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

Optimal. Leaf size=108 \[ \frac{3 \sin (e+f x)}{8 f (a+b)^2 \left (-a \sin ^2(e+f x)+a+b\right )}+\frac{\sin (e+f x)}{4 f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )^2}+\frac{3 \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{8 \sqrt{a} f (a+b)^{5/2}} \]

[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))

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Rubi [A]  time = 0.0993723, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {4147, 199, 208} \[ \frac{3 \sin (e+f x)}{8 f (a+b)^2 \left (-a \sin ^2(e+f x)+a+b\right )}+\frac{\sin (e+f x)}{4 f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )^2}+\frac{3 \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{8 \sqrt{a} f (a+b)^{5/2}} \]

Antiderivative was successfully verified.

[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 4147

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]

Rule 199

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] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\sec ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{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 \operatorname{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 \operatorname{Subst}\left (\int \frac{1}{a+b-a x^2} \, dx,x,\sin (e+f x)\right )}{8 (a+b)^2 f}\\ &=\frac{3 \tanh ^{-1}\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]  time = 0.534259, size = 128, normalized size = 1.19 \[ \frac{\sec ^6(e+f x) (a \cos (2 (e+f x))+a+2 b)^3 \left (\frac{3 \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{\sqrt{a} (a+b)^{5/2}}+\frac{4 \sin (e+f x) \left (5 (a+b)-3 a \sin ^2(e+f x)\right )}{(a+b)^2 (a \cos (2 (e+f x))+a+2 b)^2}\right )}{64 f \left (a+b \sec ^2(e+f x)\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])^3*Sec[e + f*x]^6*((3*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(Sqrt[a]*(a
+ b)^(5/2)) + (4*Sin[e + f*x]*(5*(a + b) - 3*a*Sin[e + f*x]^2))/((a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)])^2)))
/(64*f*(a + b*Sec[e + f*x]^2)^3)

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Maple [A]  time = 0.074, size = 108, normalized size = 1. \begin{align*}{\frac{1}{f} \left ({\frac{\sin \left ( fx+e \right ) }{ \left ( 4\,a+4\,b \right ) \left ( -a-b+a \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{3}{4\,a+4\,b} \left ( -{\frac{\sin \left ( fx+e \right ) }{ \left ( 2\,a+2\,b \right ) \left ( -a-b+a \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{1}{2\,a+2\,b}{\it Artanh} \left ({\sin \left ( fx+e \right ) a{\frac{1}{\sqrt{ \left ( a+b \right ) a}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.608319, size = 1053, normalized size = 9.75 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.35633, size = 165, normalized size = 1.53 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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