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

Optimal. Leaf size=225 \[ \frac{2 a^2 (33 A+28 C) \sin (c+d x) \cos ^3(c+d x)}{231 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (143 A+112 C) \sin (c+d x)}{165 d \sqrt{a \cos (c+d x)+a}}+\frac{2 (143 A+112 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{385 d}-\frac{4 a (143 A+112 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{1155 d}+\frac{2 C \sin (c+d x) \cos ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{11 d}+\frac{2 a C \sin (c+d x) \cos ^3(c+d x) \sqrt{a \cos (c+d x)+a}}{33 d} \]

[Out]

(2*a^2*(143*A + 112*C)*Sin[c + d*x])/(165*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(33*A + 28*C)*Cos[c + d*x]^3*Si
n[c + d*x])/(231*d*Sqrt[a + a*Cos[c + d*x]]) - (4*a*(143*A + 112*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(11
55*d) + (2*a*C*Cos[c + d*x]^3*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(33*d) + (2*(143*A + 112*C)*(a + a*Cos[c
+ d*x])^(3/2)*Sin[c + d*x])/(385*d) + (2*C*Cos[c + d*x]^3*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(11*d)

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Rubi [A]  time = 0.638437, antiderivative size = 225, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {3046, 2976, 2981, 2759, 2751, 2646} \[ \frac{2 a^2 (33 A+28 C) \sin (c+d x) \cos ^3(c+d x)}{231 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (143 A+112 C) \sin (c+d x)}{165 d \sqrt{a \cos (c+d x)+a}}+\frac{2 (143 A+112 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{385 d}-\frac{4 a (143 A+112 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{1155 d}+\frac{2 C \sin (c+d x) \cos ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{11 d}+\frac{2 a C \sin (c+d x) \cos ^3(c+d x) \sqrt{a \cos (c+d x)+a}}{33 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(143*A + 112*C)*Sin[c + d*x])/(165*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(33*A + 28*C)*Cos[c + d*x]^3*Si
n[c + d*x])/(231*d*Sqrt[a + a*Cos[c + d*x]]) - (4*a*(143*A + 112*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(11
55*d) + (2*a*C*Cos[c + d*x]^3*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(33*d) + (2*(143*A + 112*C)*(a + a*Cos[c
+ d*x])^(3/2)*Sin[c + d*x])/(385*d) + (2*C*Cos[c + d*x]^3*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(11*d)

Rule 3046

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (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))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b
, c, d, e, f, A, 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 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 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 ^2(c+d x) (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac{2 C \cos ^3(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{11 d}+\frac{2 \int \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} \left (\frac{1}{2} a (11 A+6 C)+\frac{3}{2} a C \cos (c+d x)\right ) \, dx}{11 a}\\ &=\frac{2 a C \cos ^3(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{33 d}+\frac{2 C \cos ^3(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{11 d}+\frac{4 \int \cos ^2(c+d x) \sqrt{a+a \cos (c+d x)} \left (\frac{9}{4} a^2 (11 A+8 C)+\frac{3}{4} a^2 (33 A+28 C) \cos (c+d x)\right ) \, dx}{99 a}\\ &=\frac{2 a^2 (33 A+28 C) \cos ^3(c+d x) \sin (c+d x)}{231 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a C \cos ^3(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{33 d}+\frac{2 C \cos ^3(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{11 d}+\frac{1}{77} (a (143 A+112 C)) \int \cos ^2(c+d x) \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{2 a^2 (33 A+28 C) \cos ^3(c+d x) \sin (c+d x)}{231 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a C \cos ^3(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{33 d}+\frac{2 (143 A+112 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{385 d}+\frac{2 C \cos ^3(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{11 d}+\frac{1}{385} (2 (143 A+112 C)) \int \left (\frac{3 a}{2}-a \cos (c+d x)\right ) \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{2 a^2 (33 A+28 C) \cos ^3(c+d x) \sin (c+d x)}{231 d \sqrt{a+a \cos (c+d x)}}-\frac{4 a (143 A+112 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{1155 d}+\frac{2 a C \cos ^3(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{33 d}+\frac{2 (143 A+112 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{385 d}+\frac{2 C \cos ^3(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{11 d}+\frac{1}{165} (a (143 A+112 C)) \int \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{2 a^2 (143 A+112 C) \sin (c+d x)}{165 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (33 A+28 C) \cos ^3(c+d x) \sin (c+d x)}{231 d \sqrt{a+a \cos (c+d x)}}-\frac{4 a (143 A+112 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{1155 d}+\frac{2 a C \cos ^3(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{33 d}+\frac{2 (143 A+112 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{385 d}+\frac{2 C \cos ^3(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{11 d}\\ \end{align*}

