3.6.60 \(\int \frac {(a+b \cos (c+d x))^2 (A+B \cos (c+d x))}{\sqrt {\sec (c+d x)}} \, dx\) [560]

Optimal. Leaf size=213 \[ \frac {2 \left (5 a^2 A+3 A b^2+6 a b B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (14 a A b+7 a^2 B+5 b^2 B\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 b^2 B \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 b (A b+2 a B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (14 a A b+7 a^2 B+5 b^2 B\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}} \]

[Out]

2/7*b^2*B*sin(d*x+c)/d/sec(d*x+c)^(5/2)+2/5*b*(A*b+2*B*a)*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/21*(14*A*a*b+7*B*a^2
+5*B*b^2)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+2/5*(5*A*a^2+3*A*b^2+6*B*a*b)*(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+2/21*(14*A*a*b+7*B*a^2+5*B*b
^2)*(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 = 0.23, antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3039, 4109, 4132, 3854, 3856, 2720, 4130, 2719} \begin {gather*} \frac {2 \left (7 a^2 B+14 a A b+5 b^2 B\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 \left (7 a^2 B+14 a A b+5 b^2 B\right ) \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 {2 \left (5 a^2 A+6 a b B+3 A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 b (2 a B+A b) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 b^2 B \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(5*a^2*A + 3*A*b^2 + 6*a*b*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*(
14*a*A*b + 7*a^2*B + 5*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*b^2
*B*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (2*b*(A*b + 2*a*B)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (2*(14
*a*A*b + 7*a^2*B + 5*b^2*B)*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]])

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3039

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(
d + c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 3854

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

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 4109

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2*(csc[(e_.) + (f_.)*(x_)]*(B_
.) + (A_)), x_Symbol] :> Simp[a^2*A*Cos[e + f*x]*((d*Csc[e + f*x])^(n + 1)/(d*f*n)), x] + Dist[1/(d*n), Int[(d
*Csc[e + f*x])^(n + 1)*(a*(2*A*b + a*B)*n + (2*a*b*B*n + A*(b^2*n + a^2*(n + 1)))*Csc[e + f*x] + b^2*B*n*Csc[e
+ f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rule 4130

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e
+ f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rule 4132

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rubi steps

\begin {align*} \int \frac {(a+b \cos (c+d x))^2 (A+B \cos (c+d x))}{\sqrt {\sec (c+d x)}} \, dx &=\int \frac {(b+a \sec (c+d x))^2 (B+A \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 b^2 B \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}-\frac {2}{7} \int \frac {-\frac {7}{2} b (A b+2 a B)+\left (-7 a A b+\left (-\frac {7 a^2}{2}-\frac {5 b^2}{2}\right ) B\right ) \sec (c+d x)-\frac {7}{2} a^2 A \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 b^2 B \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}-\frac {2}{7} \int \frac {-\frac {7}{2} b (A b+2 a B)-\frac {7}{2} a^2 A \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x)} \, dx-\frac {1}{7} \left (-14 a A b-7 a^2 B-5 b^2 B\right ) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 b^2 B \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 b (A b+2 a B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (14 a A b+7 a^2 B+5 b^2 B\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}-\frac {1}{5} \left (-5 a^2 A-3 A b^2-6 a b B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx-\frac {1}{21} \left (-14 a A b-7 a^2 B-5 b^2 B\right ) \int \sqrt {\sec (c+d x)} \, dx\\ &=\frac {2 b^2 B \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 b (A b+2 a B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (14 a A b+7 a^2 B+5 b^2 B\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}-\frac {1}{5} \left (\left (-5 a^2 A-3 A b^2-6 a b B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx-\frac {1}{21} \left (\left (-14 a A b-7 a^2 B-5 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 \left (5 a^2 A+3 A b^2+6 a b B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (14 a A b+7 a^2 B+5 b^2 B\right ) \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 b^2 B \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 b (A b+2 a B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (14 a A b+7 a^2 B+5 b^2 B\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 1.47, size = 161, normalized size = 0.76 \begin {gather*} \frac {\sqrt {\sec (c+d x)} \left (84 \left (5 a^2 A+3 A b^2+6 a b B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 \left (14 a A b+7 a^2 B+5 b^2 B\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\left (42 b (A b+2 a B) \cos (c+d x)+5 \left (28 a A b+14 a^2 B+13 b^2 B+3 b^2 B \cos (2 (c+d x))\right )\right ) \sin (2 (c+d x))\right )}{210 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sec[c + d*x]]*(84*(5*a^2*A + 3*A*b^2 + 6*a*b*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 20*(14*a*
A*b + 7*a^2*B + 5*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + (42*b*(A*b + 2*a*B)*Cos[c + d*x] + 5*(
28*a*A*b + 14*a^2*B + 13*b^2*B + 3*b^2*B*Cos[2*(c + d*x)]))*Sin[2*(c + d*x)]))/(210*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(547\) vs. \(2(241)=482\).
time = 0.43, size = 548, normalized size = 2.57

