\(\int \cos ^3(e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\) [245]

Optimal result

Integrand size = 25, antiderivative size = 245 \[ \int \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {a \cos ^2(e+f x) \sin (e+f x) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}{3 f}+\frac {2 (a+2 b) \sqrt {\cos ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |\frac {a}{a+b}\right .\right ) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}{3 f \sqrt {\frac {a+b-a \sin ^2(e+f x)}{a+b}}}-\frac {b (a+b) \sqrt {\cos ^2(e+f x)} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),\frac {a}{a+b}\right ) \sqrt {\frac {a+b-a \sin ^2(e+f x)}{a+b}} \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}{3 f \left (a+b-a \sin ^2(e+f x)\right )} \] Output:

1/3*a*cos(f*x+e)^2*sin(f*x+e)*(sec(f*x+e)^2*(a+b-a*sin(f*x+e)^2))^(1/2)/f+ 
2/3*(a+2*b)*(cos(f*x+e)^2)^(1/2)*EllipticE(sin(f*x+e),(a/(a+b))^(1/2))*(se 
c(f*x+e)^2*(a+b-a*sin(f*x+e)^2))^(1/2)/f/((a+b-a*sin(f*x+e)^2)/(a+b))^(1/2 
)-1/3*b*(a+b)*(cos(f*x+e)^2)^(1/2)*EllipticF(sin(f*x+e),(a/(a+b))^(1/2))*( 
(a+b-a*sin(f*x+e)^2)/(a+b))^(1/2)*(sec(f*x+e)^2*(a+b-a*sin(f*x+e)^2))^(1/2 
)/f/(a+b-a*sin(f*x+e)^2)
 

Mathematica [A] (verified)

Time = 1.84 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.73 \[ \int \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)} \left (4 \sqrt {2} \left (a^2+3 a b+2 b^2\right ) \sqrt {\frac {a+2 b+a \cos (2 (e+f x))}{a+b}} E\left (e+f x\left |\frac {a}{a+b}\right .\right )-2 \sqrt {2} b (a+b) \sqrt {\frac {a+2 b+a \cos (2 (e+f x))}{a+b}} \operatorname {EllipticF}\left (e+f x,\frac {a}{a+b}\right )+a (a+2 b+a \cos (2 (e+f x))) \sin (2 (e+f x))\right )}{6 f (a+2 b+a \cos (2 (e+f x)))} \] Input:

Integrate[Cos[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(3/2),x]
 

Output:

(Cos[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2]*(4*Sqrt[2]*(a^2 + 3*a*b + 2*b^2)* 
Sqrt[(a + 2*b + a*Cos[2*(e + f*x)])/(a + b)]*EllipticE[e + f*x, a/(a + b)] 
 - 2*Sqrt[2]*b*(a + b)*Sqrt[(a + 2*b + a*Cos[2*(e + f*x)])/(a + b)]*Ellipt 
icF[e + f*x, a/(a + b)] + a*(a + 2*b + a*Cos[2*(e + f*x)])*Sin[2*(e + f*x) 
]))/(6*f*(a + 2*b + a*Cos[2*(e + f*x)]))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4636, 2057, 2058, 318, 399, 323, 321, 330, 327}

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.

\(\displaystyle \int \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \sec (e+f x)^2\right )^{3/2}}{\sec (e+f x)^3}dx\)

\(\Big \downarrow \) 4636

\(\displaystyle \frac {\int \left (1-\sin ^2(e+f x)\right ) \left (a+\frac {b}{1-\sin ^2(e+f x)}\right )^{3/2}d\sin (e+f x)}{f}\)

\(\Big \downarrow \) 2057

\(\displaystyle \frac {\int \left (1-\sin ^2(e+f x)\right ) \left (\frac {-a \sin ^2(e+f x)+a+b}{1-\sin ^2(e+f x)}\right )^{3/2}d\sin (e+f x)}{f}\)

\(\Big \downarrow \) 2058

\(\displaystyle \frac {\sqrt {1-\sin ^2(e+f x)} \sqrt {\frac {-a \sin ^2(e+f x)+a+b}{1-\sin ^2(e+f x)}} \int \frac {\left (-a \sin ^2(e+f x)+a+b\right )^{3/2}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{f \sqrt {-a \sin ^2(e+f x)+a+b}}\)

