\(\int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx\) [852]

Optimal result

Integrand size = 34, antiderivative size = 353 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {2 \left (4 a b B-a^2 C-3 b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^2 (a+b)^{3/2} d}+\frac {2 (3 a B-b B+a C-3 b C) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b (a+b)^{3/2} d}-\frac {2 (b B-a C) \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (4 a b B-a^2 C-3 b^2 C\right ) \tan (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \] Output:

-2/3*(4*B*a*b-C*a^2-3*C*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^2/(a+b)^(3/2)/d+2/3*(3*B*a-B*b+C*a-3*C*b)*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)/(a-b)/b/(a+ 
b)^(3/2)/d-2/3*(B*b-C*a)*tan(d*x+c)/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)-2/3 
*(4*B*a*b-C*a^2-3*C*b^2)*tan(d*x+c)/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 15.17 (sec) , antiderivative size = 559, normalized size of antiderivative = 1.58 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {(b+a \cos (c+d x))^3 \sec ^3(c+d x) \left (-\frac {2 \left (-4 a b B+a^2 C+3 b^2 C\right ) \sin (c+d x)}{3 b \left (-a^2+b^2\right )^2}+\frac {2 \left (b^2 B \sin (c+d x)-a b C \sin (c+d x)\right )}{3 a \left (a^2-b^2\right ) (b+a \cos (c+d x))^2}+\frac {2 \left (-5 a^2 b B \sin (c+d x)+b^3 B \sin (c+d x)+2 a^3 C \sin (c+d x)+2 a b^2 C \sin (c+d x)\right )}{3 a \left (a^2-b^2\right )^2 (b+a \cos (c+d x))}\right )}{d (a+b \sec (c+d x))^{5/2}}+\frac {2 (b+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 (a+b) \left (-4 a b B+a^2 C+3 b^2 C\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+2 b (a+b) (3 a B+b B-a C-3 b C) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (-4 a b B+a^2 C+3 b^2 C\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 b \left (a^2-b^2\right )^2 d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} (a+b \sec (c+d x))^{5/2}} \] Input:

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

Output:

((b + a*Cos[c + d*x])^3*Sec[c + d*x]^3*((-2*(-4*a*b*B + a^2*C + 3*b^2*C)*S 
in[c + d*x])/(3*b*(-a^2 + b^2)^2) + (2*(b^2*B*Sin[c + d*x] - a*b*C*Sin[c + 
 d*x]))/(3*a*(a^2 - b^2)*(b + a*Cos[c + d*x])^2) + (2*(-5*a^2*b*B*Sin[c + 
d*x] + b^3*B*Sin[c + d*x] + 2*a^3*C*Sin[c + d*x] + 2*a*b^2*C*Sin[c + d*x]) 
)/(3*a*(a^2 - b^2)^2*(b + a*Cos[c + d*x]))))/(d*(a + b*Sec[c + d*x])^(5/2) 
) + (2*(b + a*Cos[c + d*x])^2*Sec[c + d*x]^(5/2)*Sqrt[Cos[(c + d*x)/2]^2*S 
ec[c + d*x]]*(2*(a + b)*(-4*a*b*B + a^2*C + 3*b^2*C)*Sqrt[Cos[c + d*x]/(1 
+ Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*E 
llipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(a + b)*(3*a*B + 
 b*B - a*C - 3*b*C)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[ 
c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]] 
, (a - b)/(a + b)] + (-4*a*b*B + a^2*C + 3*b^2*C)*Cos[c + d*x]*(b + a*Cos[ 
c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*b*(a^2 - b^2)^2*d*Sqrt[ 
Sec[(c + d*x)/2]^2]*(a + b*Sec[c + d*x])^(5/2))
 

Rubi [A] (verified)

Time = 1.34 (sec) , antiderivative size = 385, normalized size of antiderivative = 1.09, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {3042, 4548, 27, 3042, 4548, 27, 3042, 4546, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4548

