\(\int \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {11}{2}}(c+d x)} \, dx\) [500]

Optimal result

Integrand size = 45, antiderivative size = 234 \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {2 a^3 (8 A+10 B+11 C) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {2 a^3 (584 A+690 B+903 C) \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (64 A+90 B+63 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a (5 A+9 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{63 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)} \]

[Out]

2/63*a*(5*A+9*B)*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d/cos(d*x+c)^(7/2)+2/9*A*(a+a*cos(d*x+c))^(5/2)*sin(d*x+c)/
d/cos(d*x+c)^(9/2)+2/15*a^3*(8*A+10*B+11*C)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2)+2/315*a^3*(58
4*A+690*B+903*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2)+2/315*a^2*(64*A+90*B+63*C)*sin(d*x+c)*(a
+a*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(5/2)

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {3122, 3054, 3059, 2850} \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {2 a^3 (8 A+10 B+11 C) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {2 a^3 (584 A+690 B+903 C) \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 (64 A+90 B+63 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{315 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a (5 A+9 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{63 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)} \]

[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 2850

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]

Rule 3054

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
*Sin[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 3059

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 3122

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])

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 1.21 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.68 \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} (2908 A+2790 B+2961 C+2 (1396 A+1215 B+882 C) \cos (c+d x)+4 (803 A+870 B+966 C) \cos (2 (c+d x))+584 A \cos (3 (c+d x))+690 B \cos (3 (c+d x))+588 C \cos (3 (c+d x))+584 A \cos (4 (c+d x))+690 B \cos (4 (c+d x))+903 C \cos (4 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{1260 d \cos ^{\frac {9}{2}}(c+d x)} \]

[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))

Maple [A] (verified)

Time = 14.55 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.71

method	result	size
default	\(\frac {2 a^{2} \sin \left (d x +c \right ) \left (584 A \left (\cos ^{4}\left (d x +c \right )\right )+690 B \left (\cos ^{4}\left (d x +c \right )\right )+903 C \left (\cos ^{4}\left (d x +c \right )\right )+292 A \left (\cos ^{3}\left (d x +c \right )\right )+345 B \left (\cos ^{3}\left (d x +c \right )\right )+294 C \left (\cos ^{3}\left (d x +c \right )\right )+219 A \left (\cos ^{2}\left (d x +c \right )\right )+180 B \left (\cos ^{2}\left (d x +c \right )\right )+63 C \left (\cos ^{2}\left (d x +c \right )\right )+130 A \cos \left (d x +c \right )+45 B \cos \left (d x +c \right )+35 A \right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{315 d \left (1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{\frac {9}{2}}}\)	\(166\)
parts	\(\frac {2 A \sin \left (d x +c \right ) \left (584 \left (\cos ^{4}\left (d x +c \right )\right )+292 \left (\cos ^{3}\left (d x +c \right )\right )+219 \left (\cos ^{2}\left (d x +c \right )\right )+130 \cos \left (d x +c \right )+35\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, a^{2}}{315 d \left (1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{\frac {9}{2}}}+\frac {2 B \sin \left (d x +c \right ) \left (46 \left (\cos ^{3}\left (d x +c \right )\right )+23 \left (\cos ^{2}\left (d x +c \right )\right )+12 \cos \left (d x +c \right )+3\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, a^{2}}{21 d \left (1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{\frac {7}{2}}}+\frac {2 C \sin \left (d x +c \right ) \left (43 \left (\cos ^{2}\left (d x +c \right )\right )+14 \cos \left (d x +c \right )+3\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, a^{2}}{15 d \left (1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{\frac {5}{2}}}\)	\(227\)

[In]

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

[Out]

2/315*a^2/d*sin(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)/(1+cos(d*x+c))/cos(d*x+c)^(9/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.61 \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {2 \, {\left ({\left (584 \, A + 690 \, B + 903 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + {\left (292 \, A + 345 \, B + 294 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (73 \, A + 60 \, B + 21 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 5 \, {\left (26 \, A + 9 \, B\right )} a^{2} \cos \left (d x + c\right ) + 35 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\text {Timed out} \]

[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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 682 vs. \(2 (204) = 408\).

Time = 0.41 (sec) , antiderivative size = 682, normalized size of antiderivative = 2.91 \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\text {Too large to display} \]

[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

Giac [F(-1)]

Timed out. \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\text {Timed out} \]

[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]

Timed out

Mupad [B] (verification not implemented)

Time = 9.60 (sec) , antiderivative size = 713, normalized size of antiderivative = 3.05 \[ \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\text {Too large to display} \]

[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]

((a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((a^2*(584*A + 690*B + 903*C)*2i)/(315*d) - (C*
a^2*exp(c*1i + d*x*1i)*2i)/d + (C*a^2*exp(c*8i + d*x*8i)*2i)/d - (a^2*exp(c*3i + d*x*3i)*(2*A + 5*B + 10*C)*4i
)/(3*d) + (a^2*exp(c*6i + d*x*6i)*(2*A + 5*B + 10*C)*4i)/(3*d) + (a^2*exp(c*4i + d*x*4i)*(24*A + 25*B + 33*C)*
4i)/(5*d) - (a^2*exp(c*5i + d*x*5i)*(24*A + 25*B + 33*C)*4i)/(5*d) + (a^2*exp(c*2i + d*x*2i)*(146*A + 155*B +
182*C)*4i)/(35*d) - (a^2*exp(c*7i + d*x*7i)*(146*A + 155*B + 182*C)*4i)/(35*d) - (a^2*exp(c*9i + d*x*9i)*(584*
A + 690*B + 903*C)*2i)/(315*d)))/((exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2) + exp(c*1i + d*x*1i)*(
exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2) + 4*exp(c*2i + d*x*2i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i
+ d*x*1i)/2)^(1/2) + 4*exp(c*3i + d*x*3i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2) + 6*exp(c*4i
+ d*x*4i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2) + 6*exp(c*5i + d*x*5i)*(exp(- c*1i - d*x*1i)/2
+ exp(c*1i + d*x*1i)/2)^(1/2) + 4*exp(c*6i + d*x*6i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2) +
4*exp(c*7i + d*x*7i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2) + exp(c*8i + d*x*8i)*(exp(- c*1i -
d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2) + exp(c*9i + d*x*9i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(
1/2))