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

Optimal. Leaf size=55 \[ \frac{\tanh ^{-1}(\sin (e+f x))}{b f}-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{b f \sqrt{a+b}} \]

[Out]

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

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Rubi [A]  time = 0.0717949, antiderivative size = 55, 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, 391, 206, 208} \[ \frac{\tanh ^{-1}(\sin (e+f x))}{b f}-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{b f \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 391

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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

Mathematica [C]  time = 2.02713, size = 1022, normalized size = 18.58 \[ \frac{(\cos (2 (e+f x)) a+a+2 b) \sec ^2(e+f x) \left (-4 \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-\sqrt{a} \cos (e) \log \left (-\cos (2 (e+f x)) a-2 i \sin (2 e) a+a+2 \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (f x) \sqrt{a}+2 \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (2 e+f x) \sqrt{a}+2 (a+b) \cos (2 e)-2 i b \sin (2 e)\right )+\sqrt{a} \cos (e) \log \left (\cos (2 (e+f x)) a+2 i \sin (2 e) a-a+2 \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (f x) \sqrt{a}+2 \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (2 e+f x) \sqrt{a}-2 (a+b) \cos (2 e)+2 i b \sin (2 e)\right )-2 i \sqrt{a} \tan ^{-1}\left (\frac{2 \sin (e) \left (\sin (2 e) a+i a-i \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (f x) \sqrt{a}-i \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (2 e+f x) \sqrt{a}+\sqrt{a+b} \cos (f x) \sqrt{(\cos (e)-i \sin (e))^2} \sqrt{a}-\sqrt{a+b} \cos (2 e+f x) \sqrt{(\cos (e)-i \sin (e))^2} \sqrt{a}+i b+i (a+b) \cos (2 e)+b \sin (2 e)\right )}{i (a+3 b) \cos (e)+i (a+b) \cos (3 e)+i a \cos (e+2 f x)+i a \cos (3 e+2 f x)+3 a \sin (e)+b \sin (e)+a \sin (3 e)+b \sin (3 e)+a \sin (e+2 f x)-a \sin (3 e+2 f x)}\right ) (\cos (e)-i \sin (e))+i \sqrt{a} \log \left (-\cos (2 (e+f x)) a-2 i \sin (2 e) a+a+2 \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (f x) \sqrt{a}+2 \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (2 e+f x) \sqrt{a}+2 (a+b) \cos (2 e)-2 i b \sin (2 e)\right ) \sin (e)-i \sqrt{a} \log \left (\cos (2 (e+f x)) a+2 i \sin (2 e) a-a+2 \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (f x) \sqrt{a}+2 \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (2 e+f x) \sqrt{a}-2 (a+b) \cos (2 e)+2 i b \sin (2 e)\right ) \sin (e)+2 \sqrt{a} \tan ^{-1}\left (\frac{(a+b) \sin (e)}{(a+b) \cos (e)-\sqrt{a} \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} (\cos (2 e)+i \sin (2 e)) \sin (e+f x)}\right ) (i \cos (e)+\sin (e))+4 \sqrt{a+b} \log \left (\cos \left (\frac{1}{2} (e+f x)\right )+\sin \left (\frac{1}{2} (e+f x)\right )\right ) \sqrt{(\cos (e)-i \sin (e))^2}\right )}{8 b \sqrt{a+b} f \left (b \sec ^2(e+f x)+a\right ) \sqrt{(\cos (e)-i \sin (e))^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^2*(-(Sqrt[a]*Cos[e]*Log[a + 2*(a + b)*Cos[2*e] - a*Cos[2*(e + f*x
)] - (2*I)*a*Sin[2*e] - (2*I)*b*Sin[2*e] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Sin[f*x] + 2*Sqrt
[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Sin[2*e + f*x]]) + Sqrt[a]*Cos[e]*Log[-a - 2*(a + b)*Cos[2*e] + a*
Cos[2*(e + f*x)] + (2*I)*a*Sin[2*e] + (2*I)*b*Sin[2*e] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Sin
[f*x] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Sin[2*e + f*x]] - (2*I)*Sqrt[a]*ArcTan[(2*Sin[e]*(I*
a + I*b + I*(a + b)*Cos[2*e] + Sqrt[a]*Sqrt[a + b]*Cos[f*x]*Sqrt[(Cos[e] - I*Sin[e])^2] - Sqrt[a]*Sqrt[a + b]*
Cos[2*e + f*x]*Sqrt[(Cos[e] - I*Sin[e])^2] + a*Sin[2*e] + b*Sin[2*e] - I*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*
Sin[e])^2]*Sin[f*x] - I*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Sin[2*e + f*x]))/(I*(a + 3*b)*Cos[e] +
I*(a + b)*Cos[3*e] + I*a*Cos[e + 2*f*x] + I*a*Cos[3*e + 2*f*x] + 3*a*Sin[e] + b*Sin[e] + a*Sin[3*e] + b*Sin[3
*e] + a*Sin[e + 2*f*x] - a*Sin[3*e + 2*f*x])]*(Cos[e] - I*Sin[e]) - 4*Sqrt[a + b]*Log[Cos[(e + f*x)/2] - Sin[(
e + f*x)/2]]*Sqrt[(Cos[e] - I*Sin[e])^2] + 4*Sqrt[a + b]*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]*Sqrt[(Cos[e]
- I*Sin[e])^2] + I*Sqrt[a]*Log[a + 2*(a + b)*Cos[2*e] - a*Cos[2*(e + f*x)] - (2*I)*a*Sin[2*e] - (2*I)*b*Sin[2
*e] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Sin[f*x] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[
e])^2]*Sin[2*e + f*x]]*Sin[e] - I*Sqrt[a]*Log[-a - 2*(a + b)*Cos[2*e] + a*Cos[2*(e + f*x)] + (2*I)*a*Sin[2*e]
+ (2*I)*b*Sin[2*e] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Sin[f*x] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(
Cos[e] - I*Sin[e])^2]*Sin[2*e + f*x]]*Sin[e] + 2*Sqrt[a]*ArcTan[((a + b)*Sin[e])/((a + b)*Cos[e] - Sqrt[a]*Sqr
t[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*(Cos[2*e] + I*Sin[2*e])*Sin[e + f*x])]*(I*Cos[e] + Sin[e])))/(8*b*Sqrt[a
+ b]*f*(a + b*Sec[e + f*x]^2)*Sqrt[(Cos[e] - I*Sin[e])^2])

