\(\int \frac {\sec ^4(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\) [519]

Optimal result

Integrand size = 43, antiderivative size = 277 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {(11 A-15 B+19 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(455 A-651 B+799 C) \tan (c+d x)}{105 a d \sqrt {a+a \sec (c+d x)}}-\frac {(35 A-63 B+67 C) \sec ^2(c+d x) \tan (c+d x)}{70 a d \sqrt {a+a \sec (c+d x)}}+\frac {(7 A-7 B+11 C) \sec ^3(c+d x) \tan (c+d x)}{14 a d \sqrt {a+a \sec (c+d x)}}+\frac {(245 A-273 B+397 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{210 a^2 d} \] Output:

1/4*(11*A-15*B+19*C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c) 
)^(1/2))*2^(1/2)/a^(3/2)/d-1/2*(A-B+C)*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec( 
d*x+c))^(3/2)-1/105*(455*A-651*B+799*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1 
/2)-1/70*(35*A-63*B+67*C)*sec(d*x+c)^2*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/ 
2)+1/14*(7*A-7*B+11*C)*sec(d*x+c)^3*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)+ 
1/210*(245*A-273*B+397*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a^2/d
 

Mathematica [A] (verified)

Time = 2.96 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.86 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\left (210 \sqrt {2} (11 A-15 B+19 C) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)-\frac {1}{4} (1435 A-2751 B+2339 C+24 (105 A-217 B+213 C) \cos (c+d x)+60 (35 A-63 B+67 C) \cos (2 (c+d x))+840 A \cos (3 (c+d x))-1512 B \cos (3 (c+d x))+1608 C \cos (3 (c+d x))+665 A \cos (4 (c+d x))-1029 B \cos (4 (c+d x))+1201 C \cos (4 (c+d x))) \sqrt {1-\sec (c+d x)} \sec ^4(c+d x)\right ) \tan (c+d x)}{420 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{3/2}} \] Input:

Integrate[(Sec[c + d*x]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a* 
Sec[c + d*x])^(3/2),x]
 

Output:

((210*Sqrt[2]*(11*A - 15*B + 19*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]] 
*Cos[(c + d*x)/2]^2*Sec[c + d*x] - ((1435*A - 2751*B + 2339*C + 24*(105*A 
- 217*B + 213*C)*Cos[c + d*x] + 60*(35*A - 63*B + 67*C)*Cos[2*(c + d*x)] + 
 840*A*Cos[3*(c + d*x)] - 1512*B*Cos[3*(c + d*x)] + 1608*C*Cos[3*(c + d*x) 
] + 665*A*Cos[4*(c + d*x)] - 1029*B*Cos[4*(c + d*x)] + 1201*C*Cos[4*(c + d 
*x)])*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^4)/4)*Tan[c + d*x])/(420*d*Sqrt[ 
1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3/2))
 

Rubi [A] (verified)

Time = 1.87 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.09, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.395, Rules used = {3042, 4572, 27, 3042, 4509, 27, 3042, 4509, 27, 3042, 4498, 27, 3042, 4489, 3042, 4282, 216}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\)

\(\Big \downarrow \) 4572

\(\displaystyle \frac {\int -\frac {\sec ^4(c+d x) (4 a (A-2 B+2 C)-a (7 A-7 B+11 C) \sec (c+d x))}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {\sec ^4(c+d x) (4 a (A-2 B+2 C)-a (7 A-7 B+11 C) \sec (c+d x))}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (4 a (A-2 B+2 C)-a (7 A-7 B+11 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4509

\(\displaystyle -\frac {\frac {2 \int -\frac {\sec ^3(c+d x) \left (6 a^2 (7 A-7 B+11 C)-a^2 (35 A-63 B+67 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{7 a}-\frac {2 a (7 A-7 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {\int \frac {\sec ^3(c+d x) \left (6 a^2 (7 A-7 B+11 C)-a^2 (35 A-63 B+67 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{7 a}-\frac {2 a (7 A-7 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (6 a^2 (7 A-7 B+11 C)-a^2 (35 A-63 B+67 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{7 a}-\frac {2 a (7 A-7 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4509

