3.1279 \(\int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\sqrt {\sec (c+d x)}} \, dx\)

Optimal. Leaf size=255 \[ \frac {2 a^2 (21 A+27 B+19 C) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (7 A+6 B+5 C) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {4 a^2 (7 A+6 B+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {4 a^2 (12 A+9 B+8 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 (9 B+4 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{63 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{9 d \sec ^{\frac {3}{2}}(c+d x)} \]

[Out]

2/105*a^2*(21*A+27*B+19*C)*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/9*C*(a+a*cos(d*x+c))^2*sin(d*x+c)/d/sec(d*x+c)^(3/2
)+2/63*(9*B+4*C)*(a^2+a^2*cos(d*x+c))*sin(d*x+c)/d/sec(d*x+c)^(3/2)+4/21*a^2*(7*A+6*B+5*C)*sin(d*x+c)/d/sec(d*
x+c)^(1/2)+4/15*a^2*(12*A+9*B+8*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+4/21*a^2*(7*A+6*B+5*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 = 0.61, antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.209, Rules used = {4221, 3045, 2976, 2968, 3023, 2748, 2639, 2635, 2641} \[ \frac {2 a^2 (21 A+27 B+19 C) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (7 A+6 B+5 C) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {4 a^2 (7 A+6 B+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {4 a^2 (12 A+9 B+8 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 (9 B+4 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{63 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^2}{9 d \sec ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^2*(12*A + 9*B + 8*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (4*a^2*(7*
A + 6*B + 5*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*a^2*(21*A + 27*B +
19*C)*Sin[c + d*x])/(105*d*Sec[c + d*x]^(3/2)) + (2*C*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(9*d*Sec[c + d*x]^
(3/2)) + (2*(9*B + 4*C)*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(63*d*Sec[c + d*x]^(3/2)) + (4*a^2*(7*A + 6*B +
5*C)*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]])

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

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 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 2976

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*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 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 3045

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] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(b*d*(m + n + 2)), Int[(a + b*Sin[e + f*x
])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + (C*(a*d*m - b*c*(m + 1)) + b*B*
d*(m + n + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-1)] && NeQ[m + n + 2, 0]

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 \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2 \left (\frac {3}{2} a (3 A+C)+\frac {1}{2} a (9 B+4 C) \cos (c+d x)\right ) \, dx}{9 a}\\ &=\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (9 B+4 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x)) \left (\frac {3}{4} a^2 (21 A+9 B+11 C)+\frac {3}{4} a^2 (21 A+27 B+19 C) \cos (c+d x)\right ) \, dx}{63 a}\\ &=\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (9 B+4 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \left (\frac {3}{4} a^3 (21 A+9 B+11 C)+\left (\frac {3}{4} a^3 (21 A+9 B+11 C)+\frac {3}{4} a^3 (21 A+27 B+19 C)\right ) \cos (c+d x)+\frac {3}{4} a^3 (21 A+27 B+19 C) \cos ^2(c+d x)\right ) \, dx}{63 a}\\ &=\frac {2 a^2 (21 A+27 B+19 C) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (9 B+4 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \left (\frac {21}{4} a^3 (12 A+9 B+8 C)+\frac {45}{4} a^3 (7 A+6 B+5 C) \cos (c+d x)\right ) \, dx}{315 a}\\ &=\frac {2 a^2 (21 A+27 B+19 C) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (9 B+4 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{7} \left (2 a^2 (7 A+6 B+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \cos ^{\frac {3}{2}}(c+d x) \, dx+\frac {1}{15} \left (2 a^2 (12 A+9 B+8 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=\frac {4 a^2 (12 A+9 B+8 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{15 d}+\frac {2 a^2 (21 A+27 B+19 C) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (9 B+4 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (7 A+6 B+5 C) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {1}{21} \left (2 a^2 (7 A+6 B+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {4 a^2 (12 A+9 B+8 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{15 d}+\frac {4 a^2 (7 A+6 B+5 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 a^2 (21 A+27 B+19 C) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+a \cos (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (9 B+4 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{63 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (7 A+6 B+5 C) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.95, size = 151, normalized size = 0.59 \[ \frac {a^2 \sqrt {\sec (c+d x)} \left (2 \sin (2 (c+d x)) (7 (36 A+72 B+79 C) \cos (c+d x)+840 A+90 (B+2 C) \cos (2 (c+d x))+810 B+35 C \cos (3 (c+d x))+780 C)+480 (7 A+6 B+5 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+672 (12 A+9 B+8 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{2520 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sqrt[Sec[c + d*x]]*(672*(12*A + 9*B + 8*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 480*(7*A + 6*B
+ 5*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 2*(840*A + 810*B + 780*C + 7*(36*A + 72*B + 79*C)*Cos[c
+ d*x] + 90*(B + 2*C)*Cos[2*(c + d*x)] + 35*C*Cos[3*(c + d*x)])*Sin[2*(c + d*x)]))/(2520*d)

