3.12.77 \(\int \sqrt {\cos (c+d x)} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [1177]

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

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4149, 2827, 2716, 2719, 2720} \begin {gather*} \frac {2 B F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 C \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[Cos[c + d*x]]*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(-2*C*EllipticE[(c + d*x)/2, 2])/d + (2*B*EllipticF[(c + d*x)/2, 2])/d + (2*C*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 4149

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(m_)*((A_.) + (B_.)*sec[(e_.) + (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)
]^2), x_Symbol] :> Dist[b^2, Int[(b*Cos[e + f*x])^(m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; F
reeQ[{b, e, f, A, B, C, m}, x] &&  !IntegerQ[m]

Rubi steps

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

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Mathematica [A]
time = 0.15, size = 51, normalized size = 0.89 \begin {gather*} \frac {2 \left (-C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+B F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\frac {C \sin (c+d x)}{\sqrt {\cos (c+d x)}}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Cos[c + d*x]]*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

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

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Maple [A]
time = 0.10, size = 148, normalized size = 2.60

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-2*si
n(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*C+C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipt
icE(cos(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(cos(d*x+c)^(1/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*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 = 0.76, size = 156, normalized size = 2.74 \begin {gather*} \frac {-i \, \sqrt {2} B \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} B \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} C \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} C \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 \, C \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(cos(d*x+c)^(1/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

(-I*sqrt(2)*B*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + I*sqrt(2)*B*cos(d*x + c
)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - I*sqrt(2)*C*cos(d*x + c)*weierstrassZeta(-4, 0,
weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + I*sqrt(2)*C*cos(d*x + c)*weierstrassZeta(-4, 0, w
eierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*C*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*} \int \left (B + C \sec {\left (c + d x \right )}\right ) \sqrt {\cos {\left (c + d x \right )}} \sec {\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*sqrt(cos(d*x + c)), x)

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Mupad [B]
time = 4.39, size = 60, normalized size = 1.05 \begin {gather*} \frac {2\,B\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,\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(cos(c + d*x)^(1/2)*(B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

(2*B*ellipticF(c/2 + (d*x)/2, 2))/d + (2*C*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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