\(\int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {\sec (c+d x)}} \, dx\) [589]

Optimal result
Mathematica [A] (warning: unable to verify)
Rubi [A] (verified)
Maple [B] (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 = 183 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {a^{3/2} (8 A+12 B+7 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}+\frac {a^2 (8 A-4 B-5 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {a (4 B+3 C) \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d} \] Output:

1/4*a^(3/2)*(8*A+12*B+7*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/ 
2))/d+1/4*a^2*(8*A-4*B-5*C)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*sec(d*x+c)) 
^(1/2)+1/4*a*(4*B+3*C)*sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(1/2)*sin(d*x+c)/ 
d+1/2*C*sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d
 

Mathematica [A] (warning: unable to verify)

Time = 2.16 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.78 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {a^2 \left ((8 A+7 C) \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)-12 B \arcsin \left (\sqrt {\sec (c+d x)}\right ) \tan (c+d x)+\sqrt {-((-1+\sec (c+d x)) \sec (c+d x))} (8 A \sin (c+d x)+(4 B+7 C+2 C \sec (c+d x)) \tan (c+d x))\right )}{4 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(a^2*((8*A + 7*C)*ArcSin[Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x] - 12*B*ArcSi 
n[Sqrt[Sec[c + d*x]]]*Tan[c + d*x] + Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d* 
x])]*(8*A*Sin[c + d*x] + (4*B + 7*C + 2*C*Sec[c + d*x])*Tan[c + d*x])))/(4 
*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 1.08 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.244, Rules used = {3042, 4576, 27, 3042, 4506, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4576

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4506

\(\displaystyle \frac {\int \frac {\sqrt {\sec (c+d x) a+a} \left ((8 A-4 B-5 C) a^2+(8 A+12 B+7 C) \sec (c+d x) a^2\right )}{2 \sqrt {\sec (c+d x)}}dx+\frac {a^2 (4 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{2} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((8 A-4 B-5 C) a^2+(8 A+12 B+7 C) \sec (c+d x) a^2\right )}{\sqrt {\sec (c+d x)}}dx+\frac {a^2 (4 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((8 A-4 B-5 C) a^2+(8 A+12 B+7 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a^2 (4 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\)

\(\Big \downarrow \) 4503

\(\displaystyle \frac {\frac {1}{2} \left (a^2 (8 A+12 B+7 C) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^3 (8 A-4 B-5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (4 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} \left (a^2 (8 A+12 B+7 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^3 (8 A-4 B-5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (4 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\)

\(\Big \downarrow \) 4288

\(\displaystyle \frac {\frac {1}{2} \left (\frac {2 a^3 (8 A-4 B-5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {2 a^2 (8 A+12 B+7 C) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a^2 (4 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {\frac {a^2 (4 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}+\frac {1}{2} \left (\frac {2 a^{5/2} (8 A+12 B+7 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^3 (8 A-4 B-5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )}{4 a}+\frac {C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d}\)

Input:

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

Output:

(C*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(2*d) + ((a 
^2*(4*B + 3*C)*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/d 
 + ((2*a^(5/2)*(8*A + 12*B + 7*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + 
a*Sec[c + d*x]]])/d + (2*a^3*(8*A - 4*B - 5*C)*Sqrt[Sec[c + d*x]]*Sin[c + 
d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/2)/(4*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 4506
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* 
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), 
 x] + Simp[1/(d*(m + n))   Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] 
)^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*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] && GtQ[m, 1/2] &&  !LtQ[n, -1]
 

rule 4576
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[(-C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Cs 
c[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1))   Int[(a + b*Cs 
c[e + f*x])^m*(d*Csc[e + f*x])^n*Simp[A*b*(m + n + 1) + b*C*n + (a*C*m + b* 
B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m 
, n}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] &&  !LtQ[n, -2^(-1)] && 
NeQ[m + n + 1, 0]
 
Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(446\) vs. \(2(157)=314\).

Time = 12.52 (sec) , antiderivative size = 447, normalized size of antiderivative = 2.44

method result size
default \(\frac {a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (C \left (28 \tan \left (d x +c \right )+8 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )+32 A \sin \left (d x +c \right )+16 B \tan \left (d x +c \right )+8 \sqrt {2}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) A +12 \sqrt {2}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) B +7 \sqrt {2}\, \left (\cos \left (d x +c \right )+1\right ) C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+8 \sqrt {2}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) A +12 \sqrt {2}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) B +7 \sqrt {2}\, \left (\cos \left (d x +c \right )+1\right ) C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )\right )}{16 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\sec \left (d x +c \right )}}\) \(447\)
parts \(\frac {A \left (\frac {\sqrt {2}\, \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}}{2}-\frac {\sqrt {2}\, \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}}{2}-2 \cot \left (d x +c \right )+2 \csc \left (d x +c \right )\right ) a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \sqrt {\sec \left (d x +c \right )}}+\frac {B \left (\sqrt {2}\, \sin \left (d x +c \right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}-3 \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+3 \cos \left (d x +c \right ) \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )\right ) a \sqrt {\sec \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{2 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}-\frac {C a \left (-7 \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+7 \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{2}+\sin \left (d x +c \right ) \left (-7 \cos \left (d x +c \right )-2\right ) \sqrt {2}\, \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {3}{2}}}{8 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\) \(479\)

Input:

int((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2 
),x,method=_RETURNVERBOSE)
 

Output:

