3.3.73 \(\int \frac {\cos (e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\) [273]

3.3.73.1 Optimal result
3.3.73.2 Mathematica [F]
3.3.73.3 Rubi [A] (verified)
3.3.73.4 Maple [C] (warning: unable to verify)
3.3.73.5 Fricas [F]
3.3.73.6 Sympy [F]
3.3.73.7 Maxima [F]
3.3.73.8 Giac [F]
3.3.73.9 Mupad [F(-1)]

3.3.73.1 Optimal result

Integrand size = 23, antiderivative size = 240 \[ \int \frac {\cos (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {b \sin (e+f x)}{a (a+b) f \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}+\frac {(a+2 b) E\left (\arcsin (\sin (e+f x))\left |\frac {a}{a+b}\right .\right ) \left (a+b-a \sin ^2(e+f x)\right )}{a^2 (a+b) f \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}-\frac {2 b \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),\frac {a}{a+b}\right ) \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}{a^2 f \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}} \]

output
-b*sin(f*x+e)/a/(a+b)/f/(sec(f*x+e)^2*(a+b-a*sin(f*x+e)^2))^(1/2)+(a+2*b)* 
EllipticE(sin(f*x+e),(a/(a+b))^(1/2))*(a+b-a*sin(f*x+e)^2)/a^2/(a+b)/f/(co 
s(f*x+e)^2)^(1/2)/(sec(f*x+e)^2*(a+b-a*sin(f*x+e)^2))^(1/2)/(1-a*sin(f*x+e 
)^2/(a+b))^(1/2)-2*b*EllipticF(sin(f*x+e),(a/(a+b))^(1/2))*(1-a*sin(f*x+e) 
^2/(a+b))^(1/2)/a^2/f/(cos(f*x+e)^2)^(1/2)/(sec(f*x+e)^2*(a+b-a*sin(f*x+e) 
^2))^(1/2)
 
3.3.73.2 Mathematica [F]

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

input
Integrate[Cos[e + f*x]/(a + b*Sec[e + f*x]^2)^(3/2),x]
 
output
Integrate[Cos[e + f*x]/(a + b*Sec[e + f*x]^2)^(3/2), x]
 
3.3.73.3 Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.09, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 4636, 2057, 2058, 315, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4636

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

\(\Big \downarrow \) 2057

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

\(\Big \downarrow \) 2058

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

\(\Big \downarrow \) 315

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 399

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

\(\Big \downarrow \) 323

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

\(\Big \downarrow \) 321

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

\(\Big \downarrow \) 330

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

\(\Big \downarrow \) 327

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

input
Int[Cos[e + f*x]/(a + b*Sec[e + f*x]^2)^(3/2),x]
 
output
(Sqrt[a + b - a*Sin[e + f*x]^2]*(-((b*Sin[e + f*x]*Sqrt[1 - Sin[e + f*x]^2 
])/(a*(a + b)*Sqrt[a + b - a*Sin[e + f*x]^2])) + (((a + 2*b)*EllipticE[Arc 
Sin[Sin[e + f*x]], a/(a + b)]*Sqrt[a + b - a*Sin[e + f*x]^2])/(a*Sqrt[1 - 
(a*Sin[e + f*x]^2)/(a + b)]) - (2*b*(a + b)*EllipticF[ArcSin[Sin[e + f*x]] 
, a/(a + b)]*Sqrt[1 - (a*Sin[e + f*x]^2)/(a + b)])/(a*Sqrt[a + b - a*Sin[e 
 + f*x]^2]))/(a*(a + b))))/(f*Sqrt[1 - Sin[e + f*x]^2]*Sqrt[(a + b - a*Sin 
[e + f*x]^2)/(1 - Sin[e + f*x]^2)])
 

3.3.73.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 315
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), 
x] - Simp[1/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S 
imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) 
*x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 
1] && GtQ[q, 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]
 
3.3.73.4 Maple [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 7.97 (sec) , antiderivative size = 6126, normalized size of antiderivative = 25.52

method result size
default \(\text {Expression too large to display}\) \(6126\)

input
int(cos(f*x+e)/(a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 
output
result too large to display
 
3.3.73.5 Fricas [F]

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

input
integrate(cos(f*x+e)/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")
 
output
integral(sqrt(b*sec(f*x + e)^2 + a)*cos(f*x + e)/(b^2*sec(f*x + e)^4 + 2*a 
*b*sec(f*x + e)^2 + a^2), x)
 
3.3.73.6 Sympy [F]

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

input
integrate(cos(f*x+e)/(a+b*sec(f*x+e)**2)**(3/2),x)
 
output
Integral(cos(e + f*x)/(a + b*sec(e + f*x)**2)**(3/2), x)
 
3.3.73.7 Maxima [F]

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

input
integrate(cos(f*x+e)/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")
 
output
integrate(cos(f*x + e)/(b*sec(f*x + e)^2 + a)^(3/2), x)
 
3.3.73.8 Giac [F]

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

input
integrate(cos(f*x+e)/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")
 
output
sage0*x
 
3.3.73.9 Mupad [F(-1)]

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

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