\(\int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^2} \, dx\) [565]

Optimal result

Integrand size = 43, antiderivative size = 251 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {(A-4 B+7 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}+\frac {(2 A-5 B+10 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}-\frac {(A-4 B+7 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 d}+\frac {(2 A-5 B+10 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d}-\frac {(A-4 B+7 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2} \] Output:

(A-4*B+7*C)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d*x 
+c)^(1/2)/a^2/d+1/3*(2*A-5*B+10*C)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d* 
x+1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/a^2/d-(A-4*B+7*C)*sec(d*x+c)^(1/2)*sin(d 
*x+c)/a^2/d+1/3*(2*A-5*B+10*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/a^2/d-1/3*(A-4* 
B+7*C)*sec(d*x+c)^(5/2)*sin(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3*(A-B+C)*sec(d* 
x+c)^(7/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^2
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 8.67 (sec) , antiderivative size = 1347, normalized size of antiderivative = 5.37 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(-2*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^( 
(2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + 
 d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4 
, -E^((2*I)*(c + d*x))])*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)) 
/(3*d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*S 
ec[c + d*x])^2) + (8*Sqrt[2]*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x 
)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*(-3*Sqrt[ 
1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric 
2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*(A + B*Sec[c + d*x] + C 
*Sec[c + d*x]^2))/(3*d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 
 2*d*x])*(a + a*Sec[c + d*x])^2) - (14*Sqrt[2]*C*Sqrt[E^(I*(c + d*x))/(1 + 
 E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^4* 
Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)* 
c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*(A + 
B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(3*d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + 
d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) + (8*A*Cos[c/2 + (d*x)/ 
2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sqrt[S 
ec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sin[c])/(3*d*(A + 2*C 
 + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) - (20*B* 
Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2,...
 

Rubi [A] (verified)

Time = 1.30 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.96, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.349, Rules used = {3042, 4572, 27, 3042, 4507, 27, 3042, 4274, 3042, 4255, 3042, 4258, 3042, 3119, 3120}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4572

\(\displaystyle \frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) (a (A+5 B-5 C)+3 a (A-B+3 C) \sec (c+d x))}{2 (\sec (c+d x) a+a)}dx}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) (a (A+5 B-5 C)+3 a (A-B+3 C) \sec (c+d x))}{\sec (c+d x) a+a}dx}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a (A+5 B-5 C)+3 a (A-B+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 4507

\(\displaystyle \frac {\frac {\int -3 \sec ^{\frac {3}{2}}(c+d x) \left (a^2 (A-4 B+7 C)-a^2 (2 A-5 B+10 C) \sec (c+d x)\right )dx}{a^2}-\frac {2 (A-4 B+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {3 \int \sec ^{\frac {3}{2}}(c+d x) \left (a^2 (A-4 B+7 C)-a^2 (2 A-5 B+10 C) \sec (c+d x)\right )dx}{a^2}-\frac {2 (A-4 B+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {3 \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a^2 (A-4 B+7 C)-a^2 (2 A-5 B+10 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}-\frac {2 (A-4 B+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {-\frac {3 \left (a^2 (A-4 B+7 C) \int \sec ^{\frac {3}{2}}(c+d x)dx-a^2 (2 A-5 B+10 C) \int \sec ^{\frac {5}{2}}(c+d x)dx\right )}{a^2}-\frac {2 (A-4 B+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {3 \left (a^2 (A-4 B+7 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx-a^2 (2 A-5 B+10 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx\right )}{a^2}-\frac {2 (A-4 B+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {-\frac {3 \left (a^2 (A-4 B+7 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )-a^2 (2 A-5 B+10 C) \left (\frac {1}{3} \int \sqrt {\sec (c+d x)}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{a^2}-\frac {2 (A-4 B+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {3 \left (a^2 (A-4 B+7 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-a^2 (2 A-5 B+10 C) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{a^2}-\frac {2 (A-4 B+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {-\frac {3 \left (a^2 (A-4 B+7 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )-a^2 (2 A-5 B+10 C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{a^2}-\frac {2 (A-4 B+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {3 \left (a^2 (A-4 B+7 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-a^2 (2 A-5 B+10 C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{a^2}-\frac {2 (A-4 B+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {-\frac {3 \left (a^2 (A-4 B+7 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-a^2 (2 A-5 B+10 C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{a^2}-\frac {2 (A-4 B+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {-\frac {3 \left (a^2 (A-4 B+7 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-a^2 (2 A-5 B+10 C) \left (\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )\right )}{a^2}-\frac {2 (A-4 B+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