\(\Big \downarrow \) 318

\(\displaystyle \frac {\sqrt {1-\sin ^2(e+f x)} \sqrt {\frac {-a \sin ^2(e+f x)+a+b}{1-\sin ^2(e+f x)}} \left (\frac {1}{3} a \sin (e+f x) \sqrt {1-\sin ^2(e+f x)} \sqrt {-a \sin ^2(e+f x)+a+b}-\frac {1}{3} \int \frac {2 a (a+2 b) \sin ^2(e+f x)+(a+b) (a-3 (a+b))}{\sqrt {1-\sin ^2(e+f x)} \sqrt {-a \sin ^2(e+f x)+a+b}}d\sin (e+f x)\right )}{f \sqrt {-a \sin ^2(e+f x)+a+b}}\)

\(\Big \downarrow \) 399

\(\displaystyle \frac {\sqrt {1-\sin ^2(e+f x)} \sqrt {\frac {-a \sin ^2(e+f x)+a+b}{1-\sin ^2(e+f x)}} \left (\frac {1}{3} \left (2 (a+2 b) \int \frac {\sqrt {-a \sin ^2(e+f x)+a+b}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)-b (a+b) \int \frac {1}{\sqrt {1-\sin ^2(e+f x)} \sqrt {-a \sin ^2(e+f x)+a+b}}d\sin (e+f x)\right )+\frac {1}{3} a \sqrt {1-\sin ^2(e+f x)} \sin (e+f x) \sqrt {-a \sin ^2(e+f x)+a+b}\right )}{f \sqrt {-a \sin ^2(e+f x)+a+b}}\)

\(\Big \downarrow \) 323

\(\displaystyle \frac {\sqrt {1-\sin ^2(e+f x)} \sqrt {\frac {-a \sin ^2(e+f x)+a+b}{1-\sin ^2(e+f x)}} \left (\frac {1}{3} \left (2 (a+2 b) \int \frac {\sqrt {-a \sin ^2(e+f x)+a+b}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)-\frac {b (a+b) \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}} \int \frac {1}{\sqrt {1-\sin ^2(e+f x)} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}d\sin (e+f x)}{\sqrt {-a \sin ^2(e+f x)+a+b}}\right )+\frac {1}{3} a \sqrt {1-\sin ^2(e+f x)} \sin (e+f x) \sqrt {-a \sin ^2(e+f x)+a+b}\right )}{f \sqrt {-a \sin ^2(e+f x)+a+b}}\)

\(\Big \downarrow \) 321

\(\displaystyle \frac {\sqrt {1-\sin ^2(e+f x)} \sqrt {\frac {-a \sin ^2(e+f x)+a+b}{1-\sin ^2(e+f x)}} \left (\frac {1}{3} \left (2 (a+2 b) \int \frac {\sqrt {-a \sin ^2(e+f x)+a+b}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)-\frac {b (a+b) \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),\frac {a}{a+b}\right )}{\sqrt {-a \sin ^2(e+f x)+a+b}}\right )+\frac {1}{3} a \sqrt {1-\sin ^2(e+f x)} \sin (e+f x) \sqrt {-a \sin ^2(e+f x)+a+b}\right )}{f \sqrt {-a \sin ^2(e+f x)+a+b}}\)

