3.9.33 \(\int \cos ^2(c+d x) (a+b \sec (c+d x))^{5/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [833]

3.9.33.1 Optimal result

Integrand size = 42, antiderivative size = 433 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {(a-b) \sqrt {a+b} \left (3 a^2 B-6 b^2 B-14 a b C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b d}+\frac {\sqrt {a+b} \left (2 a b (9 B-7 C)-2 b^2 (3 B-C)+3 a^2 (B+6 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 d}-\frac {a \sqrt {a+b} (5 b B+2 a C) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}+\frac {a B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{d}-\frac {b (3 a B-2 b C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 d} \]

output

a*B*(a+b*sec(d*x+c))^(3/2)*sin(d*x+c)/d+1/3*(a-b)*(3*B*a^2-6*B*b^2-14*C*a* 
b)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^( 
1/2))*(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/d+1/3*(2*a*b*(9*B-7*C)-2*b^2*(3*B-C)+3*a^2*(B+6*C))*cot(d*x+c)*El 
lipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2 
)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/d-a*(5*B* 
b+2*C*a)*cot(d*x+c)*EllipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a, 
((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec 
(d*x+c))/(a-b))^(1/2)/d-1/3*b*(3*B*a-2*C*b)*(a+b*sec(d*x+c))^(1/2)*tan(d*x 
+c)/d
 

3.9.33.2 Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(7745\) vs. \(2(433)=866\).

Time = 28.77 (sec) , antiderivative size = 7745, normalized size of antiderivative = 17.89 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \]

input

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

output

Result too large to show
 

3.9.33.3 Rubi [A] (verified)

Time = 1.93 (sec) , antiderivative size = 439, normalized size of antiderivative = 1.01, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 4560, 3042, 4513, 27, 3042, 4544, 27, 3042, 4546, 3042, 4409, 3042, 4271, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4560

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4513

\(\displaystyle \frac {a B \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}-\int -\frac {1}{2} \sqrt {a+b \sec (c+d x)} \left (-b (3 a B-2 b C) \sec ^2(c+d x)+2 b (b B+2 a C) \sec (c+d x)+a (5 b B+2 a C)\right )dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \int \sqrt {a+b \sec (c+d x)} \left (-b (3 a B-2 b C) \sec ^2(c+d x)+2 b (b B+2 a C) \sec (c+d x)+a (5 b B+2 a C)\right )dx+\frac {a B \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \int \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (-b (3 a B-2 b C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 b (b B+2 a C) \csc \left (c+d x+\frac {\pi }{2}\right )+a (5 b B+2 a C)\right )dx+\frac {a B \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 4544

\(\displaystyle \frac {1}{2} \left (\frac {2}{3} \int \frac {3 (5 b B+2 a C) a^2-b \left (3 B a^2-14 b C a-6 b^2 B\right ) \sec ^2(c+d x)+2 b \left (9 C a^2+9 b B a+b^2 C\right ) \sec (c+d x)}{2 \sqrt {a+b \sec (c+d x)}}dx-\frac {2 b (3 a B-2 b C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {a B \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \int \frac {3 (5 b B+2 a C) a^2-b \left (3 B a^2-14 b C a-6 b^2 B\right ) \sec ^2(c+d x)+2 b \left (9 C a^2+9 b B a+b^2 C\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-\frac {2 b (3 a B-2 b C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {a B \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \int \frac {3 (5 b B+2 a C) a^2-b \left (3 B a^2-14 b C a-6 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 b \left (9 C a^2+9 b B a+b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b (3 a B-2 b C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {a B \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 4546

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\int \frac {3 (5 b B+2 a C) a^2+\left (b \left (3 B a^2-14 b C a-6 b^2 B\right )+2 b \left (9 C a^2+9 b B a+b^2 C\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-b \left (3 a^2 B-14 a b C-6 b^2 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\frac {2 b (3 a B-2 b C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {a B \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\int \frac {3 (5 b B+2 a C) a^2+\left (b \left (3 B a^2-14 b C a-6 b^2 B\right )+2 b \left (9 C a^2+9 b B a+b^2 C\right )\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b \left (3 a^2 B-14 a b C-6 b^2 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 b (3 a B-2 b C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {a B \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 4409

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (-b \left (3 a^2 B-14 a b C-6 b^2 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (3 a^2 (B+6 C)+2 a b (9 B-7 C)-2 b^2 (3 B-C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+3 a^2 (2 a C+5 b B) \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx\right )-\frac {2 b (3 a B-2 b C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {a B \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (b \left (3 a^2 (B+6 C)+2 a b (9 B-7 C)-2 b^2 (3 B-C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b \left (3 a^2 B-14 a b C-6 b^2 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a^2 (2 a C+5 b B) \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 b (3 a B-2 b C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {a B \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 4271

