3.500 \(\int \frac{(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac{11}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=234 \[ \frac{2 a^3 (8 A+10 B+11 C) \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (64 A+90 B+63 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{315 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 a^3 (584 A+690 B+903 C) \sin (c+d x)}{315 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{2 a (5 A+9 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{63 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{9 d \cos ^{\frac{9}{2}}(c+d x)} \]

[Out]

(2*a^3*(8*A + 10*B + 11*C)*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + (2*a^3*(584*A +
690*B + 903*C)*Sin[c + d*x])/(315*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(64*A + 90*B + 63*C)
*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(315*d*Cos[c + d*x]^(5/2)) + (2*a*(5*A + 9*B)*(a + a*Cos[c + d*x])^(3/
2)*Sin[c + d*x])/(63*d*Cos[c + d*x]^(7/2)) + (2*A*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(9*d*Cos[c + d*x]^(
9/2))

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Rubi [A]  time = 0.808779, antiderivative size = 234, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.089, Rules used = {3043, 2975, 2980, 2771} \[ \frac{2 a^3 (8 A+10 B+11 C) \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (64 A+90 B+63 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{315 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 a^3 (584 A+690 B+903 C) \sin (c+d x)}{315 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{2 a (5 A+9 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{63 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{9 d \cos ^{\frac{9}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(11/2),x]

[Out]

(2*a^3*(8*A + 10*B + 11*C)*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + (2*a^3*(584*A +
690*B + 903*C)*Sin[c + d*x])/(315*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(64*A + 90*B + 63*C)
*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(315*d*Cos[c + d*x]^(5/2)) + (2*a*(5*A + 9*B)*(a + a*Cos[c + d*x])^(3/
2)*Sin[c + d*x])/(63*d*Cos[c + d*x]^(7/2)) + (2*A*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(9*d*Cos[c + d*x]^(
9/2))

Rule 3043

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

Rule 2975

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^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, 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 2980

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

Rule 2771

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Sim
p[(-2*b^2*Cos[e + f*x])/(f*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*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]

Rubi steps

\begin{align*} \int \frac{(a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac{11}{2}}(c+d x)} \, dx &=\frac{2 A (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{2 \int \frac{(a+a \cos (c+d x))^{5/2} \left (\frac{1}{2} a (5 A+9 B)+\frac{1}{2} a (2 A+9 C) \cos (c+d x)\right )}{\cos ^{\frac{9}{2}}(c+d x)} \, dx}{9 a}\\ &=\frac{2 a (5 A+9 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{63 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{4 \int \frac{(a+a \cos (c+d x))^{3/2} \left (\frac{1}{4} a^2 (64 A+90 B+63 C)+\frac{3}{4} a^2 (8 A+6 B+21 C) \cos (c+d x)\right )}{\cos ^{\frac{7}{2}}(c+d x)} \, dx}{63 a}\\ &=\frac{2 a^2 (64 A+90 B+63 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{315 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 a (5 A+9 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{63 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{8 \int \frac{\sqrt{a+a \cos (c+d x)} \left (\frac{63}{8} a^3 (8 A+10 B+11 C)+\frac{1}{8} a^3 (248 A+270 B+441 C) \cos (c+d x)\right )}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{315 a}\\ &=\frac{2 a^3 (8 A+10 B+11 C) \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (64 A+90 B+63 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{315 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 a (5 A+9 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{63 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{1}{315} \left (a^2 (584 A+690 B+903 C)\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a^3 (8 A+10 B+11 C) \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{2 a^3 (584 A+690 B+903 C) \sin (c+d x)}{315 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (64 A+90 B+63 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{315 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 a (5 A+9 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{63 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 1.21635, size = 158, normalized size = 0.68 \[ \frac{a^2 \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)} (2 (1396 A+1215 B+882 C) \cos (c+d x)+4 (803 A+870 B+966 C) \cos (2 (c+d x))+584 A \cos (3 (c+d x))+584 A \cos (4 (c+d x))+2908 A+690 B \cos (3 (c+d x))+690 B \cos (4 (c+d x))+2790 B+588 C \cos (3 (c+d x))+903 C \cos (4 (c+d x))+2961 C)}{1260 d \cos ^{\frac{9}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(11/2),x]

