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

Optimal. Leaf size=187 \[ \frac{2 a (9 B+8 C) \tan (c+d x) \sec ^3(c+d x)}{63 d \sqrt{a \sec (c+d x)+a}}+\frac{4 (9 B+8 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 a d}-\frac{8 (9 B+8 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{315 d}+\frac{4 a (9 B+8 C) \tan (c+d x)}{45 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt{a \sec (c+d x)+a}} \]

[Out]

(4*a*(9*B + 8*C)*Tan[c + d*x])/(45*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(9*B + 8*C)*Sec[c + d*x]^3*Tan[c + d*x])
/(63*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*C*Sec[c + d*x]^4*Tan[c + d*x])/(9*d*Sqrt[a + a*Sec[c + d*x]]) - (8*(9*
B + 8*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(315*d) + (4*(9*B + 8*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*
x])/(105*a*d)

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Rubi [A]  time = 0.418851, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4072, 4016, 3803, 3800, 4001, 3792} \[ \frac{2 a (9 B+8 C) \tan (c+d x) \sec ^3(c+d x)}{63 d \sqrt{a \sec (c+d x)+a}}+\frac{4 (9 B+8 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 a d}-\frac{8 (9 B+8 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{315 d}+\frac{4 a (9 B+8 C) \tan (c+d x)}{45 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*(9*B + 8*C)*Tan[c + d*x])/(45*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(9*B + 8*C)*Sec[c + d*x]^3*Tan[c + d*x])
/(63*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*C*Sec[c + d*x]^4*Tan[c + d*x])/(9*d*Sqrt[a + a*Sec[c + d*x]]) - (8*(9*
B + 8*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(315*d) + (4*(9*B + 8*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*
x])/(105*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 3803

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

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 (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} (B+C \sec (c+d x)) \, dx\\ &=\frac{2 a C \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt{a+a \sec (c+d x)}}+\frac{1}{9} (9 B+8 C) \int \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a (9 B+8 C) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a C \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt{a+a \sec (c+d x)}}+\frac{1}{21} (2 (9 B+8 C)) \int \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a (9 B+8 C) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a C \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt{a+a \sec (c+d x)}}+\frac{4 (9 B+8 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 a d}+\frac{(4 (9 B+8 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 (9 B+8 C) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a C \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt{a+a \sec (c+d x)}}-\frac{8 (9 B+8 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac{4 (9 B+8 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 a d}+\frac{1}{45} (2 (9 B+8 C)) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{4 a (9 B+8 C) \tan (c+d x)}{45 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (9 B+8 C) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a C \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt{a+a \sec (c+d x)}}-\frac{8 (9 B+8 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac{4 (9 B+8 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 a d}\\ \end{align*}

Mathematica [A]  time = 0.536767, size = 98, normalized size = 0.52 \[ \frac{2 a \tan (c+d x) \left (5 (9 B+8 C) \sec ^3(c+d x)+6 (9 B+8 C) \sec ^2(c+d x)+8 (9 B+8 C) \sec (c+d x)+16 (9 B+8 C)+35 C \sec ^4(c+d x)\right )}{315 d \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(16*(9*B + 8*C) + 8*(9*B + 8*C)*Sec[c + d*x] + 6*(9*B + 8*C)*Sec[c + d*x]^2 + 5*(9*B + 8*C)*Sec[c + d*x]^
3 + 35*C*Sec[c + d*x]^4)*Tan[c + d*x])/(315*d*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [A]  time = 0.398, size = 138, normalized size = 0.7 \begin{align*} -{\frac{ \left ( -2+2\,\cos \left ( dx+c \right ) \right ) \left ( 144\,B \left ( \cos \left ( dx+c \right ) \right ) ^{4}+128\,C \left ( \cos \left ( dx+c \right ) \right ) ^{4}+72\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+64\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+54\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+48\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+45\,B\cos \left ( dx+c \right ) +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*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x)

[Out]

-2/315/d*(-1+cos(d*x+c))*(144*B*cos(d*x+c)^4+128*C*cos(d*x+c)^4+72*B*cos(d*x+c)^3+64*C*cos(d*x+c)^3+54*B*cos(d
*x+c)^2+48*C*cos(d*x+c)^2+45*B*cos(d*x+c)+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*(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 = 0.505987, size = 308, normalized size = 1.65 \begin{align*} \frac{2 \,{\left (16 \,{\left (9 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{4} + 8 \,{\left (9 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{3} + 6 \,{\left (9 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{2} + 5 \,{\left (9 \, B + 8 \, C\right )} \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*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/315*(16*(9*B + 8*C)*cos(d*x + c)^4 + 8*(9*B + 8*C)*cos(d*x + c)^3 + 6*(9*B + 8*C)*cos(d*x + c)^2 + 5*(9*B +
8*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 (B + C \sec{\left (c + d x \right )}\right ) \sec ^{4}{\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*(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)**4, x)

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Giac [A]  time = 4.56227, size = 362, normalized size = 1.94 \begin{align*} \frac{2 \,{\left (315 \, \sqrt{2} B 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 (630 \, \sqrt{2} B 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 (756 \, \sqrt{2} B 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 (522 \, \sqrt{2} B 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 (81 \, \sqrt{2} B 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*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

2/315*(315*sqrt(2)*B*a^5*sgn(cos(d*x + c)) + 315*sqrt(2)*C*a^5*sgn(cos(d*x + c)) - (630*sqrt(2)*B*a^5*sgn(cos(
d*x + c)) + 420*sqrt(2)*C*a^5*sgn(cos(d*x + c)) - (756*sqrt(2)*B*a^5*sgn(cos(d*x + c)) + 882*sqrt(2)*C*a^5*sgn
(cos(d*x + c)) - (522*sqrt(2)*B*a^5*sgn(cos(d*x + c)) + 324*sqrt(2)*C*a^5*sgn(cos(d*x + c)) - (81*sqrt(2)*B*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)*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)^4*sqrt(-a*ta
n(1/2*d*x + 1/2*c)^2 + a)*d)