\(\Big \downarrow \) 330

\(\displaystyle \frac {\sqrt {1-\sin ^2(e+f x)} \sqrt {\frac {-a \sin ^2(e+f x)+a+b}{1-\sin ^2(e+f x)}} \left (\frac {1}{3} \left (\frac {2 (a+2 b) \sqrt {-a \sin ^2(e+f x)+a+b} \int \frac {\sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{\sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}-\frac {b (a+b) \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),\frac {a}{a+b}\right )}{\sqrt {-a \sin ^2(e+f x)+a+b}}\right )+\frac {1}{3} a \sqrt {1-\sin ^2(e+f x)} \sin (e+f x) \sqrt {-a \sin ^2(e+f x)+a+b}\right )}{f \sqrt {-a \sin ^2(e+f x)+a+b}}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {\sqrt {1-\sin ^2(e+f x)} \sqrt {\frac {-a \sin ^2(e+f x)+a+b}{1-\sin ^2(e+f x)}} \left (\frac {1}{3} \left (\frac {2 (a+2 b) \sqrt {-a \sin ^2(e+f x)+a+b} E\left (\arcsin (\sin (e+f x))\left |\frac {a}{a+b}\right .\right )}{\sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}-\frac {b (a+b) \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),\frac {a}{a+b}\right )}{\sqrt {-a \sin ^2(e+f x)+a+b}}\right )+\frac {1}{3} a \sqrt {1-\sin ^2(e+f x)} \sin (e+f x) \sqrt {-a \sin ^2(e+f x)+a+b}\right )}{f \sqrt {-a \sin ^2(e+f x)+a+b}}\)

Input:

Int[Cos[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(3/2),x]
 

Output:

(Sqrt[1 - Sin[e + f*x]^2]*Sqrt[(a + b - a*Sin[e + f*x]^2)/(1 - Sin[e + f*x 
]^2)]*((a*Sin[e + f*x]*Sqrt[1 - Sin[e + f*x]^2]*Sqrt[a + b - a*Sin[e + f*x 
]^2])/3 + ((2*(a + 2*b)*EllipticE[ArcSin[Sin[e + f*x]], a/(a + b)]*Sqrt[a 
+ b - a*Sin[e + f*x]^2])/Sqrt[1 - (a*Sin[e + f*x]^2)/(a + b)] - (b*(a + b) 
*EllipticF[ArcSin[Sin[e + f*x]], a/(a + b)]*Sqrt[1 - (a*Sin[e + f*x]^2)/(a 
 + b)])/Sqrt[a + b - a*Sin[e + f*x]^2])/3))/(f*Sqrt[a + b - a*Sin[e + f*x] 
^2])
 

Defintions of rubi rules used

rule 318

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S 
imp[1/(b*(2*(p + q) + 1))   Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b 
*c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 
 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G 
tQ[q, 1] && NeQ[2*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntBinomialQ[a, b, c, 
d, 2, p, q, x]
 

rule 321

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c 
/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 
0] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])
 

rule 323

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[Sqrt[1 + (d/c)*x^2]/Sqrt[c + d*x^2]   Int[1/(Sqrt[a + b*x^2]*Sqrt[1 + ( 
d/c)*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] &&  !GtQ[c, 0]
 

rule 327

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

rule 330

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
Sqrt[a + b*x^2]/Sqrt[1 + (b/a)*x^2]   Int[Sqrt[1 + (b/a)*x^2]/Sqrt[c + d*x^ 
2], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &&  !GtQ[a, 
0]
 

rule 399

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_) 
^2]), x_Symbol] :> Simp[f/b   Int[Sqrt[a + b*x^2]/Sqrt[c + d*x^2], x], x] + 
 Simp[(b*e - a*f)/b   Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; Fr 
eeQ[{a, b, c, d, e, f}, x] &&  !((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && 
(PosQ[d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-b/a, -d/c])))))
 

rule 2057

Int[(u_.)*((a_) + (b_.)/((c_) + (d_.)*(x_)^(n_)))^(p_), x_Symbol] :> Int[u* 
((b + a*c + a*d*x^n)/(c + d*x^n))^p, x] /; FreeQ[{a, b, c, d, n, p}, x]
 

rule 2058

Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))^(q_.)*((c_) + (d_.)*(x_)^(n_))^ 
(r_.))^(p_), x_Symbol] :> Simp[Simp[(e*(a + b*x^n)^q*(c + d*x^n)^r)^p/((a + 
 b*x^n)^(p*q)*(c + d*x^n)^(p*r))]   Int[u*(a + b*x^n)^(p*q)*(c + d*x^n)^(p* 
r), x], x] /; FreeQ[{a, b, c, d, e, n, p, q, r}, x]
 

rule 3042

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

rule 4636

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_ 
))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f 
 Subst[Int[(a + b/(1 - ff^2*x^2)^(n/2))^p/(1 - ff^2*x^2)^((m + 1)/2), x], x 
, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2] 
 && IntegerQ[n/2] &&  !IntegerQ[p]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 4.51 (sec) , antiderivative size = 2625, normalized size of antiderivative = 10.71

method	result	size
default	\(\text {Expression too large to display}\)	\(2625\)

