\(\int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [590]

Optimal result

Integrand size = 45, antiderivative size = 177 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a^{3/2} (2 B+3 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {a^2 (8 A+6 B-3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}-\frac {a (2 A-3 C) \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \] Output:

a^(3/2)*(2*B+3*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+1/3 
*a^2*(8*A+6*B-3*C)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-1/ 
3*a*(2*A-3*C)*sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(1/2)*sin(d*x+c)/d+2/3*A*( 
a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 5.93 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.89 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a^2 \left (6 B \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)-9 C \arcsin \left (\sqrt {\sec (c+d x)}\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)+\sqrt {1-\sec (c+d x)} (2 A \sin (c+d x)+(10 A+6 B+3 C \sec (c+d x)) \tan (c+d x))\right )}{3 d \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(a^2*(6*B*ArcSin[Sqrt[1 - Sec[c + d*x]]]*Sec[c + d*x]^(3/2)*Sin[c + d*x] - 
 9*C*ArcSin[Sqrt[Sec[c + d*x]]]*Sec[c + d*x]^(3/2)*Sin[c + d*x] + Sqrt[1 - 
 Sec[c + d*x]]*(2*A*Sin[c + d*x] + (10*A + 6*B + 3*C*Sec[c + d*x])*Tan[c + 
 d*x])))/(3*d*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])]*Sqrt[a*(1 + Sec[c 
+ d*x])])
 

Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.244, Rules used = {3042, 4574, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4574

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4506

\(\displaystyle \frac {\int \frac {\sqrt {\sec (c+d x) a+a} \left ((8 A+6 B-3 C) a^2+3 (2 B+3 C) \sec (c+d x) a^2\right )}{2 \sqrt {\sec (c+d x)}}dx-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{2} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((8 A+6 B-3 C) a^2+3 (2 B+3 C) \sec (c+d x) a^2\right )}{\sqrt {\sec (c+d x)}}dx-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((8 A+6 B-3 C) a^2+3 (2 B+3 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 4503

\(\displaystyle \frac {\frac {1}{2} \left (3 a^2 (2 B+3 C) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^3 (8 A+6 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} \left (3 a^2 (2 B+3 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^3 (8 A+6 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 4288

\(\displaystyle \frac {\frac {1}{2} \left (\frac {2 a^3 (8 A+6 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {6 a^2 (2 B+3 C) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {\frac {1}{2} \left (\frac {6 a^{5/2} (2 B+3 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^3 (8 A+6 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )-\frac {a^2 (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}\)

Input:

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

Output:

(2*A*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + ( 
-((a^2*(2*A - 3*C)*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x 
])/d) + ((6*a^(5/2)*(2*B + 3*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a* 
Sec[c + d*x]]])/d + (2*a^3*(8*A + 6*B - 3*C)*Sqrt[Sec[c + d*x]]*Sin[c + d* 
x])/(d*Sqrt[a + a*Sec[c + d*x]]))/2)/(3*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 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 [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(328\) vs. \(2(153)=306\).

Time = 10.95 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.86

method	result	size
default	\(\frac {a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sqrt {2}\, B \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (6+6 \sec \left (d x +c \right )\right )+\sqrt {2}\, C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (9+9 \sec \left (d x +c \right )\right )+\sqrt {2}\, B \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (6+6 \sec \left (d x +c \right )\right )+\sqrt {2}\, C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (9+9 \sec \left (d x +c \right )\right )+A \left (8 \sin \left (d x +c \right )+40 \tan \left (d x +c \right )\right )+24 B \tan \left (d x +c \right )+12 C \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{12 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\)	\(329\)
parts	\(\frac {A \left (2 \sin \left (d x +c \right )+10 \tan \left (d x +c \right )\right ) a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \left (3 \cos \left (d x +c \right )+3\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {B \left (\frac {\sqrt {2}\, \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}}{2}-\frac {\sqrt {2}\, \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}}{2}-2 \cot \left (d x +c \right )+2 \csc \left (d x +c \right )\right ) a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \sqrt {\sec \left (d x +c \right )}}+\frac {C \left (\sqrt {2}\, \sin \left (d x +c \right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}-3 \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+3 \cos \left (d x +c \right ) \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )\right ) a \sqrt {\sec \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{2 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(362\)

