3.2.84 \(\int \frac {\sec ^3(c+d x) (A+C \sec ^2(c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\) [184]

3.2.84.1 Optimal result
3.2.84.2 Mathematica [A] (verified)
3.2.84.3 Rubi [A] (verified)
3.2.84.4 Maple [B] (warning: unable to verify)
3.2.84.5 Fricas [A] (verification not implemented)
3.2.84.6 Sympy [F]
3.2.84.7 Maxima [F]
3.2.84.8 Giac [A] (verification not implemented)
3.2.84.9 Mupad [F(-1)]

3.2.84.1 Optimal result

Integrand size = 35, antiderivative size = 193 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {2} (A+C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}-\frac {4 (35 A+37 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}-\frac {2 C \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (35 A+31 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 a d} \]

output
(A+C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*2^(1/2 
)/d/a^(1/2)-4/105*(35*A+37*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-2/35*C*s 
ec(d*x+c)^2*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/7*C*sec(d*x+c)^3*tan(d*x 
+c)/d/(a+a*sec(d*x+c))^(1/2)+2/105*(35*A+31*C)*(a+a*sec(d*x+c))^(1/2)*tan( 
d*x+c)/a/d
 
3.2.84.2 Mathematica [A] (verified)

Time = 2.07 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.60 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\left (105 \sqrt {2} (A+C) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )-2 (1-\sec (c+d x))^{3/2} \left (35 A+43 C+12 C \sec (c+d x)+15 C \sec ^2(c+d x)\right )\right ) \tan (c+d x)}{105 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]

input
Integrate[(Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2))/Sqrt[a + a*Sec[c + d*x]] 
,x]
 
output
((105*Sqrt[2]*(A + C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]] - 2*(1 - Sec 
[c + d*x])^(3/2)*(35*A + 43*C + 12*C*Sec[c + d*x] + 15*C*Sec[c + d*x]^2))* 
Tan[c + d*x])/(105*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
 
3.2.84.3 Rubi [A] (verified)

Time = 1.24 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.13, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4577, 27, 3042, 4509, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4577

\(\displaystyle \frac {2 \int \frac {\sec ^3(c+d x) (a (7 A+6 C)-a C \sec (c+d x))}{2 \sqrt {\sec (c+d x) a+a}}dx}{7 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sec ^3(c+d x) (a (7 A+6 C)-a C \sec (c+d x))}{\sqrt {\sec (c+d x) a+a}}dx}{7 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a (7 A+6 C)-a C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{7 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4509

\(\displaystyle \frac {\frac {2 \int -\frac {\sec ^2(c+d x) \left (4 a^2 C-a^2 (35 A+31 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{5 a}-\frac {2 a C \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\int \frac {\sec ^2(c+d x) \left (4 a^2 C-a^2 (35 A+31 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{5 a}-\frac {2 a C \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (4 a^2 C-a^2 (35 A+31 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}-\frac {2 a C \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4498

\(\displaystyle \frac {-\frac {\frac {2 \int -\frac {\sec (c+d x) \left (a^3 (35 A+31 C)-2 a^3 (35 A+37 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 a (35 A+31 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a C \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {-\frac {\int \frac {\sec (c+d x) \left (a^3 (35 A+31 C)-2 a^3 (35 A+37 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 a (35 A+31 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a C \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a^3 (35 A+31 C)-2 a^3 (35 A+37 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}-\frac {2 a (35 A+31 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a C \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4489

\(\displaystyle \frac {-\frac {-\frac {105 a^3 (A+C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx-\frac {4 a^3 (35 A+37 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a (35 A+31 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a C \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {-\frac {105 a^3 (A+C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {4 a^3 (35 A+37 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a (35 A+31 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a C \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {-\frac {-\frac {-\frac {210 a^3 (A+C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {4 a^3 (35 A+37 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a (35 A+31 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a C \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {-\frac {-\frac {\frac {105 \sqrt {2} a^{5/2} (A+C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {4 a^3 (35 A+37 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a (35 A+31 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a C \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{7 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\)

input
Int[(Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2))/Sqrt[a + a*Sec[c + d*x]],x]
 
output
(2*C*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt[a + a*Sec[c + d*x]]) + ((-2*a* 
C*Sec[c + d*x]^2*Tan[c + d*x])/(5*d*Sqrt[a + a*Sec[c + d*x]]) - ((-2*a*(35 
*A + 31*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d) - ((105*Sqrt[2]*a^ 
(5/2)*(A + C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d* 
x]])])/d - (4*a^3*(35*A + 37*C)*Tan[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]) 
)/(3*a))/(5*a))/(7*a)
 

3.2.84.3.1 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 4509
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*d* 
Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(f*(m + n))), 
 x] + Simp[d/(b*(m + n))   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 
 1)*Simp[b*B*(n - 1) + (A*b*(m + n) + a*B*m)*Csc[e + f*x], x], x], x] /; Fr 
eeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] 
&& GtQ[n, 1]
 

rule 4577
Int[((A_.) + 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*Csc[e + f*x])^n/(f*(m + n + 1))), 
x] + Simp[1/(b*(m + n + 1))   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n 
*Simp[A*b*(m + n + 1) + b*C*n + a*C*m*Csc[e + f*x], x], x], x] /; FreeQ[{a, 
 b, d, e, f, A, C, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && 
!LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]
 
3.2.84.4 Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(340\) vs. \(2(168)=336\).

