3.2.53 \(\int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\) [153]

3.2.53.1 Optimal result
3.2.53.2 Mathematica [A] (verified)
3.2.53.3 Rubi [A] (verified)
3.2.53.4 Maple [B] (warning: unable to verify)
3.2.53.5 Fricas [A] (verification not implemented)
3.2.53.6 Sympy [F]
3.2.53.7 Maxima [F]
3.2.53.8 Giac [A] (verification not implemented)
3.2.53.9 Mupad [F(-1)]

3.2.53.1 Optimal result

Integrand size = 33, antiderivative size = 171 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {(7 A-11 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(9 A-13 B) \tan (c+d x)}{3 a d \sqrt {a+a \sec (c+d x)}}-\frac {(3 A-7 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{6 a^2 d} \]

output
-1/4*(7*A-11*B)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/ 
2))/a^(3/2)/d*2^(1/2)+1/2*(A-B)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*sec(d*x+c)) 
^(3/2)+1/3*(9*A-13*B)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)-1/6*(3*A-7*B)* 
(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a^2/d
 
3.2.53.2 Mathematica [A] (verified)

Time = 1.04 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.82 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\left (-3 \sqrt {2} (7 A-11 B) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)+\sqrt {1-\sec (c+d x)} \left (15 A-19 B+12 (A-B) \sec (c+d x)+4 B \sec ^2(c+d x)\right )\right ) \tan (c+d x)}{6 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{3/2}} \]

input
Integrate[(Sec[c + d*x]^3*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^(3/2) 
,x]
 
output
((-3*Sqrt[2]*(7*A - 11*B)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Cos[(c + 
 d*x)/2]^2*Sec[c + d*x] + Sqrt[1 - Sec[c + d*x]]*(15*A - 19*B + 12*(A - B) 
*Sec[c + d*x] + 4*B*Sec[c + d*x]^2))*Tan[c + d*x])/(6*d*Sqrt[1 - Sec[c + d 
*x]]*(a*(1 + Sec[c + d*x]))^(3/2))
 
3.2.53.3 Rubi [A] (verified)

Time = 0.97 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4507, 27, 3042, 4498, 27, 3042, 4489, 3042, 4282, 216}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4507

\(\displaystyle \frac {\int \frac {\sec ^2(c+d x) (4 a (A-B)-a (3 A-7 B) \sec (c+d x))}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}+\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sec ^2(c+d x) (4 a (A-B)-a (3 A-7 B) \sec (c+d x))}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}+\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4498

\(\displaystyle \frac {\frac {2 \int -\frac {\sec (c+d x) \left (a^2 (3 A-7 B)-2 a^2 (9 A-13 B) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 (3 A-7 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}+\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\int \frac {\sec (c+d x) \left (a^2 (3 A-7 B)-2 a^2 (9 A-13 B) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 (3 A-7 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}+\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a^2 (3 A-7 B)-2 a^2 (9 A-13 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}-\frac {2 (3 A-7 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}+\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4489

\(\displaystyle \frac {-\frac {3 a^2 (7 A-11 B) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx-\frac {4 a^2 (9 A-13 B) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 (3 A-7 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}+\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {3 a^2 (7 A-11 B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {4 a^2 (9 A-13 B) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 (3 A-7 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}+\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {-\frac {-\frac {6 a^2 (7 A-11 B) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {4 a^2 (9 A-13 B) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 (3 A-7 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}+\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {-\frac {\frac {3 \sqrt {2} a^{3/2} (7 A-11 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {4 a^2 (9 A-13 B) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 (3 A-7 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}+\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

input
Int[(Sec[c + d*x]^3*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^(3/2),x]
 
output
((A - B)*Sec[c + d*x]^2*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) + ( 
(-2*(3*A - 7*B)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d) - ((3*Sqrt[2] 
*a^(3/2)*(7*A - 11*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Se 
c[c + d*x]])])/d - (4*a^2*(9*A - 13*B)*Tan[c + d*x])/(d*Sqrt[a + a*Sec[c + 
 d*x]]))/(3*a))/(4*a^2)
 

3.2.53.3.1 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 216
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

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

rule 4282
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2/f   Subst[Int[1/(2*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[ 
a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
 

rule 4489
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs 
c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( 
a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[(a*B*m + A*b*(m + 1))/(b*(m + 
 1))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B 
, e, f, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b 
*(m + 1), 0] &&  !LtQ[m, -2^(-1)]
 

rule 4498
Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*( 
csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]* 
((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2))   Int 
[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B) 
*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a 
*B, 0] &&  !LtQ[m, -1]
 

rule 4507
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b 
- a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(a*f*( 
2*m + 1))), x] - Simp[1/(a*b*(2*m + 1))   Int[(a + b*Csc[e + f*x])^(m + 1)* 
(d*Csc[e + f*x])^(n - 1)*Simp[A*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m 
 - n + 1) + 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] && LtQ[m, -2^(-1)] && G 
tQ[n, 0]
 
3.2.53.4 Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(330\) vs. \(2(148)=296\).

