\(\int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\) [601]

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

Optimal result

Integrand size = 45, antiderivative size = 222 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 a^{5/2} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a^3 (160 A+224 B+245 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (40 A+56 B+35 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}+\frac {2 a (5 A+7 B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)} \] Output:

2*a^(5/2)*C*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+2/105*a^3 
*(160*A+224*B+245*C)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+ 
2/105*a^2*(40*A+56*B+35*C)*(a+a*sec(d*x+c))^(1/2)*sin(d*x+c)/d/sec(d*x+c)^ 
(1/2)+2/35*a*(5*A+7*B)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(3/2 
)+2/7*A*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d/sec(d*x+c)^(5/2)
 

Mathematica [A] (warning: unable to verify)

Time = 7.13 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.65 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {a^3 \left ((1040 A+1246 B+1120 C+(505 A+392 B+140 C) \cos (c+d x)+6 (20 A+7 B) \cos (2 (c+d x))+15 A \cos (3 (c+d x))) \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))} \sin (c+d x)+420 C \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)\right )}{210 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

Integrate[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x] 
^2))/Sec[c + d*x]^(7/2),x]
 

Output:

(a^3*((1040*A + 1246*B + 1120*C + (505*A + 392*B + 140*C)*Cos[c + d*x] + 6 
*(20*A + 7*B)*Cos[2*(c + d*x)] + 15*A*Cos[3*(c + d*x)])*Sqrt[-((-1 + Sec[c 
 + d*x])*Sec[c + d*x])]*Sin[c + d*x] + 420*C*ArcSin[Sqrt[1 - Sec[c + d*x]] 
]*Tan[c + d*x]))/(210*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 1.45 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.08, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.311, Rules used = {3042, 4574, 27, 3042, 4505, 27, 3042, 4505, 27, 3042, 4503, 3042, 4288, 222}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4574

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4505

\(\displaystyle \frac {\frac {2}{5} \int \frac {(\sec (c+d x) a+a)^{3/2} \left ((40 A+56 B+35 C) a^2+35 C \sec (c+d x) a^2\right )}{2 \sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a^2 (5 A+7 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} \int \frac {(\sec (c+d x) a+a)^{3/2} \left ((40 A+56 B+35 C) a^2+35 C \sec (c+d x) a^2\right )}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a^2 (5 A+7 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left ((40 A+56 B+35 C) a^2+35 C \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a^2 (5 A+7 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 4505

\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((160 A+224 B+245 C) a^3+105 C \sec (c+d x) a^3\right )}{2 \sqrt {\sec (c+d x)}}dx+\frac {2 a^3 (40 A+56 B+35 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a^2 (5 A+7 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((160 A+224 B+245 C) a^3+105 C \sec (c+d x) a^3\right )}{\sqrt {\sec (c+d x)}}dx+\frac {2 a^3 (40 A+56 B+35 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a^2 (5 A+7 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((160 A+224 B+245 C) a^3+105 C \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^3 (40 A+56 B+35 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a^2 (5 A+7 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 4503

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (105 a^3 C \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^4 (160 A+224 B+245 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (40 A+56 B+35 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a^2 (5 A+7 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (105 a^3 C \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^4 (160 A+224 B+245 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (40 A+56 B+35 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a^2 (5 A+7 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 4288

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (\frac {2 a^4 (160 A+224 B+245 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {210 a^3 C \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {2 a^3 (40 A+56 B+35 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a^2 (5 A+7 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {\frac {2 a^2 (5 A+7 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{5} \left (\frac {2 a^3 (40 A+56 B+35 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}+\frac {1}{3} \left (\frac {210 a^{7/2} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^4 (160 A+224 B+245 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )\right )}{7 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\)

Input:

Int[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/S 
ec[c + d*x]^(7/2),x]
 

Output:

(2*A*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + ( 
(2*a^2*(5*A + 7*B)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(5*d*Sec[c + d 
*x]^(3/2)) + ((2*a^3*(40*A + 56*B + 35*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + 
 d*x])/(3*d*Sqrt[Sec[c + d*x]]) + ((210*a^(7/2)*C*ArcSinh[(Sqrt[a]*Tan[c + 
 d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^4*(160*A + 224*B + 245*C)*Sqrt[ 
Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/3)/5)/(7*a)
 

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 222
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt 
[a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
 

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

rule 4288
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
+ (a_)], x_Symbol] :> Simp[-2*(a/(b*f))*Sqrt[a*(d/b)]   Subst[Int[1/Sqrt[1 
+ x^2/a], x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; FreeQ[{a 
, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[a*(d/b), 0]
 

rule 4503
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*b^2*Co 
t[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp 
[(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n)   Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[ 
e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a 
*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && LtQ[n, 0]
 

