\(\int \frac {\sec (c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\) [972]

Optimal result

Integrand size = 41, antiderivative size = 416 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {2 \left (4 a A b^2-a^2 b B-3 b^3 B-2 a^3 C+6 a b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^3 (a+b)^{3/2} d}+\frac {2 \left (2 a^2 C+a b (3 A+B+3 C)-b^2 (A+3 (B+C))\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^2 \sqrt {a+b} \left (a^2-b^2\right ) d}-\frac {2 \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (a^2 b B+3 b^3 B+2 a^3 C-2 a b^2 (2 A+3 C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \] Output:

-2/3*(4*A*a*b^2-B*a^2*b-3*B*b^3-2*C*a^3+6*C*a*b^2)*cot(d*x+c)*EllipticE((a 
+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a 
+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/(a-b)/b^3/(a+b)^(3/2)/d+2/3*(2* 
C*a^2+a*b*(3*A+B+3*C)-b^2*(A+3*B+3*C))*cot(d*x+c)*EllipticF((a+b*sec(d*x+c 
))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*( 
-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^2/(a+b)^(1/2)/(a^2-b^2)/d-2/3*(A*b^2-a*(B 
*b-C*a))*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)+2/3*(B*a^2*b+3*B* 
b^3+2*a^3*C-2*a*b^2*(2*A+3*C))*tan(d*x+c)/b/(a^2-b^2)^2/d/(a+b*sec(d*x+c)) 
^(1/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(3980\) vs. \(2(416)=832\).

Time = 24.20 (sec) , antiderivative size = 3980, normalized size of antiderivative = 9.57 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Result too large to show} \] Input:

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

Output:

((b + a*Cos[c + d*x])^3*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^ 
2)*((-4*(-4*a*A*b^2 + a^2*b*B + 3*b^3*B + 2*a^3*C - 6*a*b^2*C)*Sin[c + d*x 
])/(3*b^2*(a^2 - b^2)^2) + (4*(A*b^2*Sin[c + d*x] - a*b*B*Sin[c + d*x] + a 
^2*C*Sin[c + d*x]))/(3*a*(a^2 - b^2)*(b + a*Cos[c + d*x])^2) + (4*(-5*a^2* 
A*b^2*Sin[c + d*x] + A*b^4*Sin[c + d*x] + 2*a^3*b*B*Sin[c + d*x] + 2*a*b^3 
*B*Sin[c + d*x] + a^4*C*Sin[c + d*x] - 5*a^2*b^2*C*Sin[c + d*x]))/(3*a*b*( 
-a^2 + b^2)^2*(b + a*Cos[c + d*x]))))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*C 
os[2*c + 2*d*x])*(a + b*Sec[c + d*x])^(5/2)) + (4*(b + a*Cos[c + d*x])^2*( 
(-8*a*A*b)/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) 
+ (2*a^2*B)/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) 
 + (2*b^2*B)/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) 
+ (4*a^3*C)/(3*b*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x] 
]) - (4*a*b*C)/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]] 
) - (2*a^2*A*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x] 
]) + (2*A*b^2*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x 
]]) + (2*a^3*B*Sqrt[Sec[c + d*x]])/(3*b*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + 
d*x]]) - (2*a*b*B*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + 
 d*x]]) - (10*a^2*C*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c 
 + d*x]]) + (4*a^4*C*Sqrt[Sec[c + d*x]])/(3*b^2*(-a^2 + b^2)^2*Sqrt[b + a* 
Cos[c + d*x]]) + (2*b^2*C*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)^2*Sqrt[b + ...
 

Rubi [A] (verified)

Time = 1.52 (sec) , antiderivative size = 429, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.268, Rules used = {3042, 4568, 27, 3042, 4491, 27, 3042, 4493, 3042, 4319, 4492}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4568

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4491

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4493

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4319

\(\displaystyle -\frac {\frac {\left (2 a^3 C+a^2 b B-2 a b^2 (2 A+3 C)+3 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (2 a^2 C+a b (3 A+B+3 C)-b^2 (A+3 (B+C))\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{a^2-b^2}-\frac {2 \tan (c+d x) \left (2 a^3 C+a^2 b B-2 a b^2 (2 A+3 C)+3 b^3 B\right )}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4492

