3.4.55 \(\int \frac {a+a \sec (c+d x)}{\sqrt {\cos (c+d x)}} \, dx\) [355]

Optimal. Leaf size=57 \[ -\frac {2 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {4310, 2827, 2716, 2719, 2720} \begin {gather*} \frac {2 a F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {2 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*EllipticE[(c + d*x)/2, 2])/d + (2*a*EllipticF[(c + d*x)/2, 2])/d + (2*a*Sin[c + d*x])/(d*Sqrt[Cos[c + d*
x]])

Rule 2716

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
))), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2719

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

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 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 4310

Int[(csc[(a_.) + (b_.)*(x_)]*(B_.) + (A_))*(u_), x_Symbol] :> Int[ActivateTrig[u]*((B + A*Sin[a + b*x])/Sin[a
+ b*x]), x] /; FreeQ[{a, b, A, B}, x] && KnownSineIntegrandQ[u, x]

Rubi steps

\begin {align*} \int \frac {a+a \sec (c+d x)}{\sqrt {\cos (c+d x)}} \, dx &=\int \frac {a+a \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=a \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx+a \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 a F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-a \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {2 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 5.04, size = 209, normalized size = 3.67 \begin {gather*} \frac {a (1+\cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (4 \cos (d x) \csc (c)-\frac {(3 \cos (c-d x-\text {ArcTan}(\tan (c)))+\cos (c+d x+\text {ArcTan}(\tan (c)))) \csc (c) \sec (c)}{\sqrt {\sec ^2(c)}}-4 \cos (c+d x) \sqrt {\cos ^2(d x-\text {ArcTan}(\cot (c)))} \sqrt {\csc ^2(c)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\text {ArcTan}(\cot (c)))\right ) \sec (d x-\text {ArcTan}(\cot (c))) \sin (c)+2 \cos (c) \csc (d x+\text {ArcTan}(\tan (c))) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\text {ArcTan}(\tan (c)))\right ) \sqrt {\sec ^2(c)} \sqrt {\sin ^2(d x+\text {ArcTan}(\tan (c)))}\right )}{4 d \sqrt {\cos (c+d x)}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*(1 + Cos[c + d*x])*Sec[(c + d*x)/2]^2*(4*Cos[d*x]*Csc[c] - ((3*Cos[c - d*x - ArcTan[Tan[c]]] + Cos[c + d*x
+ ArcTan[Tan[c]]])*Csc[c]*Sec[c])/Sqrt[Sec[c]^2] - 4*Cos[c + d*x]*Sqrt[Cos[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Csc[c
]^2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[d*x - ArcTan[Cot[c]]]*Sin[c] + 2*Co
s[c]*Csc[d*x + ArcTan[Tan[c]]]*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sqrt[Sec[c]
^2]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2]))/(4*d*Sqrt[Cos[c + d*x]])

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Maple [A]
time = 0.08, size = 148, normalized size = 2.60

method	result	size
default	\(\frac {2 a \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(148\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*E
llipticF(cos(1/2*d*x+1/2*c),2^(1/2))-(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(c
os(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+a*sec(d*x+c))/cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.97, size = 156, normalized size = 2.74 \begin {gather*} \frac {-i \, \sqrt {2} a \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} a \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - i \, \sqrt {2} a \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + i \, \sqrt {2} a \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, a \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-I*sqrt(2)*a*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + I*sqrt(2)*a*cos(d*x + c
)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - I*sqrt(2)*a*cos(d*x + c)*weierstrassZeta(-4, 0,
weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + I*sqrt(2)*a*cos(d*x + c)*weierstrassZeta(-4, 0, w
eierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*a*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x +
c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a \left (\int \frac {\sec {\left (c + d x \right )}}{\sqrt {\cos {\left (c + d x \right )}}}\, dx + \int \frac {1}{\sqrt {\cos {\left (c + d x \right )}}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [B]
time = 1.03, size = 60, normalized size = 1.05 \begin {gather*} \frac {2\,a\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,a\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*ellipticF(c/2 + (d*x)/2, 2))/d + (2*a*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c
+ d*x)^(1/2)*(sin(c + d*x)^2)^(1/2))

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