\(\int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} (A+C \sec ^2(c+d x)) \, dx\) [174]

Optimal result

Integrand size = 35, antiderivative size = 211 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {64 a^3 (33 A+25 C) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (33 A+25 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{693 d}+\frac {2 a (33 A+25 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{231 d}+\frac {2 (99 A+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {10 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d} \]

[Out]

2/231*a*(33*A+25*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/693*(99*A+26*C)*(a+a*sec(d*x+c))^(5/2)*tan(d*x+c)/d+
2/11*C*sec(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*tan(d*x+c)/d+10/99*C*(a+a*sec(d*x+c))^(7/2)*tan(d*x+c)/a/d+64/693*a
^3*(33*A+25*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+16/693*a^2*(33*A+25*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4174, 4095, 4086, 3878, 3877} \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {64 a^3 (33 A+25 C) \tan (c+d x)}{693 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 (33 A+25 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{693 d}+\frac {2 (99 A+26 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{693 d}+\frac {2 a (33 A+25 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{231 d}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d}+\frac {10 C \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{99 a d} \]

[In]

Int[Sec[c + d*x]^2*(a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2),x]

[Out]

(64*a^3*(33*A + 25*C)*Tan[c + d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(33*A + 25*C)*Sqrt[a + a*Sec[c
+ d*x]]*Tan[c + d*x])/(693*d) + (2*a*(33*A + 25*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(231*d) + (2*(99*A
+ 26*C)*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(693*d) + (2*C*Sec[c + d*x]^2*(a + a*Sec[c + d*x])^(5/2)*Tan
[c + d*x])/(11*d) + (10*C*(a + a*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(99*a*d)

Rule 3877

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]

Rule 3878

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

Rule 4086

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 4095

Int[csc[(e_.) + (f_.)*(x_)]^2*(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 + 1)/(b*f*(m + 2))), x] + Dist[1/(b*(m + 2)),
Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B)*Csc[e + f*x], x], x], x] /; Fr
eeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rule 4174

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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {2 \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac {1}{2} a (11 A+4 C)+\frac {5}{2} a C \sec (c+d x)\right ) \, dx}{11 a} \\ & = \frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {10 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}+\frac {4 \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac {35 a^2 C}{4}+\frac {1}{4} a^2 (99 A+26 C) \sec (c+d x)\right ) \, dx}{99 a^2} \\ & = \frac {2 (99 A+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {10 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}+\frac {1}{231} (5 (33 A+25 C)) \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \, dx \\ & = \frac {2 a (33 A+25 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{231 d}+\frac {2 (99 A+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {10 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}+\frac {1}{231} (8 a (33 A+25 C)) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx \\ & = \frac {16 a^2 (33 A+25 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{693 d}+\frac {2 a (33 A+25 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{231 d}+\frac {2 (99 A+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {10 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d}+\frac {1}{693} \left (32 a^2 (33 A+25 C)\right ) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx \\ & = \frac {64 a^3 (33 A+25 C) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (33 A+25 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{693 d}+\frac {2 a (33 A+25 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{231 d}+\frac {2 (99 A+26 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{693 d}+\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{11 d}+\frac {10 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{99 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.28 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.68 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^3 (2673 A+3628 C+2 (4983 A+5014 C) \cos (c+d x)+52 (66 A+71 C) \cos (2 (c+d x))+4587 A \cos (3 (c+d x))+3692 C \cos (3 (c+d x))+759 A \cos (4 (c+d x))+568 C \cos (4 (c+d x))+759 A \cos (5 (c+d x))+568 C \cos (5 (c+d x))) \sec ^5(c+d x) \tan (c+d x)}{2772 d \sqrt {a (1+\sec (c+d x))}} \]

[In]

Integrate[Sec[c + d*x]^2*(a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^3*(2673*A + 3628*C + 2*(4983*A + 5014*C)*Cos[c + d*x] + 52*(66*A + 71*C)*Cos[2*(c + d*x)] + 4587*A*Cos[3*(c
+ d*x)] + 3692*C*Cos[3*(c + d*x)] + 759*A*Cos[4*(c + d*x)] + 568*C*Cos[4*(c + d*x)] + 759*A*Cos[5*(c + d*x)]
+ 568*C*Cos[5*(c + d*x)])*Sec[c + d*x]^5*Tan[c + d*x])/(2772*d*Sqrt[a*(1 + Sec[c + d*x])])

Maple [A] (verified)

