(7/2*d*x + 7/2*c) + 378*sqrt(2)*a*sin(5/2*d*x + 5/2*c) + 1050*sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 378
0*sqrt(2)*a*sin(1/2*d*x + 1/2*c))*C*sqrt(a))/d

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Fricas [A]  time = 1.89954, size = 320, normalized size = 1.71 \begin{align*} \frac{2 \,{\left (35 \, C a \cos \left (d x + c\right )^{4} + 5 \,{\left (9 \, B + 17 \, C\right )} a \cos \left (d x + c\right )^{3} + 3 \,{\left (21 \, A + 39 \, B + 34 \, C\right )} a \cos \left (d x + c\right )^{2} +{\left (189 \, A + 156 \, B + 136 \, C\right )} a \cos \left (d x + c\right ) + 2 \,{\left (189 \, A + 156 \, B + 136 \, C\right )} a\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{315 \,{\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))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

2/315*(35*C*a*cos(d*x + c)^4 + 5*(9*B + 17*C)*a*cos(d*x + c)^3 + 3*(21*A + 39*B + 34*C)*a*cos(d*x + c)^2 + (18
9*A + 156*B + 136*C)*a*cos(d*x + c) + 2*(189*A + 156*B + 136*C)*a)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/(d*co
s(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))**(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )}{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \cos \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

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