3.1319 \(\int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac{11}{2}}(c+d x) \, dx\)
Optimal. Leaf size=232 \[ \frac{2 a^2 (52 A+72 B+63 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{315 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (136 A+156 B+189 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{315 d \sqrt{a \cos (c+d x)+a}}+\frac{4 a^2 (136 A+156 B+189 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{315 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a (A+3 B) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{21 d}+\frac{2 A \sin (c+d x) \sec ^{\frac{9}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{9 d} \]
[Out]
(4*a^2*(136*A + 156*B + 189*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(136
*A + 156*B + 189*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(52*A + 72*B +
63*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(A + 3*B)*Sqrt[a + a*Cos[c + d*
x]]*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(21*d) + (2*A*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(9/2)*Sin[c + d*x])
/(9*d)
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Rubi [A] time = 0.809637, antiderivative size = 232, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used =
{4221, 3043, 2975, 2980, 2772, 2771} \[ \frac{2 a^2 (52 A+72 B+63 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{315 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (136 A+156 B+189 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{315 d \sqrt{a \cos (c+d x)+a}}+\frac{4 a^2 (136 A+156 B+189 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{315 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a (A+3 B) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{21 d}+\frac{2 A \sin (c+d x) \sec ^{\frac{9}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{9 d} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(11/2),x]
[Out]
(4*a^2*(136*A + 156*B + 189*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(136
*A + 156*B + 189*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(52*A + 72*B +
63*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(A + 3*B)*Sqrt[a + a*Cos[c + d*
x]]*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(21*d) + (2*A*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(9/2)*Sin[c + d*x])
/(9*d)
Rule 4221
Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u,
x]
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 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 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 (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac{11}{2}}(c+d x) \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \cos (c+d x))^{3/2} \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))^{3/2} \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \cos (c+d x))^{3/2} \left (\frac{3}{2} a (A+3 B)+\frac{1}{2} a (4 A+9 C) \cos (c+d x)\right )}{\cos ^{\frac{9}{2}}(c+d x)} \, dx}{9 a}\\ &=\frac{2 a (A+3 B) \sqrt{a+a \cos (c+d x)} \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac{2 A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)} \left (\frac{1}{4} a^2 (52 A+72 B+63 C)+\frac{1}{4} a^2 (40 A+36 B+63 C) \cos (c+d x)\right )}{\cos ^{\frac{7}{2}}(c+d x)} \, dx}{63 a}\\ &=\frac{2 a^2 (52 A+72 B+63 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{315 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a (A+3 B) \sqrt{a+a \cos (c+d x)} \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac{2 A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac{1}{105} \left (a (136 A+156 B+189 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 (136 A+156 B+189 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{315 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (52 A+72 B+63 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{315 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a (A+3 B) \sqrt{a+a \cos (c+d x)} \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac{2 A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac{1}{315} \left (2 a (136 A+156 B+189 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{4 a^2 (136 A+156 B+189 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{315 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (136 A+156 B+189 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{315 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (52 A+72 B+63 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{315 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a (A+3 B) \sqrt{a+a \cos (c+d x)} \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac{2 A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{9 d}\\ \end{align*}
Mathematica [A] time = 1.05458, size = 157, normalized size = 0.68 \[ \frac{a \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{9}{2}}(c+d x) \sqrt{a (\cos (c+d x)+1)} ((748 A+81 (8 B+7 C)) \cos (c+d x)+(748 A+858 B+882 C) \cos (2 (c+d x))+136 A \cos (3 (c+d x))+136 A \cos (4 (c+d x))+752 A+156 B \cos (3 (c+d x))+156 B \cos (4 (c+d x))+702 B+189 C \cos (3 (c+d x))+189 C \cos (4 (c+d x))+693 C)}{630 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(11/2),x]
[Out]
(a*Sqrt[a*(1 + Cos[c + d*x])]*(752*A + 702*B + 693*C + (748*A + 81*(8*B + 7*C))*Cos[c + d*x] + (748*A + 858*B
