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

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

[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)

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Rubi [A]  time = 0.0539407, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {4046, 3771, 2641} \[ \frac{2 (3 A+C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \sec (c+d x)}}{3 d}+\frac{2 C \tan (c+d x) \sqrt{b \sec (c+d x)}}{3 d} \]

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 4046

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]

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 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]

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*}

Mathematica [A]  time = 0.374326, size = 58, normalized size = 0.81 \[ \frac{2 (b \sec (c+d x))^{3/2} \left ((3 A+C) \cos ^{\frac{3}{2}}(c+d x) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+C \sin (c+d x)\right )}{3 b d} \]

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]  time = 0.251, size = 201, normalized size = 2.8 \begin{align*} -{\frac{ \left ( -2+2\,\cos \left ( dx+c \right ) \right ) \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}\cos \left ( dx+c \right ) }\sqrt{{\frac{b}{\cos \left ( dx+c \right ) }}} \left ( 3\,iA\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }},i \right ) \cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +iC\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }},i \right ) \cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -C\cos \left ( dx+c \right ) +C \right ) } \end{align*}

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)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt{b \sec \left (d x + c\right )}\,{d x} \end{align*}

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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt{b \sec \left (d x + c\right )}, x\right ) \end{align*}

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]

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

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

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., size = 0, normalized size = 0. \begin{align*} \int{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt{b \sec \left (d x + c\right )}\,{d x} \end{align*}

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)