\(\int \sec (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [502]

Optimal result

Integrand size = 41, antiderivative size = 184 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {64 a^3 (21 A+15 B+13 C) \tan (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (21 A+15 B+13 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac {2 a (21 A+15 B+13 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}+\frac {2 (9 B-2 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac {2 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d} \]

[Out]

2/105*a*(21*A+15*B+13*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/63*(9*B-2*C)*(a+a*sec(d*x+c))^(5/2)*tan(d*x+c)/
d+2/9*C*(a+a*sec(d*x+c))^(7/2)*tan(d*x+c)/a/d+64/315*a^3*(21*A+15*B+13*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+
16/315*a^2*(21*A+15*B+13*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {4167, 4086, 3878, 3877} \[ \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {64 a^3 (21 A+15 B+13 C) \tan (c+d x)}{315 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 (21 A+15 B+13 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{315 d}+\frac {2 a (21 A+15 B+13 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d}+\frac {2 (9 B-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{63 d}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d} \]

[In]

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

[Out]

(64*a^3*(21*A + 15*B + 13*C)*Tan[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(21*A + 15*B + 13*C)*Sqr
t[a + a*Sec[c + d*x]]*Tan[c + d*x])/(315*d) + (2*a*(21*A + 15*B + 13*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x
])/(105*d) + (2*(9*B - 2*C)*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(63*d) + (2*C*(a + a*Sec[c + d*x])^(7/2)*
Tan[c + d*x])/(9*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 4167

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_
.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-C)*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*A*(m + 2) + b*C*(m + 1) + (b*
B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {2 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac {2 \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac {1}{2} a (9 A+7 C)+\frac {1}{2} a (9 B-2 C) \sec (c+d x)\right ) \, dx}{9 a} \\ & = \frac {2 (9 B-2 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac {2 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac {1}{21} (21 A+15 B+13 C) \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \, dx \\ & = \frac {2 a (21 A+15 B+13 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}+\frac {2 (9 B-2 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac {2 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac {1}{105} (8 a (21 A+15 B+13 C)) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx \\ & = \frac {16 a^2 (21 A+15 B+13 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac {2 a (21 A+15 B+13 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}+\frac {2 (9 B-2 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac {2 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d}+\frac {1}{315} \left (32 a^2 (21 A+15 B+13 C)\right ) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx \\ & = \frac {64 a^3 (21 A+15 B+13 C) \tan (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (21 A+15 B+13 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac {2 a (21 A+15 B+13 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}+\frac {2 (9 B-2 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac {2 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.49 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.57 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a^3 \left (903 A+690 B+584 C+(294 A+345 B+292 C) \sec (c+d x)+3 (21 A+60 B+73 C) \sec ^2(c+d x)+5 (9 B+26 C) \sec ^3(c+d x)+35 C \sec ^4(c+d x)\right ) \tan (c+d x)}{315 d \sqrt {a (1+\sec (c+d x))}} \]

[In]

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

[Out]

(2*a^3*(903*A + 690*B + 584*C + (294*A + 345*B + 292*C)*Sec[c + d*x] + 3*(21*A + 60*B + 73*C)*Sec[c + d*x]^2 +
5*(9*B + 26*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])])

Maple [A] (verified)

Time = 73.12 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.90

method	result	size
default	\(\frac {2 a^{2} \left (903 A \cos \left (d x +c \right )^{4}+690 B \cos \left (d x +c \right )^{4}+584 C \cos \left (d x +c \right )^{4}+294 A \cos \left (d x +c \right )^{3}+345 B \cos \left (d x +c \right )^{3}+292 C \cos \left (d x +c \right )^{3}+63 A \cos \left (d x +c \right )^{2}+180 B \cos \left (d x +c \right )^{2}+219 C \cos \left (d x +c \right )^{2}+45 B \cos \left (d x +c \right )+130 C \cos \left (d x +c \right )+35 C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{315 d \left (\cos \left (d x +c \right )+1\right )}\)	\(166\)
parts	\(\frac {2 A \,a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (43 \sin \left (d x +c \right )+14 \tan \left (d x +c \right )+3 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{15 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 B \,a^{2} \left (46 \cos \left (d x +c \right )^{3}+23 \cos \left (d x +c \right )^{2}+12 \cos \left (d x +c \right )+3\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{21 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 C \,a^{2} \left (584 \cos \left (d x +c \right )^{4}+292 \cos \left (d x +c \right )^{3}+219 \cos \left (d x +c \right )^{2}+130 \cos \left (d x +c \right )+35\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{315 d \left (\cos \left (d x +c \right )+1\right )}\)	\(224\)

