\(\int \frac {(a+b \cos (c+d x))^2 (B \cos (c+d x)+C \cos ^2(c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) [862]

Optimal result

Integrand size = 42, antiderivative size = 182 \[ \int \frac {(a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 \left (5 a^2 B+3 b^2 B+6 a b C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (5 b^2 C+7 a (2 b B+a C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 \left (5 b^2 C+7 a (2 b B+a C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 b (7 b B+9 a C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 b C \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{7 d} \]

[Out]

2/5*(5*B*a^2+3*B*b^2+6*C*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))/d+2/21*(5*C*b^2+7*a*(2*B*b+C*a))*(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))/d+2/35*b*(7*B*b+9*C*a)*cos(d*x+c)^(3/2)*sin(d*x+c)/d+2/7*b*C*cos(d*x+c)^(3/2)*(a+b*cos(d*x+c))*
sin(d*x+c)/d+2/21*(5*C*b^2+7*a*(2*B*b+C*a))*sin(d*x+c)*cos(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3108, 3069, 3102, 2827, 2719, 2715, 2720} \[ \int \frac {(a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 \left (5 a^2 B+6 a b C+3 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (7 a (a C+2 b B)+5 b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 \left (7 a (a C+2 b B)+5 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{21 d}+\frac {2 b (9 a C+7 b B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{35 d}+\frac {2 b C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}{7 d} \]

[In]

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

[Out]

(2*(5*a^2*B + 3*b^2*B + 6*a*b*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(5*b^2*C + 7*a*(2*b*B + a*C))*EllipticF
[(c + d*x)/2, 2])/(21*d) + (2*(5*b^2*C + 7*a*(2*b*B + a*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(21*d) + (2*b*(7*
b*B + 9*a*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(35*d) + (2*b*C*Cos[c + d*x]^(3/2)*(a + b*Cos[c + d*x])*Sin[c +
d*x])/(7*d)

Rule 2715

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] && Integ
erQ[2*n]

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 2827

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 3069

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 - 2)*(c + d*Sin[e + f
*x])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*(m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c
- b*d*(m + n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*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] && GtQ[m
, 1] &&  !(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3102

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 3108

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]

Rubi steps \begin{align*} \text {integral}& = \int \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2 (B+C \cos (c+d x)) \, dx \\ & = \frac {2 b C \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{7 d}+\frac {2}{7} \int \sqrt {\cos (c+d x)} \left (\frac {1}{2} a (7 a B+3 b C)+\frac {1}{2} \left (5 b^2 C+7 a (2 b B+a C)\right ) \cos (c+d x)+\frac {1}{2} b (7 b B+9 a C) \cos ^2(c+d x)\right ) \, dx \\ & = \frac {2 b (7 b B+9 a C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 b C \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{7 d}+\frac {4}{35} \int \sqrt {\cos (c+d x)} \left (\frac {7}{4} \left (5 a^2 B+3 b^2 B+6 a b C\right )+\frac {5}{4} \left (5 b^2 C+7 a (2 b B+a C)\right ) \cos (c+d x)\right ) \, dx \\ & = \frac {2 b (7 b B+9 a C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 b C \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{7 d}+\frac {1}{5} \left (5 a^2 B+3 b^2 B+6 a b C\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{7} \left (5 b^2 C+7 a (2 b B+a C)\right ) \int \cos ^{\frac {3}{2}}(c+d x) \, dx \\ & = \frac {2 \left (5 a^2 B+3 b^2 B+6 a b C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (5 b^2 C+7 a (2 b B+a C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 b (7 b B+9 a C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 b C \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{7 d}+\frac {1}{21} \left (5 b^2 C+7 a (2 b B+a C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {2 \left (5 a^2 B+3 b^2 B+6 a b C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (5 b^2 C+7 a (2 b B+a C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 \left (5 b^2 C+7 a (2 b B+a C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 b (7 b B+9 a C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 b C \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{7 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.31 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.76 \[ \int \frac {(a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {42 \left (5 a^2 B+3 b^2 B+6 a b C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 \left (14 a b B+7 a^2 C+5 b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\sqrt {\cos (c+d x)} \left (42 b (b B+2 a C) \cos (c+d x)+5 \left (28 a b B+14 a^2 C+13 b^2 C+3 b^2 C \cos (2 (c+d x))\right )\right ) \sin (c+d x)}{105 d} \]

[In]

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

[Out]

(42*(5*a^2*B + 3*b^2*B + 6*a*b*C)*EllipticE[(c + d*x)/2, 2] + 10*(14*a*b*B + 7*a^2*C + 5*b^2*C)*EllipticF[(c +
d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(42*b*(b*B + 2*a*C)*Cos[c + d*x] + 5*(28*a*b*B + 14*a^2*C + 13*b^2*C + 3*b^2*
C*Cos[2*(c + d*x)]))*Sin[c + d*x])/(105*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(547\) vs. \(2(218)=436\).

