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

Optimal. Leaf size=45 \[ \frac{(a+b) \log \left (a \cos ^2(e+f x)+b\right )}{2 a b f}-\frac{\log (\cos (e+f x))}{b f} \]

[Out]

-(Log[Cos[e + f*x]]/(b*f)) + ((a + b)*Log[b + a*Cos[e + f*x]^2])/(2*a*b*f)

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Rubi [A]  time = 0.0772854, antiderivative size = 45, 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 = {4138, 446, 72} \[ \frac{(a+b) \log \left (a \cos ^2(e+f x)+b\right )}{2 a b f}-\frac{\log (\cos (e+f x))}{b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Log[Cos[e + f*x]]/(b*f)) + ((a + b)*Log[b + a*Cos[e + f*x]^2])/(2*a*b*f)

Rule 4138

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

Rule 446

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

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

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

Mathematica [A]  time = 0.114396, size = 41, normalized size = 0.91 \[ \frac{(a+b) \log \left (a \cos ^2(e+f x)+b\right )-2 a \log (\cos (e+f x))}{2 a b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*Log[Cos[e + f*x]] + (a + b)*Log[b + a*Cos[e + f*x]^2])/(2*a*b*f)

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Maple [A]  time = 0.058, size = 59, normalized size = 1.3 \begin{align*}{\frac{\ln \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }{2\,fb}}+{\frac{\ln \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }{2\,af}}-{\frac{\ln \left ( \cos \left ( fx+e \right ) \right ) }{fb}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.02529, size = 68, normalized size = 1.51 \begin{align*} \frac{\frac{{\left (a + b\right )} \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a b} - \frac{\log \left (\sin \left (f x + e\right )^{2} - 1\right )}{b}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.596589, size = 100, normalized size = 2.22 \begin{align*} \frac{{\left (a + b\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) - 2 \, a \log \left (-\cos \left (f x + e\right )\right )}{2 \, a b f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a + b)*log(a*cos(f*x + e)^2 + b) - 2*a*log(-cos(f*x + e)))/(a*b*f)

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

[Out]

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

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Giac [B]  time = 1.66071, size = 319, normalized size = 7.09 \begin{align*} \frac{\frac{{\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left ({\left | -a{\left (\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} - b{\left (\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} - 2 \, a + 2 \, b \right |}\right )}{a^{2} b + a b^{2}} - \frac{\log \left (-\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 2\right )}{a} - \frac{\log \left (-\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 2\right )}{b}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^2 + 2*a*b + b^2)*log(abs(-a*((cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(cos(f*x + e)
+ 1)) - b*((cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(cos(f*x + e) + 1)) - 2*a + 2*b))/(a^2*b
+ a*b^2) - log(-(cos(f*x + e) + 1)/(cos(f*x + e) - 1) - (cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 2)/a - log(-(
cos(f*x + e) + 1)/(cos(f*x + e) - 1) - (cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 2)/b)/f