\(\int \frac {\sec ^5(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\) [206]

Optimal result

Integrand size = 23, antiderivative size = 108 \[ \int \frac {\sec ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {3 \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{8 \sqrt {a} (a+b)^{5/2} f}+\frac {\sin (e+f x)}{4 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )^2}+\frac {3 \sin (e+f x)}{8 (a+b)^2 f \left (a+b-a \sin ^2(e+f x)\right )} \]

[Out]

1/4*sin(f*x+e)/(a+b)/f/(a+b-a*sin(f*x+e)^2)^2+3/8*sin(f*x+e)/(a+b)^2/f/(a+b-a*sin(f*x+e)^2)+3/8*arctanh(sin(f*
x+e)*a^(1/2)/(a+b)^(1/2))/(a+b)^(5/2)/f/a^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4232, 205, 214} \[ \int \frac {\sec ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {3 \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{8 \sqrt {a} f (a+b)^{5/2}}+\frac {3 \sin (e+f x)}{8 f (a+b)^2 \left (-a \sin ^2(e+f x)+a+b\right )}+\frac {\sin (e+f x)}{4 f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )^2} \]

[In]

Int[Sec[e + f*x]^5/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

(3*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(8*Sqrt[a]*(a + b)^(5/2)*f) + Sin[e + f*x]/(4*(a + b)*f*(a + b
- a*Sin[e + f*x]^2)^2) + (3*Sin[e + f*x])/(8*(a + b)^2*f*(a + b - a*Sin[e + f*x]^2))

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 214

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

Rule 4232

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\left (a+b-a x^2\right )^3} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\sin (e+f x)}{4 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )^2}+\frac {3 \text {Subst}\left (\int \frac {1}{\left (a+b-a x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{4 (a+b) f} \\ & = \frac {\sin (e+f x)}{4 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )^2}+\frac {3 \sin (e+f x)}{8 (a+b)^2 f \left (a+b-a \sin ^2(e+f x)\right )}+\frac {3 \text {Subst}\left (\int \frac {1}{a+b-a x^2} \, dx,x,\sin (e+f x)\right )}{8 (a+b)^2 f} \\ & = \frac {3 \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{8 \sqrt {a} (a+b)^{5/2} f}+\frac {\sin (e+f x)}{4 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )^2}+\frac {3 \sin (e+f x)}{8 (a+b)^2 f \left (a+b-a \sin ^2(e+f x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.76 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.83 \[ \int \frac {\sec ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {3 \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{\sqrt {a} (a+b)^{5/2}}+\frac {2 (7 a+10 b+3 a \cos (2 (e+f x))) \sin (e+f x)}{(a+b)^2 (a+2 b+a \cos (2 (e+f x)))^2}}{8 f} \]

[In]

Integrate[Sec[e + f*x]^5/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

((3*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(Sqrt[a]*(a + b)^(5/2)) + (2*(7*a + 10*b + 3*a*Cos[2*(e + f*x
)])*Sin[e + f*x])/((a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)])^2))/(8*f)

Maple [A] (verified)

Time = 1.38 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {\frac {\sin \left (f x +e \right )}{4 \left (a +b \right ) \left (a \sin \left (f x +e \right )^{2}-a -b \right )^{2}}+\frac {-\frac {3 \sin \left (f x +e \right )}{8 \left (a +b \right ) \left (a \sin \left (f x +e \right )^{2}-a -b \right )}+\frac {3 \,\operatorname {arctanh}\left (\frac {a \sin \left (f x +e \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 \left (a +b \right ) \sqrt {a \left (a +b \right )}}}{a +b}}{f}\)	\(108\)
default	\(\frac {\frac {\sin \left (f x +e \right )}{4 \left (a +b \right ) \left (a \sin \left (f x +e \right )^{2}-a -b \right )^{2}}+\frac {-\frac {3 \sin \left (f x +e \right )}{8 \left (a +b \right ) \left (a \sin \left (f x +e \right )^{2}-a -b \right )}+\frac {3 \,\operatorname {arctanh}\left (\frac {a \sin \left (f x +e \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 \left (a +b \right ) \sqrt {a \left (a +b \right )}}}{a +b}}{f}\)	\(108\)
risch	\(-\frac {i \left (3 a \,{\mathrm e}^{7 i \left (f x +e \right )}+11 a \,{\mathrm e}^{5 i \left (f x +e \right )}+20 b \,{\mathrm e}^{5 i \left (f x +e \right )}-11 a \,{\mathrm e}^{3 i \left (f x +e \right )}-20 b \,{\mathrm e}^{3 i \left (f x +e \right )}-3 a \,{\mathrm e}^{i \left (f x +e \right )}\right )}{4 \left (a +b \right )^{2} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )^{2}}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{16 \sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{16 \sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f}\)	\(235\)