Time = 6.36 (sec) , antiderivative size = 548, normalized size of antiderivative = 3.01

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 C \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 B \,b^{2}-336 C a b -360 b^{2} 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 (280 B a b +168 B \,b^{2}+140 a^{2} C +336 C a b +280 b^{2} 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 (-140 B a b -42 B \,b^{2}-70 a^{2} C -84 C a b -80 b^{2} 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 )+70 B 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}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-105 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}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}-63 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}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}+35 a^{2} 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}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+25 b^{2} 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}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-126 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}\, E\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}\)	\(548\)
parts	\(\text {Expression too large to display}\)	\(741\)

[In]

int((a+cos(d*x+c)*b)^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(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*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8*
b^2+(-168*B*b^2-336*C*a*b-360*C*b^2)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(280*B*a*b+168*B*b^2+140*C*a^2+33
6*C*a*b+280*C*b^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-140*B*a*b-42*B*b^2-70*C*a^2-84*C*a*b-80*C*b^2)*si
n(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+70*B*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*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)*Ellip
ticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2-63*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)*Ellipt
icE(cos(1/2*d*x+1/2*c),2^(1/2))*b^2+35*a^2*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)*Ell
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))+25*b^2*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)*Elli
pticF(cos(1/2*d*x+1/2*c),2^(1/2))-126*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)*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

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.12 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.34 \[ \int \frac {(a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 \, {\left (15 \, C b^{2} \cos \left (d x + c\right )^{2} + 35 \, C a^{2} + 70 \, B a b + 25 \, C b^{2} + 21 \, {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 5 \, \sqrt {2} {\left (7 i \, C a^{2} + 14 i \, B a b + 5 i \, C b^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 \, \sqrt {2} {\left (-7 i \, C a^{2} - 14 i \, B a b - 5 i \, C b^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 21 \, \sqrt {2} {\left (-5 i \, B a^{2} - 6 i \, C a b - 3 i \, B b^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 \, \sqrt {2} {\left (5 i \, B a^{2} + 6 i \, C a b + 3 i \, B b^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{105 \, d} \]

[In]

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

[Out]

1/105*(2*(15*C*b^2*cos(d*x + c)^2 + 35*C*a^2 + 70*B*a*b + 25*C*b^2 + 21*(2*C*a*b + B*b^2)*cos(d*x + c))*sqrt(c
os(d*x + c))*sin(d*x + c) - 5*sqrt(2)*(7*I*C*a^2 + 14*I*B*a*b + 5*I*C*b^2)*weierstrassPInverse(-4, 0, cos(d*x
+ c) + I*sin(d*x + c)) - 5*sqrt(2)*(-7*I*C*a^2 - 14*I*B*a*b - 5*I*C*b^2)*weierstrassPInverse(-4, 0, cos(d*x +
c) - I*sin(d*x + c)) - 21*sqrt(2)*(-5*I*B*a^2 - 6*I*C*a*b - 3*I*B*b^2)*weierstrassZeta(-4, 0, weierstrassPInve
rse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 21*sqrt(2)*(5*I*B*a^2 + 6*I*C*a*b + 3*I*B*b^2)*weierstrassZeta(-4
, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/d

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {(a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 2.92 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.26 \[ \int \frac {(a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {2\,C\,a^2\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,B\,a^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,a\,b\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}-\frac {2\,B\,b^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,b^2\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {4\,C\,a\,b\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

[In]

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

[Out]

(2*C*a^2*(cos(c + d*x)^(1/2)*sin(c + d*x) + ellipticF(c/2 + (d*x)/2, 2)))/(3*d) + (2*B*a^2*ellipticE(c/2 + (d*
x)/2, 2))/d + (2*B*a*b*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2, 2))/3))/d - (2*B*b
^2*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2)) -
(2*C*b^2*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(9*d*(sin(c + d*x)^2)^(
1/2)) - (4*C*a*b*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*
x)^2)^(1/2))