3.4.84 \(\int \frac {\sec ^4(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\) [384]

3.4.84.1 Optimal result
3.4.84.2 Mathematica [A] (verified)
3.4.84.3 Rubi [A] (verified)
3.4.84.4 Maple [A] (warning: unable to verify)
3.4.84.5 Fricas [A] (verification not implemented)
3.4.84.6 Sympy [F]
3.4.84.7 Maxima [F]
3.4.84.8 Giac [A] (verification not implemented)
3.4.84.9 Mupad [F(-1)]

3.4.84.1 Optimal result

Integrand size = 42, antiderivative size = 243 \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {2} (B-C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}-\frac {4 (111 B-143 C) \tan (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}-\frac {2 (3 B-19 C) \sec ^2(c+d x) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (9 B-C) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (93 B-29 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 a d} \]

output
(B-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 
)/d/a^(1/2)-4/315*(111*B-143*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-2/105* 
(3*B-19*C)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/63*(9*B-C)*s 
ec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/9*C*sec(d*x+c)^4*tan(d*x 
+c)/d/(a+a*sec(d*x+c))^(1/2)+2/315*(93*B-29*C)*(a+a*sec(d*x+c))^(1/2)*tan( 
d*x+c)/a/d
 
3.4.84.2 Mathematica [A] (verified)

Time = 0.85 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.75 \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\left (315 \sqrt {2} (B-C) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )+\frac {1}{4} (-423 B+1279 C+(918 B-214 C) \cos (c+d x)-8 (69 B-157 C) \cos (2 (c+d x))+186 B \cos (3 (c+d x))-58 C \cos (3 (c+d x))-129 B \cos (4 (c+d x))+257 C \cos (4 (c+d x))) \sqrt {1-\sec (c+d x)} \sec ^4(c+d x)\right ) \tan (c+d x)}{315 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]

input
Integrate[(Sec[c + d*x]^4*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + a* 
Sec[c + d*x]],x]
 
output
((315*Sqrt[2]*(B - C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]] + ((-423*B + 
 1279*C + (918*B - 214*C)*Cos[c + d*x] - 8*(69*B - 157*C)*Cos[2*(c + d*x)] 
 + 186*B*Cos[3*(c + d*x)] - 58*C*Cos[3*(c + d*x)] - 129*B*Cos[4*(c + d*x)] 
 + 257*C*Cos[4*(c + d*x)])*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^4)/4)*Tan[c 
 + d*x])/(315*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
 
3.4.84.3 Rubi [A] (verified)

Time = 1.85 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.16, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {3042, 4560, 3042, 4509, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4560

\(\displaystyle \int \frac {\sec ^5(c+d x) (B+C \sec (c+d x))}{\sqrt {a \sec (c+d x)+a}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^5 \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a}}dx\)

\(\Big \downarrow \) 4509

\(\displaystyle \frac {2 \int \frac {\sec ^4(c+d x) (8 a C+a (9 B-C) \sec (c+d x))}{2 \sqrt {\sec (c+d x) a+a}}dx}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sec ^4(c+d x) (8 a C+a (9 B-C) \sec (c+d x))}{\sqrt {\sec (c+d x) a+a}}dx}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (8 a C+a (9 B-C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4509

\(\displaystyle \frac {\frac {2 \int \frac {3 \sec ^3(c+d x) \left (2 a^2 (9 B-C)-a^2 (3 B-19 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 \int \frac {\sec ^3(c+d x) \left (2 a^2 (9 B-C)-a^2 (3 B-19 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (2 a^2 (9 B-C)-a^2 (3 B-19 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4509

\(\displaystyle \frac {\frac {3 \left (\frac {2 \int -\frac {\sec ^2(c+d x) \left (4 a^3 (3 B-19 C)-a^3 (93 B-29 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{5 a}-\frac {2 a^2 (3 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\right )}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 \left (-\frac {\int \frac {\sec ^2(c+d x) \left (4 a^3 (3 B-19 C)-a^3 (93 B-29 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{5 a}-\frac {2 a^2 (3 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\right )}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 \left (-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (4 a^3 (3 B-19 C)-a^3 (93 B-29 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}-\frac {2 a^2 (3 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\right )}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4498

\(\displaystyle \frac {\frac {3 \left (-\frac {\frac {2 \int -\frac {\sec (c+d x) \left (a^4 (93 B-29 C)-2 a^4 (111 B-143 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 a^2 (93 B-29 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (3 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\right )}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 \left (-\frac {-\frac {\int \frac {\sec (c+d x) \left (a^4 (93 B-29 C)-2 a^4 (111 B-143 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 a^2 (93 B-29 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (3 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\right )}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 \left (-\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a^4 (93 B-29 C)-2 a^4 (111 B-143 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}-\frac {2 a^2 (93 B-29 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (3 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\right )}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4489

