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

Optimal. Leaf size=234 \[ \frac{2 a^2 (11 B+12 C) \tan (c+d x) \sec ^4(c+d x)}{99 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (187 B+168 C) \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt{a \sec (c+d x)+a}}+\frac{4 a^2 (187 B+168 C) \tan (c+d x)}{495 d \sqrt{a \sec (c+d x)+a}}+\frac{4 (187 B+168 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 d}-\frac{8 a (187 B+168 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{3465 d}+\frac{2 a C \tan (c+d x) \sec ^4(c+d x) \sqrt{a \sec (c+d x)+a}}{11 d} \]

[Out]

(4*a^2*(187*B + 168*C)*Tan[c + d*x])/(495*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(187*B + 168*C)*Sec[c + d*x]^3*
Tan[c + d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(11*B + 12*C)*Sec[c + d*x]^4*Tan[c + d*x])/(99*d*Sqrt[
a + a*Sec[c + d*x]]) - (8*a*(187*B + 168*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3465*d) + (2*a*C*Sec[c + d
*x]^4*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(11*d) + (4*(187*B + 168*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*
x])/(1155*d)

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Rubi [A]  time = 0.637845, antiderivative size = 234, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {4072, 4018, 4016, 3803, 3800, 4001, 3792} \[ \frac{2 a^2 (11 B+12 C) \tan (c+d x) \sec ^4(c+d x)}{99 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (187 B+168 C) \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt{a \sec (c+d x)+a}}+\frac{4 a^2 (187 B+168 C) \tan (c+d x)}{495 d \sqrt{a \sec (c+d x)+a}}+\frac{4 (187 B+168 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 d}-\frac{8 a (187 B+168 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{3465 d}+\frac{2 a C \tan (c+d x) \sec ^4(c+d x) \sqrt{a \sec (c+d x)+a}}{11 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^2*(187*B + 168*C)*Tan[c + d*x])/(495*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(187*B + 168*C)*Sec[c + d*x]^3*
Tan[c + d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(11*B + 12*C)*Sec[c + d*x]^4*Tan[c + d*x])/(99*d*Sqrt[
a + a*Sec[c + d*x]]) - (8*a*(187*B + 168*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3465*d) + (2*a*C*Sec[c + d
*x]^4*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(11*d) + (4*(187*B + 168*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*
x])/(1155*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 4018

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

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) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sec ^4(c+d x) (a+a \sec (c+d x))^{3/2} (B+C \sec (c+d x)) \, dx\\ &=\frac{2 a C \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac{2}{11} \int \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{1}{2} a (11 B+8 C)+\frac{1}{2} a (11 B+12 C) \sec (c+d x)\right ) \, dx\\ &=\frac{2 a^2 (11 B+12 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a C \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac{1}{99} (a (187 B+168 C)) \int \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a^2 (187 B+168 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (11 B+12 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a C \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac{1}{231} (2 a (187 B+168 C)) \int \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a^2 (187 B+168 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (11 B+12 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a C \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac{4 (187 B+168 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac{(4 (187 B+168 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}\\ &=\frac{2 a^2 (187 B+168 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (11 B+12 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt{a+a \sec (c+d x)}}-\frac{8 a (187 B+168 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac{2 a C \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac{4 (187 B+168 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac{1}{495} (2 a (187 B+168 C)) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{4 a^2 (187 B+168 C) \tan (c+d x)}{495 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (187 B+168 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (11 B+12 C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt{a+a \sec (c+d x)}}-\frac{8 a (187 B+168 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac{2 a C \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac{4 (187 B+168 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}\\ \end{align*}

