3.30 \(\int (A+C \cos ^2(c+d x)) \sqrt {b \sec (c+d x)} \, dx\)

Optimal. Leaf size=75 \[ \frac {2 (3 A+C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 b^2 C \tan (c+d x)}{3 d (b \sec (c+d x))^{3/2}} \]

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

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Rubi [A]  time = 0.10, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3238, 4045, 3771, 2641} \[ \frac {2 (3 A+C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 b^2 C \tan (c+d x)}{3 d (b \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

Rule 2641

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

Rule 3238

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Csc[e + f*x])^(m - n*p)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rule 3771

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 4045

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

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 58, normalized size = 0.77 \[ \frac {\sqrt {b \sec (c+d x)} \left (2 (3 A+C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+C \sin (2 (c+d x))\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt {b \sec \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt {b \sec \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.34, size = 190, normalized size = 2.53 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (3 i A \sin \left (d x +c \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+i C \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-C \left (\cos ^{2}\left (d x +c \right )\right )+C \cos \left (d x +c \right )\right ) \left (1+\cos \left (d x +c \right )\right )^{2} \sqrt {\frac {b}{\cos \left (d x +c \right )}}}{3 d \sin \left (d x +c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/d*(-1+cos(d*x+c))*(3*I*A*(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)*sin(d*x+c)+I*C*sin(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*Elli
pticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)-C*cos(d*x+c)^2+C*cos(d*x+c))*(1+cos(d*x+c))^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 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt {b \sec \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*cos(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 \[ \int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,\sqrt {\frac {b}{\cos \left (c+d\,x\right )}} \,d x \]

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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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \sec {\left (c + d x \right )}} \left (A + C \cos ^{2}{\left (c + d x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(b*sec(c + d*x))*(A + C*cos(c + d*x)**2), x)

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