3.6.81 \(\int \sqrt {\sec (c+d x)} (a+b \sec (c+d x)) \, dx\) [581]

Optimal. Leaf size=97 \[ -\frac {2 b \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {2 a \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {2 b \sqrt {\sec (c+d x)} \sin (c+d x)}{d} \]

[Out]

2*b*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2*b*(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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+2*a*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elliptic
F(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d

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Rubi [A]
time = 0.05, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3872, 3856, 2720, 3853, 2719} \begin {gather*} \frac {2 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 b \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3853

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

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rubi steps

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

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Mathematica [A]
time = 0.08, size = 71, normalized size = 0.73 \begin {gather*} \frac {2 \sqrt {\sec (c+d x)} \left (-b \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+a \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+b \sin (c+d x)\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.12, size = 150, normalized size = 1.55

method	result	size
default	\(\frac {4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -2 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, a -2 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, b}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(150\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate((b*sec(d*x + c) + a)*sqrt(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 = 0.96, size = 124, normalized size = 1.28 \begin {gather*} \frac {-i \, \sqrt {2} a {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} a {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - i \, \sqrt {2} b {\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} b {\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 ) + \frac {2 \, b \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate((b*sec(d*x + c) + a)*sqrt(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 {b}{\cos \left (c+d\,x\right )}\right )\,\sqrt {\frac {1}{\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))*(1/cos(c + d*x))^(1/2),x)

[Out]

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

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