\(\int \frac {\sec ^4(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) [264]

Optimal result

Integrand size = 25, antiderivative size = 81 \[ \int \frac {\sec ^4(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {(a-b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 b^{3/2} f}+\frac {\tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 b f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {4231, 396, 223, 212} \[ \int \frac {\sec ^4(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 b f}-\frac {(a-b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 b^{3/2} f} \]

[In]

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

[Out]

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

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 396

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

Rule 4231

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fre
eFactors[Tan[e + f*x], x]}, Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1+x^2}{\sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 b f}-\frac {(a-b) \text {Subst}\left (\int \frac {1}{\sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 b f} \\ & = \frac {\tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 b f}-\frac {(a-b) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 b f} \\ & = -\frac {(a-b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 b^{3/2} f}+\frac {\tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 b f} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

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

Time = 7.00 (sec) , antiderivative size = 326, normalized size of antiderivative = 4.02 \[ \int \frac {\sec ^4(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {\sqrt {a+2 b+a \cos (2 e+2 f x)} \sec ^4(e+f x) \left (1-\frac {a \sin ^2(e+f x)}{a+b}\right ) \tan (e+f x) \left (\frac {16 b^2 \left (b+a \cos ^2(e+f x)\right ) \operatorname {Hypergeometric2F1}\left (2,3,\frac {7}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \tan ^4(e+f x) \sqrt {-\frac {b \sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{(a+b)^2}}}{(a+b)^3}+\frac {15 \left (3 b+a \left (3-2 \sin ^2(e+f x)\right )\right ) \left (\arcsin \left (\sqrt {-\frac {b \tan ^2(e+f x)}{a+b}}\right )-\sqrt {-\frac {b \sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{(a+b)^2}}\right )}{a+b}\right )}{30 \sqrt {2} f \sqrt {a+b \sec ^2(e+f x)} \sqrt {\frac {a+b \sec ^2(e+f x)}{a+b}} \sqrt {a+b-a \sin ^2(e+f x)} \left (-\frac {b \tan ^2(e+f x)}{a+b}\right )^{3/2}} \]

[In]

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

[Out]

(Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]]*Sec[e + f*x]^4*(1 - (a*Sin[e + f*x]^2)/(a + b))*Tan[e + f*x]*((16*b^2*(b +
a*Cos[e + f*x]^2)*Hypergeometric2F1[2, 3, 7/2, -((b*Tan[e + f*x]^2)/(a + b))]*Tan[e + f*x]^4*Sqrt[-((b*Sec[e
+ f*x]^2*(a + b - a*Sin[e + f*x]^2)*Tan[e + f*x]^2)/(a + b)^2)])/(a + b)^3 + (15*(3*b + a*(3 - 2*Sin[e + f*x]^
2))*(ArcSin[Sqrt[-((b*Tan[e + f*x]^2)/(a + b))]] - Sqrt[-((b*Sec[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)*Tan[e +
f*x]^2)/(a + b)^2)]))/(a + b)))/(30*Sqrt[2]*f*Sqrt[a + b*Sec[e + f*x]^2]*Sqrt[(a + b*Sec[e + f*x]^2)/(a + b)]
*Sqrt[a + b - a*Sin[e + f*x]^2]*(-((b*Tan[e + f*x]^2)/(a + b)))^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1038\) vs. \(2(69)=138\).

Time = 11.29 (sec) , antiderivative size = 1039, normalized size of antiderivative = 12.83

method	result	size
default	\(\text {Expression too large to display}\)	\(1039\)

[In]

int(sec(f*x+e)^4/(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 146 vs. \(2 (69) = 138\).

Time = 0.32 (sec) , antiderivative size = 324, normalized size of antiderivative = 4.00 \[ \int \frac {\sec ^4(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\left [-\frac {{\left (a - b\right )} \sqrt {b} \cos \left (f x + e\right ) \log \left (\frac {{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) - 4 \, b \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{8 \, b^{2} f \cos \left (f x + e\right )}, -\frac {{\left (a - b\right )} \sqrt {-b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right ) - 2 \, b \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{4 \, b^{2} f \cos \left (f x + e\right )}\right ] \]

[In]

integrate(sec(f*x+e)^4/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

integrate(sec(f*x+e)**4/(a+b*sec(f*x+e)**2)**(1/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.91 \[ \int \frac {\sec ^4(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\frac {a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} - \frac {\operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} - \frac {\sqrt {b \tan \left (f x + e\right )^{2} + a + b} \tan \left (f x + e\right )}{b}}{2 \, f} \]

[In]

integrate(sec(f*x+e)^4/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(sec(f*x+e)^4/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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