[In]

int(sec(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x,method=_RETURNVERBOSE)

[Out]

1/f*(1/4*sin(f*x+e)/(a+b)/(a*sin(f*x+e)^2-a-b)^2+3/4/(a+b)*(-1/2*sin(f*x+e)/(a+b)/(a*sin(f*x+e)^2-a-b)+1/2/(a+
b)/(a*(a+b))^(1/2)*arctanh(a*sin(f*x+e)/(a*(a+b))^(1/2))))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (100) = 200\).

Time = 0.30 (sec) , antiderivative size = 472, normalized size of antiderivative = 4.37 \[ \int \frac {\sec ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\left [\frac {3 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {a^{2} + a b} \log \left (-\frac {a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a^{2} + a b} \sin \left (f x + e\right ) - 2 \, a - b}{a \cos \left (f x + e\right )^{2} + b}\right ) + 2 \, {\left (2 \, a^{3} + 7 \, a^{2} b + 5 \, a b^{2} + 3 \, {\left (a^{3} + a^{2} b\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{16 \, {\left ({\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b^{2} + 3 \, a^{3} b^{3} + 3 \, a^{2} b^{4} + a b^{5}\right )} f\right )}}, -\frac {3 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-a^{2} - a b} \arctan \left (\frac {\sqrt {-a^{2} - a b} \sin \left (f x + e\right )}{a + b}\right ) - {\left (2 \, a^{3} + 7 \, a^{2} b + 5 \, a b^{2} + 3 \, {\left (a^{3} + a^{2} b\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{8 \, {\left ({\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b^{2} + 3 \, a^{3} b^{3} + 3 \, a^{2} b^{4} + a b^{5}\right )} f\right )}}\right ] \]

[In]

integrate(sec(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

[1/16*(3*(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)*sqrt(a^2 + a*b)*log(-(a*cos(f*x + e)^2 - 2*sqrt(a^2
+ a*b)*sin(f*x + e) - 2*a - b)/(a*cos(f*x + e)^2 + b)) + 2*(2*a^3 + 7*a^2*b + 5*a*b^2 + 3*(a^3 + a^2*b)*cos(f
*x + e)^2)*sin(f*x + e))/((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*f*cos(f*x + e)^4 + 2*(a^5*b + 3*a^4*b^2 + 3*a^
3*b^3 + a^2*b^4)*f*cos(f*x + e)^2 + (a^4*b^2 + 3*a^3*b^3 + 3*a^2*b^4 + a*b^5)*f), -1/8*(3*(a^2*cos(f*x + e)^4
+ 2*a*b*cos(f*x + e)^2 + b^2)*sqrt(-a^2 - a*b)*arctan(sqrt(-a^2 - a*b)*sin(f*x + e)/(a + b)) - (2*a^3 + 7*a^2*
b + 5*a*b^2 + 3*(a^3 + a^2*b)*cos(f*x + e)^2)*sin(f*x + e))/((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*f*cos(f*x +
e)^4 + 2*(a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*f*cos(f*x + e)^2 + (a^4*b^2 + 3*a^3*b^3 + 3*a^2*b^4 + a*b^
5)*f)]

