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

Optimal. Leaf size=101 \[ \frac{2 (5 B-2 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{15 d}+\frac{2 a (5 B+7 C) \tan (c+d x)}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d} \]

[Out]

(2*a*(5*B + 7*C)*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*(5*B - 2*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c
+ d*x])/(15*d) + (2*C*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*a*d)

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Rubi [A]  time = 0.276975, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {4072, 4010, 4001, 3792} \[ \frac{2 (5 B-2 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{15 d}+\frac{2 a (5 B+7 C) \tan (c+d x)}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(5*B + 7*C)*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*(5*B - 2*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c
+ d*x])/(15*d) + (2*C*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*a*d)

Rule 4072

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

Rule 4010

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

Rule 4001

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

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 \sec (c+d x) \sqrt{a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sec ^2(c+d x) \sqrt{a+a \sec (c+d x)} (B+C \sec (c+d x)) \, dx\\ &=\frac{2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 a d}+\frac{2 \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{3 a C}{2}+\frac{1}{2} a (5 B-2 C) \sec (c+d x)\right ) \, dx}{5 a}\\ &=\frac{2 (5 B-2 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac{2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 a d}+\frac{1}{15} (5 B+7 C) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a (5 B+7 C) \tan (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (5 B-2 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac{2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 a d}\\ \end{align*}

Mathematica [A]  time = 0.305914, size = 80, normalized size = 0.79 \[ \frac{2 \tan (c+d x) \sec (c+d x) \sqrt{a (\sec (c+d x)+1)} ((5 B+4 C) \cos (c+d x)+(5 B+4 C) \cos (2 (c+d x))+5 B+7 C)}{15 d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(5*B + 7*C + (5*B + 4*C)*Cos[c + d*x] + (5*B + 4*C)*Cos[2*(c + d*x)])*Sec[c + d*x]*Sqrt[a*(1 + Sec[c + d*x]
)]*Tan[c + d*x])/(15*d*(1 + Cos[c + d*x]))

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Maple [A]  time = 0.285, size = 94, normalized size = 0.9 \begin{align*} -{\frac{ \left ( -2+2\,\cos \left ( dx+c \right ) \right ) \left ( 10\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+8\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+5\,B\cos \left ( dx+c \right ) +4\,C\cos \left ( dx+c \right ) +3\,C \right ) }{15\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \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(sec(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x)

[Out]

-2/15/d*(-1+cos(d*x+c))*(10*B*cos(d*x+c)^2+8*C*cos(d*x+c)^2+5*B*cos(d*x+c)+4*C*cos(d*x+c)+3*C)*(a*(cos(d*x+c)+
1)/cos(d*x+c))^(1/2)/cos(d*x+c)^2/sin(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(sec(d*x+c)*(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 = 0.492669, size = 217, normalized size = 2.15 \begin{align*} \frac{2 \,{\left (2 \,{\left (5 \, B + 4 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (5 \, B + 4 \, C\right )} \cos \left (d x + c\right ) + 3 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \,{\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(2*(5*B + 4*C)*cos(d*x + c)^2 + (5*B + 4*C)*cos(d*x + c) + 3*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*s
in(d*x + c)/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)

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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 ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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)**2, x)

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Giac [A]  time = 4.41245, size = 238, normalized size = 2.36 \begin{align*} \frac{2 \,{\left (15 \, \sqrt{2} B a^{3} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 15 \, \sqrt{2} C a^{3} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (20 \, \sqrt{2} B a^{3} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 10 \, \sqrt{2} C a^{3} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (5 \, \sqrt{2} B a^{3} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 7 \, \sqrt{2} C a^{3} \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 )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{15 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{2} \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(sec(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

2/15*(15*sqrt(2)*B*a^3*sgn(cos(d*x + c)) + 15*sqrt(2)*C*a^3*sgn(cos(d*x + c)) - (20*sqrt(2)*B*a^3*sgn(cos(d*x
+ c)) + 10*sqrt(2)*C*a^3*sgn(cos(d*x + c)) - (5*sqrt(2)*B*a^3*sgn(cos(d*x + c)) + 7*sqrt(2)*C*a^3*sgn(cos(d*x
+ c)))*tan(1/2*d*x + 1/2*c)^2)*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)^2*
sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)