3.386 \(\int \frac {a B+b B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx\)

Optimal. Leaf size=17 \[ \frac {2 B F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {21, 2641} \[ \frac {2 B F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*B*EllipticF[(c + d*x)/2, 2])/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 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]

Rubi steps

\begin {align*} \int \frac {a B+b B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx &=B \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 B F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 17, normalized size = 1.00 \[ \frac {2 B F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*B*EllipticF[(c + d*x)/2, 2])/d

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fricas [F]  time = 1.26, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {B}{\sqrt {\cos \left (d x + c\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(B/sqrt(cos(d*x + c)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.01, size = 19, normalized size = 1.12 \[ \frac {2 B \,\mathrm {am}^{-1}\left (\frac {d x}{2}+\frac {c}{2}| \sqrt {2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*B/d*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.06 \[ \int \frac {B\,a+B\,b\,\cos \left (c+d\,x\right )}{\sqrt {\cos \left (c+d\,x\right )}\,\left (a+b\,\cos \left (c+d\,x\right )\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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