\(\int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) [190]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [C] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 33, antiderivative size = 166 \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {4 a^2 (4 A+5 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^2 (A+2 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 a^2 (7 A+5 B) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 A \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)} \] Output:

4/5*a^2*(4*A+5*B)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*s 
ec(d*x+c)^(1/2)/d+4/3*a^2*(A+2*B)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x 
+1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/d+2/15*a^2*(7*A+5*B)*sin(d*x+c)/d/sec(d*x 
+c)^(1/2)+2/5*A*(a^2+a^2*sec(d*x+c))*sin(d*x+c)/d/sec(d*x+c)^(3/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 2.84 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.92 \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {a^2 \sqrt {\sec (c+d x)} \left (20 (A+2 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-4 i (4 A+5 B) e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+\cos (c+d x) (48 i A+60 i B+10 (2 A+B) \sin (c+d x)+3 A \sin (2 (c+d x)))\right )}{15 d} \] Input:

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

Output:

(a^2*Sqrt[Sec[c + d*x]]*(20*(A + 2*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d* 
x)/2, 2] - (4*I)*(4*A + 5*B)*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))] 
*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))] + Cos[c + d*x]*((4 
8*I)*A + (60*I)*B + 10*(2*A + B)*Sin[c + d*x] + 3*A*Sin[2*(c + d*x)])))/(1 
5*d)
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.04, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3042, 4505, 27, 3042, 4484, 25, 3042, 4274, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4505

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4484

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 4484
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*a*Cot[e + 
f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Simp[1/(d*n)   Int[(d*Csc[e + f*x])^( 
n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] 
/; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]
 

rule 4505
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot 
[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] - Sim 
p[b/(a*d*n)   Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Sim 
p[a*A*(m - n - 1) - b*B*n - (a*B*n + 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 
] && GtQ[m, 1/2] && LtQ[n, -1]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(356\) vs. \(2(149)=298\).

Time = 6.87 (sec) , antiderivative size = 357, normalized size of antiderivative = 2.15

method result size
default \(-\frac {4 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, a^{2} \left (-12 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (32 A +10 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-13 A -5 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}-12 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}+10 B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}-15 B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\right )}{15 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(357\)
parts \(\frac {2 \left (A \,a^{2}+2 B \,a^{2}\right ) \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}}\, \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 {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {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}}\, \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}-\frac {2 \left (2 A \,a^{2}+B \,a^{2}\right ) \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 (4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\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}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\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}}\, \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}-\frac {2 A \,a^{2} \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 (-8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 \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}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\right )}{5 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \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}-\frac {2 B \,a^{2} \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}}\, \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 )}{\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \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}\) \(677\)

Input:

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

Output:

-4/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(-12*A*c 
os(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+(32*A+10*B)*sin(1/2*d*x+1/2*c)^4*co 
s(1/2*d*x+1/2*c)+(-13*A-5*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+5*A*E 
llipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1 
/2*d*x+1/2*c)^2-1)^(1/2)-12*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1 
/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)+10*B*EllipticF(cos 
(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c 
)^2-1)^(1/2)-15*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c 
)^2)^(1/2)*(2*sin(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)/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 = 0.09 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.13 \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (A + 2 \, B\right )} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (A + 2 \, B\right )} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} {\left (4 \, A + 5 \, B\right )} a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} {\left (4 \, A + 5 \, B\right )} a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (3 \, A a^{2} \cos \left (d x + c\right )^{2} + 5 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{15 \, d} \] Input:

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

Output:

-2/15*(5*I*sqrt(2)*(A + 2*B)*a^2*weierstrassPInverse(-4, 0, cos(d*x + c) + 
 I*sin(d*x + c)) - 5*I*sqrt(2)*(A + 2*B)*a^2*weierstrassPInverse(-4, 0, co 
s(d*x + c) - I*sin(d*x + c)) - 3*I*sqrt(2)*(4*A + 5*B)*a^2*weierstrassZeta 
(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*I*s 
qrt(2)*(4*A + 5*B)*a^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, c 
os(d*x + c) - I*sin(d*x + c))) - (3*A*a^2*cos(d*x + c)^2 + 5*(2*A + B)*a^2 
*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/d
 

Sympy [F]

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

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

Output:

a**2*(Integral(A/sec(c + d*x)**(5/2), x) + Integral(2*A/sec(c + d*x)**(3/2 
), x) + Integral(A/sqrt(sec(c + d*x)), x) + Integral(B/sec(c + d*x)**(3/2) 
, x) + Integral(2*B/sqrt(sec(c + d*x)), x) + Integral(B*sqrt(sec(c + d*x)) 
, x))
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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