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4548

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4546

\(\displaystyle \frac {\frac {a^2 \left (a^2 (-C)+4 a b B-3 b^2 C\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx+\int \frac {\left (a^2 \left (3 B a^2-4 b C a+b^2 B\right )-a^2 \left (-C a^2+4 b B a-3 b^2 C\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 a \left (a^2 (-C)+4 a b B-3 b^2 C\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 (b B-a C) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {a^2 \left (a^2 (-C)+4 a b B-3 b^2 C\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx+a^2 (a-b) (a (3 B+C)-b (B+3 C)) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 a \left (a^2 (-C)+4 a b B-3 b^2 C\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 (b B-a C) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a^2 \left (a^2 (-C)+4 a b B-3 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+a^2 (a-b) (a (3 B+C)-b (B+3 C)) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}-\frac {2 a \left (a^2 (-C)+4 a b B-3 b^2 C\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 (b B-a C) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {\frac {a^2 \left (a^2 (-C)+4 a b B-3 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^2 (a-b) \sqrt {a+b} (a (3 B+C)-b (B+3 C)) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{a \left (a^2-b^2\right )}-\frac {2 a \left (a^2 (-C)+4 a b B-3 b^2 C\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 (b B-a C) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {\frac {\frac {2 a^2 (a-b) \sqrt {a+b} (a (3 B+C)-b (B+3 C)) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 a^2 (a-b) \sqrt {a+b} \left (a^2 (-C)+4 a b B-3 b^2 C\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}}{a \left (a^2-b^2\right )}-\frac {2 a \left (a^2 (-C)+4 a b B-3 b^2 C\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 (b B-a C) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

Input:

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

Output:

(-2*(b*B - a*C)*Tan[c + d*x])/(3*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) 
 + (((-2*a^2*(a - b)*Sqrt[a + b]*(4*a*b*B - a^2*C - 3*b^2*C)*Cot[c + d*x]* 
EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*S 
qrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b)) 
])/(b^2*d) + (2*a^2*(a - b)*Sqrt[a + b]*(a*(3*B + C) - b*(B + 3*C))*Cot[c 
+ d*x]*EllipticF[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*d))/(a*(a^2 - b^2)) - (2*a*(4*a*b*B - a^2*C - 3*b^2*C)*Tan[c 
+ d*x])/((a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]))/(3*a*(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 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 4546

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

rule 4548

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2042\) vs. \(2(323)=646\).

Time = 28.30 (sec) , antiderivative size = 2043, normalized size of antiderivative = 5.79

method	result	size
default	\(\text {Expression too large to display}\)	\(2043\)
parts	\(\text {Expression too large to display}\)	\(2099\)

Input:

int((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x,method=_RETURNV 
ERBOSE)
 

Output:

-2/3/d/(a-b)^2/(a+b)^2/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^4*EllipticE(-csc(d*x+c)+cot(d*x+c),( 
(a-b)/(a+b))^(1/2))*(cos(d*x+c)^3+2*cos(d*x+c)^2+cos(d*x+c))-4*B*a^3*b*cos 
(d*x+c)^2*sin(d*x+c)-2*C*a^3*b*cos(d*x+c)^2*sin(d*x+c)+sin(d*x+c)*cos(d*x+ 
c)*(-cos(d*x+c)-1)*B*b^4+(-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^4*El 
lipticF(-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)/(c 
os(d*x+c)+1))^(1/2)*b^4*EllipticE(-csc(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)*B*a^2*b^2+sin(d*x+c)*cos(d*x+c) 
*(3*cos(d*x+c)-1)*C*a^2*b^2+sin(d*x+c)*cos(d*x+c)*(-2*cos(d*x+c)+4)*C*a*b^ 
3+(cos(d*x+c)^2+2*cos(d*x+c)+1)*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^4*EllipticF(-csc(d*x+c)+cot(d* 
x+c),((a-b)/(a+b))^(1/2))+(-4*cos(d*x+c)^3-9*cos(d*x+c)^2-6*cos(d*x+c)-1)* 
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^2*b^2*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(- 
3*cos(d*x+c)^3-10*cos(d*x+c)^2-11*cos(d*x+c)-4)*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*b^3*EllipticF( 
-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(-4*cos(d*x+c)^3-12*cos(d*x+c) 
^2-12*cos(d*x+c)-4)*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(...
 

Fricas [F]

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

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

Output:

integral((C*sec(d*x + c)^2 + B*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]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int((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)**2)/(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))/(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