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (b \left (3 a^2 (B+6 C)+2 a b (9 B-7 C)-2 b^2 (3 B-C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b \left (3 a^2 B-14 a b C-6 b^2 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a \sqrt {a+b} (2 a C+5 b B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )-\frac {2 b (3 a B-2 b C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {a B \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (-b \left (3 a^2 B-14 a b C-6 b^2 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sqrt {a+b} \left (3 a^2 (B+6 C)+2 a b (9 B-7 C)-2 b^2 (3 B-C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {6 a \sqrt {a+b} (2 a C+5 b B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )-\frac {2 b (3 a B-2 b C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {a B \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {2 \sqrt {a+b} \left (3 a^2 (B+6 C)+2 a b (9 B-7 C)-2 b^2 (3 B-C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}+\frac {2 (a-b) \sqrt {a+b} \left (3 a^2 B-14 a b C-6 b^2 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}-\frac {6 a \sqrt {a+b} (2 a C+5 b B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )-\frac {2 b (3 a B-2 b C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {a B \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\)

input

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

output

(a*B*(a + b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/d + (((2*(a - b)*Sqrt[a + b] 
*(3*a^2*B - 6*b^2*B - 14*a*b*C)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*S 
ec[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) + (2*Sqrt[a + b]*(2* 
a*b*(9*B - 7*C) - 2*b^2*(3*B - C) + 3*a^2*(B + 6*C))*Cot[c + d*x]*Elliptic 
F[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))])/d - ( 
6*a*Sqrt[a + b]*(5*b*B + 2*a*C)*Cot[c + d*x]*EllipticPi[(a + b)/a, 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))])/d)/3 - (2*b*(3 
*a*B - 2*b*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*d))/2
 

3.9.33.3.1 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 3042

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

rule 4271

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a 
 + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) 
*((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ 
c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && 
NeQ[a^2 - b^2, 0]
 

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 4409

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ 
.) + (a_)], x_Symbol] :> Simp[c   Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] + 
Simp[d   Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, 
c, d, e, f}, x] && NeQ[b*c - a*d, 0] && 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 4513

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

rule 4544

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Simp[(-C)*Cot 
[e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(m + 1)   Int[( 
a + b*Csc[e + f*x])^(m - 1)*Simp[a*A*(m + 1) + ((A*b + a*B)*(m + 1) + b*C*m 
)*Csc[e + f*x] + (b*B*(m + 1) + a*C*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[ 
{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 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 4560

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

3.9.33.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(4118\) vs. \(2(396)=792\).

Time = 19.83 (sec) , antiderivative size = 4119, normalized size of antiderivative = 9.51

method	result	size
default	\(\text {Expression too large to display}\)	\(4119\)

input

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

output

1/3/d*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+1)*(2*C*b^3*tan( 
d*x+c)-12*B*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(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 
^3*cos(d*x+c)-3*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Elliptic 
E(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^( 
1/2)*a^2*b+6*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE(c 
ot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2 
)*a*b^2-6*B*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(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 
^3*cos(d*x+c)^2+18*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Ellip 
ticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1) 
)^(1/2)*a^2*b-18*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Ellipti 
cF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^ 
(1/2)*a*b^2+2*C*sin(d*x+c)*b^3+6*B*b^3*sin(d*x+c)+3*B*a^3*cos(d*x+c)^2*sin 
(d*x+c)-36*C*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x 
+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)* 
a^2*b*cos(d*x+c)+28*C*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2)) 
*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1 
))^(1/2)*a^2*b*cos(d*x+c)+14*C*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b 
))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(...
 

3.9.33.5 Fricas [F]

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

input

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

output

integral((C*b^2*cos(d*x + c)^2*sec(d*x + c)^4 + B*a^2*cos(d*x + c)^2*sec(d 
*x + c) + (2*C*a*b + B*b^2)*cos(d*x + c)^2*sec(d*x + c)^3 + (C*a^2 + 2*B*a 
*b)*cos(d*x + c)^2*sec(d*x + c)^2)*sqrt(b*sec(d*x + c) + a), x)
 

3.9.33.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.9.33.7 Maxima [F]

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

input

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

output

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

3.9.33.8 Giac [F]

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

input

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

output

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

3.9.33.9 Mupad [F(-1)]

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

input

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

output

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