Mathematica [A]  time = 0.912158, size = 115, normalized size = 0.51 \[ \frac{a \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)} (2 (5566 A+5789 C) \cos (c+d x)+8 (429 A+581 C) \cos (2 (c+d x))+660 A \cos (3 (c+d x))+21736 A+1645 C \cos (3 (c+d x))+490 C \cos (4 (c+d x))+105 C \cos (5 (c+d x))+18494 C)}{9240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[a*(1 + Cos[c + d*x])]*(21736*A + 18494*C + 2*(5566*A + 5789*C)*Cos[c + d*x] + 8*(429*A + 581*C)*Cos[2*
(c + d*x)] + 660*A*Cos[3*(c + d*x)] + 1645*C*Cos[3*(c + d*x)] + 490*C*Cos[4*(c + d*x)] + 105*C*Cos[5*(c + d*x)
])*Tan[(c + d*x)/2])/(9240*d)

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Maple [A]  time = 0.044, size = 137, normalized size = 0.6 \begin{align*}{\frac{4\,{a}^{2}\sqrt{2}}{1155\,d}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( -1680\,C \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{10}+6160\,C \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}+ \left ( -660\,A-9240\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{6}+ \left ( 1848\,A+7392\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+ \left ( -1925\,A-3465\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1155\,A+1155\,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)^2*(a+a*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2),x)

[Out]

4/1155*cos(1/2*d*x+1/2*c)*a^2*sin(1/2*d*x+1/2*c)*(-1680*C*sin(1/2*d*x+1/2*c)^10+6160*C*sin(1/2*d*x+1/2*c)^8+(-
660*A-9240*C)*sin(1/2*d*x+1/2*c)^6+(1848*A+7392*C)*sin(1/2*d*x+1/2*c)^4+(-1925*A-3465*C)*sin(1/2*d*x+1/2*c)^2+
1155*A+1155*C)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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Maxima [A]  time = 2.04439, size = 230, normalized size = 1.02 \begin{align*} \frac{44 \,{\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 )} A \sqrt{a} + 7 \,{\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 )} C \sqrt{a}}{18480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18480*(44*(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))*A*sqrt(a) + 7*(15*sqrt(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*sqrt(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))*C*sqrt(a))/d

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Fricas [A]  time = 1.40677, size = 332, normalized size = 1.48 \begin{align*} \frac{2 \,{\left (105 \, C a \cos \left (d x + c\right )^{5} + 245 \, C a \cos \left (d x + c\right )^{4} + 5 \,{\left (33 \, A + 56 \, C\right )} a \cos \left (d x + c\right )^{3} + 3 \,{\left (143 \, A + 112 \, C\right )} a \cos \left (d x + c\right )^{2} + 4 \,{\left (143 \, A + 112 \, C\right )} a \cos \left (d x + c\right ) + 8 \,{\left (143 \, A + 112 \, C\right )} a\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{1155 \,{\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)^2*(a+a*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

2/1155*(105*C*a*cos(d*x + c)^5 + 245*C*a*cos(d*x + c)^4 + 5*(33*A + 56*C)*a*cos(d*x + c)^3 + 3*(143*A + 112*C)
*a*cos(d*x + c)^2 + 4*(143*A + 112*C)*a*cos(d*x + c) + 8*(143*A + 112*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(cos(d*x+c)**2*(a+a*cos(d*x+c))**(3/2)*(A+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} + A\right )}{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \cos \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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