Input:

int(cos(f*x+e)^3*(a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

1/3/f/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/(2*I*a^(1/2)*b^(1/2)-a+b)*co 
s(f*x+e)^3*((2*cos(f*x+e)^2+4*cos(f*x+e)+2)*(-1/(a+b)*(I*a^(1/2)*b^(1/2)*c 
os(f*x+e)-I*a^(1/2)*b^(1/2)-cos(f*x+e)*a-b)/(1+cos(f*x+e)))^(1/2)*(1/(a+b) 
*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+cos(f*x+e)*a+b)/(1+cos(f* 
x+e)))^(1/2)*a^3*EllipticE(((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*(cot(f* 
x+e)-csc(f*x+e)),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2) 
/(a+b)^2)^(1/2))+(8*cos(f*x+e)^2+16*cos(f*x+e)+8)*(1/(a+b)*(I*a^(1/2)*b^(1 
/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+cos(f*x+e)*a+b)/(1+cos(f*x+e)))^(1/2)*(-1 
/(a+b)*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-cos(f*x+e)*a-b)/(1+ 
cos(f*x+e)))^(1/2)*a^2*b*EllipticE(((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2) 
*(cot(f*x+e)-csc(f*x+e)),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6* 
a*b-b^2)/(a+b)^2)^(1/2))+(10*cos(f*x+e)^2+20*cos(f*x+e)+10)*(1/(a+b)*(I*a^ 
(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+cos(f*x+e)*a+b)/(1+cos(f*x+e))) 
^(1/2)*(-1/(a+b)*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-cos(f*x+e 
)*a-b)/(1+cos(f*x+e)))^(1/2)*a*b^2*EllipticE(((2*I*a^(1/2)*b^(1/2)+a-b)/(a 
+b))^(1/2)*(cot(f*x+e)-csc(f*x+e)),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3 
/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))+(4*cos(f*x+e)^2+8*cos(f*x+e)+4)*(1/(a+b 
)*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+cos(f*x+e)*a+b)/(1+cos(f 
*x+e)))^(1/2)*(-1/(a+b)*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-co 
s(f*x+e)*a-b)/(1+cos(f*x+e)))^(1/2)*b^3*EllipticE(((2*I*a^(1/2)*b^(1/2)...
 

Fricas [F]

\[ \int \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} \,d x } \] Input:

integrate(cos(f*x+e)^3*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

integral((b*cos(f*x + e)^3*sec(f*x + e)^2 + a*cos(f*x + e)^3)*sqrt(b*sec(f 
*x + e)^2 + a), x)
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \] Input:

integrate(cos(f*x+e)**3*(a+b*sec(f*x+e)**2)**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} \,d x } \] Input:

integrate(cos(f*x+e)^3*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^(3/2)*cos(f*x + e)^3, x)
 

Giac [F]

\[ \int \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} \,d x } \] Input:

integrate(cos(f*x+e)^3*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^(3/2)*cos(f*x + e)^3, x)
 

Mupad [F(-1)]

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

int(cos(e + f*x)^3*(a + b/cos(e + f*x)^2)^(3/2),x)
 

Output:

int(cos(e + f*x)^3*(a + b/cos(e + f*x)^2)^(3/2), x)
 

Reduce [F]

\[ \int \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\left (\int \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \cos \left (f x +e \right )^{3} \sec \left (f x +e \right )^{2}d x \right ) b +\left (\int \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \cos \left (f x +e \right )^{3}d x \right ) a \] Input:

int(cos(f*x+e)^3*(a+b*sec(f*x+e)^2)^(3/2),x)
 

Output:

int(sqrt(sec(e + f*x)**2*b + a)*cos(e + f*x)**3*sec(e + f*x)**2,x)*b + int 
(sqrt(sec(e + f*x)**2*b + a)*cos(e + f*x)**3,x)*a