\(\int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2} \, dx\) [717]

Optimal result

Integrand size = 35, antiderivative size = 214 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2} \, dx=-\frac {\left (A b^2+a^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a b \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a^2 C+2 b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^4+a^4 C-3 a^2 b^2 (A+C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a (a-b) b^2 (a+b)^2 d}+\frac {\left (A b^2+a^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \]

[Out]

-(A*b^2+C*a^2)*(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/b/(a^2-
b^2)/d-(A*b^2-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-(A*b^4+a^4*C-3*a^2*b^2*(A+C))*(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/(a-b)/b^2/(a+b)^2/d+(A*b^2+C*a^2)*sin(d*x+c)*cos(d*x+c)^(1/2)/a/(a^2-b
^2)/d/(a+b*cos(d*x+c))

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3135, 3138, 2719, 3081, 2720, 2884} \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2} \, dx=-\frac {\left (a^2 (-C)+A b^2+2 b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b^2 d \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a b d \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\left (a^4 C-3 a^2 b^2 (A+C)+A b^4\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a b^2 d (a-b) (a+b)^2} \]

[In]

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

[Out]

-(((A*b^2 + a^2*C)*EllipticE[(c + d*x)/2, 2])/(a*b*(a^2 - b^2)*d)) - ((A*b^2 - a^2*C + 2*b^2*C)*EllipticF[(c +
d*x)/2, 2])/(b^2*(a^2 - b^2)*d) - ((A*b^4 + a^4*C - 3*a^2*b^2*(A + C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2,
2])/(a*(a - b)*b^2*(a + b)^2*d) + ((A*b^2 + a^2*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*C
os[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 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 3135

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c
+ d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C
)*(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n +
3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2,
0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && L
tQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 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}& = \frac {\left (A b^2+a^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\int \frac {\frac {1}{2} \left (-A b^2+a^2 (2 A+C)\right )-a b (A+C) \cos (c+d x)-\frac {1}{2} \left (A b^2+a^2 C\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a \left (a^2-b^2\right )} \\ & = \frac {\left (A b^2+a^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {\int \frac {\frac {1}{2} b \left (A b^2-a^2 (2 A+C)\right )+\frac {1}{2} a \left (A b^2-\left (a^2-2 b^2\right ) C\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a b \left (a^2-b^2\right )}-\frac {\left (A b^2+a^2 C\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 a b \left (a^2-b^2\right )} \\ & = -\frac {\left (A b^2+a^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a b \left (a^2-b^2\right ) d}+\frac {\left (A b^2+a^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {\left (A b^2-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 b^4+a^4 C-3 a^2 b^2 (A+C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 a b^2 \left (a^2-b^2\right )} \\ & = -\frac {\left (A b^2+a^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a b \left (a^2-b^2\right ) d}-\frac {\left (A b^2-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 b^4+a^4 C-3 a^2 b^2 (A+C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a (a-b) b^2 (a+b)^2 d}+\frac {\left (A b^2+a^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.95 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.27 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2} \, dx=\frac {\frac {4 \left (A b^2+a^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\frac {2 \left (-3 A b^2+a^2 (4 A+C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}-\frac {8 a (A+C) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}-\frac {2 \left (A b^2+a^2 C\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b^2 \sqrt {\sin ^2(c+d x)}}}{(a-b) (a+b)}}{4 a d} \]

[In]

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

[Out]

((4*(A*b^2 + a^2*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*(a + b*Cos[c + d*x])) + ((2*(-3*A*b^2 + a^2*
(4*A + C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) - (8*a*(A + C)*((a + b)*EllipticF[(c + d*x)/2, 2
] - a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]))/(a + b) - (2*(A*b^2 + a^2*C)*(-2*a*b*EllipticE[ArcSin[Sqrt[C
os[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*a*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(803\) vs. \(2(290)=580\).

Time = 10.49 (sec) , antiderivative size = 804, normalized size of antiderivative = 3.76

method	result	size
default	\(\text {Expression too large to display}\)	\(804\)

[In]

int((A+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)
)+8/b*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/b^2*(A*b^2+C*a^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. \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]

[In]

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

[In]

integrate((A+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]

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

[In]

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

Giac [F]

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

[In]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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