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)^(3/2 
),x,method=_RETURNVERBOSE)
 

Output:

1/12/d*a*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)/sec(d*x+c)^(3/2)*(2^(1/2) 
*B*(-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))*(6+6*sec(d*x+c))+2^(1/2)*C*(-2/(cos(d*x+c)+1))^(1/2)*a 
rctan(1/2*(-cot(d*x+c)+csc(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))*(9+9*sec(d 
*x+c))+2^(1/2)*B*(-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))*(6+6*sec(d*x+c))+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))*(9+9*sec(d*x+c))+A*(8*sin(d*x+c)+40*tan(d*x+c))+24*B*tan(d*x+c)+12*C*s 
ec(d*x+c)*tan(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 401, normalized size of antiderivative = 2.27 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\left [\frac {3 \, {\left ({\left (2 \, B + 3 \, C\right )} a \cos \left (d x + c\right ) + {\left (2 \, B + 3 \, C\right )} a\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (2 \, A a \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A + 3 \, B\right )} a \cos \left (d x + c\right ) + 3 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{12 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, \frac {3 \, {\left ({\left (2 \, B + 3 \, C\right )} a \cos \left (d x + c\right ) + {\left (2 \, B + 3 \, C\right )} a\right )} \sqrt {-a} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{2 \, a \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left (2 \, A a \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A + 3 \, B\right )} a \cos \left (d x + c\right ) + 3 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \] 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 
)^(3/2),x, algorithm="fricas")
 

Output:

[1/12*(3*((2*B + 3*C)*a*cos(d*x + c) + (2*B + 3*C)*a)*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*(2*A*a*cos(d*x + c)^2 + 2*( 
5*A + 3*B)*a*cos(d*x + c) + 3*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) + d), 1/6*(3*((2*B + 3*C 
)*a*cos(d*x + c) + (2*B + 3*C)*a)*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*(2*A*a*cos(d*x + c)^2 + 2*(5*A + 3*B)*a*cos( 
d*x + c) + 3*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqr 
t(cos(d*x + c)))/(d*cos(d*x + c) + d)]
 

Sympy [F(-1)]

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1455 vs. \(2 (153) = 306\).

Time = 0.35 (sec) , antiderivative size = 1455, normalized size of antiderivative = 8.22 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \] 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 
)^(3/2),x, algorithm="maxima")
 

Output:

1/12*(3*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))*B*sqrt(a) + 4*(sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 9*sqrt(2)*a*sin(1/2 
*d*x + 1/2*c))*A*sqrt(a) + 3*(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*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))*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*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 414 vs. \(2 (153) = 306\).

Time = 2.20 (sec) , antiderivative size = 414, normalized size of antiderivative = 2.34 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx =\text {Too large to display} \] 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 
)^(3/2),x, algorithm="giac")
 

Output:

1/6*(3*(2*B*a^(3/2)*sgn(cos(d*x + c)) + 3*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))) - 3*(2*B*a^(3/2)*sgn(cos(d*x + c)) + 3*C*a^(3/2)*sg 
n(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*(6*sqrt(2)*A*a^3*sgn(cos(d* 
x + c)) + 3*sqrt(2)*B*a^3*sgn(cos(d*x + c)) + (4*sqrt(2)*A*a^3*sgn(cos(d*x 
 + c)) + 3*sqrt(2)*B*a^3*sgn(cos(d*x + c)))*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)^(3/2) + 12*(3*sqrt(2)*(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^(5/2)* 
sgn(cos(d*x + c)) - sqrt(2)*C*a^(7/2)*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))/d
 

Mupad [F(-1)]

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

Reduce [F]

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

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

Output:

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