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]
 

rule 4574
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*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e 
 + f*x])^n/(f*n)), x] - Simp[1/(b*d*n)   Int[(a + b*Csc[e + f*x])^m*(d*Csc[ 
e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x] 
, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] 
&&  !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
 
Maple [A] (warning: unable to verify)

Time = 6.78 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.12

method result size
default \(\frac {a^{2} \left (\left (105 \cos \left (d x +c \right )+105\right ) \sqrt {2}\, C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\left (105 \cos \left (d x +c \right )+105\right ) \sqrt {2}\, C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\left (60 \cos \left (d x +c \right )^{3}+240 \cos \left (d x +c \right )^{2}+460 \cos \left (d x +c \right )+920\right ) \sin \left (d x +c \right ) A +\left (84 \cos \left (d x +c \right )^{2}+392 \cos \left (d x +c \right )+1204\right ) \sin \left (d x +c \right ) B +\left (140 \cos \left (d x +c \right )+1120\right ) \sin \left (d x +c \right ) C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{210 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\sec \left (d x +c \right )}}\) \(248\)
parts \(\frac {2 A \,a^{2} \left (3 \cos \left (d x +c \right )^{3}+12 \cos \left (d x +c \right )^{2}+23 \cos \left (d x +c \right )+46\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{21 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {B \left (6 \sin \left (d x +c \right )+28 \tan \left (d x +c \right )+86 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right ) a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \left (15 \cos \left (d x +c \right )+15\right ) \sec \left (d x +c \right )^{\frac {5}{2}}}+\frac {C \,a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (4 \sin \left (d x +c \right )+32 \tan \left (d x +c \right )+\sqrt {2}\, \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (-3-3 \sec \left (d x +c \right )\right )+\sqrt {2}\, \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \left (-3-3 \sec \left (d x +c \right )\right )\right )}{6 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\) \(326\)

Input:

int((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2 
),x,method=_RETURNVERBOSE)
 

Output:

1/210/d*a^2*((105*cos(d*x+c)+105)*2^(1/2)*C*(-2/(cos(d*x+c)+1))^(1/2)*arct 
an(1/2*(-cot(d*x+c)+csc(d*x+c)-1)/(-1/(cos(d*x+c)+1))^(1/2))+(105*cos(d*x+ 
c)+105)*2^(1/2)*C*(-2/(cos(d*x+c)+1))^(1/2)*arctan(1/2*(-cot(d*x+c)+csc(d* 
x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))+(60*cos(d*x+c)^3+240*cos(d*x+c)^2+460*c 
os(d*x+c)+920)*sin(d*x+c)*A+(84*cos(d*x+c)^2+392*cos(d*x+c)+1204)*sin(d*x+ 
c)*B+(140*cos(d*x+c)+1120)*sin(d*x+c)*C)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x 
+c)+1)/sec(d*x+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 473, normalized size of antiderivative = 2.13 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\left [\frac {105 \, {\left (C a^{2} \cos \left (d x + c\right ) + C a^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \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 )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (15 \, A a^{2} \cos \left (d x + c\right )^{4} + 3 \, {\left (20 \, A + 7 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (115 \, A + 98 \, B + 35 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (230 \, A + 301 \, B + 280 \, C\right )} a^{2} \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 )}}}{210 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, \frac {105 \, {\left (C a^{2} \cos \left (d x + c\right ) + C a^{2}\right )} \sqrt {-a} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{2 \, a \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left (15 \, A a^{2} \cos \left (d x + c\right )^{4} + 3 \, {\left (20 \, A + 7 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (115 \, A + 98 \, B + 35 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (230 \, A + 301 \, B + 280 \, C\right )} a^{2} \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 )}}}{105 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \] Input:

integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c 
)^(7/2),x, algorithm="fricas")
 

Output:

[1/210*(105*(C*a^2*cos(d*x + c) + C*a^2)*sqrt(a)*log((a*cos(d*x + c)^3 - 7 
*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt(a)*sqrt((a*co 
s(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos( 
d*x + c)^3 + cos(d*x + c)^2)) + 4*(15*A*a^2*cos(d*x + c)^4 + 3*(20*A + 7*B 
)*a^2*cos(d*x + c)^3 + (115*A + 98*B + 35*C)*a^2*cos(d*x + c)^2 + (230*A + 
 301*B + 280*C)*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))* 
sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d), 1/105*(105*(C*a^2*c 
os(d*x + c) + C*a^2)*sqrt(-a)*arctan(1/2*(cos(d*x + c)^2 - 2*cos(d*x + c)) 
*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))/(a*sqrt(cos(d*x + c))*si 
n(d*x + c))) + 2*(15*A*a^2*cos(d*x + c)^4 + 3*(20*A + 7*B)*a^2*cos(d*x + c 
)^3 + (115*A + 98*B + 35*C)*a^2*cos(d*x + c)^2 + (230*A + 301*B + 280*C)*a 
^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt 
(cos(d*x + c)))/(d*cos(d*x + c) + d)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x 
+c)**(7/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 976 vs. \(2 (192) = 384\).