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(2908*A + 2790*B + 2961*C + 2*(1396*A + 1215*B + 882*C)*Cos[c + d*x] + 4*(803*
A + 870*B + 966*C)*Cos[2*(c + d*x)] + 584*A*Cos[3*(c + d*x)] + 690*B*Cos[3*(c + d*x)] + 588*C*Cos[3*(c + d*x)]
+ 584*A*Cos[4*(c + d*x)] + 690*B*Cos[4*(c + d*x)] + 903*C*Cos[4*(c + d*x)])*Tan[(c + d*x)/2])/(1260*d*Cos[c +
d*x]^(9/2))

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Maple [A]  time = 0.123, size = 166, normalized size = 0.7 \begin{align*} -{\frac{2\,{a}^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 584\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+690\,B \left ( \cos \left ( dx+c \right ) \right ) ^{4}+903\,C \left ( \cos \left ( dx+c \right ) \right ) ^{4}+292\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+345\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+294\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+219\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+180\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+63\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+130\,A\cos \left ( dx+c \right ) +45\,B\cos \left ( dx+c \right ) +35\,A \right ) }{315\,d\sin \left ( dx+c \right ) }\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x)

[Out]

-2/315/d*a^2*(-1+cos(d*x+c))*(584*A*cos(d*x+c)^4+690*B*cos(d*x+c)^4+903*C*cos(d*x+c)^4+292*A*cos(d*x+c)^3+345*
B*cos(d*x+c)^3+294*C*cos(d*x+c)^3+219*A*cos(d*x+c)^2+180*B*cos(d*x+c)^2+63*C*cos(d*x+c)^2+130*A*cos(d*x+c)+45*
B*cos(d*x+c)+35*A)*(a*(1+cos(d*x+c)))^(1/2)/sin(d*x+c)/cos(d*x+c)^(9/2)

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Maxima [B]  time = 1.73398, size = 921, normalized size = 3.94 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x, algorithm="maxima")

[Out]

8/315*(21*(15*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 35*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c
) + 1)^3 + 28*sqrt(2)*a^(5/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 8*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(cos(d*x
+ c) + 1)^7)*C/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)) + 15
*(21*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 56*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3
+ 63*sqrt(2)*a^(5/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 36*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(cos(d*x + c) + 1
)^7 + 8*sqrt(2)*a^(5/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*B*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^2/((s
in(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(2*sin(d*x + c)^2/(cos(
d*x + c) + 1)^2 + sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1)) + (315*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x + c)
+ 1) - 945*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 1449*sqrt(2)*a^(5/2)*sin(d*x + c)^5/(cos(d*x
+ c) + 1)^5 - 1287*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 572*sqrt(2)*a^(5/2)*sin(d*x + c)^9/(c
os(d*x + c) + 1)^9 - 104*sqrt(2)*a^(5/2)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11)*A*(sin(d*x + c)^2/(cos(d*x + c
) + 1)^2 + 1)^3/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(11/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(11/2)*(3
*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + sin(d*x + c)^6/(cos(d*x + c) +
1)^6 + 1)))/d

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Fricas [A]  time = 1.82521, size = 379, normalized size = 1.62 \begin{align*} \frac{2 \,{\left ({\left (584 \, A + 690 \, B + 903 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} +{\left (292 \, A + 345 \, B + 294 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \,{\left (73 \, A + 60 \, B + 21 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 5 \,{\left (26 \, A + 9 \, B\right )} a^{2} \cos \left (d x + c\right ) + 35 \, A a^{2}\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{315 \,{\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

2/315*((584*A + 690*B + 903*C)*a^2*cos(d*x + c)^4 + (292*A + 345*B + 294*C)*a^2*cos(d*x + c)^3 + 3*(73*A + 60*
B + 21*C)*a^2*cos(d*x + c)^2 + 5*(26*A + 9*B)*a^2*cos(d*x + c) + 35*A*a^2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d
*x + c))*sin(d*x + c)/(d*cos(d*x + c)^6 + d*cos(d*x + c)^5)

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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))**(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(11/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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{5}{2}}}{\cos \left (d x + c\right )^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^(5/2)/cos(d*x + c)^(11/2), x)