3.883 \(\int \frac{B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac{9}{2}}(c+d x) (a+b \cos (c+d x))} \, dx\)
Optimal. Leaf size=217 \[ -\frac{2 \left (3 a^2 B-5 a b C+5 b^2 B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^3 d}-\frac{2 b^2 (b B-a C) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d (a+b)}+\frac{2 \left (3 a^2 B-5 a b C+5 b^2 B\right ) \sin (c+d x)}{5 a^3 d \sqrt{\cos (c+d x)}}-\frac{2 (b B-a C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac{2 (b B-a C) \sin (c+d x)}{3 a^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 B \sin (c+d x)}{5 a d \cos ^{\frac{5}{2}}(c+d x)} \]
[Out]
(-2*(3*a^2*B + 5*b^2*B - 5*a*b*C)*EllipticE[(c + d*x)/2, 2])/(5*a^3*d) - (2*(b*B - a*C)*EllipticF[(c + d*x)/2,
2])/(3*a^2*d) - (2*b^2*(b*B - a*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a^3*(a + b)*d) + (2*B*Sin[c +
d*x])/(5*a*d*Cos[c + d*x]^(5/2)) - (2*(b*B - a*C)*Sin[c + d*x])/(3*a^2*d*Cos[c + d*x]^(3/2)) + (2*(3*a^2*B + 5
*b^2*B - 5*a*b*C)*Sin[c + d*x])/(5*a^3*d*Sqrt[Cos[c + d*x]])
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Rubi [A] time = 1.23606, antiderivative size = 217, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 8, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used =
{3029, 3000, 3055, 3059, 2639, 3002, 2641, 2805} \[ -\frac{2 \left (3 a^2 B-5 a b C+5 b^2 B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^3 d}-\frac{2 b^2 (b B-a C) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d (a+b)}+\frac{2 \left (3 a^2 B-5 a b C+5 b^2 B\right ) \sin (c+d x)}{5 a^3 d \sqrt{\cos (c+d x)}}-\frac{2 (b B-a C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac{2 (b B-a C) \sin (c+d x)}{3 a^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 B \sin (c+d x)}{5 a d \cos ^{\frac{5}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Int[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(9/2)*(a + b*Cos[c + d*x])),x]
[Out]
(-2*(3*a^2*B + 5*b^2*B - 5*a*b*C)*EllipticE[(c + d*x)/2, 2])/(5*a^3*d) - (2*(b*B - a*C)*EllipticF[(c + d*x)/2,
2])/(3*a^2*d) - (2*b^2*(b*B - a*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a^3*(a + b)*d) + (2*B*Sin[c +
d*x])/(5*a*d*Cos[c + d*x]^(5/2)) - (2*(b*B - a*C)*Sin[c + d*x])/(3*a^2*d*Cos[c + d*x]^(3/2)) + (2*(3*a^2*B + 5
*b^2*B - 5*a*b*C)*Sin[c + d*x])/(5*a^3*d*Sqrt[Cos[c + d*x]])
Rule 3029
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Rule 3000
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((A*b^2 - a*b*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*
Sin[e + f*x])^(1 + n))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int
[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m
+ n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B)*(m + n + 3)*Sin[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^
2, 0] && RationalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n,
-1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
Rule 3055
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[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))
Rule 3059
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0]
Rule 2639
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]
Rule 3002
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Rule 2641
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]
Rule 2805
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Rubi steps
\begin{align*} \int \frac{B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac{9}{2}}(c+d x) (a+b \cos (c+d x))} \, dx &=\int \frac{B+C \cos (c+d x)}{\cos ^{\frac{7}{2}}(c+d x) (a+b \cos (c+d x))} \, dx\\ &=\frac{2 B \sin (c+d x)}{5 a d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \int \frac{-\frac{5}{2} (b B-a C)+\frac{3}{2} a B \cos (c+d x)+\frac{3}{2} b B \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{5 a}\\ &=\frac{2 B \sin (c+d x)}{5 a d \cos ^{\frac{5}{2}}(c+d x)}-\frac{2 (b B-a C) \sin (c+d x)}{3 a^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 \int \frac{\frac{3}{4} \left (3 a^2 B+5 b^2 B-5 a b C\right )+\frac{1}{4} a (4 b B+5 a C) \cos (c+d x)-\frac{5}{4} b (b B-a C) \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{15 a^2}\\ &=\frac{2 B \sin (c+d x)}{5 a d \cos ^{\frac{5}{2}}(c+d x)}-\frac{2 (b B-a C) \sin (c+d x)}{3 a^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (3 a^2 B+5 b^2 B-5 a b C\right ) \sin (c+d x)}{5 a^3 d \sqrt{\cos (c+d x)}}+\frac{8 \int \frac{-\frac{5}{8} \left (a^2+3 b^2\right ) (b B-a C)-\frac{1}{8} a \left (9 a^2 B+20 b^2 B-20 a b C\right ) \cos (c+d x)-\frac{3}{8} b \left (3 a^2 B+5 b^2 B-5 a b C\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{15 a^3}\\ &=\frac{2 B \sin (c+d x)}{5 a d \cos ^{\frac{5}{2}}(c+d x)}-\frac{2 (b B-a C) \sin (c+d x)}{3 a^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (3 a^2 B+5 b^2 B-5 a b C\right ) \sin (c+d x)}{5 a^3 d \sqrt{\cos (c+d x)}}-\frac{8 \int \frac{\frac{5}{8} b \left (a^2+3 b^2\right ) (b B-a C)+\frac{5}{8} a b^2 (b B-a C) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{15 a^3 b}-\frac{\left (3 a^2 B+5 b^2 B-5 a b C\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 a^3}\\ &=-\frac{2 \left (3 a^2 B+5 b^2 B-5 a b C\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^3 d}+\frac{2 B \sin (c+d x)}{5 a d \cos ^{\frac{5}{2}}(c+d x)}-\frac{2 (b B-a C) \sin (c+d x)}{3 a^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (3 a^2 B+5 b^2 B-5 a b C\right ) \sin (c+d x)}{5 a^3 d \sqrt{\cos (c+d x)}}-\frac{(b B-a C) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 a^2}-\frac{\left (b^2 (b B-a C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a^3}\\ &=-\frac{2 \left (3 a^2 B+5 b^2 B-5 a b C\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^3 d}-\frac{2 (b B-a C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac{2 b^2 (b B-a C) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 (a+b) d}+\frac{2 B \sin (c+d x)}{5 a d \cos ^{\frac{5}{2}}(c+d x)}-\frac{2 (b B-a C) \sin (c+d x)}{3 a^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (3 a^2 B+5 b^2 B-5 a b C\right ) \sin (c+d x)}{5 a^3 d \sqrt{\cos (c+d x)}}\\ \end{align*}
Mathematica [A] time = 3.86291, size = 313, normalized size = 1.44 \[ \frac{\frac{2 \left (-19 a^2 b B+10 a^3 C+45 a b^2 C-45 b^3 B\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}-\frac{2 a \left (9 a^2 B-20 a b C+20 b^2 B\right ) \left (2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-\frac{2 a \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}\right )}{b}+\frac{2 \left (3 \left (3 a^2 B-5 a b C+5 b^2 B\right ) \sin (2 (c+d x))+6 a^2 B \tan (c+d x)+10 a (a C-b B) \sin (c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)}-\frac{6 \left (3 a^2 B-5 a b C+5 b^2 B\right ) \sin (c+d x) \left (\left (2 a^2-b^2\right ) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a b \sqrt{\sin ^2(c+d x)}}}{30 a^3 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(9/2)*(a + b*Cos[c + d*x])),x]
[Out]
((2*(-19*a^2*b*B - 45*b^3*B + 10*a^3*C + 45*a*b^2*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) - (2*a
*(9*a^2*B + 20*b^2*B - 20*a*b*C)*(2*EllipticF[(c + d*x)/2, 2] - (2*a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]
)/(a + b)))/b - (6*(3*a^2*B + 5*b^2*B - 5*a*b*C)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a +
b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (2*a^2 - b^2)*EllipticPi[-(b/a), -ArcSin[Sqrt[Cos[c + d*x]]], -
1])*Sin[c + d*x])/(a*b*Sqrt[Sin[c + d*x]^2]) + (2*(10*a*(-(b*B) + a*C)*Sin[c + d*x] + 3*(3*a^2*B + 5*b^2*B - 5
*a*b*C)*Sin[2*(c + d*x)] + 6*a^2*B*Tan[c + d*x]))/Cos[c + d*x]^(3/2))/(30*a^3*d)
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Maple [B] time = 2.697, size = 787, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(a+b*cos(d*x+c)),x)
[Out]
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*b^3*(B*b-C*a)/a^3/(-2*a*b+2*b^2)*(sin(1/2*d*x+1/
2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticP
i(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))-2/5*B/a/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x
+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si
n(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*EllipticE(cos(1
/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+24*s
in(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2
*sin(1/2*d*x+1/2*c)^2-1)^(1/2)-8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x
+1/2*c)^2)^(1/2)+2*(-B*b+C*a)/a^2*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2
)/(-1/2+cos(1/2*d*x+1/2*c)^2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2
*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+2*(B*b-C*a)/a^3*b*(-(sin(1/2*
d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Elli
pticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)*si
n(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^
2-1)^(1/2)/d
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Maxima [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((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(a+b*cos(d*x+c)),x, algorithm="maxima")
[Out]
Timed out
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Fricas [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((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(a+b*cos(d*x+c)),x, algorithm="fricas")
[Out]
Timed out
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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((B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(9/2)/(a+b*cos(d*x+c)),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac{9}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(a+b*cos(d*x+c)),x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))/((b*cos(d*x + c) + a)*cos(d*x + c)^(9/2)), x)