\(\int \frac {\sec ^4(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\) [277]

Optimal result

Integrand size = 25, antiderivative size = 77 \[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{b^{3/2} f}-\frac {a \tan (e+f x)}{b (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}} \] Output:

arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/b^(3/2)/f-a*tan(f*x 
+e)/b/(a+b)/f/(a+b+b*tan(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 = 7.36 (sec) , antiderivative size = 405, normalized size of antiderivative = 5.26 \[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {(a+2 b+a \cos (2 e+2 f x))^{3/2} \sec ^4(e+f x) \sqrt {1-\frac {2 a \sin ^2(e+f x)}{2 a+2 b}} \tan (e+f x) \left (15 \arcsin \left (\sqrt {-\frac {b \tan ^2(e+f x)}{a+b}}\right ) \sec ^2(e+f x) \left (3 b^2+a b \left (6-5 \sin ^2(e+f x)\right )+a^2 \left (3-5 \sin ^2(e+f x)+2 \sin ^4(e+f x)\right )\right )+15 (a+b) \left (-3 b+a \left (-3+2 \sin ^2(e+f x)\right )\right ) \sqrt {-\frac {b \sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{(a+b)^2}}+4 b (a+b) \operatorname {Hypergeometric2F1}\left (2,2,\frac {7}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^2(e+f x) \left (-\frac {b \sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{(a+b)^2}\right )^{3/2}\right )}{15 (a+b)^2 (2 a+2 b) f \left (a+b \sec ^2(e+f x)\right )^{3/2} \sqrt {\frac {a+b \sec ^2(e+f x)}{a+b}} \sqrt {2 a+2 b-2 a \sin ^2(e+f x)} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}} \left (-\frac {b \tan ^2(e+f x)}{a+b}\right )^{3/2}} \] Input:

Integrate[Sec[e + f*x]^4/(a + b*Sec[e + f*x]^2)^(3/2),x]
 

Output:

-1/15*((a + 2*b + a*Cos[2*e + 2*f*x])^(3/2)*Sec[e + f*x]^4*Sqrt[1 - (2*a*S 
in[e + f*x]^2)/(2*a + 2*b)]*Tan[e + f*x]*(15*ArcSin[Sqrt[-((b*Tan[e + f*x] 
^2)/(a + b))]]*Sec[e + f*x]^2*(3*b^2 + a*b*(6 - 5*Sin[e + f*x]^2) + a^2*(3 
 - 5*Sin[e + f*x]^2 + 2*Sin[e + f*x]^4)) + 15*(a + b)*(-3*b + a*(-3 + 2*Si 
n[e + f*x]^2))*Sqrt[-((b*Sec[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)*Tan[e + 
 f*x]^2)/(a + b)^2)] + 4*b*(a + b)*Hypergeometric2F1[2, 2, 7/2, -((b*Tan[e 
 + f*x]^2)/(a + b))]*Sin[e + f*x]^2*(-((b*Sec[e + f*x]^2*(a + b - a*Sin[e 
+ f*x]^2)*Tan[e + f*x]^2)/(a + b)^2))^(3/2)))/((a + b)^2*(2*a + 2*b)*f*(a 
+ b*Sec[e + f*x]^2)^(3/2)*Sqrt[(a + b*Sec[e + f*x]^2)/(a + b)]*Sqrt[2*a + 
2*b - 2*a*Sin[e + f*x]^2]*Sqrt[1 - (a*Sin[e + f*x]^2)/(a + b)]*(-((b*Tan[e 
 + f*x]^2)/(a + b)))^(3/2))
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4634, 298, 224, 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.

\(\displaystyle \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (e+f x)^4}{\left (a+b \sec (e+f x)^2\right )^{3/2}}dx\)

\(\Big \downarrow \) 4634

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

\(\Big \downarrow \) 298

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

Int[Sec[e + f*x]^4/(a + b*Sec[e + f*x]^2)^(3/2),x]
 

Output:

(ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]]/b^(3/2) - 
(a*Tan[e + f*x])/(b*(a + b)*Sqrt[a + b + b*Tan[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 224

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

rule 298

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( 
b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 
2*p + 3))/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
 

rule 3042

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

rule 4634

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(889\) vs. \(2(69)=138\).

Time = 9.94 (sec) , antiderivative size = 890, normalized size of antiderivative = 11.56

method	result	size
default	\(\text {Expression too large to display}\)	\(890\)

Input:

int(sec(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (69) = 138\).

Time = 0.15 (sec) , antiderivative size = 410, normalized size of antiderivative = 5.32 \[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\left [-\frac {4 \, a b \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left ({\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{2} + a b + b^{2}\right )} \sqrt {b} \log \left (\frac {{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right )}{4 \, {\left ({\left (a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a b^{3} + b^{4}\right )} f\right )}}, -\frac {2 \, a b \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left ({\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{2} + a b + b^{2}\right )} \sqrt {-b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right )}{2 \, {\left ({\left (a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a b^{3} + b^{4}\right )} f\right )}}\right ] \] Input:

integrate(sec(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

[-1/4*(4*a*b*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)*sin( 
f*x + e) - ((a^2 + a*b)*cos(f*x + e)^2 + a*b + b^2)*sqrt(b)*log(((a^2 - 6* 
a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b^2)*cos(f*x + e)^2 + 4*((a - b)*cos( 
f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x 
 + e)^2)*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4))/((a^2*b^2 + a*b^3)*f*cos(f 
*x + e)^2 + (a*b^3 + b^4)*f), -1/2*(2*a*b*sqrt((a*cos(f*x + e)^2 + b)/cos( 
f*x + e)^2)*cos(f*x + e)*sin(f*x + e) - ((a^2 + a*b)*cos(f*x + e)^2 + a*b 
+ b^2)*sqrt(-b)*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sq 
rt(-b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a*b*cos(f*x + e)^2 + 
b^2)*sin(f*x + e))))/((a^2*b^2 + a*b^3)*f*cos(f*x + e)^2 + (a*b^3 + b^4)*f 
)]
 

Sympy [F]

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

integrate(sec(f*x+e)**4/(a+b*sec(f*x+e)**2)**(3/2),x)
 

Output:

Integral(sec(e + f*x)**4/(a + b*sec(e + f*x)**2)**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.01 \[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\frac {\frac {\operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} + \frac {\tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}} - \frac {\tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} b}}{f} \] Input:

integrate(sec(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

(arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/b^(3/2) + tan(f*x + e)/(sqrt(b*ta 
n(f*x + e)^2 + a + b)*(a + b)) - tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a + 
 b)*b))/f
 

Giac [F]

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

integrate(sec(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

integrate(sec(f*x + e)^4/(b*sec(f*x + e)^2 + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {1}{{\cos \left (e+f\,x\right )}^4\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}} \,d x \] Input:

int(1/(cos(e + f*x)^4*(a + b/cos(e + f*x)^2)^(3/2)),x)
 

Output:

int(1/(cos(e + f*x)^4*(a + b/cos(e + f*x)^2)^(3/2)), x)
 

Reduce [F]

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

int(sec(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x)
 

Output:

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