3.156 \(\int \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=180 \[ \frac{2 (21 A+16 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 a d}-\frac{4 (21 A+16 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{315 d}+\frac{2 a (21 A+16 C) \tan (c+d x)}{45 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) \sec ^3(c+d x) \sqrt{a \sec (c+d x)+a}}{9 d}+\frac{2 a C \tan (c+d x) \sec ^3(c+d x)}{63 d \sqrt{a \sec (c+d x)+a}} \]

[Out]

(2*a*(21*A + 16*C)*Tan[c + d*x])/(45*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*C*Sec[c + d*x]^3*Tan[c + d*x])/(63*d*S
qrt[a + a*Sec[c + d*x]]) - (4*(21*A + 16*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(315*d) + (2*C*Sec[c + d*x]
^3*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(9*d) + (2*(21*A + 16*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(1
05*a*d)

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Rubi [A]  time = 0.445481, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4089, 4016, 3800, 4001, 3792} \[ \frac{2 (21 A+16 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 a d}-\frac{4 (21 A+16 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{315 d}+\frac{2 a (21 A+16 C) \tan (c+d x)}{45 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) \sec ^3(c+d x) \sqrt{a \sec (c+d x)+a}}{9 d}+\frac{2 a C \tan (c+d x) \sec ^3(c+d x)}{63 d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*(A + C*Sec[c + d*x]^2),x]

[Out]

(2*a*(21*A + 16*C)*Tan[c + d*x])/(45*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*C*Sec[c + d*x]^3*Tan[c + d*x])/(63*d*S
qrt[a + a*Sec[c + d*x]]) - (4*(21*A + 16*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(315*d) + (2*C*Sec[c + d*x]
^3*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(9*d) + (2*(21*A + 16*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(1
05*a*d)

Rule 4089

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> -Simp[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*(m + n + 1)
), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n*Simp[A*b*(m + n + 1) + b*C*n + a
*C*m*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1
)] &&  !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 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 ^3(c+d x) \sqrt{a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{2 C \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{9 d}+\frac{2 \int \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{3}{2} a (3 A+2 C)+\frac{1}{2} a C \sec (c+d x)\right ) \, dx}{9 a}\\ &=\frac{2 a C \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{9 d}+\frac{1}{21} (21 A+16 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)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{9 d}+\frac{2 (21 A+16 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 a d}+\frac{(2 (21 A+16 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}{105 a}\\ &=\frac{2 a C \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}-\frac{4 (21 A+16 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac{2 C \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{9 d}+\frac{2 (21 A+16 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 a d}+\frac{1}{45} (21 A+16 C) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a (21 A+16 C) \tan (c+d x)}{45 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a C \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}-\frac{4 (21 A+16 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac{2 C \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{9 d}+\frac{2 (21 A+16 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 a d}\\ \end{align*}

Mathematica [A]  time = 0.98639, size = 122, normalized size = 0.68 \[ \frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^4(c+d x) \sqrt{a (\sec (c+d x)+1)} (2 (63 A+88 C) \cos (c+d x)+11 (21 A+16 C) \cos (2 (c+d x))+42 A \cos (3 (c+d x))+42 A \cos (4 (c+d x))+189 A+32 C \cos (3 (c+d x))+32 C \cos (4 (c+d x))+214 C)}{315 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*(A + C*Sec[c + d*x]^2),x]

[Out]

((189*A + 214*C + 2*(63*A + 88*C)*Cos[c + d*x] + 11*(21*A + 16*C)*Cos[2*(c + d*x)] + 42*A*Cos[3*(c + d*x)] + 3
2*C*Cos[3*(c + d*x)] + 42*A*Cos[4*(c + d*x)] + 32*C*Cos[4*(c + d*x)])*Sec[c + d*x]^4*Sqrt[a*(1 + Sec[c + d*x])
]*Tan[(c + d*x)/2])/(315*d)

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Maple [A]  time = 0.352, size = 129, normalized size = 0.7 \begin{align*} -{\frac{ \left ( -2+2\,\cos \left ( dx+c \right ) \right ) \left ( 168\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+128\,C \left ( \cos \left ( dx+c \right ) \right ) ^{4}+84\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+64\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+63\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+48\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+40\,C\cos \left ( dx+c \right ) +35\,C \right ) }{315\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}\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)^3*(A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x)

[Out]

-2/315/d*(-1+cos(d*x+c))*(168*A*cos(d*x+c)^4+128*C*cos(d*x+c)^4+84*A*cos(d*x+c)^3+64*C*cos(d*x+c)^3+63*A*cos(d
*x+c)^2+48*C*cos(d*x+c)^2+40*C*cos(d*x+c)+35*C)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/cos(d*x+c)^4/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)^3*(A+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 = 0.496488, size = 302, normalized size = 1.68 \begin{align*} \frac{2 \,{\left (8 \,{\left (21 \, A + 16 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \,{\left (21 \, A + 16 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (21 \, A + 16 \, C\right )} \cos \left (d x + c\right )^{2} + 40 \, C \cos \left (d x + c\right ) + 35 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \,{\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/315*(8*(21*A + 16*C)*cos(d*x + c)^4 + 4*(21*A + 16*C)*cos(d*x + c)^3 + 3*(21*A + 16*C)*cos(d*x + c)^2 + 40*C
*cos(d*x + c) + 35*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^5 + d*cos(d*x + c)^
4)

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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 (A + C \sec ^{2}{\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)**3*(A+C*sec(d*x+c)**2)*(a+a*sec(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a*(sec(c + d*x) + 1))*(A + C*sec(c + d*x)**2)*sec(c + d*x)**3, x)

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Giac [A]  time = 4.62029, size = 362, normalized size = 2.01 \begin{align*} \frac{2 \,{\left (315 \, \sqrt{2} A a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 315 \, \sqrt{2} C a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (840 \, \sqrt{2} A a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 420 \, \sqrt{2} C a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (882 \, \sqrt{2} A a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 882 \, \sqrt{2} C a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (504 \, \sqrt{2} A a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 324 \, \sqrt{2} C a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (147 \, \sqrt{2} A a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 107 \, \sqrt{2} C a^{5} \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 )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{315 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{4} \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)^3*(A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

2/315*(315*sqrt(2)*A*a^5*sgn(cos(d*x + c)) + 315*sqrt(2)*C*a^5*sgn(cos(d*x + c)) - (840*sqrt(2)*A*a^5*sgn(cos(
d*x + c)) + 420*sqrt(2)*C*a^5*sgn(cos(d*x + c)) - (882*sqrt(2)*A*a^5*sgn(cos(d*x + c)) + 882*sqrt(2)*C*a^5*sgn
(cos(d*x + c)) - (504*sqrt(2)*A*a^5*sgn(cos(d*x + c)) + 324*sqrt(2)*C*a^5*sgn(cos(d*x + c)) - (147*sqrt(2)*A*a
^5*sgn(cos(d*x + c)) + 107*sqrt(2)*C*a^5*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*ta
n(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)^4*sqrt(-a*t
an(1/2*d*x + 1/2*c)^2 + a)*d)