\(\int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\) [135]

Optimal result

Integrand size = 33, antiderivative size = 237 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {2 a^3 (803 A+710 B) \tan (c+d x)}{495 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (209 A+194 B) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}-\frac {4 a^2 (803 A+710 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a^2 (11 A+14 B) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{99 d}+\frac {2 a (803 A+710 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 a B \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d} \]

[Out]

2/1155*a*(803*A+710*B)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/11*a*B*sec(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*tan(d*
x+c)/d+2/495*a^3*(803*A+710*B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/693*a^3*(209*A+194*B)*sec(d*x+c)^3*tan(d*
x+c)/d/(a+a*sec(d*x+c))^(1/2)-4/3465*a^2*(803*A+710*B)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/99*a^2*(11*A+14*B
)*sec(d*x+c)^3*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {4103, 4101, 3885, 4086, 3877} \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {2 a^3 (209 A+194 B) \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^3 (803 A+710 B) \tan (c+d x)}{495 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (11 A+14 B) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{99 d}-\frac {4 a^2 (803 A+710 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3465 d}+\frac {2 a (803 A+710 B) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 d}+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d} \]

[In]

Int[Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]

[Out]

(2*a^3*(803*A + 710*B)*Tan[c + d*x])/(495*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(209*A + 194*B)*Sec[c + d*x]^3*
Tan[c + d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) - (4*a^2*(803*A + 710*B)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])
/(3465*d) + (2*a^2*(11*A + 14*B)*Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(99*d) + (2*a*(803*A +
710*B)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(1155*d) + (2*a*B*Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2)*Ta
n[c + d*x])/(11*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 3885

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 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 4101

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 4103

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] &&
NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {2 a B \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac {2}{11} \int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {1}{2} a (11 A+6 B)+\frac {1}{2} a (11 A+14 B) \sec (c+d x)\right ) \, dx \\ & = \frac {2 a^2 (11 A+14 B) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{99 d}+\frac {2 a B \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac {4}{99} \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {3}{4} a^2 (55 A+46 B)+\frac {1}{4} a^2 (209 A+194 B) \sec (c+d x)\right ) \, dx \\ & = \frac {2 a^3 (209 A+194 B) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (11 A+14 B) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{99 d}+\frac {2 a B \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac {1}{231} \left (a^2 (803 A+710 B)\right ) \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx \\ & = \frac {2 a^3 (209 A+194 B) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (11 A+14 B) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{99 d}+\frac {2 a (803 A+710 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 a B \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac {(2 a (803 A+710 B)) \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^3 (209 A+194 B) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}-\frac {4 a^2 (803 A+710 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a^2 (11 A+14 B) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{99 d}+\frac {2 a (803 A+710 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 a B \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac {1}{495} \left (a^2 (803 A+710 B)\right ) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx \\ & = \frac {2 a^3 (803 A+710 B) \tan (c+d x)}{495 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (209 A+194 B) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}-\frac {4 a^2 (803 A+710 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 a^2 (11 A+14 B) \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{99 d}+\frac {2 a (803 A+710 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac {2 a B \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.16 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.49 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {2 a^3 \left (8 (803 A+710 B)+4 (803 A+710 B) \sec (c+d x)+3 (803 A+710 B) \sec ^2(c+d x)+5 (286 A+355 B) \sec ^3(c+d x)+35 (11 A+32 B) \sec ^4(c+d x)+315 B \sec ^5(c+d x)\right ) \tan (c+d x)}{3465 d \sqrt {a (1+\sec (c+d x))}} \]

[In]

Integrate[Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]

[Out]

(2*a^3*(8*(803*A + 710*B) + 4*(803*A + 710*B)*Sec[c + d*x] + 3*(803*A + 710*B)*Sec[c + d*x]^2 + 5*(286*A + 355
*B)*Sec[c + d*x]^3 + 35*(11*A + 32*B)*Sec[c + d*x]^4 + 315*B*Sec[c + d*x]^5)*Tan[c + d*x])/(3465*d*Sqrt[a*(1 +
Sec[c + d*x])])

Maple [A] (verified)

