3.374 \(\int \frac{A+B \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=200 \[ -\frac{(A b-a B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b d \left (a^2-b^2\right )}-\frac{(A b-a B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d \left (a^2-b^2\right )}+\frac{\left (3 a^2 A b+a^3 (-B)-a b^2 B-A b^3\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a b d (a-b) (a+b)^2}+\frac{b (A b-a B) \sin (c+d x) \sqrt{\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]

[Out]

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

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Rubi [A]  time = 0.618138, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3000, 3059, 2639, 3002, 2641, 2805} \[ -\frac{(A b-a B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b d \left (a^2-b^2\right )}-\frac{(A b-a B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d \left (a^2-b^2\right )}+\frac{\left (3 a^2 A b+a^3 (-B)-a b^2 B-A b^3\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a b d (a-b) (a+b)^2}+\frac{b (A b-a B) \sin (c+d x) \sqrt{\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3000

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)), Int
[(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 3059

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]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 3002

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 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[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]

Rubi steps

\begin{align*} \int \frac{A+B \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))^2} \, dx &=\frac{b (A b-a B) \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 (2 a^2 A-A b^2-a b B\right )-a (A b-a B) \cos (c+d x)-\frac{1}{2} b (A b-a B) \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{b (A b-a B) \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 (2 a^2 A-A b^2-a b B\right )+\frac{1}{2} a b (A b-a B) \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{(A b-a B) \int \sqrt{\cos (c+d x)} \, dx}{2 a \left (a^2-b^2\right )}\\ &=-\frac{(A b-a B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \sqrt{\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{(A b-a B) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{2 b \left (a^2-b^2\right )}+\frac{\left (3 a^2 A b-A b^3-a^3 B-a b^2 B\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 a b \left (a^2-b^2\right )}\\ &=-\frac{(A b-a B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a \left (a^2-b^2\right ) d}-\frac{(A b-a B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b \left (a^2-b^2\right ) d}+\frac{\left (3 a^2 A b-A b^3-a^3 B-a b^2 B\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a (a-b) b (a+b)^2 d}+\frac{b (A b-a B) \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]  time = 2.55624, size = 276, normalized size = 1.38 \[ \frac{\frac{4 b (A b-a B) \sin (c+d x) \sqrt{\cos (c+d x)}}{\left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{\frac{2 \left (4 a^2 A-a b B-3 A b^2\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}+\frac{2 (A b-a B) \sin (c+d x) \left (\left (b^2-2 a^2\right ) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a b \sqrt{\sin ^2(c+d x)}}+\frac{4 a (a B-A b) \left (2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-\frac{2 a \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}\right )}{b}}{(a-b) (a+b)}}{4 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 7.84, size = 721, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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