\(\int \frac {\cos ^2(c+d x) (A+C \cos ^2(c+d x))}{\sqrt {b \cos (c+d x)}} \, dx\) [63]

Optimal result

Integrand size = 33, antiderivative size = 112 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {b \cos (c+d x)}} \, dx=\frac {2 (7 A+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {2 (7 A+5 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 b d}+\frac {2 C (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^3 d} \]

[Out]

2/7*C*(b*cos(d*x+c))^(5/2)*sin(d*x+c)/b^3/d+2/21*(7*A+5*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ell
ipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/d/(b*cos(d*x+c))^(1/2)+2/21*(7*A+5*C)*sin(d*x+c)*(b*cos(d*
x+c))^(1/2)/b/d

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {16, 3093, 2715, 2721, 2720} \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {b \cos (c+d x)}} \, dx=\frac {2 (7 A+5 C) \sin (c+d x) \sqrt {b \cos (c+d x)}}{21 b d}+\frac {2 (7 A+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b^3 d} \]

[In]

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

[Out]

(2*(7*A + 5*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*d*Sqrt[b*Cos[c + d*x]]) + (2*(7*A + 5*C)*Sqrt
[b*Cos[c + d*x]]*Sin[c + d*x])/(21*b*d) + (2*C*(b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*b^3*d)

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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 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 2721

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

Rule 3093

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos
[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e +
f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\int (b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx}{b^2} \\ & = \frac {2 C (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^3 d}+\frac {(7 A+5 C) \int (b \cos (c+d x))^{3/2} \, dx}{7 b^2} \\ & = \frac {2 (7 A+5 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 b d}+\frac {2 C (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^3 d}+\frac {1}{21} (7 A+5 C) \int \frac {1}{\sqrt {b \cos (c+d x)}} \, dx \\ & = \frac {2 (7 A+5 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 b d}+\frac {2 C (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^3 d}+\frac {\left ((7 A+5 C) \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 \sqrt {b \cos (c+d x)}} \\ & = \frac {2 (7 A+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {2 (7 A+5 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 b d}+\frac {2 C (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.69 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {b \cos (c+d x)}} \, dx=\frac {4 (7 A+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+(14 A+13 C+3 C \cos (2 (c+d x))) \sin (2 (c+d x))}{42 d \sqrt {b \cos (c+d x)}} \]

[In]

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

[Out]

(4*(7*A + 5*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + (14*A + 13*C + 3*C*Cos[2*(c + d*x)])*Sin[2*(c +
d*x)])/(42*d*Sqrt[b*Cos[c + d*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(292\) vs. \(2(124)=248\).

Time = 10.06 (sec) , antiderivative size = 293, normalized size of antiderivative = 2.62

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (48 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-72 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) C \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (28 A +56 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-14 A -16 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+7 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+5 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{21 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\)	\(293\)
parts	\(-\frac {2 A \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\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 )\right )}{3 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}-\frac {2 C \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (48 \left (\cos ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-120 \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+128 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-72 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{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 )+16 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{21 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\)	\(396\)

[In]

int(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(cos(d*x+c)*b)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/21*((2*cos(1/2*d*x+1/2*c)^2-1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)*(48*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-
72*cos(1/2*d*x+1/2*c)*C*sin(1/2*d*x+1/2*c)^6+(28*A+56*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-14*A-16*C)*
sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+7*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)*Elli
pticF(cos(1/2*d*x+1/2*c),2^(1/2))+5*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)))/(-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*c
os(1/2*d*x+1/2*c)^2-1)*b)^(1/2)/d

Fricas [C] (verification not implemented)

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

Time = 0.10 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.97 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {b \cos (c+d x)}} \, dx=\frac {\sqrt {2} {\left (-7 i \, A - 5 i \, C\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (7 i \, A + 5 i \, C\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (3 \, C \cos \left (d x + c\right )^{2} + 7 \, A + 5 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{21 \, b d} \]

[In]

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

[Out]

1/21*(sqrt(2)*(-7*I*A - 5*I*C)*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(2)*(7*
I*A + 5*I*C)*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(3*C*cos(d*x + c)^2 + 7*A +
5*C)*sqrt(b*cos(d*x + c))*sin(d*x + c))/(b*d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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