3.74 \(\int \cos ^3(c+d x) \sqrt{a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx\)

Optimal. Leaf size=187 \[ \frac{2 a (9 A+8 B) \sin (c+d x) \cos ^3(c+d x)}{63 d \sqrt{a \cos (c+d x)+a}}+\frac{4 (9 A+8 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{105 a d}-\frac{8 (9 A+8 B) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{315 d}+\frac{4 a (9 A+8 B) \sin (c+d x)}{45 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a B \sin (c+d x) \cos ^4(c+d x)}{9 d \sqrt{a \cos (c+d x)+a}} \]

[Out]

(4*a*(9*A + 8*B)*Sin[c + d*x])/(45*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(9*A + 8*B)*Cos[c + d*x]^3*Sin[c + d*x])
/(63*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*B*Cos[c + d*x]^4*Sin[c + d*x])/(9*d*Sqrt[a + a*Cos[c + d*x]]) - (8*(9*
A + 8*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(315*d) + (4*(9*A + 8*B)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*
x])/(105*a*d)

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Rubi [A]  time = 0.303999, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {2981, 2770, 2759, 2751, 2646} \[ \frac{2 a (9 A+8 B) \sin (c+d x) \cos ^3(c+d x)}{63 d \sqrt{a \cos (c+d x)+a}}+\frac{4 (9 A+8 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{105 a d}-\frac{8 (9 A+8 B) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{315 d}+\frac{4 a (9 A+8 B) \sin (c+d x)}{45 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a B \sin (c+d x) \cos ^4(c+d x)}{9 d \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*(9*A + 8*B)*Sin[c + d*x])/(45*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(9*A + 8*B)*Cos[c + d*x]^3*Sin[c + d*x])
/(63*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*B*Cos[c + d*x]^4*Sin[c + d*x])/(9*d*Sqrt[a + a*Cos[c + d*x]]) - (8*(9*
A + 8*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(315*d) + (4*(9*A + 8*B)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*
x])/(105*a*d)

Rule 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

Rule 2770

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

Rule 2759

Int[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(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*(b*(m + 1) - a*
Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-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 ^3(c+d x) \sqrt{a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx &=\frac{2 a B \cos ^4(c+d x) \sin (c+d x)}{9 d \sqrt{a+a \cos (c+d x)}}+\frac{1}{9} (9 A+8 B) \int \cos ^3(c+d x) \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{2 a (9 A+8 B) \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a B \cos ^4(c+d x) \sin (c+d x)}{9 d \sqrt{a+a \cos (c+d x)}}+\frac{1}{21} (2 (9 A+8 B)) \int \cos ^2(c+d x) \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{2 a (9 A+8 B) \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a B \cos ^4(c+d x) \sin (c+d x)}{9 d \sqrt{a+a \cos (c+d x)}}+\frac{4 (9 A+8 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 a d}+\frac{(4 (9 A+8 B)) \int \left (\frac{3 a}{2}-a \cos (c+d x)\right ) \sqrt{a+a \cos (c+d x)} \, dx}{105 a}\\ &=\frac{2 a (9 A+8 B) \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a B \cos ^4(c+d x) \sin (c+d x)}{9 d \sqrt{a+a \cos (c+d x)}}-\frac{8 (9 A+8 B) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac{4 (9 A+8 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 a d}+\frac{1}{45} (2 (9 A+8 B)) \int \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{4 a (9 A+8 B) \sin (c+d x)}{45 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a (9 A+8 B) \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a B \cos ^4(c+d x) \sin (c+d x)}{9 d \sqrt{a+a \cos (c+d x)}}-\frac{8 (9 A+8 B) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac{4 (9 A+8 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 a d}\\ \end{align*}

Mathematica [A]  time = 0.628738, size = 103, normalized size = 0.55 \[ \frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)} (94 (9 A+8 B) \cos (c+d x)+4 (54 A+83 B) \cos (2 (c+d x))+90 A \cos (3 (c+d x))+1368 A+80 B \cos (3 (c+d x))+35 B \cos (4 (c+d x))+1321 B)}{1260 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a*(1 + Cos[c + d*x])]*(1368*A + 1321*B + 94*(9*A + 8*B)*Cos[c + d*x] + 4*(54*A + 83*B)*Cos[2*(c + d*x)]
+ 90*A*Cos[3*(c + d*x)] + 80*B*Cos[3*(c + d*x)] + 35*B*Cos[4*(c + d*x)])*Tan[(c + d*x)/2])/(1260*d)

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Maple [A]  time = 1.147, size = 121, normalized size = 0.7 \begin{align*}{\frac{2\,a\sqrt{2}}{315\,d}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 560\,B \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}+ \left ( -360\,A-1440\,B \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{6}+ \left ( 756\,A+1512\,B \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+ \left ( -630\,A-840\,B \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+315\,A+315\,B \right ){\frac{1}{\sqrt{ \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+cos(d*x+c)*a)^(1/2)*(A+B*cos(d*x+c)),x)

[Out]

2/315*cos(1/2*d*x+1/2*c)*a*sin(1/2*d*x+1/2*c)*(560*B*sin(1/2*d*x+1/2*c)^8+(-360*A-1440*B)*sin(1/2*d*x+1/2*c)^6
+(756*A+1512*B)*sin(1/2*d*x+1/2*c)^4+(-630*A-840*B)*sin(1/2*d*x+1/2*c)^2+315*A+315*B)*2^(1/2)/(cos(1/2*d*x+1/2
*c)^2*a)^(1/2)/d

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Maxima [A]  time = 1.90054, size = 196, normalized size = 1.05 \begin{align*} \frac{18 \,{\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 )} A \sqrt{a} +{\left (35 \, \sqrt{2} \sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ) + 45 \, \sqrt{2} \sin \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ) + 252 \, \sqrt{2} \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 420 \, \sqrt{2} \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 1890 \, \sqrt{2} \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} B \sqrt{a}}{2520 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x, algorithm="maxima")

[Out]

1/2520*(18*(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))*A*sqrt(a) + (35*sqrt(2)*sin(9/2*d*x + 9/2*c) + 45*sqrt(2)*sin(7/2*d*x + 7/
2*c) + 252*sqrt(2)*sin(5/2*d*x + 5/2*c) + 420*sqrt(2)*sin(3/2*d*x + 3/2*c) + 1890*sqrt(2)*sin(1/2*d*x + 1/2*c)
)*B*sqrt(a))/d

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Fricas [A]  time = 1.33308, size = 263, normalized size = 1.41 \begin{align*} \frac{2 \,{\left (35 \, B \cos \left (d x + c\right )^{4} + 5 \,{\left (9 \, A + 8 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \,{\left (9 \, A + 8 \, B\right )} \cos \left (d x + c\right )^{2} + 8 \,{\left (9 \, A + 8 \, B\right )} \cos \left (d x + c\right ) + 144 \, A + 128 \, B\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)^3*(a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x, algorithm="fricas")

[Out]

2/315*(35*B*cos(d*x + c)^4 + 5*(9*A + 8*B)*cos(d*x + c)^3 + 6*(9*A + 8*B)*cos(d*x + c)^2 + 8*(9*A + 8*B)*cos(d
*x + c) + 144*A + 128*B)*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)**3*(a+a*cos(d*x+c))**(1/2)*(A+B*cos(d*x+c)),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*sqrt(a*cos(d*x + c) + a)*cos(d*x + c)^3, x)