3.482 \(\int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=178 \[ \frac {2 a (24 A+28 B+35 C) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {4 a (24 A+28 B+35 C) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}+\frac {2 a (A+7 B) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {2 A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)} \]

[Out]

2/35*a*(A+7*B)*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2)+2/105*a*(24*A+28*B+35*C)*sin(d*x+c)/d/cos(
d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2)+4/105*a*(24*A+28*B+35*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1
/2)+2/7*A*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(7/2)

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Rubi [A]  time = 0.44, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {3043, 2980, 2772, 2771} \[ \frac {2 a (24 A+28 B+35 C) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {4 a (24 A+28 B+35 C) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}+\frac {2 a (A+7 B) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {2 A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(9/2),x]

[Out]

(2*a*(A + 7*B)*Sin[c + d*x])/(35*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(24*A + 28*B + 35*C)*Si
n[c + d*x])/(105*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + (4*a*(24*A + 28*B + 35*C)*Sin[c + d*x])/(105
*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]) + (2*A*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(7*d*Cos[c + d*x
]^(7/2))

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 2772

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

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 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])

Rubi steps

\begin {align*} \int \frac {\sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx &=\frac {2 A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 \int \frac {\sqrt {a+a \cos (c+d x)} \left (\frac {1}{2} a (A+7 B)+\frac {1}{2} a (4 A+7 C) \cos (c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx}{7 a}\\ &=\frac {2 a (A+7 B) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {1}{35} (24 A+28 B+35 C) \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 a (A+7 B) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {2 a (24 A+28 B+35 C) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {1}{105} (2 (24 A+28 B+35 C)) \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a (A+7 B) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {2 a (24 A+28 B+35 C) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {4 a (24 A+28 B+35 C) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 0.60, size = 121, normalized size = 0.68 \[ \frac {\tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\cos (c+d x)+1)} (3 (36 A+42 B+35 C) \cos (c+d x)+(24 A+28 B+35 C) \cos (2 (c+d x))+24 A \cos (3 (c+d x))+54 A+28 B \cos (3 (c+d x))+28 B+35 C \cos (3 (c+d x))+35 C)}{105 d \cos ^{\frac {7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(9/2),x]

[Out]

(Sqrt[a*(1 + Cos[c + d*x])]*(54*A + 28*B + 35*C + 3*(36*A + 42*B + 35*C)*Cos[c + d*x] + (24*A + 28*B + 35*C)*C
os[2*(c + d*x)] + 24*A*Cos[3*(c + d*x)] + 28*B*Cos[3*(c + d*x)] + 35*C*Cos[3*(c + d*x)])*Tan[(c + d*x)/2])/(10
5*d*Cos[c + d*x]^(7/2))

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fricas [A]  time = 0.58, size = 109, normalized size = 0.61 \[ \frac {2 \, {\left (2 \, {\left (24 \, A + 28 \, B + 35 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (24 \, A + 28 \, B + 35 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (6 \, A + 7 \, B\right )} \cos \left (d x + c\right ) + 15 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*(a+a*cos(d*x+c))^(1/2)/cos(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

2/105*(2*(24*A + 28*B + 35*C)*cos(d*x + c)^3 + (24*A + 28*B + 35*C)*cos(d*x + c)^2 + 3*(6*A + 7*B)*cos(d*x + c
) + 15*A)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^5 + d*cos(d*x + c)^4)

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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+B*cos(d*x+c)+C*cos(d*x+c)^2)*(a+a*cos(d*x+c))^(1/2)/cos(d*x+c)^(9/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.37, size = 130, normalized size = 0.73 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (48 A \left (\cos ^{3}\left (d x +c \right )\right )+56 B \left (\cos ^{3}\left (d x +c \right )\right )+70 C \left (\cos ^{3}\left (d x +c \right )\right )+24 A \left (\cos ^{2}\left (d x +c \right )\right )+28 B \left (\cos ^{2}\left (d x +c \right )\right )+35 C \left (\cos ^{2}\left (d x +c \right )\right )+18 A \cos \left (d x +c \right )+21 B \cos \left (d x +c \right )+15 A \right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{105 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*(a+a*cos(d*x+c))^(1/2)/cos(d*x+c)^(9/2),x)

