3.7.22 \(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx\) [622]

3.7.22.1 Optimal result
3.7.22.2 Mathematica [B] (verified)
3.7.22.3 Rubi [A] (verified)
3.7.22.4 Maple [B] (warning: unable to verify)
3.7.22.5 Fricas [A] (verification not implemented)
3.7.22.6 Sympy [F(-1)]
3.7.22.7 Maxima [B] (verification not implemented)
3.7.22.8 Giac [F]
3.7.22.9 Mupad [F(-1)]

3.7.22.1 Optimal result

Integrand size = 45, antiderivative size = 211 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=-\frac {(75 A-19 B-5 C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(13 A-5 B-3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(49 A-9 B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}} \]

output
-1/32*(75*A-19*B-5*C)*arctanh(1/2*sin(d*x+c)*a^(1/2)*sec(d*x+c)^(1/2)*2^(1 
/2)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)-1/4*(A-B+C)*sin(d*x+c)*sec(d 
*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(5/2)-1/16*(13*A-5*B-3*C)*sin(d*x+c)*sec(d* 
x+c)^(1/2)/a/d/(a+a*sec(d*x+c))^(3/2)+1/16*(49*A-9*B+C)*sin(d*x+c)*sec(d*x 
+c)^(1/2)/a^2/d/(a+a*sec(d*x+c))^(1/2)
 
3.7.22.2 Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(653\) vs. \(2(211)=422\).

Time = 7.45 (sec) , antiderivative size = 653, normalized size of antiderivative = 3.09 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=-\frac {A \sqrt {1+\sec (c+d x)} \left (\frac {8 \sqrt {\sec (c+d x)} \sin (c+d x)}{d (1+\sec (c+d x))^{5/2}}+\frac {26 \sqrt {\sec (c+d x)} \sin (c+d x)}{d (1+\sec (c+d x))^{3/2}}-\frac {98 \sqrt {\sec (c+d x)} \sin (c+d x)}{d \sqrt {1+\sec (c+d x)}}-\frac {75 \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}\right )}{32 a^2 \sqrt {a (1+\sec (c+d x))}}-\frac {B \sqrt {1+\sec (c+d x)} \left (\frac {8 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{d (1+\sec (c+d x))^{5/2}}+\frac {18 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{d (1+\sec (c+d x))^{3/2}}+\frac {19 \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}\right )}{32 a^2 \sqrt {a (1+\sec (c+d x))}}-\frac {C \sqrt {1+\sec (c+d x)} \left (\frac {8 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (1+\sec (c+d x))^{5/2}}+5 \left (\frac {2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (1+\sec (c+d x))^{3/2}}-\frac {\left (2 \arcsin \left (\sqrt {\sec (c+d x)}\right )-\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}-\frac {2 \left (\arcsin \left (\sqrt {1-\sec (c+d x)}\right )+\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)}\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}\right )\right )}{32 a^2 \sqrt {a (1+\sec (c+d x))}} \]

input
Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sqrt[Sec[c + d*x]]*(a + 
 a*Sec[c + d*x])^(5/2)),x]
 
output
-1/32*(A*Sqrt[1 + Sec[c + d*x]]*((8*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(1 
 + Sec[c + d*x])^(5/2)) + (26*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(1 + Sec 
[c + d*x])^(3/2)) - (98*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[1 + Sec[c 
 + d*x]]) - (75*Sqrt[2]*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c 
 + d*x]]]*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]])) 
)/(a^2*Sqrt[a*(1 + Sec[c + d*x])]) - (B*Sqrt[1 + Sec[c + d*x]]*((8*Sec[c + 
 d*x]^(3/2)*Sin[c + d*x])/(d*(1 + Sec[c + d*x])^(5/2)) + (18*Sec[c + d*x]^ 
(3/2)*Sin[c + d*x])/(d*(1 + Sec[c + d*x])^(3/2)) + (19*Sqrt[2]*ArcTan[(Sqr 
t[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x])/(d*Sqrt[1 - 
 Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]])))/(32*a^2*Sqrt[a*(1 + Sec[c + d*x]) 
]) - (C*Sqrt[1 + Sec[c + d*x]]*((8*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(d*(1 
+ Sec[c + d*x])^(5/2)) + 5*((2*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(d*(1 + Se 
c[c + d*x])^(3/2)) - ((2*ArcSin[Sqrt[Sec[c + d*x]]] - Sqrt[2]*ArcTan[(Sqrt 
[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]])*Tan[c + d*x])/(d*Sqrt[1 - 
 Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]) - (2*(ArcSin[Sqrt[1 - Sec[c + d*x]] 
] + Sqrt[1 - Sec[c + d*x]]*Sqrt[Sec[c + d*x]])*Tan[c + d*x])/(d*Sqrt[1 - S 
ec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]))))/(32*a^2*Sqrt[a*(1 + Sec[c + d*x])] 
)
 
3.7.22.3 Rubi [A] (verified)

Time = 1.15 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.244, Rules used = {3042, 4572, 27, 3042, 4508, 27, 3042, 4501, 3042, 4295, 219}

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.

