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

Optimal. Leaf size=125 \[ \frac{(4 a+b) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{8 a^{3/2} f (a+b)^{5/2}}+\frac{(4 a+b) \sin (e+f x)}{8 a f (a+b)^2 \left (-a \sin ^2(e+f x)+a+b\right )}-\frac{b \sin (e+f x)}{4 a f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.106025, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {4147, 385, 199, 208} \[ \frac{(4 a+b) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{8 a^{3/2} f (a+b)^{5/2}}+\frac{(4 a+b) \sin (e+f x)}{8 a f (a+b)^2 \left (-a \sin ^2(e+f x)+a+b\right )}-\frac{b \sin (e+f x)}{4 a f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a + b)*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(8*a^(3/2)*(a + b)^(5/2)*f) - (b*Sin[e + f*x])/(4*a*(a
+ b)*f*(a + b - a*Sin[e + f*x]^2)^2) + ((4*a + b)*Sin[e + f*x])/(8*a*(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 385

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

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 ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{\left (a+b-a x^2\right )^3} \, dx,x,\sin (e+f x)\right )}{f}\\ &=-\frac{b \sin (e+f x)}{4 a (a+b) f \left (a+b-a \sin ^2(e+f x)\right )^2}+\frac{(4 a+b) \operatorname{Subst}\left (\int \frac{1}{\left (a+b-a x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{4 a (a+b) f}\\ &=-\frac{b \sin (e+f x)}{4 a (a+b) f \left (a+b-a \sin ^2(e+f x)\right )^2}+\frac{(4 a+b) \sin (e+f x)}{8 a (a+b)^2 f \left (a+b-a \sin ^2(e+f x)\right )}+\frac{(4 a+b) \operatorname{Subst}\left (\int \frac{1}{a+b-a x^2} \, dx,x,\sin (e+f x)\right )}{8 a (a+b)^2 f}\\ &=\frac{(4 a+b) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{8 a^{3/2} (a+b)^{5/2} f}-\frac{b \sin (e+f x)}{4 a (a+b) f \left (a+b-a \sin ^2(e+f x)\right )^2}+\frac{(4 a+b) \sin (e+f x)}{8 a (a+b)^2 f \left (a+b-a \sin ^2(e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.619619, size = 163, normalized size = 1.3 \[ -\frac{\sec ^6(e+f x) (a \cos (2 (e+f x))+a+2 b)^3 \left (\frac{8 \sin (e+f x)}{\left (-a \sin ^2(e+f x)+a+b\right )^2}-(4 a+b) \left (\frac{3 \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{\sqrt{a} (a+b)^{5/2}}+\frac{4 \sin (e+f x) \left (5 (a+b)-3 a \sin ^2(e+f x)\right )}{(a+b)^2 (a \cos (2 (e+f x))+a+2 b)^2}\right )\right )}{192 a f \left (a+b \sec ^2(e+f x)\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^3/(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*((8*Sin[e + f*x])/(a + b - a*Sin[e + f*x]^2)^2 - (4*a + b)*(
(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))))/(192*a*f*(a + b*Sec[e + f*x]^2)^3)

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Maple [A]  time = 0.078, size = 124, normalized size = 1. \begin{align*}{\frac{1}{f} \left ({\frac{1}{ \left ( -a-b+a \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}} \left ( -{\frac{ \left ( 4\,a+b \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{8\,{a}^{2}+16\,ab+8\,{b}^{2}}}+{\frac{ \left ( 4\,a-b \right ) \sin \left ( fx+e \right ) }{ \left ( 8\,a+8\,b \right ) a}} \right ) }+{\frac{4\,a+b}{ \left ( 8\,{a}^{2}+16\,ab+8\,{b}^{2} \right ) a}{\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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*((-1/8*(4*a+b)/(a^2+2*a*b+b^2)*sin(f*x+e)^3+1/8*(4*a-b)/(a+b)/a*sin(f*x+e))/(-a-b+a*sin(f*x+e)^2)^2+1/8*(4
*a+b)/(a^2+2*a*b+b^2)/a/((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)^3/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.651157, size = 1183, normalized size = 9.46 \begin{align*} \left [\frac{{\left ({\left (4 \, a^{3} + a^{2} b\right )} \cos \left (f x + e\right )^{4} + 4 \, a b^{2} + b^{3} + 2 \,{\left (4 \, a^{2} b + a b^{2}\right )} \cos \left (f x + e\right )^{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} b + a^{2} b^{2} - a b^{3} +{\left (4 \, a^{4} + 5 \, a^{3} b + a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{16 \,{\left ({\left (a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \,{\left (a^{6} b + 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} + a^{3} b^{4}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + a^{2} b^{5}\right )} f\right )}}, -\frac{{\left ({\left (4 \, a^{3} + a^{2} b\right )} \cos \left (f x + e\right )^{4} + 4 \, a b^{2} + b^{3} + 2 \,{\left (4 \, a^{2} b + a b^{2}\right )} \cos \left (f x + e\right )^{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} b + a^{2} b^{2} - a b^{3} +{\left (4 \, a^{4} + 5 \, a^{3} b + a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{8 \,{\left ({\left (a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \,{\left (a^{6} b + 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} + a^{3} b^{4}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + a^{2} b^{5}\right )} f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.30061, size = 219, normalized size = 1.75 \begin{align*} -\frac{\frac{{\left (4 \, a + b\right )} \arctan \left (\frac{a \sin \left (f x + e\right )}{\sqrt{-a^{2} - a b}}\right )}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sqrt{-a^{2} - a b}} + \frac{4 \, a^{2} \sin \left (f x + e\right )^{3} + a b \sin \left (f x + e\right )^{3} - 4 \, a^{2} \sin \left (f x + e\right ) - 3 \, a b \sin \left (f x + e\right ) + b^{2} \sin \left (f x + e\right )}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )}{\left (a \sin \left (f x + e\right )^{2} - a - b\right )}^{2}}}{8 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((4*a + b)*arctan(a*sin(f*x + e)/sqrt(-a^2 - a*b))/((a^3 + 2*a^2*b + a*b^2)*sqrt(-a^2 - a*b)) + (4*a^2*si
n(f*x + e)^3 + a*b*sin(f*x + e)^3 - 4*a^2*sin(f*x + e) - 3*a*b*sin(f*x + e) + b^2*sin(f*x + e))/((a^3 + 2*a^2*
b + a*b^2)*(a*sin(f*x + e)^2 - a - b)^2))/f