Input:

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

Output:

-1/3*((A - B + C)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]) 
^2) + ((-2*(A - 4*B + 7*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(d*(1 + Sec[c 
+ d*x])) - (3*(a^2*(A - 4*B + 7*C)*((-2*Sqrt[Cos[c + d*x]]*EllipticE[(c + 
d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d) 
- a^2*(2*A - 5*B + 10*C)*((2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]* 
Sqrt[Sec[c + d*x]])/(3*d) + (2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d))))/a 
^2)/(6*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 3042

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

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 4255

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

rule 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

rule 4274

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

rule 4507

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(723\) vs. \(2(230)=460\).

Time = 30.70 (sec) , antiderivative size = 724, normalized size of antiderivative = 2.88

method	result	size
default	\(\text {Expression too large to display}\)	\(724\)

Input:

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

Output:

-1/2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a^2*((-2*B+ 
4*C)*(cos(1/2*d*x+1/2*c)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c 
)^2-1)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+ 
1/2*c),2^(1/2)))-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)/cos(1/2*d*x+ 
1/2*c)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+(4*B-8*C)/sin( 
1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1 
/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(2*sin(1 
/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d* 
x+1/2*c)^2)^(1/2))+1/3*(A-B+C)*(2*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/ 
2*d*x+1/2*c)^2)^(1/2)*(2*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE 
(cos(1/2*d*x+1/2*c),2^(1/2)))*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-2*(2 
*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*EllipticF(c 
os(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/ 
2*d*x+1/2*c)-12*sin(1/2*d*x+1/2*c)^6+20*sin(1/2*d*x+1/2*c)^4-7*sin(1/2*d*x 
+1/2*c)^2)/cos(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^ 
2)^(1/2)/(sin(1/2*d*x+1/2*c)^2-1)+4*C*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2 
*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3 
*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/ 
2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^ 
(1/2))))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 464, normalized size of antiderivative = 1.85 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

1/6*((sqrt(2)*(-2*I*A + 5*I*B - 10*I*C)*cos(d*x + c)^3 - 2*sqrt(2)*(2*I*A 
- 5*I*B + 10*I*C)*cos(d*x + c)^2 + sqrt(2)*(-2*I*A + 5*I*B - 10*I*C)*cos(d 
*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt 
(2)*(2*I*A - 5*I*B + 10*I*C)*cos(d*x + c)^3 - 2*sqrt(2)*(-2*I*A + 5*I*B - 
10*I*C)*cos(d*x + c)^2 + sqrt(2)*(2*I*A - 5*I*B + 10*I*C)*cos(d*x + c))*we 
ierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*(sqrt(2)*(-I*A 
 + 4*I*B - 7*I*C)*cos(d*x + c)^3 + 2*sqrt(2)*(-I*A + 4*I*B - 7*I*C)*cos(d* 
x + c)^2 + sqrt(2)*(-I*A + 4*I*B - 7*I*C)*cos(d*x + c))*weierstrassZeta(-4 
, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*(sqrt( 
2)*(I*A - 4*I*B + 7*I*C)*cos(d*x + c)^3 + 2*sqrt(2)*(I*A - 4*I*B + 7*I*C)* 
cos(d*x + c)^2 + sqrt(2)*(I*A - 4*I*B + 7*I*C)*cos(d*x + c))*weierstrassZe 
ta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*( 
3*(A - 4*B + 7*C)*cos(d*x + c)^3 + (4*A - 19*B + 32*C)*cos(d*x + c)^2 - 2* 
(3*B - 4*C)*cos(d*x + c) - 2*C)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*co 
s(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))
 

Sympy [F(-1)]

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

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

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(int((sqrt(sec(c + d*x))*sec(c + d*x)**4)/(sec(c + d*x)**2 + 2*sec(c + d*x 
) + 1),x)*c + int((sqrt(sec(c + d*x))*sec(c + d*x)**3)/(sec(c + d*x)**2 + 
2*sec(c + d*x) + 1),x)*b + int((sqrt(sec(c + d*x))*sec(c + d*x)**2)/(sec(c 
 + d*x)**2 + 2*sec(c + d*x) + 1),x)*a)/a**2