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C a^{2} \cos \left (d x + c\right )^{4} + {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (A + 2 \, B + C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + A a^{2}}{\sqrt {\sec \left (d x + c\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*a^2*cos(d*x + c)^4 + (B + 2*C)*a^2*cos(d*x + c)^3 + (A + 2*B + C)*a^2*cos(d*x + c)^2 + (2*A + B)*a
^2*cos(d*x + c) + A*a^2)/sqrt(sec(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2}}{\sqrt {\sec \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^2/sqrt(sec(d*x + c)), x)

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maple [A]  time = 2.85, size = 514, normalized size = 2.02 \[ -\frac {4 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a^{2} \left (-560 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (360 B +1840 C \right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-252 A -1044 B -2368 C \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (672 A +1134 B +1568 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-273 A -351 B -387 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+105 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-252 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+90 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-189 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+75 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-168 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{315 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/315*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(-560*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*
c)^10+(360*B+1840*C)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-252*A-1044*B-2368*C)*sin(1/2*d*x+1/2*c)^6*cos(1
/2*d*x+1/2*c)+(672*A+1134*B+1568*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-273*A-351*B-387*C)*sin(1/2*d*x+1
/2*c)^2*cos(1/2*d*x+1/2*c)+105*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1
/2*d*x+1/2*c),2^(1/2))-252*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d
*x+1/2*c),2^(1/2))+90*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)*EllipticF(cos(1/2*d*x+1/
2*c),2^(1/2))-189*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)*EllipticE(cos(1/2*d*x+1/2*c)
,2^(1/2))+75*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1
/2))-168*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))
)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/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]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2}}{\sqrt {\sec \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^2/sqrt(sec(d*x + c)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+a\,\cos \left (c+d\,x\right )\right )}^2\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int \frac {A}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {2 A \cos {\left (c + d x \right )}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {A \cos ^{2}{\left (c + d x \right )}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {B \cos {\left (c + d x \right )}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {2 B \cos ^{2}{\left (c + d x \right )}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {B \cos ^{3}{\left (c + d x \right )}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {2 C \cos ^{3}{\left (c + d x \right )}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {C \cos ^{4}{\left (c + d x \right )}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/sec(d*x+c)**(1/2),x)

[Out]

a**2*(Integral(A/sqrt(sec(c + d*x)), x) + Integral(2*A*cos(c + d*x)/sqrt(sec(c + d*x)), x) + Integral(A*cos(c
+ d*x)**2/sqrt(sec(c + d*x)), x) + Integral(B*cos(c + d*x)/sqrt(sec(c + d*x)), x) + Integral(2*B*cos(c + d*x)*
*2/sqrt(sec(c + d*x)), x) + Integral(B*cos(c + d*x)**3/sqrt(sec(c + d*x)), x) + Integral(C*cos(c + d*x)**2/sqr
t(sec(c + d*x)), x) + Integral(2*C*cos(c + d*x)**3/sqrt(sec(c + d*x)), x) + Integral(C*cos(c + d*x)**4/sqrt(se
c(c + d*x)), x))

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