\(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4572

\(\displaystyle \frac {\int \frac {a (9 A-B+C)-4 a (A-B-C) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (9 A-B+C)-4 a (A-B-C) \sec (c+d x)}{\sqrt {\sec (c+d x)} (\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (9 A-B+C)-4 a (A-B-C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4508

\(\displaystyle \frac {\frac {\int \frac {a^2 (49 A-9 B+C)-2 a^2 (13 A-5 B-3 C) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {a^2 (49 A-9 B+C)-2 a^2 (13 A-5 B-3 C) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {a^2 (49 A-9 B+C)-2 a^2 (13 A-5 B-3 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4501

\(\displaystyle \frac {\frac {\frac {2 a^2 (49 A-9 B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-a^2 (75 A-19 B-5 C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 a^2 (49 A-9 B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-a^2 (75 A-19 B-5 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4295

\(\displaystyle \frac {\frac {\frac {2 a^2 (75 A-19 B-5 C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^2 (49 A-9 B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {2 a^2 (49 A-9 B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {\sqrt {2} a^{3/2} (75 A-19 B-5 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\)

input
Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sqrt[Sec[c + d*x]]*(a + a*Sec 
[c + d*x])^(5/2)),x]
 
output
-1/4*((A - B + C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]) 
^(5/2)) + (-1/2*(a*(13*A - 5*B - 3*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d* 
(a + a*Sec[c + d*x])^(3/2)) + (-((Sqrt[2]*a^(3/2)*(75*A - 19*B - 5*C)*ArcT 
anh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + 
d*x]])])/d) + (2*a^2*(49*A - 9*B + C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d* 
Sqrt[a + a*Sec[c + d*x]]))/(4*a^2))/(8*a^2)
 

3.7.22.3.1 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 219
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4295
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
+ (a_)], x_Symbol] :> Simp[-2*b*(d/(a*f))   Subst[Int[1/(2*b - d*x^2), x], 
x, b*(Cot[e + f*x]/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]]))], x] /; 
 FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0]
 

rule 4501
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e 
 + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[(a*A*m 
 - b*B*n)/(b*d*n)   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x] 
, x] /; FreeQ[{a, b, d, e, f, A, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a 
^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -1]
 

rule 4508
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b 
- a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(2*m + 
 1))), x] - Simp[1/(a^2*(2*m + 1))   Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Cs 
c[e + f*x])^n*Simp[b*B*n - a*A*(2*m + n + 1) + (A*b - a*B)*(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] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]
 

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)]
 
3.7.22.4 Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(499\) vs. \(2(180)=360\).

Time = 1.54 (sec) , antiderivative size = 500, normalized size of antiderivative = 2.37

method result size
default \(-\frac {\left (2 A \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-2 B \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+2 C \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-17 A \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+9 B \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-C \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+75 A \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )}{\sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right ) \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}-19 \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )}{\sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right ) \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, B -5 C \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )}{\sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right ) \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}-83 A \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+11 B \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-3 C \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right ) \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}}{32 a^{3} d \sqrt {-\frac {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+1}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}}\) \(500\)
parts \(-\frac {A \left (2 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-17 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+75 \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \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 ) \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}+83 \cot \left (d x +c \right )-83 \csc \left (d x +c \right )\right ) \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}}{32 d \,a^{3} \sqrt {-\frac {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+1}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}}+\frac {B \sqrt {-\frac {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+1}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right ) \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (-2 \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+11 \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+19 \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \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 )\right )}{32 d \,a^{3} \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}-\frac {C \left (-\frac {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+1}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )^{\frac {3}{2}} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{2} \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (2 \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-3 \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+5 \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \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 )\right )}{32 d \,a^{3} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+1\right ) \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\) \(806\)

input
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2 
),x,method=_RETURNVERBOSE)
 
output
-1/32/a^3/d*(2*A*(1-cos(d*x+c))^5*csc(d*x+c)^5-2*B*(1-cos(d*x+c))^5*csc(d* 
x+c)^5+2*C*(1-cos(d*x+c))^5*csc(d*x+c)^5-17*A*(1-cos(d*x+c))^3*csc(d*x+c)^ 
3+9*B*(1-cos(d*x+c))^3*csc(d*x+c)^3-C*(1-cos(d*x+c))^3*csc(d*x+c)^3+75*A*a 
rctan(1/(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c))) 
*(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)-19*arctan(1/(-(1-cos(d*x+c))^2*c 
sc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c)))*(-(1-cos(d*x+c))^2*csc(d*x+ 
c)^2-1)^(1/2)*B-5*C*arctan(1/(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-co 
t(d*x+c)+csc(d*x+c)))*(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)-83*A*(-cot( 
d*x+c)+csc(d*x+c))+11*B*(-cot(d*x+c)+csc(d*x+c))-3*C*(-cot(d*x+c)+csc(d*x+ 
c)))*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)/(-((1-cos(d*x+c))^2*cs 
c(d*x+c)^2+1)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)
 
