\(\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))} \, dx\) [881]

Optimal result

Integrand size = 42, antiderivative size = 86 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=-\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}-\frac {2 (b B-a C) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a (a+b) d}+\frac {2 B \sin (c+d x)}{a d \sqrt {\cos (c+d x)}} \]

[Out]

-2*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))/a/d-2*(B*b-C*a)*(co
s(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/(a+b)/d+2*B*si
n(d*x+c)/a/d/cos(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3108, 3079, 3138, 2719, 12, 2884} \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=-\frac {2 (b B-a C) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a d (a+b)}-\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {2 B \sin (c+d x)}{a d \sqrt {\cos (c+d x)}} \]

[In]

Int[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + b*Cos[c + d*x])),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 3079

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(A*b^2 - a*b*B))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c +
d*Sin[e + f*x])^(1 + n)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*
(m + n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B)*(m + n + 3)*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] && RationalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n
, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 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 {B+C \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx \\ & = \frac {2 B \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}+\frac {2 \int \frac {\frac {1}{2} (-b B+a C)-\frac {1}{2} a B \cos (c+d x)-\frac {1}{2} b B \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a} \\ & = \frac {2 B \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {2 \int \frac {b (b B-a C)}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a b}-\frac {B \int \sqrt {\cos (c+d x)} \, dx}{a} \\ & = -\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {2 B \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {(b B-a C) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a} \\ & = -\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}-\frac {2 (b B-a C) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a (a+b) d}+\frac {2 B \sin (c+d x)}{a d \sqrt {\cos (c+d x)}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(206\) vs. \(2(86)=172\).

Time = 2.68 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.40 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=\frac {\frac {2 (-3 b B+2 a C) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}-\frac {2 a B \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )}{b}+\frac {4 B \sin (c+d x)}{\sqrt {\cos (c+d x)}}-\frac {2 B \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b \sqrt {\sin ^2(c+d x)}}}{2 a d} \]

[In]

Integrate[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + b*Cos[c + d*x])),x]

[Out]

((2*(-3*b*B + 2*a*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) - (2*a*B*(2*EllipticF[(c + d*x)/2, 2]
- (2*a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b)))/b + (4*B*Sin[c + d*x])/Sqrt[Cos[c + d*x]] - (2*B*(
-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*Sqrt[Sin[c + d*x]^2]))/(2*a
*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(299\) vs. \(2(136)=272\).

Time = 5.49 (sec) , antiderivative size = 300, normalized size of antiderivative = 3.49

method	result	size
default	\(-\frac {\sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\frac {2 B \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}-\frac {4 \left (-B b +a C \right ) b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \Pi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right )}{a \left (-2 a b +2 b^{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(300\)

[In]

int((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+cos(d*x+c)*b),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*B/a/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2
-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-4/a*(-B*b+C*a)/(-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)))/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. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=\text {Timed out} \]

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=\text {Timed out} \]

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(5/2)/(a+b*cos(d*x+c)),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c)),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))/((b*cos(d*x + c) + a)*cos(d*x + c)^(5/2)), x)

Giac [F]

\[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c)),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))/((b*cos(d*x + c) + a)*cos(d*x + c)^(5/2)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{5/2}\,\left (a+b\,\cos \left (c+d\,x\right )\right )} \,d x \]

[In]

int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(a + b*cos(c + d*x))),x)

[Out]

int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(a + b*cos(c + d*x))), x)