Mathematica [A]  time = 6.10408, size = 113, normalized size = 0.48 \[ \frac{2 a^2 \tan (c+d x) \left (35 (11 B+21 C) \sec ^4(c+d x)+(935 B+840 C) \sec ^3(c+d x)+6 (187 B+168 C) \sec ^2(c+d x)+8 (187 B+168 C) \sec (c+d x)+2992 B+315 C \sec ^5(c+d x)+2688 C\right )}{3465 d \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(2992*B + 2688*C + 8*(187*B + 168*C)*Sec[c + d*x] + 6*(187*B + 168*C)*Sec[c + d*x]^2 + (935*B + 840*C)*
Sec[c + d*x]^3 + 35*(11*B + 21*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 = 0.463, size = 161, normalized size = 0.7 \begin{align*} -{\frac{2\,a \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 2992\,B \left ( \cos \left ( dx+c \right ) \right ) ^{5}+2688\,C \left ( \cos \left ( dx+c \right ) \right ) ^{5}+1496\,B \left ( \cos \left ( dx+c \right ) \right ) ^{4}+1344\,C \left ( \cos \left ( dx+c \right ) \right ) ^{4}+1122\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+1008\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+935\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+840\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+385\,B\cos \left ( dx+c \right ) +735\,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)^3*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

-2/3465/d*a*(-1+cos(d*x+c))*(2992*B*cos(d*x+c)^5+2688*C*cos(d*x+c)^5+1496*B*cos(d*x+c)^4+1344*C*cos(d*x+c)^4+1
122*B*cos(d*x+c)^3+1008*C*cos(d*x+c)^3+935*B*cos(d*x+c)^2+840*C*cos(d*x+c)^2+385*B*cos(d*x+c)+735*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)^3*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.51914, size = 394, normalized size = 1.68 \begin{align*} \frac{2 \,{\left (16 \,{\left (187 \, B + 168 \, C\right )} a \cos \left (d x + c\right )^{5} + 8 \,{\left (187 \, B + 168 \, C\right )} a \cos \left (d x + c\right )^{4} + 6 \,{\left (187 \, B + 168 \, C\right )} a \cos \left (d x + c\right )^{3} + 5 \,{\left (187 \, B + 168 \, C\right )} a \cos \left (d x + c\right )^{2} + 35 \,{\left (11 \, B + 21 \, C\right )} a \cos \left (d x + c\right ) + 315 \, C a\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)^3*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

2/3465*(16*(187*B + 168*C)*a*cos(d*x + c)^5 + 8*(187*B + 168*C)*a*cos(d*x + c)^4 + 6*(187*B + 168*C)*a*cos(d*x
+ c)^3 + 5*(187*B + 168*C)*a*cos(d*x + c)^2 + 35*(11*B + 21*C)*a*cos(d*x + c) + 315*C*a)*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)**3*(a+a*sec(d*x+c))**(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 4.91038, size = 424, normalized size = 1.81 \begin{align*} -\frac{4 \,{\left (3465 \, \sqrt{2} B a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 3465 \, \sqrt{2} C a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (9240 \, \sqrt{2} B a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 6930 \, \sqrt{2} C a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (14784 \, \sqrt{2} B a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 15246 \, \sqrt{2} C a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (13662 \, \sqrt{2} B a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 11088 \, \sqrt{2} C a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (5687 \, \sqrt{2} B a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 5313 \, \sqrt{2} C a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 2 \,{\left (517 \, \sqrt{2} B a^{7} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 483 \, \sqrt{2} C a^{7} \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)^3*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

-4/3465*(3465*sqrt(2)*B*a^7*sgn(cos(d*x + c)) + 3465*sqrt(2)*C*a^7*sgn(cos(d*x + c)) - (9240*sqrt(2)*B*a^7*sgn
(cos(d*x + c)) + 6930*sqrt(2)*C*a^7*sgn(cos(d*x + c)) - (14784*sqrt(2)*B*a^7*sgn(cos(d*x + c)) + 15246*sqrt(2)
*C*a^7*sgn(cos(d*x + c)) - (13662*sqrt(2)*B*a^7*sgn(cos(d*x + c)) + 11088*sqrt(2)*C*a^7*sgn(cos(d*x + c)) - (5
687*sqrt(2)*B*a^7*sgn(cos(d*x + c)) + 5313*sqrt(2)*C*a^7*sgn(cos(d*x + c)) - 2*(517*sqrt(2)*B*a^7*sgn(cos(d*x
+ c)) + 483*sqrt(2)*C*a^7*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*
sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)