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

Optimal. Leaf size=74 \[ \frac {\sin (e+f x)}{2 f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{2 \sqrt {a} f (a+b)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4147, 199, 208} \[ \frac {\sin (e+f x)}{2 f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{2 \sqrt {a} f (a+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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]

Rubi steps

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

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Mathematica [A]  time = 0.27, size = 88, normalized size = 1.19 \[ \frac {\sqrt {a} \sqrt {a+b} \sin (e+f x)+\left (-a \sin ^2(e+f x)+a+b\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{\sqrt {a} f (a+b)^{3/2} (a \cos (2 (e+f x))+a+2 b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a]*Sqrt[a + b]*Sin[e + f*x] + ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]]*(a + b - a*Sin[e + f*x]^2))/(S
qrt[a]*(a + b)^(3/2)*f*(a + 2*b + a*Cos[2*(e + f*x)]))

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fricas [A]  time = 1.34, size = 262, normalized size = 3.54 \[ \left [\frac {{\left (a \cos \left (f x + e\right )^{2} + b\right )} \sqrt {a^{2} + a b} \log \left (-\frac {a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a^{2} + a b} \sin \left (f x + e\right ) - 2 \, a - b}{a \cos \left (f x + e\right )^{2} + b}\right ) + 2 \, {\left (a^{2} + a b\right )} \sin \left (f x + e\right )}{4 \, {\left ({\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} f\right )}}, -\frac {{\left (a \cos \left (f x + e\right )^{2} + b\right )} \sqrt {-a^{2} - a b} \arctan \left (\frac {\sqrt {-a^{2} - a b} \sin \left (f x + e\right )}{a + b}\right ) - {\left (a^{2} + a b\right )} \sin \left (f x + e\right )}{2 \, {\left ({\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*((a*cos(f*x + e)^2 + b)*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*(a^2 + a*b)*sin(f*x + e))/((a^4 + 2*a^3*b + a^2*b^2)*f*cos(f*x + e)^2 + (a^3*b +
2*a^2*b^2 + a*b^3)*f), -1/2*((a*cos(f*x + e)^2 + b)*sqrt(-a^2 - a*b)*arctan(sqrt(-a^2 - a*b)*sin(f*x + e)/(a
+ b)) - (a^2 + a*b)*sin(f*x + e))/((a^4 + 2*a^3*b + a^2*b^2)*f*cos(f*x + e)^2 + (a^3*b + 2*a^2*b^2 + a*b^3)*f)
]

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giac [A]  time = 0.30, size = 79, normalized size = 1.07 \[ -\frac {\frac {\arctan \left (\frac {a \sin \left (f x + e\right )}{\sqrt {-a^{2} - a b}}\right )}{\sqrt {-a^{2} - a b} {\left (a + b\right )}} + \frac {\sin \left (f x + e\right )}{{\left (a \sin \left (f x + e\right )^{2} - a - b\right )} {\left (a + b\right )}}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.64, size = 68, normalized size = 0.92 \[ \frac {-\frac {\sin \left (f x +e \right )}{2 \left (a +b \right ) \left (-a -b +a \left (\sin ^{2}\left (f x +e \right )\right )\right )}+\frac {\arctanh \left (\frac {a \sin \left (f x +e \right )}{\sqrt {\left (a +b \right ) a}}\right )}{2 \left (a +b \right ) \sqrt {\left (a +b \right ) a}}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(-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 [A]  time = 0.44, size = 98, normalized size = 1.32 \[ -\frac {\frac {2 \, \sin \left (f x + e\right )}{{\left (a^{2} + a b\right )} \sin \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}} + \frac {\log \left (\frac {a \sin \left (f x + e\right ) - \sqrt {{\left (a + b\right )} a}}{a \sin \left (f x + e\right ) + \sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} {\left (a + b\right )}}}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.14, size = 62, normalized size = 0.84 \[ \frac {\sin \left (e+f\,x\right )}{2\,f\,\left (a+b\right )\,\left (-a\,{\sin \left (e+f\,x\right )}^2+a+b\right )}+\frac {\mathrm {atanh}\left (\frac {\sqrt {a}\,\sin \left (e+f\,x\right )}{\sqrt {a+b}}\right )}{2\,\sqrt {a}\,f\,{\left (a+b\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{3}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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