3.356 \(\int \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=230 \[ \frac{2 a (11 B+10 C) \tan (c+d x) \sec ^4(c+d x)}{99 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a (11 B+10 C) \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt{a \sec (c+d x)+a}}+\frac{32 (11 B+10 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 a d}-\frac{64 (11 B+10 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{3465 d}+\frac{32 a (11 B+10 C) \tan (c+d x)}{495 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a C \tan (c+d x) \sec ^5(c+d x)}{11 d \sqrt{a \sec (c+d x)+a}} \]

[Out]

(32*a*(11*B + 10*C)*Tan[c + d*x])/(495*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a*(11*B + 10*C)*Sec[c + d*x]^3*Tan[c
+ d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(11*B + 10*C)*Sec[c + d*x]^4*Tan[c + d*x])/(99*d*Sqrt[a + a*Se
c[c + d*x]]) + (2*a*C*Sec[c + d*x]^5*Tan[c + d*x])/(11*d*Sqrt[a + a*Sec[c + d*x]]) - (64*(11*B + 10*C)*Sqrt[a
+ a*Sec[c + d*x]]*Tan[c + d*x])/(3465*d) + (32*(11*B + 10*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(1155*a*
d)

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Rubi [A]  time = 0.485266, antiderivative size = 230, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{2 a (11 B+10 C) \tan (c+d x) \sec ^4(c+d x)}{99 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a (11 B+10 C) \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt{a \sec (c+d x)+a}}+\frac{32 (11 B+10 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 a d}-\frac{64 (11 B+10 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{3465 d}+\frac{32 a (11 B+10 C) \tan (c+d x)}{495 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a C \tan (c+d x) \sec ^5(c+d x)}{11 d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*Sqrt[a + a*Sec[c + d*x]]*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(32*a*(11*B + 10*C)*Tan[c + d*x])/(495*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a*(11*B + 10*C)*Sec[c + d*x]^3*Tan[c
+ d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(11*B + 10*C)*Sec[c + d*x]^4*Tan[c + d*x])/(99*d*Sqrt[a + a*Se
c[c + d*x]]) + (2*a*C*Sec[c + d*x]^5*Tan[c + d*x])/(11*d*Sqrt[a + a*Sec[c + d*x]]) - (64*(11*B + 10*C)*Sqrt[a
+ a*Sec[c + d*x]]*Tan[c + d*x])/(3465*d) + (32*(11*B + 10*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(1155*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 ^4(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 ^5(c+d x) \sqrt{a+a \sec (c+d x)} (B+C \sec (c+d x)) \, dx\\ &=\frac{2 a C \sec ^5(c+d x) \tan (c+d x)}{11 d \sqrt{a+a \sec (c+d x)}}+\frac{1}{11} (11 B+10 C) \int \sec ^5(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a (11 B+10 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a C \sec ^5(c+d x) \tan (c+d x)}{11 d \sqrt{a+a \sec (c+d x)}}+\frac{1}{99} (8 (11 B+10 C)) \int \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{16 a (11 B+10 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (11 B+10 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a C \sec ^5(c+d x) \tan (c+d x)}{11 d \sqrt{a+a \sec (c+d x)}}+\frac{1}{231} (16 (11 B+10 C)) \int \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{16 a (11 B+10 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (11 B+10 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a C \sec ^5(c+d x) \tan (c+d x)}{11 d \sqrt{a+a \sec (c+d x)}}+\frac{32 (11 B+10 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 a d}+\frac{(32 (11 B+10 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}{1155 a}\\ &=\frac{16 a (11 B+10 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (11 B+10 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a C \sec ^5(c+d x) \tan (c+d x)}{11 d \sqrt{a+a \sec (c+d x)}}-\frac{64 (11 B+10 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac{32 (11 B+10 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 a d}+\frac{1}{495} (16 (11 B+10 C)) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{32 a (11 B+10 C) \tan (c+d x)}{495 d \sqrt{a+a \sec (c+d x)}}+\frac{16 a (11 B+10 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (11 B+10 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a C \sec ^5(c+d x) \tan (c+d x)}{11 d \sqrt{a+a \sec (c+d x)}}-\frac{64 (11 B+10 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac{32 (11 B+10 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 a d}\\ \end{align*}