\(\displaystyle -\frac {-\frac {\frac {2 \int -\frac {\sec ^2(c+d x) \left (4 a^3 (35 A-63 B+67 C)-a^3 (245 A-273 B+397 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{5 a}-\frac {2 a^2 (35 A-63 B+67 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}-\frac {2 a (7 A-7 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {-\frac {\int \frac {\sec ^2(c+d x) \left (4 a^3 (35 A-63 B+67 C)-a^3 (245 A-273 B+397 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{5 a}-\frac {2 a^2 (35 A-63 B+67 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}-\frac {2 a (7 A-7 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (4 a^3 (35 A-63 B+67 C)-a^3 (245 A-273 B+397 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}-\frac {2 a^2 (35 A-63 B+67 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}-\frac {2 a (7 A-7 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4498

\(\displaystyle -\frac {-\frac {-\frac {\frac {2 \int -\frac {\sec (c+d x) \left (a^4 (245 A-273 B+397 C)-2 a^4 (455 A-651 B+799 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 a^2 (245 A-273 B+397 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (35 A-63 B+67 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}-\frac {2 a (7 A-7 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {-\frac {-\frac {\int \frac {\sec (c+d x) \left (a^4 (245 A-273 B+397 C)-2 a^4 (455 A-651 B+799 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 a^2 (245 A-273 B+397 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (35 A-63 B+67 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}-\frac {2 a (7 A-7 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {-\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a^4 (245 A-273 B+397 C)-2 a^4 (455 A-651 B+799 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}-\frac {2 a^2 (245 A-273 B+397 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (35 A-63 B+67 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}-\frac {2 a (7 A-7 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4489

\(\displaystyle -\frac {-\frac {-\frac {-\frac {105 a^4 (11 A-15 B+19 C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx-\frac {4 a^4 (455 A-651 B+799 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a^2 (245 A-273 B+397 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (35 A-63 B+67 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}-\frac {2 a (7 A-7 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {-\frac {-\frac {105 a^4 (11 A-15 B+19 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {4 a^4 (455 A-651 B+799 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a^2 (245 A-273 B+397 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (35 A-63 B+67 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}-\frac {2 a (7 A-7 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4282

\(\displaystyle -\frac {-\frac {-\frac {-\frac {-\frac {210 a^4 (11 A-15 B+19 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {4 a^4 (455 A-651 B+799 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a^2 (245 A-273 B+397 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (35 A-63 B+67 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}-\frac {2 a (7 A-7 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {-\frac {-\frac {2 a^2 (35 A-63 B+67 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}-\frac {-\frac {2 a^2 (245 A-273 B+397 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}-\frac {\frac {105 \sqrt {2} a^{7/2} (11 A-15 B+19 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {4 a^4 (455 A-651 B+799 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}}{5 a}}{7 a}-\frac {2 a (7 A-7 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

Input:

Int[(Sec[c + d*x]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c 
+ d*x])^(3/2),x]
 

Output:

-1/2*((A - B + C)*Sec[c + d*x]^4*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^(3/ 
2)) - ((-2*a*(7*A - 7*B + 11*C)*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt[a + 
 a*Sec[c + d*x]]) - ((-2*a^2*(35*A - 63*B + 67*C)*Sec[c + d*x]^2*Tan[c + d 
*x])/(5*d*Sqrt[a + a*Sec[c + d*x]]) - ((-2*a^2*(245*A - 273*B + 397*C)*Sqr 
t[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d) - ((105*Sqrt[2]*a^(7/2)*(11*A - 
15*B + 19*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x 
]])])/d - (4*a^4*(455*A - 651*B + 799*C)*Tan[c + d*x])/(d*Sqrt[a + a*Sec[c 
 + d*x]]))/(3*a))/(5*a))/(7*a))/(4*a^2)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4282

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2/f   Subst[Int[1/(2*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[ 
a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
 

rule 4489

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs 
c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( 
a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[(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 4498

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] + Simp[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] /; FreeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a 
*B, 0] &&  !LtQ[m, -1]
 

rule 4509

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

rule 4572

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

Maple [A] (warning: unable to verify)