Time = 0.54 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.77

method result size
default \(\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (105 A \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {7}{2}} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )+105 C \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {7}{2}} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )-140 A \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}-184 C \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}+280 A \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+224 C \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-140 A \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-280 C \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}\right )}{105 d a \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{3}}\) \(341\)
parts \(\frac {A \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}}-4 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}\right )}{3 d a \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )}+\frac {C \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (105 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {7}{2}}-184 \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}+224 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-280 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}\right )}{105 d a \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{3}}\) \(356\)

input
int(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x,method=_RETUR 
NVERBOSE)
 
output
1/105/d/a*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(105*A*((1-cos(d* 
x+c))^2*csc(d*x+c)^2-1)^(7/2)*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*c 
sc(d*x+c)^2-1)^(1/2))+105*C*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(7/2)*ln(csc 
(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))-140*A*(1-cos(d 
*x+c))^7*csc(d*x+c)^7-184*C*(1-cos(d*x+c))^7*csc(d*x+c)^7+280*A*(1-cos(d*x 
+c))^5*csc(d*x+c)^5+224*C*(1-cos(d*x+c))^5*csc(d*x+c)^5-140*A*(1-cos(d*x+c 
))^3*csc(d*x+c)^3-280*C*(1-cos(d*x+c))^3*csc(d*x+c)^3)/((1-cos(d*x+c))^2*c 
sc(d*x+c)^2-1)^3
 
3.2.84.5 Fricas [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 412, normalized size of antiderivative = 2.13 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [\frac {105 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{4} + {\left (A + C\right )} a \cos \left (d x + c\right )^{3}\right )} \sqrt {-\frac {1}{a}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left ({\left (35 \, A + 43 \, C\right )} \cos \left (d x + c\right )^{3} - {\left (35 \, A + 31 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) - 15 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{210 \, {\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}}, -\frac {2 \, {\left ({\left (35 \, A + 43 \, C\right )} \cos \left (d x + c\right )^{3} - {\left (35 \, A + 31 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) - 15 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) + \frac {105 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{4} + {\left (A + C\right )} a \cos \left (d x + c\right )^{3}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}}}{105 \, {\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}}\right ] \]

input
integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algori 
thm="fricas")
 
output
[1/210*(105*sqrt(2)*((A + C)*a*cos(d*x + c)^4 + (A + C)*a*cos(d*x + c)^3)* 
sqrt(-1/a)*log(-(2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1 
/a)*cos(d*x + c)*sin(d*x + c) - 3*cos(d*x + c)^2 - 2*cos(d*x + c) + 1)/(co 
s(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 4*((35*A + 43*C)*cos(d*x + c)^3 - (3 
5*A + 31*C)*cos(d*x + c)^2 + 3*C*cos(d*x + c) - 15*C)*sqrt((a*cos(d*x + c) 
 + a)/cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c)^4 + a*d*cos(d*x + c)^3 
), -1/105*(2*((35*A + 43*C)*cos(d*x + c)^3 - (35*A + 31*C)*cos(d*x + c)^2 
+ 3*C*cos(d*x + c) - 15*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x 
 + c) + 105*sqrt(2)*((A + C)*a*cos(d*x + c)^4 + (A + C)*a*cos(d*x + c)^3)* 
arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt( 
a)*sin(d*x + c)))/sqrt(a))/(a*d*cos(d*x + c)^4 + a*d*cos(d*x + c)^3)]
 
3.2.84.6 Sympy [F]

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

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

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

input
integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algori 
thm="maxima")
 
output
integrate((C*sec(d*x + c)^2 + A)*sec(d*x + c)^3/sqrt(a*sec(d*x + c) + a), 
x)
 
3.2.84.8 Giac [A] (verification not implemented)

Time = 1.43 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.13 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\frac {105 \, \sqrt {2} {\left (A + C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {4 \, {\left ({\left (\frac {\sqrt {2} {\left (35 \, A a^{3} + 46 \, C a^{3}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {14 \, \sqrt {2} {\left (5 \, A a^{3} + 4 \, C a^{3}\right )}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {35 \, \sqrt {2} {\left (A a^{3} + 2 \, C a^{3}\right )}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{105 \, d} \]

input
integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algori 
thm="giac")
 
output
-1/105*(105*sqrt(2)*(A + 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)))/(sqrt(-a)*sgn(cos(d*x + c))) + 4*((sqrt(2 
)*(35*A*a^3 + 46*C*a^3)*tan(1/2*d*x + 1/2*c)^2/sgn(cos(d*x + c)) - 14*sqrt 
(2)*(5*A*a^3 + 4*C*a^3)/sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2 + 35*sqr 
t(2)*(A*a^3 + 2*C*a^3)/sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^3/((a*tan(1 
/2*d*x + 1/2*c)^2 - a)^3*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d
 
3.2.84.9 Mupad [F(-1)]

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

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