\(\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2} \, dx\) [887]

Optimal result

Integrand size = 42, antiderivative size = 198 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2} \, dx=\frac {(b B-a C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b \left (a^2-b^2\right ) d}+\frac {\left (a b B+a^2 C-2 b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b^2 \left (a^2-b^2\right ) d}-\frac {\left (a^2 b B+b^3 B+a^3 C-3 a b^2 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{(a-b) b^2 (a+b)^2 d}-\frac {(b B-a C) \sqrt {\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))} \]

[Out]

(B*b-C*a)*(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))/b/(a^2-b^2)/d+
(B*a*b+C*a^2-2*C*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))/b^
2/(a^2-b^2)/d-(B*a^2*b+B*b^3+C*a^3-3*C*a*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1
/2*d*x+1/2*c),2*b/(a+b),2^(1/2))/(a-b)/b^2/(a+b)^2/d-(B*b-C*a)*sin(d*x+c)*cos(d*x+c)^(1/2)/(a^2-b^2)/d/(a+b*co
s(d*x+c))

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 198, 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, 3078, 3138, 2719, 3081, 2720, 2884} \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2} \, dx=\frac {\left (a^2 C+a b B-2 b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b^2 d \left (a^2-b^2\right )}+\frac {(b B-a C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d \left (a^2-b^2\right )}-\frac {(b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\left (a^3 C+a^2 b B-3 a b^2 C+b^3 B\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b^2 d (a-b) (a+b)^2} \]

[In]

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

[Out]

((b*B - a*C)*EllipticE[(c + d*x)/2, 2])/(b*(a^2 - b^2)*d) + ((a*b*B + a^2*C - 2*b^2*C)*EllipticF[(c + d*x)/2,
2])/(b^2*(a^2 - b^2)*d) - ((a^2*b*B + b^3*B + a^3*C - 3*a*b^2*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/((
a - b)*b^2*(a + b)^2*d) - ((b*B - a*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*d*(a + b*Cos[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 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rule 3078

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

Rule 3081

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

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]

Rule 3138

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*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]

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

Mathematica [A] (verified)

Time = 2.81 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.31 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2} \, dx=\frac {\frac {4 (-b B+a C) \sqrt {\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {2 (-b B+a C) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {(4 a B-4 b C) \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )}{b}+\frac {2 (b B-a C) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b^2 \sqrt {\sin ^2(c+d x)}}}{(-a+b) (a+b)}}{4 d} \]

[In]

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

[Out]

((4*(-(b*B) + a*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*(a + b*Cos[c + d*x])) - ((2*(-(b*B) + a*C)*El
lipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + ((4*a*B - 4*b*C)*(2*EllipticF[(c + d*x)/2, 2] - (2*a*Ellipt
icPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b)))/b + (2*(b*B - a*C)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]]
, -1] + 2*a*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (-2*a^2 + b^2)*EllipticPi[-(b/a), ArcSin[Sqrt[
Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b^2*Sqrt[Sin[c + d*x]^2]))/((-a + b)*(a + b)))/(4*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(807\) vs. \(2(274)=548\).

Time = 7.25 (sec) , antiderivative size = 808, normalized size of antiderivative = 4.08

method	result	size
default	\(\text {Expression too large to display}\)	\(808\)

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*C/b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d
*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)
)-4/b*(B*b-2*C*a)/(-2*a*b+2*b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*
x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))-2*a*(B*b-C*a)/b^2*(-1
/a*b^2/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*b*cos(1/2*d*x+1/2*
c)^2+a-b)-1/2/a/(a+b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+
sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/(a^2-b^2)*b/a*(sin(1/2*d*x+1/2*c)^2)^(1/
2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*
x+1/2*c),2^(1/2))+1/2/(a^2-b^2)*b/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2
*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3*a/(a^2-b^2)/(-2*a*b+2*b^2)*b
*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)
^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+1/a/(a^2-b^2)/(-2*a*b+2*b^2)*b^3*(sin(1/2*d*x+1/2*c)^
2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos
(1/2*d*x+1/2*c),-2*b/(a-b),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 [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

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

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*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 [F(-1)]

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

[In]

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

[Out]

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