3.887 \(\int \frac{B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))^2} \, dx\)
Optimal. Leaf size=198 \[ \frac{\left (a^2 C+a b B-2 b^2 C\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d \left (a^2-b^2\right )}+\frac{(b B-a C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b d \left (a^2-b^2\right )}-\frac{\left (a^2 b B+a^3 C-3 a b^2 C+b^3 B\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d (a-b) (a+b)^2}-\frac{(b B-a C) \sin (c+d x) \sqrt{\cos (c+d x)}}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]
[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]))
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Rubi [A] time = 0.63593, antiderivative size = 198, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used =
{3029, 2999, 3059, 2639, 3002, 2641, 2805} \[ \frac{\left (a^2 C+a b B-2 b^2 C\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d \left (a^2-b^2\right )}+\frac{(b B-a C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b d \left (a^2-b^2\right )}-\frac{\left (a^2 b B+a^3 C-3 a b^2 C+b^3 B\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d (a-b) (a+b)^2}-\frac{(b B-a C) \sin (c+d x) \sqrt{\cos (c+d x)}}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]
Antiderivative was successfully verified.
[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 3029
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 2999
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 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{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{\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 ) F\left (\left .\frac{1}{2} (c+d x)\right |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 ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |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] time = 2.31026, size = 262, normalized size = 1.32 \[ \frac{\frac{4 (a C-b B) \sin (c+d x) \sqrt{\cos (c+d x)}}{\left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{-\frac{2 (b B-a C) \sin (c+d x) \left (\left (b^2-2 a^2\right ) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a b^2 \sqrt{\sin ^2(c+d x)}}+\frac{2 (a C-b B) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}+\frac{(4 a B-4 b C) \left (2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-\frac{2 a \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}\right )}{b}}{(b-a) (a+b)}}{4 d} \]
Antiderivative was successfully verified.
[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)
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Maple [B] time = 1.829, size = 808, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((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]
-(-(-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*cos(1/2*d*x+1/2*c)
^2*b+a-b)-1/2/(a+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*b/(a^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)*EllipticF(cos(1/2*d*
x+1/2*c),2^(1/2))+1/2/a*b/(a^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)*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((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]
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((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
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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((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
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)