\(\int (b \sec (c+d x))^{3/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [47]

Optimal result

Integrand size = 32, antiderivative size = 169 \[ \int (b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {6 b^2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 b B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {6 b C \sqrt {b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 B (b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {2 C (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 b d} \] Output:

-6/5*b^2*C*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/cos(d*x+c)^(1/2)/(b*sec 
(d*x+c))^(1/2)+2/3*b*B*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1 
/2))*(b*sec(d*x+c))^(1/2)/d+6/5*b*C*(b*sec(d*x+c))^(1/2)*sin(d*x+c)/d+2/3* 
B*(b*sec(d*x+c))^(3/2)*sin(d*x+c)/d+2/5*C*(b*sec(d*x+c))^(5/2)*sin(d*x+c)/ 
b/d
 

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.60 \[ \int (b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {(b \sec (c+d x))^{5/2} \left (-36 C \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 B \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+21 C \sin (c+d x)+10 B \sin (2 (c+d x))+9 C \sin (3 (c+d x))\right )}{30 b d} \] Input:

Integrate[(b*Sec[c + d*x])^(3/2)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
 

Output:

((b*Sec[c + d*x])^(5/2)*(-36*C*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2 
] + 20*B*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 21*C*Sin[c + d*x] 
+ 10*B*Sin[2*(c + d*x)] + 9*C*Sin[3*(c + d*x)]))/(30*b*d)
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.07, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.406, Rules used = {3042, 4535, 27, 2030, 3042, 4255, 3042, 4255, 3042, 4258, 3042, 3119, 3120}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4535

\(\displaystyle \frac {B \int (b \sec (c+d x))^{5/2}dx}{b}+\int C \sec ^2(c+d x) (b \sec (c+d x))^{3/2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {B \int (b \sec (c+d x))^{5/2}dx}{b}+C \int \sec ^2(c+d x) (b \sec (c+d x))^{3/2}dx\)

\(\Big \downarrow \) 2030

\(\displaystyle \frac {C \int (b \sec (c+d x))^{7/2}dx}{b^2}+\frac {B \int (b \sec (c+d x))^{5/2}dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {C \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}dx}{b^2}+\frac {B \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx}{b}\)

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

\(\displaystyle \frac {B \left (\frac {2 b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )}{b}+\frac {C \left (\frac {3}{5} b^2 \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-\frac {2 b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d}\right )}{b^2}\)

Input:

Int[(b*Sec[c + d*x])^(3/2)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
 

Output:

(B*((2*b^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x 
]])/(3*d) + (2*b*(b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d)))/b + (C*((2*b 
*(b*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(5*d) + (3*b^2*((-2*b^2*EllipticE[(c 
 + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*b*Sqrt[b*S 
ec[c + d*x]]*Sin[c + d*x])/d))/5))/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 2030

Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m   Int[(b*v) 
^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
 

rule 3042

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

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 4255

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

rule 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

rule 4535

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* 
(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b   Int[(b*Cs 
c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) 
, x] /; FreeQ[{b, e, f, A, B, C, m}, x]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 3.50 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.75

method	result	size
parts	\(\frac {B \left (-\frac {2 i \left (\cos \left (d x +c \right )+1\right ) \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{3}+\frac {2 \tan \left (d x +c \right )}{3}\right ) b \sqrt {b \sec \left (d x +c \right )}}{d}+\frac {2 C \sqrt {b \sec \left (d x +c \right )}\, b \left (3 \sin \left (d x +c \right )+\tan \left (d x +c \right )+\sec \left (d x +c \right ) \tan \left (d x +c \right )+i \left (3 \cos \left (d x +c \right )^{2}+6 \cos \left (d x +c \right )+3\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+i \left (-3 \cos \left (d x +c \right )^{2}-6 \cos \left (d x +c \right )-3\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )\right )}{5 d \left (\cos \left (d x +c \right )+1\right )}\)	\(295\)
default	\(-\frac {2 b \left (9 i \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) C \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+5 i \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) B \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+9 i \left (-\cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-1\right ) C \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+5 B \left (-\sin \left (d x +c \right )-\tan \left (d x +c \right )\right )+3 C \left (-3 \sin \left (d x +c \right )-\tan \left (d x +c \right )-\sec \left (d x +c \right ) \tan \left (d x +c \right )\right )\right ) \sqrt {b \sec \left (d x +c \right )}}{15 d \left (\cos \left (d x +c \right )+1\right )}\)	\(306\)

Input:

int((b*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVER 
BOSE)
 

Output:

B/d*(-2/3*I*(cos(d*x+c)+1)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(cos( 
d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+2/3*tan(d*x+c))*b*(b*se 
c(d*x+c))^(1/2)+2/5*C/d*(b*sec(d*x+c))^(1/2)*b/(cos(d*x+c)+1)*(3*sin(d*x+c 
)+tan(d*x+c)+sec(d*x+c)*tan(d*x+c)+I*(3*cos(d*x+c)^2+6*cos(d*x+c)+3)*(1/(c 
os(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(csc(d*x 
+c)-cot(d*x+c)),I)+I*(-3*cos(d*x+c)^2-6*cos(d*x+c)-3)*(1/(cos(d*x+c)+1))^( 
1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)) 
,I))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.22 \[ \int (b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {-5 i \, \sqrt {2} B b^{\frac {3}{2}} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} B b^{\frac {3}{2}} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 9 i \, \sqrt {2} C b^{\frac {3}{2}} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 9 i \, \sqrt {2} C b^{\frac {3}{2}} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (9 \, C b \cos \left (d x + c\right )^{2} + 5 \, B b \cos \left (d x + c\right ) + 3 \, C b\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{2}} \] Input:

integrate((b*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm= 
"fricas")
 

Output:

1/15*(-5*I*sqrt(2)*B*b^(3/2)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos 
(d*x + c) + I*sin(d*x + c)) + 5*I*sqrt(2)*B*b^(3/2)*cos(d*x + c)^2*weierst 
rassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 9*I*sqrt(2)*C*b^(3/2) 
*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x 
+ c) + I*sin(d*x + c))) + 9*I*sqrt(2)*C*b^(3/2)*cos(d*x + c)^2*weierstrass 
Zeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2 
*(9*C*b*cos(d*x + c)^2 + 5*B*b*cos(d*x + c) + 3*C*b)*sqrt(b/cos(d*x + c))* 
sin(d*x + c))/(d*cos(d*x + c)^2)
 

Sympy [F]

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

integrate((b*sec(d*x+c))**(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)
 

Output:

Integral((b*sec(c + d*x))**(3/2)*(B + C*sec(c + d*x))*sec(c + d*x), x)
 

Maxima [F]

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

integrate((b*sec(d*x+c))^(3/2)*(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))^(3/2), x)
 

Giac [F]

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

integrate((b*sec(d*x+c))^(3/2)*(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))^(3/2), x)
 

Mupad [F(-1)]

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

int((B/cos(c + d*x) + C/cos(c + d*x)^2)*(b/cos(c + d*x))^(3/2),x)
 

Output:

int((B/cos(c + d*x) + C/cos(c + d*x)^2)*(b/cos(c + d*x))^(3/2), x)
 

Reduce [F]

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

int((b*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)
 

Output:

sqrt(b)*b*(int(sqrt(sec(c + d*x))*sec(c + d*x)**3,x)*c + int(sqrt(sec(c + 
d*x))*sec(c + d*x)**2,x)*b)