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

Optimal. Leaf size=234 \[ \frac{2 a^2 (11 A+12 B) \sin (c+d x) \cos ^4(c+d x)}{99 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (187 A+168 B) \sin (c+d x) \cos ^3(c+d x)}{693 d \sqrt{a \cos (c+d x)+a}}+\frac{4 a^2 (187 A+168 B) \sin (c+d x)}{495 d \sqrt{a \cos (c+d x)+a}}+\frac{4 (187 A+168 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{1155 d}-\frac{8 a (187 A+168 B) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{3465 d}+\frac{2 a B \sin (c+d x) \cos ^4(c+d x) \sqrt{a \cos (c+d x)+a}}{11 d} \]

[Out]

(4*a^2*(187*A + 168*B)*Sin[c + d*x])/(495*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(187*A + 168*B)*Cos[c + d*x]^3*
Sin[c + d*x])/(693*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(11*A + 12*B)*Cos[c + d*x]^4*Sin[c + d*x])/(99*d*Sqrt[
a + a*Cos[c + d*x]]) - (8*a*(187*A + 168*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3465*d) + (2*a*B*Cos[c + d
*x]^4*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(11*d) + (4*(187*A + 168*B)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*
x])/(1155*d)

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Rubi [A]  time = 0.529324, antiderivative size = 234, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2976, 2981, 2770, 2759, 2751, 2646} \[ \frac{2 a^2 (11 A+12 B) \sin (c+d x) \cos ^4(c+d x)}{99 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (187 A+168 B) \sin (c+d x) \cos ^3(c+d x)}{693 d \sqrt{a \cos (c+d x)+a}}+\frac{4 a^2 (187 A+168 B) \sin (c+d x)}{495 d \sqrt{a \cos (c+d x)+a}}+\frac{4 (187 A+168 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{1155 d}-\frac{8 a (187 A+168 B) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{3465 d}+\frac{2 a B \sin (c+d x) \cos ^4(c+d x) \sqrt{a \cos (c+d x)+a}}{11 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^2*(187*A + 168*B)*Sin[c + d*x])/(495*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(187*A + 168*B)*Cos[c + d*x]^3*
Sin[c + d*x])/(693*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(11*A + 12*B)*Cos[c + d*x]^4*Sin[c + d*x])/(99*d*Sqrt[
a + a*Cos[c + d*x]]) - (8*a*(187*A + 168*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3465*d) + (2*a*B*Cos[c + d
*x]^4*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(11*d) + (4*(187*A + 168*B)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*
x])/(1155*d)

Rule 2976

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], 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] && GtQ[m, 1/2] &&  !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

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) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx &=\frac{2 a B \cos ^4(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{11 d}+\frac{2}{11} \int \cos ^3(c+d x) \sqrt{a+a \cos (c+d x)} \left (\frac{1}{2} a (11 A+8 B)+\frac{1}{2} a (11 A+12 B) \cos (c+d x)\right ) \, dx\\ &=\frac{2 a^2 (11 A+12 B) \cos ^4(c+d x) \sin (c+d x)}{99 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a B \cos ^4(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{11 d}+\frac{1}{99} (a (187 A+168 B)) \int \cos ^3(c+d x) \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{2 a^2 (187 A+168 B) \cos ^3(c+d x) \sin (c+d x)}{693 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (11 A+12 B) \cos ^4(c+d x) \sin (c+d x)}{99 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a B \cos ^4(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{11 d}+\frac{1}{231} (2 a (187 A+168 B)) \int \cos ^2(c+d x) \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{2 a^2 (187 A+168 B) \cos ^3(c+d x) \sin (c+d x)}{693 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (11 A+12 B) \cos ^4(c+d x) \sin (c+d x)}{99 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a B \cos ^4(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{11 d}+\frac{4 (187 A+168 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{1155 d}+\frac{(4 (187 A+168 B)) \int \left (\frac{3 a}{2}-a \cos (c+d x)\right ) \sqrt{a+a \cos (c+d x)} \, dx}{1155}\\ &=\frac{2 a^2 (187 A+168 B) \cos ^3(c+d x) \sin (c+d x)}{693 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (11 A+12 B) \cos ^4(c+d x) \sin (c+d x)}{99 d \sqrt{a+a \cos (c+d x)}}-\frac{8 a (187 A+168 B) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{3465 d}+\frac{2 a B \cos ^4(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{11 d}+\frac{4 (187 A+168 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{1155 d}+\frac{1}{495} (2 a (187 A+168 B)) \int \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{4 a^2 (187 A+168 B) \sin (c+d x)}{495 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (187 A+168 B) \cos ^3(c+d x) \sin (c+d x)}{693 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (11 A+12 B) \cos ^4(c+d x) \sin (c+d x)}{99 d \sqrt{a+a \cos (c+d x)}}-\frac{8 a (187 A+168 B) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{3465 d}+\frac{2 a B \cos ^4(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{11 d}+\frac{4 (187 A+168 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{1155 d}\\ \end{align*}

