3.438 \(\int \frac{\sqrt{\cos (c+d x)} (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=117 \[ -\frac{2 B \sqrt{a+b} \cot (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{b};\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{b d} \]

[Out]

(-2*Sqrt[a + b]*B*Cot[c + d*x]*EllipticPi[(a + b)/b, ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c +
d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(b*d)

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Rubi [A]  time = 0.0845342, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.053, Rules used = {21, 2809} \[ -\frac{2 B \sqrt{a+b} \cot (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{b};\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{b d} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[Cos[c + d*x]]*(a*B + b*B*Cos[c + d*x]))/(a + b*Cos[c + d*x])^(3/2),x]

[Out]

(-2*Sqrt[a + b]*B*Cot[c + d*x]*EllipticPi[(a + b)/b, ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c +
d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(b*d)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 2809

Int[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[(2*b*Tan
[e + f*x]*Rt[(c + d)/b, 2]*Sqrt[(c*(1 + Csc[e + f*x]))/(c - d)]*Sqrt[(c*(1 - Csc[e + f*x]))/(c + d)]*EllipticP
i[(c + d)/d, ArcSin[Sqrt[c + d*Sin[e + f*x]]/(Sqrt[b*Sin[e + f*x]]*Rt[(c + d)/b, 2])], -((c + d)/(c - d))])/(d
*f), x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] && PosQ[(c + d)/b]

Rubi steps

\begin{align*} \int \frac{\sqrt{\cos (c+d x)} (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^{3/2}} \, dx &=B \int \frac{\sqrt{\cos (c+d x)}}{\sqrt{a+b \cos (c+d x)}} \, dx\\ &=-\frac{2 \sqrt{a+b} B \cot (c+d x) \Pi \left (\frac{a+b}{b};\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right ) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (1+\sec (c+d x))}{a-b}}}{b d}\\ \end{align*}

Mathematica [A]  time = 0.138341, size = 133, normalized size = 1.14 \[ -\frac{2 B \sqrt{\cos (c+d x)} \sqrt{\frac{a+b \cos (c+d x)}{(a+b) (\cos (c+d x)+1)}} \left (F\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{b-a}{a+b}\right )+2 \Pi \left (-1;-\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{b-a}{a+b}\right )\right )}{d \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[Cos[c + d*x]]*(a*B + b*B*Cos[c + d*x]))/(a + b*Cos[c + d*x])^(3/2),x]

[Out]

(-2*B*Sqrt[Cos[c + d*x]]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*(EllipticF[ArcSin[Tan[(c + d*
x)/2]], (-a + b)/(a + b)] + 2*EllipticPi[-1, -ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]))/(d*Sqrt[Cos[c + d*
x]/(1 + Cos[c + d*x])]*Sqrt[a + b*Cos[c + d*x]])

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Maple [A]  time = 0.479, size = 160, normalized size = 1.4 \begin{align*} -2\,{\frac{B \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{d\sqrt{a+b\cos \left ( dx+c \right ) } \left ( -1+\cos \left ( dx+c \right ) \right ) \sqrt{\cos \left ( dx+c \right ) }} \left ({\it EllipticF} \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }},\sqrt{-{\frac{a-b}{a+b}}} \right ) -2\,{\it EllipticPi} \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }},-1,\sqrt{-{\frac{a-b}{a+b}}} \right ) \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\sqrt{{\frac{a+b\cos \left ( dx+c \right ) }{ \left ( a+b \right ) \left ( 1+\cos \left ( dx+c \right ) \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(1/2)*(a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c))^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B b \cos \left (d x + c\right ) + B a\right )} \sqrt{\cos \left (d x + c\right )}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)*(a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((B*b*cos(d*x + c) + B*a)*sqrt(cos(d*x + c))/(b*cos(d*x + c) + a)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{B \sqrt{\cos \left (d x + c\right )}}{\sqrt{b \cos \left (d x + c\right ) + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)*(a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

integral(B*sqrt(cos(d*x + c))/sqrt(b*cos(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**(1/2)*(a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c))**(3/2),x)

[Out]

B*Integral(sqrt(cos(c + d*x))/sqrt(a + b*cos(c + d*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B b \cos \left (d x + c\right ) + B a\right )} \sqrt{\cos \left (d x + c\right )}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)*(a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((B*b*cos(d*x + c) + B*a)*sqrt(cos(d*x + c))/(b*cos(d*x + c) + a)^(3/2), x)