3.6.77 \(\int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))} \, dx\) [577]

Optimal. Leaf size=61 \[ \frac {2 A F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}-\frac {2 (A b-a B) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a (a+b) d} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {3033, 3081, 2720, 2884} \begin {gather*} \frac {2 A F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}-\frac {2 (A b-a B) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a d (a+b)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*A*EllipticF[(c + d*x)/2, 2])/(a*d) - (2*(A*b - a*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a*(a + b)*d
)

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(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]

Rule 3033

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(
d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 3081

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]

Rubi steps

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

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Mathematica [A]
time = 0.23, size = 58, normalized size = 0.95 \begin {gather*} \frac {2 \left (A (a+b) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+(-A b+a B) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{a (a+b) d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(A*(a + b)*EllipticF[(c + d*x)/2, 2] + (-(A*b) + a*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]))/(a*(a + b
)*d)

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Maple [A]
time = 1.82, size = 217, normalized size = 3.56

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (A \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a -A \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b +A \EllipticPi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 a}{a -b}, \sqrt {2}\right ) b -B \EllipticPi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 a}{a -b}, \sqrt {2}\right ) a \right )}{a \left (a -b \right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(217\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c))/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-2*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)
^2+1)^(1/2)*(A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a-A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b+A*EllipticPi(
cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))*b-B*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))*a)/a/(a-b)/(-2*sin(
1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right ) \sqrt {\cos {\left (c + d x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{\sqrt {\cos \left (c+d\,x\right )}\,\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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