\(\displaystyle \frac {\frac {3 \left (-\frac {-\frac {315 a^4 (B-C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx-\frac {4 a^4 (111 B-143 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a^2 (93 B-29 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (3 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\right )}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 \left (-\frac {-\frac {315 a^4 (B-C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {4 a^4 (111 B-143 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a^2 (93 B-29 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (3 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\right )}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {\frac {3 \left (-\frac {-\frac {-\frac {630 a^4 (B-C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {4 a^4 (111 B-143 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a^2 (93 B-29 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (3 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\right )}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {3 \left (-\frac {2 a^2 (3 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}-\frac {-\frac {2 a^2 (93 B-29 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}-\frac {\frac {315 \sqrt {2} a^{7/2} (B-C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {4 a^4 (111 B-143 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}}{5 a}\right )}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\)

input
Int[(Sec[c + d*x]^4*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + a*Sec[c 
+ d*x]],x]
 
output
(2*C*Sec[c + d*x]^4*Tan[c + d*x])/(9*d*Sqrt[a + a*Sec[c + d*x]]) + ((2*a*( 
9*B - C)*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt[a + a*Sec[c + d*x]]) + (3* 
((-2*a^2*(3*B - 19*C)*Sec[c + d*x]^2*Tan[c + d*x])/(5*d*Sqrt[a + a*Sec[c + 
 d*x]]) - ((-2*a^2*(93*B - 29*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3 
*d) - ((315*Sqrt[2]*a^(7/2)*(B - C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2] 
*Sqrt[a + a*Sec[c + d*x]])])/d - (4*a^4*(111*B - 143*C)*Tan[c + d*x])/(d*S 
qrt[a + a*Sec[c + d*x]]))/(3*a))/(5*a)))/(7*a))/(9*a)
 

3.4.84.3.1 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 4560
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] :> Simp[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]
 
3.4.84.4 Maple [A] (warning: unable to verify)

Time = 0.98 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.67

method result size
default \(\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (315 B \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {9}{2}}-315 C \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {9}{2}} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )-552 B \left (1-\cos \left (d x +c \right )\right )^{9} \csc \left (d x +c \right )^{9}+766 C \left (1-\cos \left (d x +c \right )\right )^{9} \csc \left (d x +c \right )^{9}+1224 B \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}-1872 C \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}-1512 B \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+3276 C \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+840 B \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-1680 C \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+630 C \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{315 d a \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{4}}\) \(405\)
parts \(\frac {B \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (105 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {7}{2}}-184 \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}+224 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-280 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}\right )}{105 d a \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{3}}-\frac {C \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (315 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {9}{2}}-766 \left (1-\cos \left (d x +c \right )\right )^{9} \csc \left (d x +c \right )^{9}+1872 \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}-3276 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+1680 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+630 \cot \left (d x +c \right )-630 \csc \left (d x +c \right )\right )}{315 d a \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{4}}\) \(438\)

input
int(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x,me 
thod=_RETURNVERBOSE)
 
output
1/315/d/a*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(315*B*ln(csc(d*x 
+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))*((1-cos(d*x+c))^2* 
csc(d*x+c)^2-1)^(9/2)-315*C*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(9/2)*ln(csc 
(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))-552*B*(1-cos(d 
*x+c))^9*csc(d*x+c)^9+766*C*(1-cos(d*x+c))^9*csc(d*x+c)^9+1224*B*(1-cos(d* 
x+c))^7*csc(d*x+c)^7-1872*C*(1-cos(d*x+c))^7*csc(d*x+c)^7-1512*B*(1-cos(d* 
x+c))^5*csc(d*x+c)^5+3276*C*(1-cos(d*x+c))^5*csc(d*x+c)^5+840*B*(1-cos(d*x 
+c))^3*csc(d*x+c)^3-1680*C*(1-cos(d*x+c))^3*csc(d*x+c)^3+630*C*(-cot(d*x+c 
)+csc(d*x+c)))/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^4
 
