\(\int \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [1190]

Optimal result
Mathematica [C] (warning: unable to verify)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [C] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 41, antiderivative size = 98 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a (A+B-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a (A+3 (B+C)) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a C \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 a A \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 6.09 (sec) , antiderivative size = 278, normalized size of antiderivative = 2.84 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a (1+\cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (-4 (A+3 (B+C)) \cos (c+d x) \sqrt {\cos ^2(d x-\arctan (\cot (c)))} \sqrt {\csc ^2(c)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec (d x-\arctan (\cot (c))) \sin (c)+\frac {\csc (c) \left (9 (A+B-C) \cos (c-d x-\arctan (\tan (c))) \sec (c)+3 (A+B-C) \cos (c+d x+\arctan (\tan (c))) \sec (c)-2 \sqrt {\sec ^2(c)} (3 (A+B-2 C) \cos (d x)+3 (A+B) \cos (2 c+d x)-A \sin (c) \sin (2 (c+d x)))\right )}{\sqrt {\sec ^2(c)}}-\frac {6 (A+B-C) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sec (c) \sin (d x+\arctan (\tan (c)))}{\sqrt {\sec ^2(c)} \sqrt {\sin ^2(d x+\arctan (\tan (c)))}}\right )}{12 d \sqrt {\cos (c+d x)}} \] Input:

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

Output:

(a*(1 + Cos[c + d*x])*Sec[(c + d*x)/2]^2*(-4*(A + 3*(B + C))*Cos[c + d*x]* 
Sqrt[Cos[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Csc[c]^2]*HypergeometricPFQ[{1/4, 1 
/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[d*x - ArcTan[Cot[c]]]*Sin[c] 
+ (Csc[c]*(9*(A + B - C)*Cos[c - d*x - ArcTan[Tan[c]]]*Sec[c] + 3*(A + B - 
 C)*Cos[c + d*x + ArcTan[Tan[c]]]*Sec[c] - 2*Sqrt[Sec[c]^2]*(3*(A + B - 2* 
C)*Cos[d*x] + 3*(A + B)*Cos[2*c + d*x] - A*Sin[c]*Sin[2*(c + d*x)])))/Sqrt 
[Sec[c]^2] - (6*(A + B - C)*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x 
 + ArcTan[Tan[c]]]^2]*Sec[c]*Sin[d*x + ArcTan[Tan[c]]])/(Sqrt[Sec[c]^2]*Sq 
rt[Sin[d*x + ArcTan[Tan[c]]]^2])))/(12*d*Sqrt[Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.03, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.317, Rules used = {3042, 4600, 3042, 3510, 27, 3042, 3502, 27, 3042, 3227, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4600

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3510

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \int \frac {a (A+3 (B+C))+3 a (A+B-C) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx+\frac {2 a A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}+\frac {2 a C \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

((6*a*(A + B - C)*EllipticE[(c + d*x)/2, 2])/d + (2*a*(A + 3*(B + C))*Elli 
pticF[(c + d*x)/2, 2])/d)/3 + (2*a*C*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) 
+ (2*a*A*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)
 

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 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 3227
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3502
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

rule 3510
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ 
e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S 
imp[1/(b^2*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( 
m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m 
 + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) 
))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F 
reeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 
 0] && LtQ[m, -1]
 

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. \(379\) vs. \(2(94)=188\).

Time = 3.35 (sec) , antiderivative size = 380, normalized size of antiderivative = 3.88

method result size
default \(-\frac {2 a \left (4 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) A +A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-6 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) C +3 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(380\)

Input:

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

Output:

-2/3*a*(4*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-2*sin(1/2*d*x+1/2*c)^2 
*cos(1/2*d*x+1/2*c)*A+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))-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))+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)*El 
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si 
n(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-6*sin(1/ 
2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*C+3*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si 
n(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*C*(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)))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2) 
/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.02 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {-i \, \sqrt {2} {\left (A + 3 \, B + 3 \, C\right )} a \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} {\left (A + 3 \, B + 3 \, C\right )} a \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (A + B - C\right )} a \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (A + B - C\right )} a \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (A a \cos \left (d x + c\right ) + 3 \, C a\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )} \] Input:

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

Output:

1/3*(-I*sqrt(2)*(A + 3*B + 3*C)*a*cos(d*x + c)*weierstrassPInverse(-4, 0, 
cos(d*x + c) + I*sin(d*x + c)) + I*sqrt(2)*(A + 3*B + 3*C)*a*cos(d*x + c)* 
weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*I*sqrt(2)*(A 
 + B - C)*a*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, 
 cos(d*x + c) + I*sin(d*x + c))) - 3*I*sqrt(2)*(A + B - C)*a*cos(d*x + c)* 
weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x 
 + c))) + 2*(A*a*cos(d*x + c) + 3*C*a)*sqrt(cos(d*x + c))*sin(d*x + c))/(d 
*cos(d*x + c))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Giac [F]

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

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

Mupad [B] (verification not implemented)

Time = 13.81 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.49 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,A\,a\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,A\,a\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,a\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,a\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,a\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,a\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

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

Output:

(2*A*a*(cos(c + d*x)^(1/2)*sin(c + d*x) + ellipticF(c/2 + (d*x)/2, 2)))/(3 
*d) + (2*A*a*ellipticE(c/2 + (d*x)/2, 2))/d + (2*B*a*ellipticE(c/2 + (d*x) 
/2, 2))/d + (2*B*a*ellipticF(c/2 + (d*x)/2, 2))/d + (2*C*a*ellipticF(c/2 + 
 (d*x)/2, 2))/d + (2*C*a*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + 
d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2))
 

Reduce [F]

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

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

Output:

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