Time = 4.10 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.94

method result size
default \(-\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (3 A \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-3 B \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+21 A \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}}-33 B \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}}-30 A \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+46 B \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+27 A \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-27 B \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{12 a^{2} d \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )}\) \(331\)
parts \(-\frac {A \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (\left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+7 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}-9 \csc \left (d x +c \right )+9 \cot \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {B \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (3 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+33 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}}-46 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+27 \csc \left (d x +c \right )-27 \cot \left (d x +c \right )\right )}{12 d \,a^{2} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )}\) \(340\)

input
int(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x,method=_RETURNV 
ERBOSE)
 
output
-1/12/a^2/d*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(3*A*(1-cos(d*x 
+c))^5*csc(d*x+c)^5-3*B*(1-cos(d*x+c))^5*csc(d*x+c)^5+21*A*ln(csc(d*x+c)-c 
ot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))*((1-cos(d*x+c))^2*csc(d 
*x+c)^2-1)^(3/2)-33*B*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c 
)^2-1)^(1/2))*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)-30*A*(1-cos(d*x+c))^ 
3*csc(d*x+c)^3+46*B*(1-cos(d*x+c))^3*csc(d*x+c)^3+27*A*(-cot(d*x+c)+csc(d* 
x+c))-27*B*(-cot(d*x+c)+csc(d*x+c)))/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)
 
3.2.53.5 Fricas [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 459, normalized size of antiderivative = 2.68 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\left [\frac {3 \, \sqrt {2} {\left ({\left (7 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (7 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (7 \, A - 11 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (15 \, A - 19 \, B\right )} \cos \left (d x + c\right )^{2} + 12 \, {\left (A - B\right )} \cos \left (d x + c\right ) + 4 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{24 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}}, \frac {3 \, \sqrt {2} {\left ({\left (7 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (7 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (7 \, A - 11 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + 2 \, {\left ({\left (15 \, A - 19 \, B\right )} \cos \left (d x + c\right )^{2} + 12 \, {\left (A - B\right )} \cos \left (d x + c\right ) + 4 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}}\right ] \]

input
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x, algorith 
m="fricas")
 
output
[1/24*(3*sqrt(2)*((7*A - 11*B)*cos(d*x + c)^3 + 2*(7*A - 11*B)*cos(d*x + c 
)^2 + (7*A - 11*B)*cos(d*x + c))*sqrt(-a)*log((2*sqrt(2)*sqrt(-a)*sqrt((a* 
cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + 3*a*cos(d*x + 
c)^2 + 2*a*cos(d*x + c) - a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*(( 
15*A - 19*B)*cos(d*x + c)^2 + 12*(A - B)*cos(d*x + c) + 4*B)*sqrt((a*cos(d 
*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*c 
os(d*x + c)^2 + a^2*d*cos(d*x + c)), 1/12*(3*sqrt(2)*((7*A - 11*B)*cos(d*x 
 + c)^3 + 2*(7*A - 11*B)*cos(d*x + c)^2 + (7*A - 11*B)*cos(d*x + c))*sqrt( 
a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sq 
rt(a)*sin(d*x + c))) + 2*((15*A - 19*B)*cos(d*x + c)^2 + 12*(A - B)*cos(d* 
x + c) + 4*B)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d 
*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))]
 
3.2.53.6 Sympy [F]

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

input
integrate(sec(d*x+c)**3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**(3/2),x)
 
output
Integral((A + B*sec(c + d*x))*sec(c + d*x)**3/(a*(sec(c + d*x) + 1))**(3/2 
), x)
 
3.2.53.7 Maxima [F]

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

input
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x, algorith 
m="maxima")
 
output
integrate((B*sec(d*x + c) + A)*sec(d*x + c)^3/(a*sec(d*x + c) + a)^(3/2), 
x)
 
3.2.53.8 Giac [A] (verification not implemented)

Time = 1.63 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.44 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {\frac {{\left ({\left (\frac {3 \, {\left (\sqrt {2} A a \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} B a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a} - \frac {2 \, {\left (15 \, \sqrt {2} A a \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 23 \, \sqrt {2} B a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {27 \, {\left (\sqrt {2} A a \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} B a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} - \frac {3 \, {\left (7 \, \sqrt {2} A - 11 \, \sqrt {2} B\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{12 \, d} \]

input
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x, algorith 
m="giac")
 
output
-1/12*(((3*(sqrt(2)*A*a*sgn(cos(d*x + c)) - sqrt(2)*B*a*sgn(cos(d*x + c))) 
*tan(1/2*d*x + 1/2*c)^2/a - 2*(15*sqrt(2)*A*a*sgn(cos(d*x + c)) - 23*sqrt( 
2)*B*a*sgn(cos(d*x + c)))/a)*tan(1/2*d*x + 1/2*c)^2 + 27*(sqrt(2)*A*a*sgn( 
cos(d*x + c)) - sqrt(2)*B*a*sgn(cos(d*x + c)))/a)*tan(1/2*d*x + 1/2*c)/((a 
*tan(1/2*d*x + 1/2*c)^2 - a)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)) - 3*(7*s 
qrt(2)*A - 11*sqrt(2)*B)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) + sqrt(-a* 
tan(1/2*d*x + 1/2*c)^2 + a)))/(sqrt(-a)*a*sgn(cos(d*x + c))))/d
 
3.2.53.9 Mupad [F(-1)]

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

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