Time = 95.81 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.65

method	result	size
default	\(\frac {2 a^{2} \left (6424 A \cos \left (d x +c \right )^{5}+5680 B \cos \left (d x +c \right )^{5}+3212 A \cos \left (d x +c \right )^{4}+2840 B \cos \left (d x +c \right )^{4}+2409 A \cos \left (d x +c \right )^{3}+2130 B \cos \left (d x +c \right )^{3}+1430 A \cos \left (d x +c \right )^{2}+1775 B \cos \left (d x +c \right )^{2}+385 A \cos \left (d x +c \right )+1120 B \cos \left (d x +c \right )+315 B \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{4}}{3465 d \left (\cos \left (d x +c \right )+1\right )}\)	\(155\)
parts	\(\frac {2 A \,a^{2} \left (584 \cos \left (d x +c \right )^{4}+292 \cos \left (d x +c \right )^{3}+219 \cos \left (d x +c \right )^{2}+130 \cos \left (d x +c \right )+35\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{315 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 B \,a^{2} \left (1136 \cos \left (d x +c \right )^{5}+568 \cos \left (d x +c \right )^{4}+426 \cos \left (d x +c \right )^{3}+355 \cos \left (d x +c \right )^{2}+224 \cos \left (d x +c \right )+63\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{4}}{693 d \left (\cos \left (d x +c \right )+1\right )}\)	\(182\)

[In]

int(sec(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

2/3465*a^2/d*(6424*A*cos(d*x+c)^5+5680*B*cos(d*x+c)^5+3212*A*cos(d*x+c)^4+2840*B*cos(d*x+c)^4+2409*A*cos(d*x+c
)^3+2130*B*cos(d*x+c)^3+1430*A*cos(d*x+c)^2+1775*B*cos(d*x+c)^2+385*A*cos(d*x+c)+1120*B*cos(d*x+c)+315*B)*(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 = 157, normalized size of antiderivative = 0.66 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {2 \, {\left (8 \, {\left (803 \, A + 710 \, B\right )} a^{2} \cos \left (d x + c\right )^{5} + 4 \, {\left (803 \, A + 710 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 3 \, {\left (803 \, A + 710 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 5 \, {\left (286 \, A + 355 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 35 \, {\left (11 \, A + 32 \, B\right )} a^{2} \cos \left (d x + c\right ) + 315 \, B a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \]

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

2/3465*(8*(803*A + 710*B)*a^2*cos(d*x + c)^5 + 4*(803*A + 710*B)*a^2*cos(d*x + c)^4 + 3*(803*A + 710*B)*a^2*co
s(d*x + c)^3 + 5*(286*A + 355*B)*a^2*cos(d*x + c)^2 + 35*(11*A + 32*B)*a^2*cos(d*x + c) + 315*B*a^2)*sqrt((a*c
os(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. \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**3*(a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)),x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 23.95 (sec) , antiderivative size = 856, normalized size of antiderivative = 3.61 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \]

[In]

int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(5/2))/cos(c + d*x)^3,x)

[Out]

((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((A*a^2*8i)/(3*d) - (a^2*exp(c*1i + d*x*1i)*(80
3*A + 710*B)*8i)/(3465*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*1i + d*x*1i)/2))^(1/2)*((A*a^2*24i)/(7*d) - exp(c*1i + d*x*1i)*((a^2*(5*A + 16*B)*8i)/(7*d) - (a^2
*(5*A + 2*B)*8i)/(7*d) + (a^2*(11*A + 50*B)*32i)/(693*d)) + (a^2*(9*A + 10*B)*8i)/(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*8i)/(11*d) + (a^2*(3*A + 4*B)*40i)/(11*d) - (a^2*(5*A + 2*B)*8i)/(11*d) - (a^2*(11*A + 10*B)*
8i)/(11*d)) + (A*a^2*8i)/(11*d) + (a^2*(3*A + 4*B)*40i)/(11*d) - (a^2*(5*A + 2*B)*8i)/(11*d) - (a^2*(11*A + 10
*B)*8i)/(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)*((A*a^2*8i)/(9*d) - exp(c*1i + d*x*1i)*((a^2*(A - 8*B)*8i)/(9*d) - (B*a^2*64i)/(99*d
) + (a^2*(5*A + 2*B)*8i)/(9*d) - (a^2*(5*A + 9*B)*16i)/(9*d)) + (a^2*(A + 2*B)*40i)/(9*d) + (B*a^2*64i)/(9*d)
- (a^2*(A + B)*80i)/(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 + 2*B)*8i)/(5*d) + (a^2*(44*A - 31*B)*16i)
/(1155*d)) - (A*a^2*8i)/(5*d) + (a^2*(4*A + 5*B)*16i)/(5*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) +
1)^2) - (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)*(803*A + 710*B)*
16i)/(3465*d*(exp(c*1i + d*x*1i) + 1))