Time = 120.66 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.69

method	result	size
default	\(\frac {2 a^{2} \left (1518 A \cos \left (d x +c \right )^{5}+1136 C \cos \left (d x +c \right )^{5}+759 A \cos \left (d x +c \right )^{4}+568 C \cos \left (d x +c \right )^{4}+396 A \cos \left (d x +c \right )^{3}+426 C \cos \left (d x +c \right )^{3}+99 A \cos \left (d x +c \right )^{2}+355 C \cos \left (d x +c \right )^{2}+224 C \cos \left (d x +c \right )+63 C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{4}}{693 d \left (\cos \left (d x +c \right )+1\right )}\)	\(146\)
parts	\(\frac {2 A \,a^{2} \left (46 \cos \left (d x +c \right )^{3}+23 \cos \left (d x +c \right )^{2}+12 \cos \left (d x +c \right )+3\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{21 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 C \,a^{2} \left (1136 \cos \left (d x +c \right )^{5}+568 \cos \left (d x +c \right )^{4}+426 \cos \left (d x +c \right )^{3}+355 \cos \left (d x +c \right )^{2}+224 \cos \left (d x +c \right )+63\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{4}}{693 d \left (\cos \left (d x +c \right )+1\right )}\)	\(172\)

[In]

int(sec(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

2/693*a^2/d*(1518*A*cos(d*x+c)^5+1136*C*cos(d*x+c)^5+759*A*cos(d*x+c)^4+568*C*cos(d*x+c)^4+396*A*cos(d*x+c)^3+
426*C*cos(d*x+c)^3+99*A*cos(d*x+c)^2+355*C*cos(d*x+c)^2+224*C*cos(d*x+c)+63*C)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d
*x+c)+1)*tan(d*x+c)*sec(d*x+c)^4

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.71 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (2 \, {\left (759 \, A + 568 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + {\left (759 \, A + 568 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 6 \, {\left (66 \, A + 71 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (99 \, A + 355 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 224 \, C a^{2} \cos \left (d x + c\right ) + 63 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{693 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \]

[In]

integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

2/693*(2*(759*A + 568*C)*a^2*cos(d*x + c)^5 + (759*A + 568*C)*a^2*cos(d*x + c)^4 + 6*(66*A + 71*C)*a^2*cos(d*x
+ c)^3 + (99*A + 355*C)*a^2*cos(d*x + c)^2 + 224*C*a^2*cos(d*x + c) + 63*C*a^2)*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)

Sympy [F(-1)]

Timed out. \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**2*(a+a*sec(d*x+c))**(5/2)*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{2} \,d x } \]

[In]

integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 27.04 (sec) , antiderivative size = 885, normalized size of antiderivative = 4.19 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]

[In]

int(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(5/2))/cos(c + d*x)^2,x)

[Out]

((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((A*a^2*24i)/(11*d) - (a^2*
(3*A + C)*16i)/(11*d) + (a^2*(13*A + 20*C)*8i)/(11*d) - (a^2*(A + C)*80i)/(11*d)) + (A*a^2*24i)/(11*d) - (a^2*
(3*A + C)*16i)/(11*d) + (a^2*(13*A + 20*C)*8i)/(11*d) - (a^2*(A + C)*80i)/(11*d)))/((exp(c*1i + d*x*1i) + 1)*(
exp(c*2i + d*x*2i) + 1)^5) - ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i
)*((A*a^2*4i)/(9*d) + (C*a^2*64i)/(99*d) - (a^2*(11*A + 4*C)*4i)/(9*d) + (a^2*(5*A + 12*C)*4i)/(3*d) - (a^2*(5
*A - 16*C)*4i)/(9*d)) + (A*a^2*20i)/(9*d) - (a^2*(A - 16*C)*4i)/(9*d) - (a^2*(3*A + 4*C)*20i)/(9*d) + (a^2*(11
*A + 20*C)*4i)/(9*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^4) + ((a + a/(exp(- c*1i - d*x*1i)/2
+ exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((a^2*(9*A + 4*C)*4i)/(5*d) - (A*a^2*4i)/(5*d) + (a^2*(33*
A - 31*C)*16i)/(1155*d)) - (A*a^2*4i)/d + (a^2*(A + 4*C)*4i)/d))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i)
+ 1)^2) + ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((A*a^2*4i)/(7*d
) + (C*a^2*1600i)/(693*d) - (a^2*(5*A + 2*C)*8i)/(7*d) + (a^2*(5*A + 32*C)*4i)/(7*d)) + (A*a^2*24i)/(7*d) - (a
^2*(A + 2*C)*40i)/(7*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^3) + ((a + a/(exp(- c*1i - d*x*1i
)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((A*a^2*4i)/(3*d) - (a^2*(66*A + 71*C)*16i)/(693*d)) +
(A*a^2*20i)/(3*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)) - (a^2*exp(c*1i + d*x*1i)*(a + a/(exp(
- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(759*A + 568*C)*4i)/(693*d*(exp(c*1i + d*x*1i) + 1))