Mathematica [A]  time = 5.88604, size = 115, normalized size = 0.5 \[ \frac{2 a \tan (c+d x) \left (35 (11 B+10 C) \sec ^4(c+d x)+40 (11 B+10 C) \sec ^3(c+d x)+48 (11 B+10 C) \sec ^2(c+d x)+64 (11 B+10 C) \sec (c+d x)+128 (11 B+10 C)+315 C \sec ^5(c+d x)\right )}{3465 d \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*Sqrt[a + a*Sec[c + d*x]]*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(2*a*(128*(11*B + 10*C) + 64*(11*B + 10*C)*Sec[c + d*x] + 48*(11*B + 10*C)*Sec[c + d*x]^2 + 40*(11*B + 10*C)*S
ec[c + d*x]^3 + 35*(11*B + 10*C)*Sec[c + d*x]^4 + 315*C*Sec[c + d*x]^5)*Tan[c + d*x])/(3465*d*Sqrt[a*(1 + Sec[
c + d*x])])

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Maple [A]  time = 0.403, size = 160, normalized size = 0.7 \begin{align*} -{\frac{ \left ( -2+2\,\cos \left ( dx+c \right ) \right ) \left ( 1408\,B \left ( \cos \left ( dx+c \right ) \right ) ^{5}+1280\,C \left ( \cos \left ( dx+c \right ) \right ) ^{5}+704\,B \left ( \cos \left ( dx+c \right ) \right ) ^{4}+640\,C \left ( \cos \left ( dx+c \right ) \right ) ^{4}+528\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+480\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+440\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+400\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+385\,B\cos \left ( dx+c \right ) +350\,C\cos \left ( dx+c \right ) +315\,C \right ) }{3465\,d \left ( \cos \left ( dx+c \right ) \right ) ^{5}\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)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x)

[Out]

-2/3465/d*(-1+cos(d*x+c))*(1408*B*cos(d*x+c)^5+1280*C*cos(d*x+c)^5+704*B*cos(d*x+c)^4+640*C*cos(d*x+c)^4+528*B
*cos(d*x+c)^3+480*C*cos(d*x+c)^3+440*B*cos(d*x+c)^2+400*C*cos(d*x+c)^2+385*B*cos(d*x+c)+350*C*cos(d*x+c)+315*C
)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/cos(d*x+c)^5/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)^4*(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 = 0.506013, size = 373, normalized size = 1.62 \begin{align*} \frac{2 \,{\left (128 \,{\left (11 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{5} + 64 \,{\left (11 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{4} + 48 \,{\left (11 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{3} + 40 \,{\left (11 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{2} + 35 \,{\left (11 \, B + 10 \, C\right )} \cos \left (d x + c\right ) + 315 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3465 \,{\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/3465*(128*(11*B + 10*C)*cos(d*x + c)^5 + 64*(11*B + 10*C)*cos(d*x + c)^4 + 48*(11*B + 10*C)*cos(d*x + c)^3 +
40*(11*B + 10*C)*cos(d*x + c)^2 + 35*(11*B + 10*C)*cos(d*x + c) + 315*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x +
c))*sin(d*x + c)/(d*cos(d*x + c)^6 + d*cos(d*x + c)^5)

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Sympy [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)**4*(B*sec(d*x+c)+C*sec(d*x+c)**2)*(a+a*sec(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 4.62683, size = 424, normalized size = 1.84 \begin{align*} -\frac{2 \,{\left (3465 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 3465 \, \sqrt{2} C a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (8085 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 5775 \, \sqrt{2} C a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (14322 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 16170 \, \sqrt{2} C a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (13266 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 8910 \, \sqrt{2} C a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (4741 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 5885 \, \sqrt{2} C a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (1177 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 755 \, \sqrt{2} C a^{6} \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 )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{3465 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{5} \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)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

-2/3465*(3465*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + 3465*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - (8085*sqrt(2)*B*a^6*sgn
(cos(d*x + c)) + 5775*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - (14322*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + 16170*sqrt(2)
*C*a^6*sgn(cos(d*x + c)) - (13266*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + 8910*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - (47
41*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + 5885*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - (1177*sqrt(2)*B*a^6*sgn(cos(d*x +
c)) + 755*sqrt(2)*C*a^6*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)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^5*sq
rt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)