\(\int \frac {\sec (e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\) [290]

Optimal result

Integrand size = 23, antiderivative size = 78 \[ \int \frac {\sec (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{(a+b)^{3/2} f}+\frac {b \sin (e+f x)}{a (a+b) f \sqrt {a+b \sin ^2(e+f x)}} \] Output:

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

Mathematica [C] (warning: unable to verify)

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

Time = 6.72 (sec) , antiderivative size = 480, normalized size of antiderivative = 6.15 \[ \int \frac {\sec (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\sec (e+f x) \tan (e+f x) \left (-45 \arcsin \left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right )-\frac {30 b \arcsin \left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right ) \sin ^2(e+f x)}{a}-\frac {45 (a+b) \arcsin \left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right ) \tan ^2(e+f x)}{a}-\frac {30 b (a+b) \arcsin \left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right ) \sin ^2(e+f x) \tan ^2(e+f x)}{a^2}+4 \operatorname {Hypergeometric2F1}\left (2,2,\frac {7}{2},-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{5/2}+\frac {4 b \operatorname {Hypergeometric2F1}\left (2,2,\frac {7}{2},-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{5/2}}{a}+45 \sqrt {-\frac {(a+b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{a^2}}+\frac {30 b \sin ^2(e+f x) \sqrt {-\frac {(a+b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{a^2}}}{a}\right )}{15 a f \sqrt {a+b \sin ^2(e+f x)} \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{3/2}} \] Input:

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

Output:

(Sec[e + f*x]*Tan[e + f*x]*(-45*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)/a - 
(45*(a + b)*ArcSin[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a)]]*Tan[e + f*x]^2)/a 
- (30*b*(a + b)*ArcSin[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a)]]*Sin[e + f*x]^2 
*Tan[e + f*x]^2)/a^2 + 4*Hypergeometric2F1[2, 2, 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) + (4*b*Hypergeometric2F1[2, 2, 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))/a + 45*Sqrt[-(((a + b)*Sec[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)])/a))/(15*a*f*Sqrt[a + b*Sin[e + f*x]^2]*Sqrt[(Sec[e + f*x]^2*(a + b*Si 
n[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(3/2))
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3669

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

\(\Big \downarrow \) 296

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(ArcTanh[(Sqrt[a + b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]]/(a + b)^(3 
/2) + (b*Sin[e + f*x])/(a*(a + b)*Sqrt[a + b*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. \(405\) vs. \(2(70)=140\).

Time = 0.45 (sec) , antiderivative size = 406, normalized size of antiderivative = 5.21

method	result	size
default	\(\frac {-\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a b \cos \left (f x +e \right )^{2}+\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a b \cos \left (f x +e \right )^{2}+2 \sqrt {a +b}\, \sqrt {-b \cos \left (f x +e \right )^{2}+\frac {b^{2} a +b^{3}}{b^{2}}}\, b \sin \left (f x +e \right )+\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{2}+\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a b -\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{2}-\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a b}{2 \sqrt {a +b}\, a \left (-\cos \left (f x +e \right )^{2} a b -\cos \left (f x +e \right )^{2} b^{2}+a^{2}+2 a b +b^{2}\right ) f}\)	\(406\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 213 vs. \(2 (70) = 140\).

Time = 0.17 (sec) , antiderivative size = 453, normalized size of antiderivative = 5.81 \[ \int \frac {\sec (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\left [\frac {{\left (a b \cos \left (f x + e\right )^{2} - a^{2} - a b\right )} \sqrt {a + b} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 8 \, {\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} \sin \left (f x + e\right ) + 8 \, a^{2} + 16 \, a b + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) - 4 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a b + b^{2}\right )} \sin \left (f x + e\right )}{4 \, {\left ({\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} f\right )}}, -\frac {{\left (a b \cos \left (f x + e\right )^{2} - a^{2} - a b\right )} \sqrt {-a - b} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{2 \, {\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sin \left (f x + e\right )}\right ) + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a b + b^{2}\right )} \sin \left (f x + e\right )}{2 \, {\left ({\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} f\right )}}\right ] \] Input:

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

Output:

[1/4*((a*b*cos(f*x + e)^2 - a^2 - a*b)*sqrt(a + b)*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)*s 
in(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 + b^2)*sin(f*x + e))/((a^3*b + 2*a^2*b^2 + a*b^3)*f*c 
os(f*x + e)^2 - (a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f), -1/2*((a*b*cos(f*x 
 + e)^2 - a^2 - a*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))) + 2*sqrt(-b*cos(f*x + e)^2 + 
 a + b)*(a*b + b^2)*sin(f*x + e))/((a^3*b + 2*a^2*b^2 + a*b^3)*f*cos(f*x + 
 e)^2 - (a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f)]
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 152 vs. \(2 (70) = 140\).

Time = 0.13 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.95 \[ \int \frac {\sec (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\frac {2 \, b \sin \left (f x + e\right )}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2} + \sqrt {b \sin \left (f x + e\right )^{2} + a} a b} + \frac {\operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}}\right )}{{\left (a + b\right )}^{\frac {3}{2}}} + \frac {\operatorname {arsinh}\left (-\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}}\right )}{{\left (a + b\right )}^{\frac {3}{2}}}}{2 \, f} \] Input:

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

Output:

1/2*(2*b*sin(f*x + e)/(sqrt(b*sin(f*x + e)^2 + a)*a^2 + sqrt(b*sin(f*x + e 
)^2 + a)*a*b) + arcsinh(b*sin(f*x + e)/(sqrt(a*b)*(sin(f*x + e) + 1)) - a/ 
(sqrt(a*b)*(sin(f*x + e) + 1)))/(a + b)^(3/2) + arcsinh(-b*sin(f*x + e)/(s 
qrt(a*b)*(sin(f*x + e) - 1)) - a/(sqrt(a*b)*(sin(f*x + e) - 1)))/(a + b)^( 
3/2))/f
                                                                                    
                                                                                    
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sec (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\sqrt {\sin \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )}{\sin \left (f x +e \right )^{4} b^{2}+2 \sin \left (f x +e \right )^{2} a b +a^{2}}d x \] Input:

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

Output:

int((sqrt(sin(e + f*x)**2*b + a)*sec(e + f*x))/(sin(e + f*x)**4*b**2 + 2*s 
in(e + f*x)**2*a*b + a**2),x)