3.4.84.5 Fricas [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 465, normalized size of antiderivative = 1.91 \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [-\frac {315 \, \sqrt {2} {\left ({\left (B - C\right )} a \cos \left (d x + c\right )^{5} + {\left (B - C\right )} a \cos \left (d x + c\right )^{4}\right )} \sqrt {-\frac {1}{a}} \log \left (\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (129 \, B - 257 \, C\right )} \cos \left (d x + c\right )^{4} - {\left (93 \, B - 29 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, B - 19 \, C\right )} \cos \left (d x + c\right )^{2} - 5 \, {\left (9 \, B - C\right )} \cos \left (d x + c\right ) - 35 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{630 \, {\left (a d \cos \left (d x + c\right )^{5} + a d \cos \left (d x + c\right )^{4}\right )}}, -\frac {2 \, {\left ({\left (129 \, B - 257 \, C\right )} \cos \left (d x + c\right )^{4} - {\left (93 \, B - 29 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, B - 19 \, C\right )} \cos \left (d x + c\right )^{2} - 5 \, {\left (9 \, B - C\right )} \cos \left (d x + c\right ) - 35 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) + \frac {315 \, \sqrt {2} {\left ({\left (B - C\right )} a \cos \left (d x + c\right )^{5} + {\left (B - C\right )} a \cos \left (d x + c\right )^{4}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}}}{315 \, {\left (a d \cos \left (d x + c\right )^{5} + a d \cos \left (d x + c\right )^{4}\right )}}\right ] \]

input
integrate(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2 
),x, algorithm="fricas")
 
output
[-1/630*(315*sqrt(2)*((B - C)*a*cos(d*x + c)^5 + (B - C)*a*cos(d*x + c)^4) 
*sqrt(-1/a)*log((2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1 
/a)*cos(d*x + c)*sin(d*x + c) + 3*cos(d*x + c)^2 + 2*cos(d*x + c) - 1)/(co 
s(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*((129*B - 257*C)*cos(d*x + c)^4 - 
(93*B - 29*C)*cos(d*x + c)^3 + 3*(3*B - 19*C)*cos(d*x + c)^2 - 5*(9*B - C) 
*cos(d*x + c) - 35*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) 
)/(a*d*cos(d*x + c)^5 + a*d*cos(d*x + c)^4), -1/315*(2*((129*B - 257*C)*co 
s(d*x + c)^4 - (93*B - 29*C)*cos(d*x + c)^3 + 3*(3*B - 19*C)*cos(d*x + c)^ 
2 - 5*(9*B - C)*cos(d*x + c) - 35*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c 
))*sin(d*x + c) + 315*sqrt(2)*((B - C)*a*cos(d*x + c)^5 + (B - C)*a*cos(d* 
x + c)^4)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + 
 c)/(sqrt(a)*sin(d*x + c)))/sqrt(a))/(a*d*cos(d*x + c)^5 + a*d*cos(d*x + c 
)^4)]
 
3.4.84.6 Sympy [F]

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

input
integrate(sec(d*x+c)**4*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**( 
1/2),x)
 
output
Integral((B + C*sec(c + d*x))*sec(c + d*x)**5/sqrt(a*(sec(c + d*x) + 1)), 
x)
 
3.4.84.7 Maxima [F]

\[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sec \left (d x + c\right )^{4}}{\sqrt {a \sec \left (d x + c\right ) + a}} \,d x } \]

input
integrate(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2 
),x, algorithm="maxima")
 
output
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*sec(d*x + c)^4/sqrt(a*sec(d* 
x + c) + a), x)
 
3.4.84.8 Giac [A] (verification not implemented)

Time = 1.43 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.32 \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\frac {315 \, {\left (\sqrt {2} B - \sqrt {2} C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {2 \, {\left (\frac {315 \, \sqrt {2} C a^{4}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + {\left (420 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 840 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (756 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 1638 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (612 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 936 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (276 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 383 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{4} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{315 \, d} \]

input
integrate(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2 
),x, algorithm="giac")
 
output
-1/315*(315*(sqrt(2)*B - 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)*sgn(cos(d*x + c))) - 2* 
(315*sqrt(2)*C*a^4/sgn(cos(d*x + c)) + (420*sqrt(2)*B*a^4*sgn(cos(d*x + c) 
) - 840*sqrt(2)*C*a^4*sgn(cos(d*x + c)) - (756*sqrt(2)*B*a^4*sgn(cos(d*x + 
 c)) - 1638*sqrt(2)*C*a^4*sgn(cos(d*x + c)) - (612*sqrt(2)*B*a^4*sgn(cos(d 
*x + c)) - 936*sqrt(2)*C*a^4*sgn(cos(d*x + c)) - (276*sqrt(2)*B*a^4*sgn(co 
s(d*x + c)) - 383*sqrt(2)*C*a^4*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)*t 
an(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^4*sqrt(-a*tan(1/2*d*x 
+ 1/2*c)^2 + a)))/d
 
3.4.84.9 Mupad [F(-1)]

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

input
int((B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^4*(a + a/cos(c + d*x 
))^(1/2)),x)
 
output
int((B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^4*(a + a/cos(c + d*x 
))^(1/2)), x)