[Out]

-2/105/d*(-1+cos(d*x+c))*(48*A*cos(d*x+c)^3+56*B*cos(d*x+c)^3+70*C*cos(d*x+c)^3+24*A*cos(d*x+c)^2+28*B*cos(d*x
+c)^2+35*C*cos(d*x+c)^2+18*A*cos(d*x+c)+21*B*cos(d*x+c)+15*A)*(a*(1+cos(d*x+c)))^(1/2)/sin(d*x+c)/cos(d*x+c)^(
7/2)

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maxima [B]  time = 1.15, size = 710, normalized size = 3.99 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*(a+a*cos(d*x+c))^(1/2)/cos(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

2/105*(35*C*(3*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 4*sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c
) + 1)^3 + sqrt(2)*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^2/((
sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)*(2*sin(d*x + c)^2/(cos
(d*x + c) + 1)^2 + sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1)) + 7*B*(15*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x +
c) + 1) - 25*sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 17*sqrt(2)*sqrt(a)*sin(d*x + c)^5/(cos(d*x
+ c) + 1)^5 - 7*sqrt(2)*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1
)^3/((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*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)) + 3
*A*(35*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 70*sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^
3 + 84*sqrt(2)*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 58*sqrt(2)*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) +
1)^7 + 9*sqrt(2)*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^4/((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)*(4*sin(d*x + c)^2/(cos(
d*x + c) + 1)^2 + 6*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + sin(d*x + c)
^8/(cos(d*x + c) + 1)^8 + 1)))/d

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mupad [B]  time = 7.20, size = 503, normalized size = 2.83 \[ \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 {\left (96\,A+112\,B+140\,C\right )\,1{}\mathrm {i}}{105\,d}-\frac {{\mathrm {e}}^{c\,7{}\mathrm {i}+d\,x\,7{}\mathrm {i}}\,\left (96\,A+112\,B+140\,C\right )\,1{}\mathrm {i}}{105\,d}+\frac {{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,\left (336\,A+392\,B+280\,C\right )\,1{}\mathrm {i}}{105\,d}-\frac {{\mathrm {e}}^{c\,5{}\mathrm {i}+d\,x\,5{}\mathrm {i}}\,\left (336\,A+392\,B+280\,C\right )\,1{}\mathrm {i}}{105\,d}-\frac {{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,\left (280\,B+140\,C\right )\,1{}\mathrm {i}}{105\,d}+\frac {{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,\left (280\,B+140\,C\right )\,1{}\mathrm {i}}{105\,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}}+3\,{\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}}+3\,{\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}}+3\,{\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}}+3\,{\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}}+{\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}}+{\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}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + a*cos(c + d*x))^(1/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^(9/2),x)

[Out]

((a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(((96*A + 112*B + 140*C)*1i)/(105*d) - (exp(c*7
i + d*x*7i)*(96*A + 112*B + 140*C)*1i)/(105*d) + (exp(c*2i + d*x*2i)*(336*A + 392*B + 280*C)*1i)/(105*d) - (ex
p(c*5i + d*x*5i)*(336*A + 392*B + 280*C)*1i)/(105*d) - (exp(c*3i + d*x*3i)*(280*B + 140*C)*1i)/(105*d) + (exp(
c*4i + d*x*4i)*(280*B + 140*C)*1i)/(105*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) + 3*exp(c*2i + d*x*2i)*(exp(- c*1i - d*x*1i)/
2 + exp(c*1i + d*x*1i)/2)^(1/2) + 3*exp(c*3i + d*x*3i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2) +
3*exp(c*4i + d*x*4i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2) + 3*exp(c*5i + d*x*5i)*(exp(- c*1i
- d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2) + exp(c*6i + d*x*6i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2
)^(1/2) + exp(c*7i + d*x*7i)*(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+B*cos(d*x+c)+C*cos(d*x+c)**2)*(a+a*cos(d*x+c))**(1/2)/cos(d*x+c)**(9/2),x)

[Out]

Timed out

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