+ 882*C)*Cos[2*(c + d*x)] + 136*A*Cos[3*(c + d*x)] + 156*B*Cos[3*(c + d*x)] + 189*C*Cos[3*(c + d*x)] + 136*A*C
os[4*(c + d*x)] + 156*B*Cos[4*(c + d*x)] + 189*C*Cos[4*(c + d*x)])*Sec[c + d*x]^(9/2)*Tan[(c + d*x)/2])/(630*d
)
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Maple [A] time = 0.198, size = 172, normalized size = 0.7 \begin{align*} -{\frac{2\,a \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 272\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+312\,B \left ( \cos \left ( dx+c \right ) \right ) ^{4}+378\,C \left ( \cos \left ( dx+c \right ) \right ) ^{4}+136\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+156\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+189\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+102\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+117\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+63\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+85\,A\cos \left ( dx+c \right ) +45\,B\cos \left ( dx+c \right ) +35\,A \right ) \cos \left ( dx+c \right ) }{315\,d\sin \left ( dx+c \right ) } \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{11}{2}}}\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) }} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(11/2),x)
[Out]
-2/315/d*a*(-1+cos(d*x+c))*(272*A*cos(d*x+c)^4+312*B*cos(d*x+c)^4+378*C*cos(d*x+c)^4+136*A*cos(d*x+c)^3+156*B*
cos(d*x+c)^3+189*C*cos(d*x+c)^3+102*A*cos(d*x+c)^2+117*B*cos(d*x+c)^2+63*C*cos(d*x+c)^2+85*A*cos(d*x+c)+45*B*c
os(d*x+c)+35*A)*cos(d*x+c)*(1/cos(d*x+c))^(11/2)*(a*(1+cos(d*x+c)))^(1/2)/sin(d*x+c)
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Maxima [B] time = 1.90508, size = 1250, normalized size = 5.39 \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))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(11/2),x, algorithm="maxima")
[Out]
4/315*((315*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 840*sqrt(2)*a^(3/2)*sin(d*x + c)^3/(cos(d*x + c)
+ 1)^3 + 1344*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 1242*sqrt(2)*a^(3/2)*sin(d*x + c)^7/(cos(
d*x + c) + 1)^7 + 517*sqrt(2)*a^(3/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 94*sqrt(2)*a^(3/2)*sin(d*x + c)^11
/(cos(d*x + c) + 1)^11)*A*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^4/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(
11/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(11/2)*(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)) + 3*(
105*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 350*sqrt(2)*a^(3/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3
+ 518*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 444*sqrt(2)*a^(3/2)*sin(d*x + c)^7/(cos(d*x + c) +
1)^7 + 209*sqrt(2)*a^(3/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 38*sqrt(2)*a^(3/2)*sin(d*x + c)^11/(cos(d*x
+ c) + 1)^11)*B*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^4/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(11/2)*(-si
n(d*x + c)/(cos(d*x + c) + 1) + 1)^(11/2)*(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)) + 63*(5*sqrt(2)
*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sqrt(2)*a^(3/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 32*sqrt(2)
*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 26*sqrt(2)*a^(3/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 11*sqr
t(2)*a^(3/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 2*sqrt(2)*a^(3/2)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11)*C*
(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^4/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(11/2)*(-sin(d*x + c)/(cos(
d*x + c) + 1) + 1)^(11/2)*(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*s
in(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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Fricas [A] time = 1.63049, size = 371, normalized size = 1.6 \begin{align*} \frac{2 \,{\left (2 \,{\left (136 \, A + 156 \, B + 189 \, C\right )} a \cos \left (d x + c\right )^{4} +{\left (136 \, A + 156 \, B + 189 \, C\right )} a \cos \left (d x + c\right )^{3} + 3 \,{\left (34 \, A + 39 \, B + 21 \, C\right )} a \cos \left (d x + c\right )^{2} + 5 \,{\left (17 \, A + 9 \, B\right )} a \cos \left (d x + c\right ) + 35 \, A a\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{315 \,{\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )} \sqrt{\cos \left (d x + c\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(11/2),x, algorithm="fricas")
[Out]
2/315*(2*(136*A + 156*B + 189*C)*a*cos(d*x + c)^4 + (136*A + 156*B + 189*C)*a*cos(d*x + c)^3 + 3*(34*A + 39*B
+ 21*C)*a*cos(d*x + c)^2 + 5*(17*A + 9*B)*a*cos(d*x + c) + 35*A*a)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/((d*c
os(d*x + c)^5 + d*cos(d*x + c)^4)*sqrt(cos(d*x + c)))
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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))**(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(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{\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{3}{2}} \sec \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))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(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)^(3/2)*sec(d*x + c)^(11/2), x)