\(\int \frac {\sec ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) [282]

Optimal result

Integrand size = 25, antiderivative size = 91 \[ \int \frac {\sec ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {(a+2 b) \text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{2 (a+b)^{3/2} f}+\frac {\sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{2 (a+b) f} \] Output:

1/2*(a+2*b)*arctanh((a+b)^(1/2)*sin(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2))/(a+b) 
^(3/2)/f+1/2*sec(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e)/(a+b)/f
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 6.42 (sec) , antiderivative size = 408, normalized size of antiderivative = 4.48 \[ \int \frac {\sec ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\sec ^3(e+f x) \left (1+\frac {b \sin ^2(e+f x)}{a}\right ) \tan (e+f x) \left (45 a \arcsin \left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right )+30 b \arcsin \left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right ) \sin ^2(e+f x)+16 a \operatorname {Hypergeometric2F1}\left (2,3,\frac {7}{2},-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{5/2}+16 b \operatorname {Hypergeometric2F1}\left (2,3,\frac {7}{2},-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{5/2}-45 a \sqrt {-\frac {(a+b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{a^2}}-30 b \sin ^2(e+f x) \sqrt {-\frac {(a+b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{a^2}}\right )}{30 a f \sqrt {a+b \sin ^2(e+f x)} \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{3/2}} \] Input:

Integrate[Sec[e + f*x]^3/Sqrt[a + b*Sin[e + f*x]^2],x]
 

Output:

(Sec[e + f*x]^3*(1 + (b*Sin[e + f*x]^2)/a)*Tan[e + f*x]*(45*a*ArcSin[Sqrt[ 
-(((a + b)*Tan[e + f*x]^2)/a)]] + 30*b*ArcSin[Sqrt[-(((a + b)*Tan[e + f*x] 
^2)/a)]]*Sin[e + f*x]^2 + 16*a*Hypergeometric2F1[2, 3, 7/2, -(((a + b)*Tan 
[e + f*x]^2)/a)]*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + 
b)*Tan[e + f*x]^2)/a))^(5/2) + 16*b*Hypergeometric2F1[2, 3, 7/2, -(((a + b 
)*Tan[e + f*x]^2)/a)]*Sin[e + f*x]^2*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f 
*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(5/2) - 45*a*Sqrt[-(((a + b)*Se 
c[e + f*x]^2*(a + b*Sin[e + f*x]^2)*Tan[e + f*x]^2)/a^2)] - 30*b*Sin[e + f 
*x]^2*Sqrt[-(((a + b)*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2)*Tan[e + f*x]^2 
)/a^2)]))/(30*a*f*Sqrt[a + b*Sin[e + f*x]^2]*Sqrt[(Sec[e + f*x]^2*(a + b*S 
in[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(3/2))
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3669, 296, 291, 219}

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.

\(\displaystyle \int \frac {\sec ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\cos (e+f x)^3 \sqrt {a+b \sin (e+f x)^2}}dx\)

\(\Big \downarrow \) 3669

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

\(\Big \downarrow \) 296

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 219

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

Input:

Int[Sec[e + f*x]^3/Sqrt[a + b*Sin[e + f*x]^2],x]
 

Output:

(((a + 2*b)*ArcTanh[(Sqrt[a + b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]] 
)/(2*(a + b)^(3/2)) + (Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(2*(a + b) 
*(1 - Sin[e + f*x]^2)))/f
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 291

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

rule 296

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[(b*c + 2*(p + 1)*(b*c - a*d))/(2*a*(p + 1)*(b*c - a*d))   Int[ 
(a + b*x^2)^(p + 1)*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, q}, x] && N 
eQ[b*c - a*d, 0] && EqQ[2*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1 
]) && NeQ[p, -1]
 

rule 3042

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

rule 3669

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(359\) vs. \(2(79)=158\).

Time = 0.47 (sec) , antiderivative size = 360, normalized size of antiderivative = 3.96

method	result	size
default	\(\frac {-\left (-\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{2}-3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a b -2 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) b^{2}+\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{2}+3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a b +2 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) b^{2}\right ) \cos \left (f x +e \right )^{2}+2 \sin \left (f x +e \right ) \sqrt {a +b -b \cos \left (f x +e \right )^{2}}\, \left (a +b \right )^{\frac {3}{2}}}{4 \left (a +b \right )^{\frac {5}{2}} \cos \left (f x +e \right )^{2} f}\)	\(360\)

Input:

int(sec(f*x+e)^3/(a+b*sin(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/4*(-(-ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin( 
f*x+e)+a))*a^2-3*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/ 
2)+b*sin(f*x+e)+a))*a*b-2*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+ 
e)^2)^(1/2)+b*sin(f*x+e)+a))*b^2+ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*c 
os(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^2+3*ln(2/(1+sin(f*x+e))*((a+b)^(1/2) 
*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a*b+2*ln(2/(1+sin(f*x+e))*((a 
+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*b^2)*cos(f*x+e)^2+2* 
sin(f*x+e)*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(3/2))/(a+b)^(5/2)/cos(f*x+e)^ 
2/f
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 167 vs. \(2 (79) = 158\).

Time = 0.16 (sec) , antiderivative size = 361, normalized size of antiderivative = 3.97 \[ \int \frac {\sec ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\left [\frac {{\left (a + 2 \, b\right )} \sqrt {a + b} \cos \left (f x + e\right )^{2} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 8 \, {\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} \sin \left (f x + e\right ) + 8 \, a^{2} + 16 \, a b + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) + 4 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \sin \left (f x + e\right )}{8 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2}}, -\frac {{\left (a + 2 \, b\right )} \sqrt {-a - b} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{2 \, {\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \sin \left (f x + e\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2}}\right ] \] Input:

integrate(sec(f*x+e)^3/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

Output:

[1/8*((a + 2*b)*sqrt(a + b)*cos(f*x + e)^2*log(((a^2 + 8*a*b + 8*b^2)*cos( 
f*x + e)^4 - 8*(a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^2 - 4*((a + 2*b)*cos(f*x 
 + e)^2 - 2*a - 2*b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a + b)*sin(f*x + 
 e) + 8*a^2 + 16*a*b + 8*b^2)/cos(f*x + e)^4) + 4*sqrt(-b*cos(f*x + e)^2 + 
 a + b)*(a + b)*sin(f*x + e))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^2), -1/4 
*((a + 2*b)*sqrt(-a - b)*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - 2*a - 2*b) 
*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a - b)/(((a*b + b^2)*cos(f*x + e)^2 
 - a^2 - 2*a*b - b^2)*sin(f*x + e)))*cos(f*x + e)^2 - 2*sqrt(-b*cos(f*x + 
e)^2 + a + b)*(a + b)*sin(f*x + e))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^2) 
]
 

Sympy [F]

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

integrate(sec(f*x+e)**3/(a+b*sin(f*x+e)**2)**(1/2),x)
 

Output:

Integral(sec(e + f*x)**3/sqrt(a + b*sin(e + f*x)**2), x)
 

Maxima [F]

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

integrate(sec(f*x+e)^3/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(sec(f*x+e)^3/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int \frac {1}{{\cos \left (e+f\,x\right )}^3\,\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}} \,d x \] Input:

int(1/(cos(e + f*x)^3*(a + b*sin(e + f*x)^2)^(1/2)),x)
 

Output:

int(1/(cos(e + f*x)^3*(a + b*sin(e + f*x)^2)^(1/2)), x)
 

Reduce [F]

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

int(sec(f*x+e)^3/(a+b*sin(f*x+e)^2)^(1/2),x)
 

Output:

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