3.360 \(\int \sqrt{a+a \sec (c+d x)} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=62 \[ \frac{2 a (3 B+C) \tan (c+d x)}{3 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{3 d} \]

[Out]

(2*a*(3*B + C)*Tan[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*C*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d
)

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Rubi [A]  time = 0.0842717, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.088, Rules used = {4054, 12, 3792} \[ \frac{2 a (3 B+C) \tan (c+d x)}{3 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(3*B + C)*Tan[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*C*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d
)

Rule 4054

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3792

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*b*Cot[e + f*x])/
(f*Sqrt[a + b*Csc[e + f*x]]), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

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

Mathematica [A]  time = 0.159898, size = 43, normalized size = 0.69 \[ \frac{2 a \tan (c+d x) (3 B+C \sec (c+d x)+2 C)}{3 d \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(3*B + 2*C + C*Sec[c + d*x])*Tan[c + d*x])/(3*d*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [A]  time = 0.275, size = 70, normalized size = 1.1 \begin{align*} -{\frac{ \left ( -2+2\,\cos \left ( dx+c \right ) \right ) \left ( 3\,B\cos \left ( dx+c \right ) +2\,C\cos \left ( dx+c \right ) +C \right ) }{3\,d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x)

[Out]

-2/3/d*(-1+cos(d*x+c))*(3*B*cos(d*x+c)+2*C*cos(d*x+c)+C)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/sin(d*x+c)/cos(d*
x+c)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.496403, size = 169, normalized size = 2.73 \begin{align*} \frac{2 \,{\left ({\left (3 \, B + 2 \, C\right )} \cos \left (d x + c\right ) + C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3 \,{\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*((3*B + 2*C)*cos(d*x + c) + C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^2 + d*
cos(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \left (\sec{\left (c + d x \right )} + 1\right )} \left (B + C \sec{\left (c + d x \right )}\right ) \sec{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*(sec(c + d*x) + 1))*(B + C*sec(c + d*x))*sec(c + d*x), x)

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Giac [B]  time = 4.31345, size = 174, normalized size = 2.81 \begin{align*} -\frac{2 \,{\left (3 \, \sqrt{2} B a^{2} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, \sqrt{2} C a^{2} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (3 \, \sqrt{2} B a^{2} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + \sqrt{2} C a^{2} \mathrm{sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{3 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

-2/3*(3*sqrt(2)*B*a^2*sgn(cos(d*x + c)) + 3*sqrt(2)*C*a^2*sgn(cos(d*x + c)) - (3*sqrt(2)*B*a^2*sgn(cos(d*x + c
)) + sqrt(2)*C*a^2*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2
- a)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)