3.4.23 \(\int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx\) [323]

Optimal. Leaf size=118 \[ \frac {2 B \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}+\frac {2 A \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}} \]

[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)*(b/(a+b))^(1/2))*((a+
b*cos(d*x+c))/(a+b))^(1/2)/d/(a+b*cos(d*x+c))^(1/2)+2*A*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipt
icPi(sin(1/2*d*x+1/2*c),2,2^(1/2)*(b/(a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b))^(1/2)/d/(a+b*cos(d*x+c))^(1/2)

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Rubi [A]
time = 0.17, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3081, 2742, 2740, 2886, 2884} \begin {gather*} \frac {2 A \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}+\frac {2 B \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2740

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*Sqrt[a + b]))*EllipticF[(1/2)*(c - P
i/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2742

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

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 2886

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/
(c + d))*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]

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

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Mathematica [A]
time = 0.21, size = 81, normalized size = 0.69 \begin {gather*} \frac {2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left (B F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )+A \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )}{d \sqrt {a+b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.28, size = 194, normalized size = 1.64

method	result	size
default	\(\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b}{a -b}}\, \left (A \EllipticPi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2, \sqrt {-\frac {2 b}{a -b}}\right )-B \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )\right )}{\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +\left (a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b}\, d}\)	\(194\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*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*b+a-b)/(a-b))^(1/2)*(A*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))-B*EllipticF(cos(1/2*d*x+1/2*c
),(-2*b/(a-b))^(1/2)))/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(-2*sin
(1/2*d*x+1/2*c)^2*b+a+b)^(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*cos(d*x+c))*sec(d*x+c)/(a+b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*sec(d*x + c)/sqrt(b*cos(d*x + c) + a), 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*cos(d*x+c))*sec(d*x+c)/(a+b*cos(d*x+c))^(1/2),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 {\left (A + B \cos {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\sqrt {a + b \cos {\left (c + d x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

[Out]

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

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