3.1010 \(\int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=61 \[ \frac {2 (b B-2 a C) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b}}+C x \]

[Out]

C*x+2*(B*b-2*C*a)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/d/(a-b)^(1/2)/(a+b)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {24, 2735, 2659, 205} \[ \frac {2 (b B-2 a C) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b}}+C x \]

Antiderivative was successfully verified.

[In]

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

[Out]

C*x + (2*(b*B - 2*a*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*d)

Rule 24

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((A_.) + (B_.)*(v_) + (C_.)*(v_)^2), x_Symbol] :> Dist[1/b^2, Int[u*(a + b*
v)^(m + 1)*Simp[b*B - a*C + b*C*v, x], x], x] /; FreeQ[{a, b, A, B, C}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0] &&
LeQ[m, -1]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 2735

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

Rubi steps

\begin {align*} \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx &=\frac {\int \frac {b^2 (b B-a C)+b^3 C \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{b^2}\\ &=C x+(b B-2 a C) \int \frac {1}{a+b \cos (c+d x)} \, dx\\ &=C x+\frac {(2 (b B-2 a C)) \operatorname {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{d}\\ &=C x+\frac {2 (b B-2 a C) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b} d}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 68, normalized size = 1.11 \[ \frac {2 (2 a C-b B) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{d \sqrt {b^2-a^2}}+\frac {C (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*(c + d*x))/d + (2*(-(b*B) + 2*a*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(Sqrt[-a^2 + b^2]*
d)

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fricas [A]  time = 1.07, size = 240, normalized size = 3.93 \[ \left [\frac {2 \, {\left (C a^{2} - C b^{2}\right )} d x + {\left (2 \, C a - B b\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right )}{2 \, {\left (a^{2} - b^{2}\right )} d}, \frac {{\left (C a^{2} - C b^{2}\right )} d x - {\left (2 \, C a - B b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right )}{{\left (a^{2} - b^{2}\right )} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*(C*a^2 - C*b^2)*d*x + (2*C*a - B*b)*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x +
c)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*
x + c) + a^2)))/((a^2 - b^2)*d), ((C*a^2 - C*b^2)*d*x - (2*C*a - B*b)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c)
+ b)/(sqrt(a^2 - b^2)*sin(d*x + c))))/((a^2 - b^2)*d)]

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giac [B]  time = 1.37, size = 318, normalized size = 5.21 \[ \frac {\frac {{\left (\sqrt {a^{2} - b^{2}} B b^{2} {\left | a - b \right |} - \sqrt {a^{2} - b^{2}} C {\left (a + b\right )} {\left | a - b \right |} {\left | b \right |} + \sqrt {a^{2} - b^{2}} B b {\left | a - b \right |} {\left | b \right |} - \sqrt {a^{2} - b^{2}} {\left (3 \, a b - b^{2}\right )} C {\left | a - b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {\frac {2 \, a + \sqrt {-4 \, {\left (a + b\right )} {\left (a - b\right )} + 4 \, a^{2}}}{a - b}}}\right )\right )}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} b^{2} + {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} {\left | b \right |}} - \frac {{\left (3 \, C a b - B b^{2} - C b^{2} - C a {\left | b \right |} + B b {\left | b \right |} - C b {\left | b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {\frac {2 \, a - \sqrt {-4 \, {\left (a + b\right )} {\left (a - b\right )} + 4 \, a^{2}}}{a - b}}}\right )\right )}}{b^{2} - a {\left | b \right |}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((sqrt(a^2 - b^2)*B*b^2*abs(a - b) - sqrt(a^2 - b^2)*C*(a + b)*abs(a - b)*abs(b) + sqrt(a^2 - b^2)*B*b*abs(a -
b)*abs(b) - sqrt(a^2 - b^2)*(3*a*b - b^2)*C*abs(a - b))*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(2*sqrt(1/2
)*tan(1/2*d*x + 1/2*c)/sqrt((2*a + sqrt(-4*(a + b)*(a - b) + 4*a^2))/(a - b))))/((a^2 - 2*a*b + b^2)*b^2 + (a^
3 - 2*a^2*b + a*b^2)*abs(b)) - (3*C*a*b - B*b^2 - C*b^2 - C*a*abs(b) + B*b*abs(b) - C*b*abs(b))*(pi*floor(1/2*
(d*x + c)/pi + 1/2) + arctan(2*sqrt(1/2)*tan(1/2*d*x + 1/2*c)/sqrt((2*a - sqrt(-4*(a + b)*(a - b) + 4*a^2))/(a
- b))))/(b^2 - a*abs(b)))/d

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maple [B]  time = 0.15, size = 108, normalized size = 1.77 \[ \frac {2 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) B b}{d \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {4 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) a C}{d \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 C \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*B*b-4/d/((a-b)*(a+b))^(1/2)*arcta
n(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*a*C+2/d*C*arctan(tan(1/2*d*x+1/2*c))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 2.50, size = 248, normalized size = 4.07 \[ \frac {2\,C\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,B^2\,b^2-4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,B\,C\,a\,b+3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,C^2\,a^2+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,C^2\,b^2}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (B^2\,b^2-4\,B\,C\,a\,b+3\,C^2\,a^2+C^2\,b^2\right )}\right )}{d}-\frac {2\,B\,b\,\mathrm {atanh}\left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}\right )}{d\,\sqrt {b^2-a^2}}+\frac {4\,C\,a\,\mathrm {atanh}\left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}\right )}{d\,\sqrt {b^2-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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