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

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Rubi [A]  time = 0.08, antiderivative size = 117, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.14, size = 131, normalized size = 1.12 \[ -\frac {2 B \sqrt {\cos (c+d x)} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (\cos (c+d x)+1)}} \left (F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )-2 \Pi \left (-1;\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )\right )}{d \sqrt {\frac {\cos (c+d x)}{\cos (c+d x)+1}} \sqrt {a+b \cos (c+d x)}} \]

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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fricas [F]  time = 1.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {B \sqrt {\cos \left (d x + c\right )}}{\sqrt {b \cos \left (d x + c\right ) + a}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B b \cos \left (d x + c\right ) + B a\right )} \sqrt {\cos \left (d x + c\right )}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

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)

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maple [A]  time = 0.33, size = 160, normalized size = 1.37 \[ -\frac {2 B \left (\EllipticF \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}, \sqrt {-\frac {a -b}{a +b}}\right )-2 \EllipticPi \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}, -1, \sqrt {-\frac {a -b}{a +b}}\right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {a +b \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \left (a +b \right )}}\, \left (\sin ^{2}\left (d x +c \right )\right )}{d \sqrt {a +b \cos \left (d x +c \right )}\, \left (-1+\cos \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right )}} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B b \cos \left (d x + c\right ) + B a\right )} \sqrt {\cos \left (d x + c\right )}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (B\,a+B\,b\,\cos \left (c+d\,x\right )\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ B \int \frac {\sqrt {\cos {\left (c + d x \right )}}}{\sqrt {a + b \cos {\left (c + d x \right )}}}\, dx \]

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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