3.7.22.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 556, normalized size of antiderivative = 2.64 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=\left [-\frac {\sqrt {2} {\left ({\left (75 \, A - 19 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (75 \, A - 19 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (75 \, A - 19 \, B - 5 \, C\right )} \cos \left (d x + c\right ) + 75 \, A - 19 \, B - 5 \, C\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - \frac {4 \, {\left (32 \, A \cos \left (d x + c\right )^{3} + {\left (85 \, A - 13 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (49 \, A - 9 \, B + C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, \frac {\sqrt {2} {\left ({\left (75 \, A - 19 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (75 \, A - 19 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (75 \, A - 19 \, B - 5 \, C\right )} \cos \left (d x + c\right ) + 75 \, A - 19 \, B - 5 \, C\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )}}{a \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left (32 \, A \cos \left (d x + c\right )^{3} + {\left (85 \, A - 13 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (49 \, A - 9 \, B + C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \]

input
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c) 
)^(5/2),x, algorithm="fricas")
 
output
[-1/64*(sqrt(2)*((75*A - 19*B - 5*C)*cos(d*x + c)^3 + 3*(75*A - 19*B - 5*C 
)*cos(d*x + c)^2 + 3*(75*A - 19*B - 5*C)*cos(d*x + c) + 75*A - 19*B - 5*C) 
*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + 
 a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a 
)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 4*(32*A*cos(d*x + c)^3 + (85*A 
- 13*B + 5*C)*cos(d*x + c)^2 + (49*A - 9*B + C)*cos(d*x + c))*sqrt((a*cos( 
d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d* 
x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), 1/32*(s 
qrt(2)*((75*A - 19*B - 5*C)*cos(d*x + c)^3 + 3*(75*A - 19*B - 5*C)*cos(d*x 
 + c)^2 + 3*(75*A - 19*B - 5*C)*cos(d*x + c) + 75*A - 19*B - 5*C)*sqrt(-a) 
*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos( 
d*x + c))/(a*sin(d*x + c))) + 2*(32*A*cos(d*x + c)^3 + (85*A - 13*B + 5*C) 
*cos(d*x + c)^2 + (49*A - 9*B + C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a) 
/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3* 
a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)]
 
3.7.22.6 Sympy [F(-1)]

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

input
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(1/2)/(a+a*sec(d*x+ 
c))**(5/2),x)
 
output
Timed out
 
3.7.22.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 264379 vs. \(2 (180) = 360\).

Time = 3.44 (sec) , antiderivative size = 264379, normalized size of antiderivative = 1252.98 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

input
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c) 
)^(5/2),x, algorithm="maxima")
 
output
-1/32*((576*(75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*si 
n(1/2*d*x + 1/2*c) + 1) - 75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/ 
2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 64*sin(1/2*d*x + 1/2*c))*cos(5/2*d* 
x + 5/2*c)^6 + 14400*(75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c) 
^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - 75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2 
*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 64*sin(1/2*d*x + 1/2*c))*c 
os(3/2*d*x + 3/2*c)^6 + 187500*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 
 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin 
(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(1/2*d*x + 1/2*c)^6 
+ 576*(75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2* 
d*x + 1/2*c) + 1) - 75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 
 - 2*sin(1/2*d*x + 1/2*c) + 1) - 64*sin(1/2*d*x + 1/2*c))*sin(5/2*d*x + 5/ 
2*c)^6 + 5184*(75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2* 
sin(1/2*d*x + 1/2*c) + 1) - 75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 
1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 64*sin(1/2*d*x + 1/2*c))*sin(3/2* 
d*x + 3/2*c)^6 + 262500*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c) 
^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d* 
x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(1/2*d*x + 1/2*c)^4*sin(1/2 
*d*x + 1/2*c)^2 + 77700*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c) 
^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2...
 
3.7.22.8 Giac [F]

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

input
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c) 
)^(5/2),x, algorithm="giac")
 
output
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^(5 
/2)*sqrt(sec(d*x + c))), x)
 
3.7.22.9 Mupad [F(-1)]

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

input
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/((a + a/cos(c + d*x))^(5/2)*(1 
/cos(c + d*x))^(1/2)),x)
 
output
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/((a + a/cos(c + d*x))^(5/2)*(1 
/cos(c + d*x))^(1/2)), x)