Sympy [F]

\[ \int \frac {\sec ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\int \frac {\sec ^{5}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{3}}\, dx \]

[In]

integrate(sec(f*x+e)**5/(a+b*sec(f*x+e)**2)**3,x)

[Out]

Integral(sec(e + f*x)**5/(a + b*sec(e + f*x)**2)**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.66 \[ \int \frac {\sec ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {2 \, {\left (3 \, a \sin \left (f x + e\right )^{3} - 5 \, {\left (a + b\right )} \sin \left (f x + e\right )\right )}}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \sin \left (f x + e\right )^{4} + a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4} - 2 \, {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \sin \left (f x + e\right )^{2}} + \frac {3 \, \log \left (\frac {a \sin \left (f x + e\right ) - \sqrt {{\left (a + b\right )} a}}{a \sin \left (f x + e\right ) + \sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} {\left (a^{2} + 2 \, a b + b^{2}\right )}}}{16 \, f} \]

[In]

integrate(sec(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

-1/16*(2*(3*a*sin(f*x + e)^3 - 5*(a + b)*sin(f*x + e))/((a^4 + 2*a^3*b + a^2*b^2)*sin(f*x + e)^4 + a^4 + 4*a^3
*b + 6*a^2*b^2 + 4*a*b^3 + b^4 - 2*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*sin(f*x + e)^2) + 3*log((a*sin(f*x + e)
- sqrt((a + b)*a))/(a*sin(f*x + e) + sqrt((a + b)*a)))/(sqrt((a + b)*a)*(a^2 + 2*a*b + b^2)))/f

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.08 \[ \int \frac {\sec ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {3 \, \arctan \left (\frac {a \sin \left (f x + e\right )}{\sqrt {-a^{2} - a b}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-a^{2} - a b}} + \frac {3 \, a \sin \left (f x + e\right )^{3} - 5 \, a \sin \left (f x + e\right ) - 5 \, b \sin \left (f x + e\right )}{{\left (a \sin \left (f x + e\right )^{2} - a - b\right )}^{2} {\left (a^{2} + 2 \, a b + b^{2}\right )}}}{8 \, f} \]

[In]

integrate(sec(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")

[Out]

-1/8*(3*arctan(a*sin(f*x + e)/sqrt(-a^2 - a*b))/((a^2 + 2*a*b + b^2)*sqrt(-a^2 - a*b)) + (3*a*sin(f*x + e)^3 -
5*a*sin(f*x + e) - 5*b*sin(f*x + e))/((a*sin(f*x + e)^2 - a - b)^2*(a^2 + 2*a*b + b^2)))/f

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.05 \[ \int \frac {\sec ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {5\,\sin \left (e+f\,x\right )}{8\,\left (a+b\right )}-\frac {3\,a\,{\sin \left (e+f\,x\right )}^3}{8\,{\left (a+b\right )}^2}}{f\,\left (2\,a\,b+a^2+b^2-{\sin \left (e+f\,x\right )}^2\,\left (2\,a^2+2\,b\,a\right )+a^2\,{\sin \left (e+f\,x\right )}^4\right )}+\frac {3\,\mathrm {atanh}\left (\frac {\sqrt {a}\,\sin \left (e+f\,x\right )}{\sqrt {a+b}}\right )}{8\,\sqrt {a}\,f\,{\left (a+b\right )}^{5/2}} \]

[In]

int(1/(cos(e + f*x)^5*(a + b/cos(e + f*x)^2)^3),x)

[Out]

((5*sin(e + f*x))/(8*(a + b)) - (3*a*sin(e + f*x)^3)/(8*(a + b)^2))/(f*(2*a*b + a^2 + b^2 - sin(e + f*x)^2*(2*
a*b + 2*a^2) + a^2*sin(e + f*x)^4)) + (3*atanh((a^(1/2)*sin(e + f*x))/(a + b)^(1/2)))/(8*a^(1/2)*f*(a + b)^(5/
2))