Mathematica [A]  time = 0.956185, size = 125, normalized size = 0.53 \[ \frac{a \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)} ((35156 A+34734 B) \cos (c+d x)+8 (1507 A+1743 B) \cos (2 (c+d x))+3740 A \cos (3 (c+d x))+770 A \cos (4 (c+d x))+59158 A+4935 B \cos (3 (c+d x))+1470 B \cos (4 (c+d x))+315 B \cos (5 (c+d x))+55482 B)}{27720 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[a*(1 + Cos[c + d*x])]*(59158*A + 55482*B + (35156*A + 34734*B)*Cos[c + d*x] + 8*(1507*A + 1743*B)*Cos[
2*(c + d*x)] + 3740*A*Cos[3*(c + d*x)] + 4935*B*Cos[3*(c + d*x)] + 770*A*Cos[4*(c + d*x)] + 1470*B*Cos[4*(c +
d*x)] + 315*B*Cos[5*(c + d*x)])*Tan[(c + d*x)/2])/(27720*d)

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Maple [A]  time = 1.046, size = 142, normalized size = 0.6 \begin{align*}{\frac{4\,{a}^{2}\sqrt{2}}{3465\,d}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( -5040\,B \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{10}+ \left ( 3080\,A+18480\,B \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{8}+ \left ( -9900\,A-27720\,B \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{6}+ \left ( 12474\,A+22176\,B \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+ \left ( -8085\,A-10395\,B \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+3465\,A+3465\,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)^(3/2)*(A+B*cos(d*x+c)),x)

[Out]

4/3465*cos(1/2*d*x+1/2*c)*a^2*sin(1/2*d*x+1/2*c)*(-5040*B*sin(1/2*d*x+1/2*c)^10+(3080*A+18480*B)*sin(1/2*d*x+1
/2*c)^8+(-9900*A-27720*B)*sin(1/2*d*x+1/2*c)^6+(12474*A+22176*B)*sin(1/2*d*x+1/2*c)^4+(-8085*A-10395*B)*sin(1/
2*d*x+1/2*c)^2+3465*A+3465*B)*2^(1/2)/(cos(1/2*d*x+1/2*c)^2*a)^(1/2)/d

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Maxima [A]  time = 2.00005, size = 250, normalized size = 1.07 \begin{align*} \frac{22 \,{\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 )} A \sqrt{a} + 21 \,{\left (15 \, \sqrt{2} a \sin \left (\frac{11}{2} \, d x + \frac{11}{2} \, c\right ) + 55 \, \sqrt{2} a \sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ) + 165 \, \sqrt{2} a \sin \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ) + 429 \, \sqrt{2} a \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 990 \, \sqrt{2} a \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 3630 \, \sqrt{2} a \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} B \sqrt{a}}{55440 \, 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))^(3/2)*(A+B*cos(d*x+c)),x, algorithm="maxima")

[Out]

1/55440*(22*(35*sqrt(2)*a*sin(9/2*d*x + 9/2*c) + 135*sqrt(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) + 3780*sqrt(2)*a*sin(1/2*d*x + 1/2*c))*A*sqrt(a) + 21*(15*sqr
t(2)*a*sin(11/2*d*x + 11/2*c) + 55*sqrt(2)*a*sin(9/2*d*x + 9/2*c) + 165*sqrt(2)*a*sin(7/2*d*x + 7/2*c) + 429*s
qrt(2)*a*sin(5/2*d*x + 5/2*c) + 990*sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 3630*sqrt(2)*a*sin(1/2*d*x + 1/2*c))*B*sq
rt(a))/d

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Fricas [A]  time = 1.33085, size = 351, normalized size = 1.5 \begin{align*} \frac{2 \,{\left (315 \, B a \cos \left (d x + c\right )^{5} + 35 \,{\left (11 \, A + 21 \, B\right )} a \cos \left (d x + c\right )^{4} + 5 \,{\left (187 \, A + 168 \, B\right )} a \cos \left (d x + c\right )^{3} + 6 \,{\left (187 \, A + 168 \, B\right )} a \cos \left (d x + c\right )^{2} + 8 \,{\left (187 \, A + 168 \, B\right )} a \cos \left (d x + c\right ) + 16 \,{\left (187 \, A + 168 \, B\right )} a\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{3465 \,{\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))^(3/2)*(A+B*cos(d*x+c)),x, algorithm="fricas")

[Out]

2/3465*(315*B*a*cos(d*x + c)^5 + 35*(11*A + 21*B)*a*cos(d*x + c)^4 + 5*(187*A + 168*B)*a*cos(d*x + c)^3 + 6*(1
87*A + 168*B)*a*cos(d*x + c)^2 + 8*(187*A + 168*B)*a*cos(d*x + c) + 16*(187*A + 168*B)*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(cos(d*x+c)**3*(a+a*cos(d*x+c))**(3/2)*(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 )}{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \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))^(3/2)*(A+B*cos(d*x+c)),x, algorithm="giac")

[Out]

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