1/16/d*a*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)/sec(d*x+c)^(1/2)*(C*(28*t 
an(d*x+c)+8*sec(d*x+c)*tan(d*x+c))+32*A*sin(d*x+c)+16*B*tan(d*x+c)+8*2^(1/ 
2)*(cos(d*x+c)+1)*(-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))*A+12*2^(1/2)*(cos(d*x+c)+1)*(-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))*B+7*2^(1/2)*(cos(d*x+c)+1)*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))+8*2^(1/2)*(cos(d*x+c)+ 
1)*(-2/(cos(d*x+c)+1))^(1/2)*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)-1)/(-1/(co 
s(d*x+c)+1))^(1/2))*A+12*2^(1/2)*(cos(d*x+c)+1)*(-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))*B+7*2^(1/ 
2)*(cos(d*x+c)+1)*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)))
 

Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 443, normalized size of antiderivative = 2.42 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\left [\frac {{\left ({\left (8 \, A + 12 \, B + 7 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (8 \, A + 12 \, B + 7 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (8 \, A a \cos \left (d x + c\right )^{2} + {\left (4 \, B + 7 \, C\right )} a \cos \left (d x + c\right ) + 2 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{16 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, \frac {{\left ({\left (8 \, A + 12 \, B + 7 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (8 \, A + 12 \, B + 7 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {-a} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{2 \, a \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left (8 \, A a \cos \left (d x + c\right )^{2} + {\left (4 \, B + 7 \, C\right )} a \cos \left (d x + c\right ) + 2 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{8 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \] Input:

integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c 
)^(1/2),x, algorithm="fricas")
 

Output:

[1/16*(((8*A + 12*B + 7*C)*a*cos(d*x + c)^2 + (8*A + 12*B + 7*C)*a*cos(d*x 
 + c))*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*cos(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* 
(8*A*a*cos(d*x + c)^2 + (4*B + 7*C)*a*cos(d*x + c) + 2*C*a)*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) 
^2 + d*cos(d*x + c)), 1/8*(((8*A + 12*B + 7*C)*a*cos(d*x + c)^2 + (8*A + 1 
2*B + 7*C)*a*cos(d*x + c))*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))*sin(d*x + c))) + 2*(8*A*a*cos(d*x + c)^2 + (4*B + 7*C)*a*cos(d*x + c) 
+ 2*C*a)*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)^2 + d*cos(d*x + c))]
 

Sympy [F(-1)]

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

integrate((a+a*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x 
+c)**(1/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3661 vs. \(2 (157) = 314\).

Time = 0.42 (sec) , antiderivative size = 3661, normalized size of antiderivative = 20.01 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c 
)^(1/2),x, algorithm="maxima")
 

Output:

1/16*(4*sqrt(2)*(sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 
1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) 
 + 2) - sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 
+ 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + s 
qrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt( 
2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a* 
log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/ 
2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 8*a*sin(1/2*d*x + 1 
/2*c))*A*sqrt(a) + 4*(3*(a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 
1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) 
 + 2) - a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt 
(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + a*log(2*c 
os(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 
 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a*log(2*cos(1/2*d*x + 1/2* 
c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt( 
2)*sin(1/2*d*x + 1/2*c) + 2))*cos(2*d*x + 2*c)^2 + 3*(a*log(2*cos(1/2*d*x 
+ 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2 
*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*si 
n(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2* 
d*x + 1/2*c) + 2) + a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 625 vs. \(2 (157) = 314\).

Time = 3.93 (sec) , antiderivative size = 625, normalized size of antiderivative = 3.42 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c 
)^(1/2),x, algorithm="giac")
 

Output:

1/8*(16*sqrt(2)*A*a^2*sgn(cos(d*x + c))*tan(1/2*d*x + 1/2*c)/sqrt(a*tan(1/ 
2*d*x + 1/2*c)^2 + a) + (8*A*a^(3/2)*sgn(cos(d*x + c)) + 12*B*a^(3/2)*sgn( 
cos(d*x + c)) + 7*C*a^(3/2)*sgn(cos(d*x + c)))*log(abs((sqrt(a)*tan(1/2*d* 
x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a*(2*sqrt(2) + 3))) - 
 (8*A*a^(3/2)*sgn(cos(d*x + c)) + 12*B*a^(3/2)*sgn(cos(d*x + c)) + 7*C*a^( 
3/2)*sgn(cos(d*x + c)))*log(abs((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan 
(1/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3))) + 4*sqrt(2)*(12*(sqrt(a) 
*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*B*a^(5/2)*sg 
n(cos(d*x + c)) + 7*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1 
/2*c)^2 + a))^6*C*a^(5/2)*sgn(cos(d*x + c)) - 76*(sqrt(a)*tan(1/2*d*x + 1/ 
2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*B*a^(7/2)*sgn(cos(d*x + c)) - 
 95*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4* 
C*a^(7/2)*sgn(cos(d*x + c)) + 36*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*ta 
n(1/2*d*x + 1/2*c)^2 + a))^2*B*a^(9/2)*sgn(cos(d*x + c)) + 53*(sqrt(a)*tan 
(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*C*a^(9/2)*sgn(co 
s(d*x + c)) - 4*B*a^(11/2)*sgn(cos(d*x + c)) - 5*C*a^(11/2)*sgn(cos(d*x + 
c)))/((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^ 
4 - 6*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^ 
2*a + a^2)^2)/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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