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

Optimal. Leaf size=219 \[ \frac {2 a^3 (8 A+11 C) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {2 a^3 (584 A+903 C) \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 (64 A+63 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{315 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {10 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{63 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)} \]

[Out]

10/63*a*A*(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+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*(584*A+903*C)*s
in(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2)+2/315*a^2*(64*A+63*C)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d/
cos(d*x+c)^(5/2)

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Rubi [A]  time = 0.72, antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {3044, 2975, 2980, 2771} \[ \frac {2 a^3 (8 A+11 C) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 (64 A+63 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{315 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a^3 (584 A+903 C) \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}+\frac {10 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{63 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{9 d \cos ^{\frac {9}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*(8*A + 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 + 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 + 63*C)*Sqrt[a + a*Cos[c + d*
x]]*Sin[c + d*x])/(315*d*Cos[c + d*x]^(5/2)) + (10*a*A*(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 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]

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 3044

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + 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*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(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, 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*} \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+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 {5 a A}{2}+\frac {1}{2} a (2 A+9 C) \cos (c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx}{9 a}\\ &=\frac {10 a A (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+63 C)+\frac {3}{4} a^2 (8 A+21 C) \cos (c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx}{63 a}\\ &=\frac {2 a^2 (64 A+63 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {10 a A (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+11 C)+\frac {1}{8} a^3 (248 A+441 C) \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{315 a}\\ &=\frac {2 a^3 (8 A+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+63 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {10 a A (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+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+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+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+63 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {10 a A (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*}

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Mathematica [A]  time = 0.92, size = 127, normalized size = 0.58 \[ \frac {a^2 \tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\cos (c+d x)+1)} (4 (698 A+441 C) \cos (c+d x)+4 (803 A+966 C) \cos (2 (c+d x))+584 A \cos (3 (c+d x))+584 A \cos (4 (c+d x))+2908 A+588 C \cos (3 (c+d x))+903 C \cos (4 (c+d x))+2961 C)}{1260 d \cos ^{\frac {9}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Cos[c + d*x])^(5/2)*(A + 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 + 2961*C + 4*(698*A + 441*C)*Cos[c + d*x] + 4*(803*A + 966*C)*Cos[2*(c
+ d*x)] + 584*A*Cos[3*(c + d*x)] + 588*C*Cos[3*(c + d*x)] + 584*A*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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fricas [A]  time = 0.42, size = 129, normalized size = 0.59 \[ \frac {2 \, {\left ({\left (584 \, A + 903 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 2 \, {\left (146 \, A + 147 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (73 \, A + 21 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 130 \, A a^{2} \cos \left (d x + c\right ) + 35 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*((584*A + 903*C)*a^2*cos(d*x + c)^4 + 2*(146*A + 147*C)*a^2*cos(d*x + c)^3 + 3*(73*A + 21*C)*a^2*cos(d*x
+ c)^2 + 130*A*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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.35, size = 124, normalized size = 0.57 \[ -\frac {2 a^{2} \left (-1+\cos \left (d x +c \right )\right ) \left (584 A \left (\cos ^{4}\left (d x +c \right )\right )+903 C \left (\cos ^{4}\left (d x +c \right )\right )+292 A \left (\cos ^{3}\left (d x +c \right )\right )+294 C \left (\cos ^{3}\left (d x +c \right )\right )+219 A \left (\cos ^{2}\left (d x +c \right )\right )+63 C \left (\cos ^{2}\left (d x +c \right )\right )+130 A \cos \left (d x +c \right )+35 A \right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{315 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^(5/2)*(A+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+903*C*cos(d*x+c)^4+292*A*cos(d*x+c)^3+294*C*cos(d*x+c)^3+219*
A*cos(d*x+c)^2+63*C*cos(d*x+c)^2+130*A*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 = 0.96, size = 441, normalized size = 2.01 \[ \frac {8 \, {\left (\frac {21 \, {\left (\frac {15 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {28 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {8 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )} C}{{\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}}} + \frac {{\left (\frac {315 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {945 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {1449 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {1287 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {572 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {104 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}}\right )} A {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{3}}{{\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {11}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {11}{2}} {\left (\frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {\sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 1\right )}}\right )}}{315 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+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)) + (3
15*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/(cos(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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mupad [B]  time = 9.77, size = 685, normalized size = 3.13 \[ \frac {\sqrt {a+a\,\left (\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}\right )}\,\left (\frac {a^2\,\left (584\,A+903\,C\right )\,2{}\mathrm {i}}{315\,d}+\frac {a^2\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,\left (8\,A+11\,C\right )\,12{}\mathrm {i}}{5\,d}-\frac {a^2\,{\mathrm {e}}^{c\,5{}\mathrm {i}+d\,x\,5{}\mathrm {i}}\,\left (8\,A+11\,C\right )\,12{}\mathrm {i}}{5\,d}+\frac {a^2\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,\left (73\,A+91\,C\right )\,8{}\mathrm {i}}{35\,d}-\frac {a^2\,{\mathrm {e}}^{c\,7{}\mathrm {i}+d\,x\,7{}\mathrm {i}}\,\left (73\,A+91\,C\right )\,8{}\mathrm {i}}{35\,d}-\frac {a^2\,{\mathrm {e}}^{c\,9{}\mathrm {i}+d\,x\,9{}\mathrm {i}}\,\left (584\,A+903\,C\right )\,2{}\mathrm {i}}{315\,d}-\frac {a^2\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,\left (A+5\,C\right )\,8{}\mathrm {i}}{3\,d}+\frac {a^2\,{\mathrm {e}}^{c\,6{}\mathrm {i}+d\,x\,6{}\mathrm {i}}\,\left (A+5\,C\right )\,8{}\mathrm {i}}{3\,d}-\frac {C\,a^2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,2{}\mathrm {i}}{d}+\frac {C\,a^2\,{\mathrm {e}}^{c\,8{}\mathrm {i}+d\,x\,8{}\mathrm {i}}\,2{}\mathrm {i}}{d}\right )}{\sqrt {\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}+{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}+4\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}+4\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}+6\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}+6\,{\mathrm {e}}^{c\,5{}\mathrm {i}+d\,x\,5{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}+4\,{\mathrm {e}}^{c\,6{}\mathrm {i}+d\,x\,6{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}+4\,{\mathrm {e}}^{c\,7{}\mathrm {i}+d\,x\,7{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}+{\mathrm {e}}^{c\,8{}\mathrm {i}+d\,x\,8{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}+{\mathrm {e}}^{c\,9{}\mathrm {i}+d\,x\,9{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^(5/2))/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 + 903*C)*2i)/(315*d) + (a^2*exp(c*
4i + d*x*4i)*(8*A + 11*C)*12i)/(5*d) - (a^2*exp(c*5i + d*x*5i)*(8*A + 11*C)*12i)/(5*d) + (a^2*exp(c*2i + d*x*2
i)*(73*A + 91*C)*8i)/(35*d) - (a^2*exp(c*7i + d*x*7i)*(73*A + 91*C)*8i)/(35*d) - (a^2*exp(c*9i + d*x*9i)*(584*
A + 903*C)*2i)/(315*d) - (a^2*exp(c*3i + d*x*3i)*(A + 5*C)*8i)/(3*d) + (a^2*exp(c*6i + d*x*6i)*(A + 5*C)*8i)/(
3*d) - (C*a^2*exp(c*1i + d*x*1i)*2i)/d + (C*a^2*exp(c*8i + d*x*8i)*2i)/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))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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