method	result	size
default	\(-\frac {2 \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 )}\, \left (240 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}+\left (-168 A \,b^{2}-336 B a b -360 B \,b^{2}\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 (280 A a b +168 A \,b^{2}+140 B \,a^{2}+336 B a b +280 B \,b^{2}\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 (-140 A a b -42 A \,b^{2}-70 B \,a^{2}-84 B a b -80 B \,b^{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+70 A a 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 )-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}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}-63 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 ) b^{2}+35 B \,a^{2} \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 )+25 B \,b^{2} \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 )-126 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 ) a b \right )}{105 \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}\)	\(548\)

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))/sec(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8*
b^2+(-168*A*b^2-336*B*a*b-360*B*b^2)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(280*A*a*b+168*A*b^2+140*B*a^2+33
6*B*a*b+280*B*b^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-140*A*a*b-42*A*b^2-70*B*a^2-84*B*a*b-80*B*b^2)*si
n(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+70*A*a*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)*E
llipticF(cos(1/2*d*x+1/2*c),2^(1/2))-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)*Ellip
ticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2-63*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)*Ellipt
icE(cos(1/2*d*x+1/2*c),2^(1/2))*b^2+35*B*a^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)*Ell
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))+25*B*b^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)*Elli
pticF(cos(1/2*d*x+1/2*c),2^(1/2))-126*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)*Elliptic
E(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)/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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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))/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.15, size = 254, normalized size = 1.19 \begin {gather*} -\frac {5 \, \sqrt {2} {\left (7 i \, B a^{2} + 14 i \, A a b + 5 i \, B b^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-7 i \, B a^{2} - 14 i \, A a b - 5 i \, B b^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, \sqrt {2} {\left (-5 i \, A a^{2} - 6 i \, B a b - 3 i \, A b^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 \, \sqrt {2} {\left (5 i \, A a^{2} + 6 i \, B a b + 3 i \, A b^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (15 \, B b^{2} \cos \left (d x + c\right )^{3} + 21 \, {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (7 \, B a^{2} + 14 \, A a b + 5 \, B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{105 \, d} \end {gather*}

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))/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

-1/105*(5*sqrt(2)*(7*I*B*a^2 + 14*I*A*a*b + 5*I*B*b^2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c
)) + 5*sqrt(2)*(-7*I*B*a^2 - 14*I*A*a*b - 5*I*B*b^2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))
+ 21*sqrt(2)*(-5*I*A*a^2 - 6*I*B*a*b - 3*I*A*b^2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x +
c) + I*sin(d*x + c))) + 21*sqrt(2)*(5*I*A*a^2 + 6*I*B*a*b + 3*I*A*b^2)*weierstrassZeta(-4, 0, weierstrassPInv
erse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(15*B*b^2*cos(d*x + c)^3 + 21*(2*B*a*b + A*b^2)*cos(d*x + c)^2
+ 5*(7*B*a^2 + 14*A*a*b + 5*B*b^2)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/d

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \left (a + b \cos {\left (c + d x \right )}\right )^{2}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx \end {gather*}

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))/sec(d*x+c)**(1/2),x)

[Out]

Integral((A + B*cos(c + d*x))*(a + b*cos(c + d*x))**2/sqrt(sec(c + d*x)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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))/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*(b*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 \begin {gather*} \int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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