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Maple [A]  time = 0.059, size = 68, normalized size = 1.2 \begin{align*}{\frac{\ln \left ( \sin \left ( fx+e \right ) +1 \right ) }{2\,fb}}-{\frac{a}{fb}{\it Artanh} \left ({a\sin \left ( fx+e \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}}}-{\frac{\ln \left ( \sin \left ( fx+e \right ) -1 \right ) }{2\,fb}} \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),x)

[Out]

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

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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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.557766, size = 405, normalized size = 7.36 \begin{align*} \left [\frac{\sqrt{\frac{a}{a + b}} \log \left (-\frac{a \cos \left (f x + e\right )^{2} + 2 \,{\left (a + b\right )} \sqrt{\frac{a}{a + b}} \sin \left (f x + e\right ) - 2 \, a - b}{a \cos \left (f x + e\right )^{2} + b}\right ) + \log \left (\sin \left (f x + e\right ) + 1\right ) - \log \left (-\sin \left (f x + e\right ) + 1\right )}{2 \, b f}, \frac{2 \, \sqrt{-\frac{a}{a + b}} \arctan \left (\sqrt{-\frac{a}{a + b}} \sin \left (f x + e\right )\right ) + \log \left (\sin \left (f x + e\right ) + 1\right ) - \log \left (-\sin \left (f x + e\right ) + 1\right )}{2 \, b f}\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),x, algorithm="fricas")

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.30815, size = 104, normalized size = 1.89 \begin{align*} \frac{\frac{2 \, a \arctan \left (\frac{a \sin \left (f x + e\right )}{\sqrt{-a^{2} - a b}}\right )}{\sqrt{-a^{2} - a b} b} + \frac{\log \left (\sin \left (f x + e\right ) + 1\right )}{b} - \frac{\log \left (-\sin \left (f x + e\right ) + 1\right )}{b}}{2 \, 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),x, algorithm="giac")

[Out]

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