Time = 3.86 (sec) , antiderivative size = 413, normalized size of antiderivative = 1.49

method	result	size
default	\(\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (1155 \cos \left (d x +c \right )^{2}+2310 \cos \left (d x +c \right )+1155\right ) \sqrt {2}\, A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\left (-1575 \cos \left (d x +c \right )^{2}-3150 \cos \left (d x +c \right )-1575\right ) \sqrt {2}\, B \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\left (1995 \cos \left (d x +c \right )^{2}+3990 \cos \left (d x +c \right )+1995\right ) \sqrt {2}\, C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\left (-1330 \cos \left (d x +c \right )^{2}-840 \cos \left (d x +c \right )+280\right ) A \tan \left (d x +c \right )+\left (2058 \cos \left (d x +c \right )^{3}+1512 \cos \left (d x +c \right )^{2}-168 \cos \left (d x +c \right )+168\right ) B \tan \left (d x +c \right ) \sec \left (d x +c \right )+\left (-2402 \cos \left (d x +c \right )^{4}-1608 \cos \left (d x +c \right )^{3}+392 \cos \left (d x +c \right )^{2}-72 \cos \left (d x +c \right )+120\right ) C \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}\right )}{420 d \,a^{2} \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right )}\)	\(413\)
parts	\(\frac {A \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (-38 \cos \left (d x +c \right )^{2}-24 \cos \left (d x +c \right )+8\right ) \tan \left (d x +c \right )+\left (33 \cos \left (d x +c \right )^{2}+66 \cos \left (d x +c \right )+33\right ) \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{12 d \,a^{2} \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right )}-\frac {B \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (-98 \cos \left (d x +c \right )^{3}-72 \cos \left (d x +c \right )^{2}+8 \cos \left (d x +c \right )-8\right ) \sec \left (d x +c \right ) \tan \left (d x +c \right )+\left (75 \cos \left (d x +c \right )^{2}+150 \cos \left (d x +c \right )+75\right ) \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{20 d \,a^{2} \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right )}+\frac {C \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (-2402 \cos \left (d x +c \right )^{4}-1608 \cos \left (d x +c \right )^{3}+392 \cos \left (d x +c \right )^{2}-72 \cos \left (d x +c \right )+120\right ) \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}+\left (1995 \cos \left (d x +c \right )^{2}+3990 \cos \left (d x +c \right )+1995\right ) \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{420 d \,a^{2} \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right )}\)	\(493\)

Input:

int(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, 
method=_RETURNVERBOSE)
 

Output:

1/420/d/a^2*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)^2+2*cos(d*x+c)+1)*((1155* 
cos(d*x+c)^2+2310*cos(d*x+c)+1155)*2^(1/2)*A*(-cos(d*x+c)/(cos(d*x+c)+1))^ 
(1/2)*ln((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-cot(d*x+c)+csc(d*x+c))+(-157 
5*cos(d*x+c)^2-3150*cos(d*x+c)-1575)*2^(1/2)*B*(-cos(d*x+c)/(cos(d*x+c)+1) 
)^(1/2)*ln((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-cot(d*x+c)+csc(d*x+c))+(19 
95*cos(d*x+c)^2+3990*cos(d*x+c)+1995)*2^(1/2)*C*(-cos(d*x+c)/(cos(d*x+c)+1 
))^(1/2)*ln((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-cot(d*x+c)+csc(d*x+c))+(- 
1330*cos(d*x+c)^2-840*cos(d*x+c)+280)*A*tan(d*x+c)+(2058*cos(d*x+c)^3+1512 
*cos(d*x+c)^2-168*cos(d*x+c)+168)*B*tan(d*x+c)*sec(d*x+c)+(-2402*cos(d*x+c 
)^4-1608*cos(d*x+c)^3+392*cos(d*x+c)^2-72*cos(d*x+c)+120)*C*tan(d*x+c)*sec 
(d*x+c)^2)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 575, normalized size of antiderivative = 2.08 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3 
/2),x, algorithm="fricas")
 

Output:

[-1/840*(105*sqrt(2)*((11*A - 15*B + 19*C)*cos(d*x + c)^5 + 2*(11*A - 15*B 
 + 19*C)*cos(d*x + c)^4 + (11*A - 15*B + 19*C)*cos(d*x + c)^3)*sqrt(-a)*lo 
g((2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c) 
*sin(d*x + c) + 3*a*cos(d*x + c)^2 + 2*a*cos(d*x + c) - a)/(cos(d*x + c)^2 
 + 2*cos(d*x + c) + 1)) + 4*((665*A - 1029*B + 1201*C)*cos(d*x + c)^4 + 12 
*(35*A - 63*B + 67*C)*cos(d*x + c)^3 - 28*(5*A - 3*B + 7*C)*cos(d*x + c)^2 
 - 12*(7*B - 3*C)*cos(d*x + c) - 60*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + 
 c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^5 + 2*a^2*d*cos(d*x + c)^4 + a^2*d* 
cos(d*x + c)^3), -1/420*(105*sqrt(2)*((11*A - 15*B + 19*C)*cos(d*x + c)^5 
+ 2*(11*A - 15*B + 19*C)*cos(d*x + c)^4 + (11*A - 15*B + 19*C)*cos(d*x + c 
)^3)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d* 
x + c)/(sqrt(a)*sin(d*x + c))) + 2*((665*A - 1029*B + 1201*C)*cos(d*x + c) 
^4 + 12*(35*A - 63*B + 67*C)*cos(d*x + c)^3 - 28*(5*A - 3*B + 7*C)*cos(d*x 
 + c)^2 - 12*(7*B - 3*C)*cos(d*x + c) - 60*C)*sqrt((a*cos(d*x + c) + a)/co 
s(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^5 + 2*a^2*d*cos(d*x + c)^4 + 
 a^2*d*cos(d*x + c)^3)]
 

Sympy [F]

\[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(sec(d*x+c)**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))* 
*(3/2),x)
 

Output:

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**4/(a*(sec( 
c + d*x) + 1))**(3/2), x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3 
/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [A] (verification not implemented)

Time = 1.19 (sec) , antiderivative size = 448, normalized size of antiderivative = 1.62 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3 
/2),x, algorithm="giac")
 

Output:

-1/420*(105*(11*sqrt(2)*A - 15*sqrt(2)*B + 19*sqrt(2)*C)*log(abs(-sqrt(-a) 
*tan(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/(sqrt(-a)*a* 
sgn(cos(d*x + c))) - ((((105*(sqrt(2)*A*a^5*sgn(cos(d*x + c)) - sqrt(2)*B* 
a^5*sgn(cos(d*x + c)) + sqrt(2)*C*a^5*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2 
*c)^2/a^3 - 4*(455*sqrt(2)*A*a^5*sgn(cos(d*x + c)) - 693*sqrt(2)*B*a^5*sgn 
(cos(d*x + c)) + 877*sqrt(2)*C*a^5*sgn(cos(d*x + c)))/a^3)*tan(1/2*d*x + 1 
/2*c)^2 + 14*(305*sqrt(2)*A*a^5*sgn(cos(d*x + c)) - 453*sqrt(2)*B*a^5*sgn( 
cos(d*x + c)) + 517*sqrt(2)*C*a^5*sgn(cos(d*x + c)))/a^3)*tan(1/2*d*x + 1/ 
2*c)^2 - 140*(25*sqrt(2)*A*a^5*sgn(cos(d*x + c)) - 39*sqrt(2)*B*a^5*sgn(co 
s(d*x + c)) + 47*sqrt(2)*C*a^5*sgn(cos(d*x + c)))/a^3)*tan(1/2*d*x + 1/2*c 
)^2 + 105*(9*sqrt(2)*A*a^5*sgn(cos(d*x + c)) - 17*sqrt(2)*B*a^5*sgn(cos(d* 
x + c)) + 17*sqrt(2)*C*a^5*sgn(cos(d*x + c)))/a^3)*tan(1/2*d*x + 1/2*c)/(( 
a*tan(1/2*d*x + 1/2*c)^2 - a)^3*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^4\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^4*(a + a/cos(c + 
 d*x))^(3/2)),x)
 

Output:

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^4*(a + a/cos(c + 
 d*x))^(3/2)), x)
 

Reduce [F]

\[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{6}}{\sec \left (d x +c \right )^{2}+2 \sec \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{5}}{\sec \left (d x +c \right )^{2}+2 \sec \left (d x +c \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}}{\sec \left (d x +c \right )^{2}+2 \sec \left (d x +c \right )+1}d x \right ) a \right )}{a^{2}} \] Input:

int(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x)
 

Output:

(sqrt(a)*(int((sqrt(sec(c + d*x) + 1)*sec(c + d*x)**6)/(sec(c + d*x)**2 + 
2*sec(c + d*x) + 1),x)*c + int((sqrt(sec(c + d*x) + 1)*sec(c + d*x)**5)/(s 
ec(c + d*x)**2 + 2*sec(c + d*x) + 1),x)*b + int((sqrt(sec(c + d*x) + 1)*se 
c(c + d*x)**4)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1),x)*a))/a**2