3.9.32 \(\int \cos (c+d x) (a+b \sec (c+d x))^{5/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [832]

3.9.32.1 Optimal result

Integrand size = 40, antiderivative size = 442 \[ \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (35 a b B+23 a^2 C+9 b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b d}+\frac {2 \sqrt {a+b} \left (a^2 b (45 B-23 C)-a b^2 (35 B-17 C)+b^3 (5 B-9 C)+15 a^3 C\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b d}-\frac {2 a^2 \sqrt {a+b} B \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}+\frac {2 b (5 b B+8 a C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 b C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \]

output

-2/15*(a-b)*(35*B*a*b+23*C*a^2+9*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))*(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+2/15*(a^2*b*(45*B-23*C)-a* 
b^2*(35*B-17*C)+b^3*(5*B-9*C)+15*a^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))*(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-2*a^2*B*cot(d*x+c)*Ellipti 
cPi((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+2/5 
*b*C*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/15*b*(5*B*b+8*C*a)*(a+b*sec(d*x 
+c))^(1/2)*tan(d*x+c)/d
 

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

Leaf count is larger than twice the leaf count of optimal. \(7094\) vs. \(2(442)=884\).

Time = 27.27 (sec) , antiderivative size = 7094, normalized size of antiderivative = 16.05 \[ \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \]

input

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

output

Result too large to show
 

3.9.32.3 Rubi [A] (verified)

Time = 1.90 (sec) , antiderivative size = 448, normalized size of antiderivative = 1.01, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4560, 3042, 4406, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4560

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4406

\(\displaystyle \frac {2}{5} \int \frac {1}{2} \sqrt {a+b \sec (c+d x)} \left (5 B a^2+b (5 b B+8 a C) \sec ^2(c+d x)+\left (5 C a^2+10 b B a+3 b^2 C\right ) \sec (c+d x)\right )dx+\frac {2 b C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \int \sqrt {a+b \sec (c+d x)} \left (5 B a^2+b (5 b B+8 a C) \sec ^2(c+d x)+\left (5 C a^2+10 b B a+3 b^2 C\right ) \sec (c+d x)\right )dx+\frac {2 b C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \int \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (5 B a^2+b (5 b B+8 a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (5 C a^2+10 b B a+3 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 b C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 4544

\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {15 B a^3+b \left (23 C a^2+35 b B a+9 b^2 C\right ) \sec ^2(c+d x)+\left (15 C a^3+45 b B a^2+17 b^2 C a+5 b^3 B\right ) \sec (c+d x)}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {2 b (8 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {15 B a^3+b \left (23 C a^2+35 b B a+9 b^2 C\right ) \sec ^2(c+d x)+\left (15 C a^3+45 b B a^2+17 b^2 C a+5 b^3 B\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+\frac {2 b (8 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {15 B a^3+b \left (23 C a^2+35 b B a+9 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (15 C a^3+45 b B a^2+17 b^2 C a+5 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b (8 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 4546

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (b \left (23 a^2 C+35 a b B+9 b^2 C\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx+\int \frac {15 B a^3+\left (15 C a^3+45 b B a^2+17 b^2 C a+5 b^3 B-b \left (23 C a^2+35 b B a+9 b^2 C\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {2 b (8 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (b \left (23 a^2 C+35 a b B+9 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\int \frac {15 B a^3+\left (15 C a^3+45 b B a^2+17 b^2 C a+5 b^3 B-b \left (23 C a^2+35 b B a+9 b^2 C\right )\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 b (8 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 4409

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (15 a^3 B \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx+b \left (23 a^2 C+35 a b B+9 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (15 a^3 C+a^2 b (45 B-23 C)-a b^2 (35 B-17 C)+b^3 (5 B-9 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {2 b (8 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (15 a^3 B \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (23 a^2 C+35 a b B+9 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (15 a^3 C+a^2 b (45 B-23 C)-a b^2 (35 B-17 C)+b^3 (5 B-9 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 b (8 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 4271

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (b \left (23 a^2 C+35 a b B+9 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (15 a^3 C+a^2 b (45 B-23 C)-a b^2 (35 B-17 C)+b^3 (5 B-9 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {30 a^2 B \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )+\frac {2 b (8 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (b \left (23 a^2 C+35 a b B+9 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {30 a^2 B \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}+\frac {2 \sqrt {a+b} \left (15 a^3 C+a^2 b (45 B-23 C)-a b^2 (35 B-17 C)+b^3 (5 B-9 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )+\frac {2 b (8 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (-\frac {2 (a-b) \sqrt {a+b} \left (23 a^2 C+35 a b B+9 b^2 C\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}-\frac {30 a^2 B \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}+\frac {2 \sqrt {a+b} \left (15 a^3 C+a^2 b (45 B-23 C)-a b^2 (35 B-17 C)+b^3 (5 B-9 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )+\frac {2 b (8 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

input

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

output

(2*b*C*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (((-2*(a - b)*Sqrt 
[a + b]*(35*a*b*B + 23*a^2*C + 9*b^2*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*d) + (2*Sqrt[a 
+ b]*(a^2*b*(45*B - 23*C) - a*b^2*(35*B - 17*C) + b^3*(5*B - 9*C) + 15*a^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) - (30*a^2*Sqrt[a + b]*B*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*(5*b*B + 8*a*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3* 
d))/5
 

3.9.32.3.1 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 4406

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d 
_.) + (c_)), x_Symbol] :> Simp[(-b)*d*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m 
 - 1)/(f*m)), x] + Simp[1/m   Int[(a + b*Csc[e + f*x])^(m - 2)*Simp[a^2*c*m 
 + (b^2*d*(m - 1) + 2*a*b*c*m + a^2*d*m)*Csc[e + f*x] + b*(b*c*m + a*d*(2*m 
 - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b* 
c - a*d, 0] && GtQ[m, 1] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m]
 

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 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]
 

3.9.32.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(4179\) vs. \(2(403)=806\).

Time = 10.51 (sec) , antiderivative size = 4180, normalized size of antiderivative = 9.46

method	result	size
default	\(\text {Expression too large to display}\)	\(4180\)

input

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

output

-2/15/d*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+1)*(-3*C*b^3*t 
an(d*x+c)+10*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)-35*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Elli 
pticE(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-35*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Ellipt 
icE(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+5*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+45*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)* 
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)*a^2*b+35*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*El 
lipticF(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^2-9*C*sin(d*x+c)*b^3-5*B*tan(d*x+c)*b^3-5*B*b^3*sin(d*x+c)+ 
46*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*co 
s(d*x+c)-46*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)-23*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+...
 

3.9.32.5 Fricas [F]

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

input

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

3.9.32.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.9.32.7 Maxima [F]

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

input

integrate(cos(d*x+c)*(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), x)
 

3.9.32.8 Giac [F]

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

input

integrate(cos(d*x+c)*(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), x)
 

3.9.32.9 Mupad [F(-1)]

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

input

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