\(\int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{a+a \sec (c+d x)} \, dx\) [559]

Optimal result

Integrand size = 43, antiderivative size = 205 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {(A-3 B+3 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {(3 A-3 B+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a d}-\frac {(A-3 B+3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{a d}+\frac {(3 A-3 B+5 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}-\frac {(A-B+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))} \] Output:

(A-3*B+3*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/d+1/3*(3*A-3*B+5*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/d-(A-3*B+3*C)*sec(d*x+c)^(1/2)*sin(d*x+c) 
/a/d+1/3*(3*A-3*B+5*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/a/d-(A-B+C)*sec(d*x+c)^ 
(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))
 

Mathematica [C] (warning: unable to verify)

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

Time = 8.34 (sec) , antiderivative size = 1261, normalized size of antiderivative = 6.15 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/3*(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]^2*Cos[c + d*x]*Csc[c/2]*(-3*Sqrt[1 
+ E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F 
1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*(A + B*Sec[c + d*x] + C*S 
ec[c + d*x]^2))/(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])) + (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]^2*Cos[c + d 
*x]*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))/(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])) - (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]^2*Cos[c + d*x]*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))/(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*A*Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d* 
x)/2, 2]*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sin[c])/(d*(A + 
2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[Sec[c + d*x]]*(a + a*Sec 
[c + d*x])) - (2*B*Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*Csc[c/2]*Ell...
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.93, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.279, Rules used = {3042, 4572, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4572

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

-(((A - B + C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]))) 
- (a*(A - 3*B + 3*C)*((-2*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqr 
t[Sec[c + d*x]])/d + (2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d) - a*(3*A - 3*B 
 + 5*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)))/(2*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 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. \(466\) vs. \(2(190)=380\).

Time = 4.28 (sec) , antiderivative size = 467, normalized size of antiderivative = 2.28

method	result	size
default	\(-\frac {\sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\frac {\left (2 B -2 C \right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}+\frac {\left (A -B +C \right ) \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \left (\operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}+2 C \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{6 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{2}}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}\right )\right )}{a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(467\)

Input:

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

Output:

-1/a*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*B-2*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+s 
in(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*s 
in(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))+(A-B+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)+2*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 = 0.10 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.63 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {{\left (\sqrt {2} {\left (-3 i \, A + 3 i \, B - 5 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-3 i \, A + 3 i \, B - 5 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (3 i \, A - 3 i \, B + 5 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (3 i \, A - 3 i \, B + 5 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, {\left (\sqrt {2} {\left (-i \, A + 3 i \, B - 3 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-i \, A + 3 i \, B - 3 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, {\left (\sqrt {2} {\left (i \, A - 3 i \, B + 3 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (i \, A - 3 i \, B + 3 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (3 \, {\left (A - 3 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (3 \, B - 2 \, C\right )} \cos \left (d x + c\right ) - 2 \, C\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}} \] Input:

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

Output:

1/6*((sqrt(2)*(-3*I*A + 3*I*B - 5*I*C)*cos(d*x + c)^2 + sqrt(2)*(-3*I*A + 
3*I*B - 5*I*C)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*s 
in(d*x + c)) + (sqrt(2)*(3*I*A - 3*I*B + 5*I*C)*cos(d*x + c)^2 + sqrt(2)*( 
3*I*A - 3*I*B + 5*I*C)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + 
c) - I*sin(d*x + c)) - 3*(sqrt(2)*(-I*A + 3*I*B - 3*I*C)*cos(d*x + c)^2 + 
sqrt(2)*(-I*A + 3*I*B - 3*I*C)*cos(d*x + c))*weierstrassZeta(-4, 0, weiers 
trassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*(sqrt(2)*(I*A - 3 
*I*B + 3*I*C)*cos(d*x + c)^2 + sqrt(2)*(I*A - 3*I*B + 3*I*C)*cos(d*x + c)) 
*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d* 
x + c))) - 2*(3*(A - 3*B + 3*C)*cos(d*x + c)^2 - 2*(3*B - 2*C)*cos(d*x + c 
) - 2*C)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos(d*x + c)^2 + a*d*cos(d* 
x + c))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

int(((1/cos(c + d*x))^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + 
a/cos(c + d*x)),x)
 

Output:

int(((1/cos(c + d*x))^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + 
a/cos(c + d*x)), x)
 

Reduce [F]

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

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

Output:

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