\(\displaystyle -\frac {2 \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {\frac {-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (2 a^2 C+a b (3 A+B+3 C)-b^2 (A+3 (B+C))\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (2 a^3 C+a^2 b B-2 a b^2 (2 A+3 C)+3 b^3 B\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}}{a^2-b^2}-\frac {2 \tan (c+d x) \left (2 a^3 C+a^2 b B-2 a b^2 (2 A+3 C)+3 b^3 B\right )}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}\)

Input:

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

Output:

(-2*(A*b^2 - a*(b*B - a*C))*Tan[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Sec[c 
+ d*x])^(3/2)) - (((-2*(a - b)*Sqrt[a + b]*(a^2*b*B + 3*b^3*B + 2*a^3*C - 
2*a*b^2*(2*A + 3*C))*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x] 
]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt 
[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b^2*d) - (2*(a - b)*Sqrt[a + b]*(2*a 
^2*C + a*b*(3*A + B + 3*C) - b^2*(A + 3*(B + C)))*Cot[c + d*x]*EllipticF[A 
rcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - 
 Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d))/( 
a^2 - b^2) - (2*(a^2*b*B + 3*b^3*B + 2*a^3*C - 2*a*b^2*(2*A + 3*C))*Tan[c 
+ d*x])/((a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]))/(3*b*(a^2 - b^2))
 

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 3042

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

rule 4319

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

rule 4491

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

rule 4492

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c 
sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a 
 + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e 
+ f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + 
 f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, 
 f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
 

rule 4493

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

rule 4568

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3275\) vs. \(2(387)=774\).

Time = 51.55 (sec) , antiderivative size = 3276, normalized size of antiderivative = 7.88

method	result	size
default	\(\text {Expression too large to display}\)	\(3276\)
parts	\(\text {Expression too large to display}\)	\(3393\)

Input:

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

Output:

-2/3/d/(a-b)^2/(a+b)^2/b^2*((cos(d*x+c)^2+2*cos(d*x+c)+1)*A*(1/(a+b)*(b+a* 
cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b^5*El 
lipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+sin(d*x+c)*cos(d*x+c)* 
(-cos(d*x+c)+3)*a^4*b*C+B*a^4*b*cos(d*x+c)^2*sin(d*x+c)-2*B*a^3*b^2*cos(d* 
x+c)^2*sin(d*x+c)+6*C*a*b^4*cos(d*x+c)*sin(d*x+c)-4*A*a^3*b^2*cos(d*x+c)^2 
*sin(d*x+c)+sin(d*x+c)*cos(d*x+c)*(-6*cos(d*x+c)-2)*C*a^3*b^2+sin(d*x+c)*c 
os(d*x+c)*(5*cos(d*x+c)-7)*C*a^2*b^3+(3*cos(d*x+c)^2+6*cos(d*x+c)+3)*B*(1/ 
(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^( 
1/2)*b^5*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(-3*cos(d*x 
+c)^2-6*cos(d*x+c)-3)*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(c 
os(d*x+c)/(cos(d*x+c)+1))^(1/2)*b^5*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b 
)/(a+b))^(1/2))+(3*cos(d*x+c)^2+6*cos(d*x+c)+3)*C*(1/(a+b)*(b+a*cos(d*x+c) 
)/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b^5*EllipticF(-c 
sc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+sin(d*x+c)*cos(d*x+c)*(5*cos(d*x 
+c)-3)*a^2*A*b^3+sin(d*x+c)*cos(d*x+c)*(3*cos(d*x+c)-1)*B*a^2*b^3+sin(d*x+ 
c)*cos(d*x+c)*(-2*cos(d*x+c)+4)*a*b^4*B+C*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d 
*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^5*EllipticE(-csc(d*x+c 
)+cot(d*x+c),((a-b)/(a+b))^(1/2))*(2*cos(d*x+c)^3+4*cos(d*x+c)^2+2*cos(d*x 
+c))+4*A*a*b^4*cos(d*x+c)*sin(d*x+c)+sin(d*x+c)*cos(d*x+c)*(-cos(d*x+c)-1) 
*A*b^5+2*C*a^5*cos(d*x+c)^2*sin(d*x+c)-3*B*b^5*cos(d*x+c)*sin(d*x+c)+(4...
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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