[In]

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

[Out]

2/315*a^2/d*(903*A*cos(d*x+c)^4+690*B*cos(d*x+c)^4+584*C*cos(d*x+c)^4+294*A*cos(d*x+c)^3+345*B*cos(d*x+c)^3+29
2*C*cos(d*x+c)^3+63*A*cos(d*x+c)^2+180*B*cos(d*x+c)^2+219*C*cos(d*x+c)^2+45*B*cos(d*x+c)+130*C*cos(d*x+c)+35*C
)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*tan(d*x+c)*sec(d*x+c)^3

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.78 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left ({\left (903 \, A + 690 \, B + 584 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + {\left (294 \, A + 345 \, B + 292 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (21 \, A + 60 \, B + 73 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 5 \, {\left (9 \, B + 26 \, C\right )} a^{2} \cos \left (d x + c\right ) + 35 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \]

[In]

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

[Out]

2/315*((903*A + 690*B + 584*C)*a^2*cos(d*x + c)^4 + (294*A + 345*B + 292*C)*a^2*cos(d*x + c)^3 + 3*(21*A + 60*
B + 73*C)*a^2*cos(d*x + c)^2 + 5*(9*B + 26*C)*a^2*cos(d*x + c) + 35*C*a^2)*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)

Sympy [F]

\[ \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \]

[In]

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

[Out]

Integral((a*(sec(c + d*x) + 1))**(5/2)*(A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x), x)

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 27.14 (sec) , antiderivative size = 885, normalized size of antiderivative = 4.81 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]

[In]

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

[Out]

(((A*a^2*2i)/d - (a^2*exp(c*1i + d*x*1i)*(903*A + 690*B + 584*C)*2i)/(315*d))*(a + a/(exp(- c*1i - d*x*1i)/2 +
exp(c*1i + d*x*1i)/2))^(1/2))/(exp(c*1i + d*x*1i) + 1) - ((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*(3*A + 4*B + 4*C)*10i)/(7*d) - (a^2*(11*A + 10*B + 20*C)*2i)/(7*d) - (a^2
*(5*A + 2*B)*2i)/(7*d) + (a^2*(9*A - 16*C)*2i)/(63*d)) + (A*a^2*2i)/(7*d) - (a^2*(5*A + 2*B - 16*C)*2i)/(7*d)
- (a^2*(11*A + 10*B + 4*C)*2i)/(7*d) + (a^2*(15*A + 20*B + 36*C)*2i)/(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^
2*(189*A + 240*B + 292*C)*2i)/(315*d) - (a^2*(5*A + 2*B)*2i)/(3*d)) + (A*a^2*2i)/(3*d) - (a^2*(9*A + 10*B + 4*
C)*2i)/(3*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)) - ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1
i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((A*a^2*2i)/(9*d) - (a^2*(6*A + 5*B + 2*C)*4i)/(9*d) - (a^2*(10*A +
11*B + 10*C)*4i)/(9*d) + (a^2*(13*A + 15*B + 20*C)*4i)/(9*d) + (a^2*(5*A + 2*B)*2i)/(9*d)) - (A*a^2*2i)/(9*d)
+ (a^2*(6*A + 5*B + 2*C)*4i)/(9*d) + (a^2*(10*A + 11*B + 10*C)*4i)/(9*d) - (a^2*(13*A + 15*B + 20*C)*4i)/(9*d)
- (a^2*(5*A + 2*B)*2i)/(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*(5*A + 9*B + 10*C)*4i)/(5*d) + (a^2*(24*B -
21*A + 32*C)*2i)/(105*d) - (a^2*(5*A + 2*B)*2i)/(5*d)) + (A*a^2*2i)/(5*d) - (a^2*(5*A + 5*B + 2*C)*4i)/(5*d)
+ (a^2*(5*A + 10*B + 32*C)*2i)/(5*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2)