\(\int \frac {\sec ^6(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\) [276]

Optimal result

Integrand size = 25, antiderivative size = 121 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {(3 a-b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 b^{5/2} f}+\frac {a^2 \tan (e+f x)}{b^2 (a+b) f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 b^2 f} \] Output:

-1/2*(3*a-b)*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/b^(5/2 
)/f+a^2*tan(f*x+e)/b^2/(a+b)/f/(a+b+b*tan(f*x+e)^2)^(1/2)+1/2*tan(f*x+e)*( 
a+b+b*tan(f*x+e)^2)^(1/2)/b^2/f
 

Mathematica [C] (warning: unable to verify)

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

Time = 6.05 (sec) , antiderivative size = 259, normalized size of antiderivative = 2.14 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^8(e+f x) \left (-7 (a+b) \cos ^2(e+f x) \left (15 b^2+10 a b (2+\cos (2 (e+f x)))+a^2 (8+6 \cos (2 (e+f x))+\cos (4 (e+f x)))\right ) \operatorname {Hypergeometric2F1}\left (1,2,\frac {7}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right )+4 b (a+2 b+a \cos (2 (e+f x)))^2 \, _3F_2\left (2,2,3;1,\frac {9}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^2(e+f x)+16 b \operatorname {Hypergeometric2F1}\left (2,3,\frac {9}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^2(e+f x) \left (4 (a+b)^2-7 a (a+b) \sin ^2(e+f x)+3 a^2 \sin ^4(e+f x)\right )\right ) \tan (e+f x)}{210 (a+b)^4 f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \] Input:

Integrate[Sec[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(3/2),x]
 

Output:

-1/210*((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^8*(-7*(a + b)*Cos[e + 
f*x]^2*(15*b^2 + 10*a*b*(2 + Cos[2*(e + f*x)]) + a^2*(8 + 6*Cos[2*(e + f*x 
)] + Cos[4*(e + f*x)]))*Hypergeometric2F1[1, 2, 7/2, -((b*Tan[e + f*x]^2)/ 
(a + b))] + 4*b*(a + 2*b + a*Cos[2*(e + f*x)])^2*HypergeometricPFQ[{2, 2, 
3}, {1, 9/2}, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]^2 + 16*b*Hyperge 
ometric2F1[2, 3, 9/2, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]^2*(4*(a 
+ b)^2 - 7*a*(a + b)*Sin[e + f*x]^2 + 3*a^2*Sin[e + f*x]^4))*Tan[e + f*x]) 
/((a + b)^4*f*(a + b*Sec[e + f*x]^2)^(3/2))
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.18, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4634, 315, 299, 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.

\(\displaystyle \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (e+f x)^6}{\left (a+b \sec (e+f x)^2\right )^{3/2}}dx\)

\(\Big \downarrow \) 4634

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

\(\Big \downarrow \) 315

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

\(\Big \downarrow \) 299

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

Int[Sec[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(3/2),x]
 

Output:

(-((a*Tan[e + f*x]*(1 + Tan[e + f*x]^2))/(b*(a + b)*Sqrt[a + b + b*Tan[e + 
 f*x]^2])) + (-1/2*((3*a - b)*(a + b)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[ 
a + b + b*Tan[e + f*x]^2]])/b^(3/2) + ((3*a + b)*Tan[e + f*x]*Sqrt[a + b + 
 b*Tan[e + f*x]^2])/(2*b))/(b*(a + b)))/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 299

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

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 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. \(1284\) vs. \(2(107)=214\).

Time = 25.35 (sec) , antiderivative size = 1285, normalized size of antiderivative = 10.62

method	result	size
default	\(\text {Expression too large to display}\)	\(1285\)

Input:

int(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/4/f/(a+b)/b^(7/2)/(a+b*sec(f*x+e)^2)^(3/2)*(((b+a*cos(f*x+e)^2)/(1+cos( 
f*x+e))^2)^(1/2)*a^3*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)+sin(f*x+e)*a-b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e)) 
^2)^(1/2)-a-b)/(sin(f*x+e)-1))*(3+3*sec(f*x+e))+((b+a*cos(f*x+e)^2)/(1+cos 
(f*x+e))^2)^(1/2)*a^2*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)+sin(f*x+e)*a-b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+ 
e))^2)^(1/2)-a-b)/(sin(f*x+e)-1))*(2+2*sec(f*x+e)+3*sec(f*x+e)^2+3*sec(f*x 
+e)^3)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*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)+sin(f*x+e)*a-b^(1/2)* 
((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-a-b)/(sin(f*x+e)-1))*(-1-sec(f 
*x+e)+2*sec(f*x+e)^2+2*sec(f*x+e)^3)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2) 
^(1/2)*b^4*ln(4*(-b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos( 
f*x+e)+sin(f*x+e)*a-b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-a- 
b)/(sin(f*x+e)-1))*(-sec(f*x+e)^2-sec(f*x+e)^3)+((b+a*cos(f*x+e)^2)/(1+cos 
(f*x+e))^2)^(1/2)*a^3*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)+sin(f*x+e)*a-b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e 
))^2)^(1/2)+a+b)/(sin(f*x+e)+1))*(3+3*sec(f*x+e))+((b+a*cos(f*x+e)^2)/(1+c 
os(f*x+e))^2)^(1/2)*a^2*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)+sin(f*x+e)*a-b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f 
*x+e))^2)^(1/2)+a+b)/(sin(f*x+e)+1))*(2+2*sec(f*x+e)+3*sec(f*x+e)^2+3*s...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 246 vs. \(2 (107) = 214\).

