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

Optimal. Leaf size=144 \[ \frac{2 (7 B+6 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 a d}-\frac{4 (7 B+6 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{105 d}+\frac{2 a (7 B+6 C) \tan (c+d x)}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt{a \sec (c+d x)+a}} \]

[Out]

(2*a*(7*B + 6*C)*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*C*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt
[a + a*Sec[c + d*x]]) - (4*(7*B + 6*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(105*d) + (2*(7*B + 6*C)*(a + a*
Sec[c + d*x])^(3/2)*Tan[c + d*x])/(35*a*d)

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Rubi [A]  time = 0.359372, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.119, Rules used = {4072, 4016, 3800, 4001, 3792} \[ \frac{2 (7 B+6 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 a d}-\frac{4 (7 B+6 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{105 d}+\frac{2 a (7 B+6 C) \tan (c+d x)}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(7*B + 6*C)*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*C*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt
[a + a*Sec[c + d*x]]) - (4*(7*B + 6*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(105*d) + (2*(7*B + 6*C)*(a + a*
Sec[c + d*x])^(3/2)*Tan[c + d*x])/(35*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 4016

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

Rule 3800

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[e + f*x]*(a
+ b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b
*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-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 ^2(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 ^3(c+d x) \sqrt{a+a \sec (c+d x)} (B+C \sec (c+d x)) \, dx\\ &=\frac{2 a C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}+\frac{1}{7} (7 B+6 C) \int \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (7 B+6 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 a d}+\frac{(2 (7 B+6 C)) \int \sec (c+d x) \left (\frac{3 a}{2}-a \sec (c+d x)\right ) \sqrt{a+a \sec (c+d x)} \, dx}{35 a}\\ &=\frac{2 a C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}-\frac{4 (7 B+6 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac{2 (7 B+6 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 a d}+\frac{1}{15} (7 B+6 C) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a (7 B+6 C) \tan (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}-\frac{4 (7 B+6 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac{2 (7 B+6 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 a d}\\ \end{align*}

Mathematica [A]  time = 0.281068, size = 81, normalized size = 0.56 \[ \frac{2 a \tan (c+d x) \left (3 (7 B+6 C) \sec ^2(c+d x)+4 (7 B+6 C) \sec (c+d x)+8 (7 B+6 C)+15 C \sec ^3(c+d x)\right )}{105 d \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(8*(7*B + 6*C) + 4*(7*B + 6*C)*Sec[c + d*x] + 3*(7*B + 6*C)*Sec[c + d*x]^2 + 15*C*Sec[c + d*x]^3)*Tan[c +
d*x])/(105*d*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [A]  time = 0.317, size = 116, normalized size = 0.8 \begin{align*} -{\frac{ \left ( -2+2\,\cos \left ( dx+c \right ) \right ) \left ( 56\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+48\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+28\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+24\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+21\,B\cos \left ( dx+c \right ) +18\,C\cos \left ( dx+c \right ) +15\,C \right ) }{105\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}\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)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x)

[Out]

-2/105/d*(-1+cos(d*x+c))*(56*B*cos(d*x+c)^3+48*C*cos(d*x+c)^3+28*B*cos(d*x+c)^2+24*C*cos(d*x+c)^2+21*B*cos(d*x
+c)+18*C*cos(d*x+c)+15*C)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/cos(d*x+c)^3/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)^2*(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 = 0.49372, size = 265, normalized size = 1.84 \begin{align*} \frac{2 \,{\left (8 \,{\left (7 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \,{\left (7 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (7 \, B + 6 \, C\right )} \cos \left (d x + c\right ) + 15 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \,{\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(8*(7*B + 6*C)*cos(d*x + c)^3 + 4*(7*B + 6*C)*cos(d*x + c)^2 + 3*(7*B + 6*C)*cos(d*x + c) + 15*C)*sqrt((
a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)

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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 ^{3}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(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)**3, x)

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Giac [A]  time = 4.45322, size = 300, normalized size = 2.08 \begin{align*} -\frac{2 \,{\left (105 \, \sqrt{2} B a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 105 \, \sqrt{2} C a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (175 \, \sqrt{2} B a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 105 \, \sqrt{2} C a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (119 \, \sqrt{2} B a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 147 \, \sqrt{2} C a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (49 \, \sqrt{2} B a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 27 \, \sqrt{2} C a^{4} \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 )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{105 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{3} \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)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

-2/105*(105*sqrt(2)*B*a^4*sgn(cos(d*x + c)) + 105*sqrt(2)*C*a^4*sgn(cos(d*x + c)) - (175*sqrt(2)*B*a^4*sgn(cos
(d*x + c)) + 105*sqrt(2)*C*a^4*sgn(cos(d*x + c)) - (119*sqrt(2)*B*a^4*sgn(cos(d*x + c)) + 147*sqrt(2)*C*a^4*sg
n(cos(d*x + c)) - (49*sqrt(2)*B*a^4*sgn(cos(d*x + c)) + 27*sqrt(2)*C*a^4*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)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^3*s
qrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)