3.1461 \(\int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx\)
Optimal. Leaf size=220 \[ \frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (A+3 C)+6 a b B+b^2 (3 A+C)\right )}{3 d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 B+2 a b (A-C)-b^2 B\right )}{d}+\frac {2 a (3 a B+4 A b) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2}{3 d}-\frac {2 b^2 (A-C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \]
[Out]
2/3*A*(a+b*cos(d*x+c))^2*sec(d*x+c)^(3/2)*sin(d*x+c)/d-2/3*b^2*(A-C)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+2/3*a*(4*A*
b+3*B*a)*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2*(a^2*B-b^2*B+2*a*b*(A-C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/
2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+2/3*(6*a*b*B+b^2*(3*A+C)+a^2*(A
+3*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*
sec(d*x+c)^(1/2)/d
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Rubi [A] time = 0.62, antiderivative size = 220, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used =
{4221, 3047, 3031, 3023, 2748, 2641, 2639} \[ \frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (A+3 C)+6 a b B+b^2 (3 A+C)\right )}{3 d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 B+2 a b (A-C)-b^2 B\right )}{d}+\frac {2 a (3 a B+4 A b) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2}{3 d}-\frac {2 b^2 (A-C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(5/2),x]
[Out]
(-2*(a^2*B - b^2*B + 2*a*b*(A - C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*(6
*a*b*B + b^2*(3*A + C) + a^2*(A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d)
- (2*b^2*(A - C)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + (2*a*(4*A*b + 3*a*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x
])/(3*d) + (2*A*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d)
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 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 2748
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Rule 3023
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Rule 3031
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]
Rule 3047
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/(d*(n + 1)*(
c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c
*C - B*d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*
d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)
))*Sin[e + f*x]^2, 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] && GtQ[m, 0] && LtQ[n, -1]
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]
Rubi steps
\begin {align*} \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 A (a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {1}{3} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x)) \left (\frac {1}{2} (4 A b+3 a B)+\frac {1}{2} (3 b B+a (A+3 C)) \cos (c+d x)-\frac {3}{2} b (A-C) \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a (4 A b+3 a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}-\frac {1}{3} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{4} \left (-4 A b^2-6 a b B-a^2 (A+3 C)\right )+\frac {3}{4} \left (a^2 B-b^2 B+2 a b (A-C)\right ) \cos (c+d x)+\frac {3}{4} b^2 (A-C) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 b^2 (A-C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 a (4 A b+3 a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}-\frac {1}{9} \left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {3}{8} \left (6 a b B+b^2 (3 A+C)+a^2 (A+3 C)\right )+\frac {9}{8} \left (a^2 B-b^2 B+2 a b (A-C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 b^2 (A-C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 a (4 A b+3 a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}-\left (\left (a^2 B-b^2 B+2 a b (A-C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{3} \left (\left (6 a b B+b^2 (3 A+C)+a^2 (A+3 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {2 \left (6 a b B+b^2 (3 A+C)+a^2 (A+3 C)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {2 b^2 (A-C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 a (4 A b+3 a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}\\ \end {align*}
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Mathematica [A] time = 1.19, size = 158, normalized size = 0.72 \[ \frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (A+3 C)+6 a b B+b^2 (3 A+C)\right )-6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 B+2 a b (A-C)-b^2 B\right )+\frac {\sin (c+d x) \left (2 a^2 A+6 a (a B+2 A b) \cos (c+d x)+b^2 C \cos (2 (c+d x))+b^2 C\right )}{\cos ^{\frac {3}{2}}(c+d x)}\right )}{3 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(5/2),x]
[Out]
(Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-6*(a^2*B - b^2*B + 2*a*b*(A - C))*EllipticE[(c + d*x)/2, 2] + 2*(6*a*
b*B + b^2*(3*A + C) + a^2*(A + 3*C))*EllipticF[(c + d*x)/2, 2] + ((2*a^2*A + b^2*C + 6*a*(2*A*b + a*B)*Cos[c +
d*x] + b^2*C*Cos[2*(c + d*x)])*Sin[c + d*x])/Cos[c + d*x]^(3/2)))/(3*d)
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fricas [F] time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{2} \cos \left (d x + c\right )^{4} + {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )^{3} + A a^{2} + {\left (C a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sec \left (d x + c\right )^{\frac {5}{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2),x, algorithm="fricas")
[Out]
integral((C*b^2*cos(d*x + c)^4 + (2*C*a*b + B*b^2)*cos(d*x + c)^3 + A*a^2 + (C*a^2 + 2*B*a*b + A*b^2)*cos(d*x
+ c)^2 + (B*a^2 + 2*A*a*b)*cos(d*x + c))*sec(d*x + c)^(5/2), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {5}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2),x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^2*sec(d*x + c)^(5/2), x)
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maple [B] time = 8.22, size = 1303, normalized size = 5.92 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2),x)
[Out]
2/3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1
)/sin(1/2*d*x+1/2*c)^3*(-8*C*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+2*A*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*
x+1/2*c)^2+6*B*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+2*C*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+8*C
*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*B*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+2*A*(2*sin(1/2*d
*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*sin(1/2*d*x+1/2*c)
^2-C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^2-A
*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2-24*A*
a*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+12*A*a*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-3*A*(2*sin(1/2*d*
x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^2-3*B*(2*sin(1/2*d*x+
1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2+3*B*(2*sin(1/2*d*x+1/
2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^2-3*C*(2*sin(1/2*d*x+1/2*
c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2-12*C*(2*sin(1/2*d*x+1/2*c
)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b*sin(1/2*d*x+1/2*c)^2+12*A*
(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b*sin(1/
2*d*x+1/2*c)^2+12*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c)
,2^(1/2))*a*b*sin(1/2*d*x+1/2*c)^2+6*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF
(cos(1/2*d*x+1/2*c),2^(1/2))*b^2*sin(1/2*d*x+1/2*c)^2+6*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)
^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*sin(1/2*d*x+1/2*c)^2-6*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^2*sin(1/2*d*x+1/2*c)^2+6*C*(2*sin(1/2*d*x
+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*sin(1/2*d*x+1/2*c)^2
+2*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^2*s
in(1/2*d*x+1/2*c)^2-6*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/
2*c),2^(1/2))*a*b-6*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*
c),2^(1/2))*a*b+6*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c)
,2^(1/2))*a*b)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {5}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2),x, algorithm="maxima")
[Out]
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^2*sec(d*x + c)^(5/2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((1/cos(c + d*x))^(5/2)*(a + b*cos(c + d*x))^2*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)
[Out]
int((1/cos(c + d*x))^(5/2)*(a + b*cos(c + d*x))^2*(A + B*cos(c + d*x) + C*cos(c + d*x)^2), x)
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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))**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**(5/2),x)
[Out]
Timed out
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