Time = 0.28 (sec) , antiderivative size = 524, normalized size of antiderivative = 4.33 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\left [-\frac {{\left ({\left (3 \, a^{3} + 2 \, a^{2} b - a b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (3 \, a^{2} b + 2 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\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 (a b^{2} + b^{3} + {\left (3 \, a^{2} b + a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \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 \, {\left ({\left (a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{3} + {\left (a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )\right )}}, -\frac {{\left ({\left (3 \, a^{3} + 2 \, a^{2} b - a b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (3 \, a^{2} b + 2 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\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 (a b^{2} + b^{3} + {\left (3 \, a^{2} b + a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{4 \, {\left ({\left (a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{3} + {\left (a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )\right )}}\right ] \] Input:

integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

[-1/8*(((3*a^3 + 2*a^2*b - a*b^2)*cos(f*x + e)^3 + (3*a^2*b + 2*a*b^2 - b^ 
3)*cos(f*x + e))*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) - 4*(a*b^2 + b^3 + (3*a^2*b + a*b^2)*cos(f*x + e)^2)*sqrt((a*c 
os(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/((a^2*b^3 + a*b^4)*f*cos( 
f*x + e)^3 + (a*b^4 + b^5)*f*cos(f*x + e)), -1/4*(((3*a^3 + 2*a^2*b - a*b^ 
2)*cos(f*x + e)^3 + (3*a^2*b + 2*a*b^2 - b^3)*cos(f*x + e))*sqrt(-b)*arcta 
n(-1/2*((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)/((a*b*cos(f*x + e)^2 + b^2)*sin(f*x + e))) - 
 2*(a*b^2 + b^3 + (3*a^2*b + a*b^2)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 
 + b)/cos(f*x + e)^2)*sin(f*x + e))/((a^2*b^3 + a*b^4)*f*cos(f*x + e)^3 + 
(a*b^4 + b^5)*f*cos(f*x + e))]
 

Sympy [F]

\[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\sec ^{6}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(sec(f*x+e)**6/(a+b*sec(f*x+e)**2)**(3/2),x)
 

Output:

Integral(sec(e + f*x)**6/(a + b*sec(e + f*x)**2)**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.33 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\frac {\frac {\tan \left (f x + e\right )^{3}}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} b} - \frac {3 \, {\left (a + b\right )} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {5}{2}}} + \frac {4 \, \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} + \frac {2 \, \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}} + \frac {3 \, {\left (a + b\right )} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} b^{2}} - \frac {4 \, \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} b}}{2 \, f} \] Input:

integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

1/2*(tan(f*x + e)^3/(sqrt(b*tan(f*x + e)^2 + a + b)*b) - 3*(a + b)*arcsinh 
(b*tan(f*x + e)/sqrt((a + b)*b))/b^(5/2) + 4*arcsinh(b*tan(f*x + e)/sqrt(( 
a + b)*b))/b^(3/2) + 2*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b 
)) + 3*(a + b)*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*b^2) - 4*tan(f 
*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*b))/f
 

Giac [F]

\[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\sec \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

integrate(sec(f*x + e)^6/(b*sec(f*x + e)^2 + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {1}{{\cos \left (e+f\,x\right )}^6\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}} \,d x \] Input:

int(1/(cos(e + f*x)^6*(a + b/cos(e + f*x)^2)^(3/2)),x)
 

Output:

int(1/(cos(e + f*x)^6*(a + b/cos(e + f*x)^2)^(3/2)), x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{6}}{\sec \left (f x +e \right )^{4} b^{2}+2 \sec \left (f x +e \right )^{2} a b +a^{2}}d x \] Input:

int(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x)
 

Output:

int((sqrt(sec(e + f*x)**2*b + a)*sec(e + f*x)**6)/(sec(e + f*x)**4*b**2 + 
2*sec(e + f*x)**2*a*b + a**2),x)