3.1.76 \(\int \cos ^2(c+d x) \sqrt {b \sec (c+d x)} \, dx\) [76]

Optimal. Leaf size=67 \[ \frac {2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 b \sin (c+d x)}{3 d \sqrt {b \sec (c+d x)}} \]

[Out]

2/3*b*sin(d*x+c)/d/(b*sec(d*x+c))^(1/2)+2/3*(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))*cos(d*x+c)^(1/2)*(b*sec(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.04, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {16, 3854, 3856, 2720} \begin {gather*} \frac {2 b \sin (c+d x)}{3 d \sqrt {b \sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*Sqrt[b*Sec[c + d*x]],x]

[Out]

(2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(3*d) + (2*b*Sin[c + d*x])/(3*d*Sqrt[b*S
ec[c + d*x]])

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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 3854

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

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 51, normalized size = 0.76 \begin {gather*} \frac {\sqrt {b \sec (c+d x)} \left (2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\sin (2 (c+d x))\right )}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*Sqrt[b*Sec[c + d*x]],x]

[Out]

(Sqrt[b*Sec[c + d*x]]*(2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + Sin[2*(c + d*x)]))/(3*d)

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Maple [C] Result contains complex when optimal does not.
time = 31.47, size = 123, normalized size = 1.84

method	result	size
default	\(\frac {2 \left (\cos \left (d x +c \right )-1\right ) \left (-i \EllipticF \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sin \left (d x +c \right )+\cos ^{2}\left (d x +c \right )-\cos \left (d x +c \right )\right ) \left (\cos \left (d x +c \right )+1\right )^{2} \sqrt {\frac {b}{\cos \left (d x +c \right )}}}{3 d \sin \left (d x +c \right )^{3}}\)	\(123\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/d*(cos(d*x+c)-1)*(-I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cos(d*x+c)-1)
/sin(d*x+c),I)*sin(d*x+c)+cos(d*x+c)^2-cos(d*x+c))*(cos(d*x+c)+1)^2*(b/cos(d*x+c))^(1/2)/sin(d*x+c)^3

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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)^2*(b*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.85, size = 84, normalized size = 1.25 \begin {gather*} \frac {2 \, \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(2*sqrt(b/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) - I*sqrt(2)*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x +
c) + I*sin(d*x + c)) + I*sqrt(2)*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)))/d

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {b \sec {\left (c + d x \right )}} \cos ^{2}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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