Time = 0.37 (sec) , antiderivative size = 976, normalized size of antiderivative = 4.40 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c 
)^(7/2),x, algorithm="maxima")
 

Output:

1/840*(5*sqrt(2)*(315*a^2*cos(6/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d* 
x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 77*a^2*cos(4/7*arctan2(sin(7/2*d*x + 7 
/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 21*a^2*cos(2/7*arctan 
2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) - 315* 
a^2*cos(7/2*d*x + 7/2*c)*sin(6/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x 
 + 7/2*c))) - 77*a^2*cos(7/2*d*x + 7/2*c)*sin(4/7*arctan2(sin(7/2*d*x + 7/ 
2*c), cos(7/2*d*x + 7/2*c))) - 21*a^2*cos(7/2*d*x + 7/2*c)*sin(2/7*arctan2 
(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 6*a^2*sin(7/2*d*x + 7/2*c) 
 + 21*a^2*sin(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 7 
7*a^2*sin(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 315*a 
^2*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))))*A*sqrt(a) 
 + 70*sqrt(2)*(30*a^2*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 
3/2*c)))*sin(3/2*d*x + 3/2*c) - 30*a^2*cos(3/2*d*x + 3/2*c)*sin(2/3*arctan 
2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 3*sqrt(2)*a^2*log(2*cos(1 
/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arct 
an2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sqrt(2)*cos(1/3*arc 
tan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*sqrt(2)*sin(1/3*arct 
an2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) - 3*sqrt(2)*a^2*log( 
2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1 
/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sqrt(2)*c...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 385 vs. \(2 (192) = 384\).

Time = 2.90 (sec) , antiderivative size = 385, normalized size of antiderivative = 1.73 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {\frac {105 \, C a^{\frac {7}{2}} \log \left (\frac {{\left | 2 \, {\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 )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\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 )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{{\left | a \right |}} + \frac {2 \, {\left (420 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 420 \, \sqrt {2} B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 315 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (700 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 980 \, \sqrt {2} B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 875 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (560 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 784 \, \sqrt {2} B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 805 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (160 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 224 \, \sqrt {2} B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 245 \, \sqrt {2} C a^{6} \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 )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {7}{2}}}}{105 \, d} \] Input:

integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c 
)^(7/2),x, algorithm="giac")
 

Output:

1/105*(105*C*a^(7/2)*log(abs(2*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan( 
1/2*d*x + 1/2*c)^2 + a))^2 - 4*sqrt(2)*abs(a) - 6*a)/abs(2*(sqrt(a)*tan(1/ 
2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 4*sqrt(2)*abs(a) 
- 6*a))*sgn(cos(d*x + c))/abs(a) + 2*(420*sqrt(2)*A*a^6*sgn(cos(d*x + c)) 
+ 420*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + 315*sqrt(2)*C*a^6*sgn(cos(d*x + c) 
) + (700*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 980*sqrt(2)*B*a^6*sgn(cos(d*x + 
 c)) + 875*sqrt(2)*C*a^6*sgn(cos(d*x + c)) + (560*sqrt(2)*A*a^6*sgn(cos(d* 
x + c)) + 784*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + 805*sqrt(2)*C*a^6*sgn(cos( 
d*x + c)) + (160*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 224*sqrt(2)*B*a^6*sgn(c 
os(d*x + c)) + 245*sqrt(2)*C*a^6*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)/(a* 
tan(1/2*d*x + 1/2*c)^2 + a)^(7/2))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}} \,d x \] Input:

int(((a + a/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/( 
1/cos(c + d*x))^(7/2),x)
                                                                                    
                                                                                    
 

Output:

int(((a + a/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/( 
1/cos(c + d*x))^(7/2), x)
 

Reduce [F]

\[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\sqrt {a}\, a^{2} \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{4}}d x \right ) a +2 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{3}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{3}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{2}}d x \right ) a +2 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{2}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{2}}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )}d x \right ) b +2 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}d x \right ) c \right ) \] Input:

int((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2 
),x)
 

Output:

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