\(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{a+b \sec (c+d x)} \, dx\) [1319]

Optimal result

Integrand size = 43, antiderivative size = 147 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=-\frac {2 (A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {2 \left (3 A b^2-3 a b B+a^2 (A+3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^3 d}-\frac {2 b \left (A b^2-a (b B-a C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a^3 (a+b) d}+\frac {2 A \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d} \] Output:

-2*(A*b-B*a)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d+2/3*(3*A*b^2-3*B* 
a*b+a^2*(A+3*C))*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^3/d-2*b*(A*b^2-a 
*(B*b-C*a))*EllipticPi(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))/a^3/(a+b)/d+2 
/3*A*cos(d*x+c)^(1/2)*sin(d*x+c)/a/d
 

Mathematica [A] (warning: unable to verify)

Time = 1.59 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.46 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {\frac {2 (-A b+3 a B) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {4 (A+3 C) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-b \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}+4 A \sqrt {\cos (c+d x)} \sin (c+d x)+\frac {6 (-A b+a B) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a^2 b \sqrt {\sin ^2(c+d x)}}}{6 a d} \] Input:

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

Output:

((2*(-(A*b) + 3*a*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + 
(4*(A + 3*C)*((a + b)*EllipticF[(c + d*x)/2, 2] - b*EllipticPi[(2*a)/(a + 
b), (c + d*x)/2, 2]))/(a + b) + 4*A*Sqrt[Cos[c + d*x]]*Sin[c + d*x] + (6*( 
-(A*b) + a*B)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + 
 b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (a^2 - 2*b^2)*EllipticPi[- 
(a/b), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a^2*b*Sqrt[Sin[c + 
d*x]^2]))/(6*a*d)
 

Rubi [A] (verified)

Time = 1.29 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.07, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.326, Rules used = {3042, 4600, 3042, 3528, 27, 3042, 3538, 25, 3042, 3119, 3481, 3042, 3120, 3284}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4600

\(\displaystyle \int \frac {\sqrt {\cos (c+d x)} \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right )}{a \cos (c+d x)+b}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

\(\displaystyle \frac {2 \int \frac {-3 (A b-a B) \cos ^2(c+d x)+a (A+3 C) \cos (c+d x)+A b}{2 \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{3 a}+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {-3 (A b-a B) \cos ^2(c+d x)+a (A+3 C) \cos (c+d x)+A b}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{3 a}+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3538

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {\int \frac {a A b+\left ((A+3 C) a^2-3 b B a+3 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {6 (A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{3 a}+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}\)

\(\Big \downarrow \) 3481

\(\displaystyle \frac {\frac {\frac {\left (a^2 (A+3 C)-3 a b B+3 A b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a}-\frac {3 b \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}}{a}-\frac {6 (A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{3 a}+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\left (a^2 (A+3 C)-3 a b B+3 A b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}-\frac {3 b \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{a}-\frac {6 (A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{3 a}+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^2 (A+3 C)-3 a b B+3 A b^2\right )}{a d}-\frac {3 b \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{a}-\frac {6 (A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{3 a}+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {\frac {\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^2 (A+3 C)-3 a b B+3 A b^2\right )}{a d}-\frac {6 b \left (A b^2-a (b B-a C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a d (a+b)}}{a}-\frac {6 (A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{3 a}+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}\)

Input:

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

Output:

((-6*(A*b - a*B)*EllipticE[(c + d*x)/2, 2])/(a*d) + ((2*(3*A*b^2 - 3*a*b*B 
 + a^2*(A + 3*C))*EllipticF[(c + d*x)/2, 2])/(a*d) - (6*b*(A*b^2 - a*(b*B 
- a*C))*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a*(a + b)*d))/a)/(3*a) 
 + (2*A*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*a*d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 3284

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 
2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[c + d, 0]
 

rule 3481

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ 
B/d   Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d   Int[(a + b* 
Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, 
 B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3528

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

rule 3538

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d)   Int[Sqrt[a + b*Sin[e + f*x]], x] 
, x] - Simp[1/(b*d)   Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ 
e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre 
eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 
] && NeQ[c^2 - d^2, 0]
 

rule 4600

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x 
_)])^(m_.)*((A_.) + (B_.)*sec[(e_.) + (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.) 
*(x_)]^2), x_Symbol] :> Simp[d^(m + 2)   Int[(b + a*Cos[e + f*x])^m*(d*Cos[ 
e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; Fr 
eeQ[{a, b, d, e, f, A, B, C, n}, x] &&  !IntegerQ[n] && IntegerQ[m]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(980\) vs. \(2(146)=292\).

Time = 6.70 (sec) , antiderivative size = 981, normalized size of antiderivative = 6.67

method	result	size
default	\(\text {Expression too large to display}\)	\(981\)

Input:

int(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x,me 
thod=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

-2/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*a^3*A*cos( 
1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-4*a^2*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x 
+1/2*c)^4*b-2*a^3*A*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+2*a^2*A*sin(1/ 
2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*b+a^3*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2* 
sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-A*(2*s 
in(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/ 
2*d*x+1/2*c)^2)^(1/2)*a^2*b+3*a*A*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin( 
1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*A*(2*sin 
(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2* 
d*x+1/2*c)^2)^(1/2)*b^3+3*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/ 
2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b-3*A*(2*sin(1/2 
*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+ 
1/2*c)^2)^(1/2)*a*b^2+3*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2* 
c)^2-1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))*b^3-3*B*a^2 
*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF 
(cos(1/2*d*x+1/2*c),2^(1/2))+3*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elliptic 
F(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a*b^2-3*B*(sin( 
1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2 
*d*x+1/2*c),2^(1/2))*a^3+3*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(co 
s(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*b-3*B*(sin(1...
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^(3/2)/(b*se 
c(d*x + c) + a), x)
 

Giac [F]

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

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

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^(3/2)/(b*se 
c(d*x + c) + a), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(cos(c + d*x))*cos(c + d*x)*sec(c + d*x)**2)/(sec(c + d*x)*b + a) 
,x)*c + int((sqrt(cos(c + d*x))*cos(c + d*x)*sec(c + d*x))/(sec(c + d*x)*b 
 + a),x)*b + int((sqrt(cos(c + d*x))*cos(c + d*x))/(sec(c + d*x)*b + a),x) 
*a