3.1.18 \(\int \sqrt {b \sec (c+d x)} (A+C \sec ^2(c+d x)) \, dx\) [18]

Optimal. Leaf size=72 \[ \frac {2 (3 A+C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 C \sqrt {b \sec (c+d x)} \tan (c+d x)}{3 d} \]

[Out]

2/3*(3*A+C)*(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+2/3*C*(b*sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]
time = 0.04, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {4131, 3856, 2720} \begin {gather*} \frac {2 (3 A+C) \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 C \tan (c+d x) \sqrt {b \sec (c+d x)}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4131

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot
[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x
] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rubi steps

\begin {align*} \int \sqrt {b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {2 C \sqrt {b \sec (c+d x)} \tan (c+d x)}{3 d}+\frac {1}{3} (3 A+C) \int \sqrt {b \sec (c+d x)} \, dx\\ &=\frac {2 C \sqrt {b \sec (c+d x)} \tan (c+d x)}{3 d}+\frac {1}{3} \left ((3 A+C) \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 (3 A+C) \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 C \sqrt {b \sec (c+d x)} \tan (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]
time = 0.26, size = 58, normalized size = 0.81 \begin {gather*} \frac {2 (b \sec (c+d x))^{3/2} \left ((3 A+C) \cos ^{\frac {3}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+C \sin (c+d x)\right )}{3 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 33.55, size = 201, normalized size = 2.79

method	result	size
default	\(-\frac {2 \sqrt {\frac {b}{\cos \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right ) \left (3 i A \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+i C \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-C \cos \left (d x +c \right )+C \right ) \left (1+\cos \left (d x +c \right )\right )^{2}}{3 d \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}\)	\(201\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.54, size = 111, normalized size = 1.54 \begin {gather*} \frac {\sqrt {2} {\left (-3 i \, A - i \, C\right )} \sqrt {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 ) + \sqrt {2} {\left (3 i \, A + i \, C\right )} \sqrt {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 ) + 2 \, C \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(sqrt(2)*(-3*I*A - I*C)*sqrt(b)*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + s
qrt(2)*(3*I*A + I*C)*sqrt(b)*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*C*sqrt
(b/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 \sqrt {b \sec {\left (c + d x \right )}} \left (A + C \sec ^{2}{\left (c + d x \right )}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

[Out]

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

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