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

Optimal. Leaf size=50 \[ \frac{a \log \left (a+b \tan ^2(e+f x)\right )}{2 b f (a-b)}+\frac{\log (\cos (e+f x))}{f (a-b)} \]

[Out]

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

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Rubi [A]  time = 0.0862229, antiderivative size = 50, 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 = {3670, 446, 72} \[ \frac{a \log \left (a+b \tan ^2(e+f x)\right )}{2 b f (a-b)}+\frac{\log (\cos (e+f x))}{f (a-b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

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

Mathematica [A]  time = 0.0328375, size = 41, normalized size = 0.82 \[ \frac{a \log \left (a+b \tan ^2(e+f x)\right )+2 b \log (\cos (e+f x))}{2 a b f-2 b^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.016, size = 54, normalized size = 1.1 \begin{align*}{\frac{a\ln \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }{ \left ( 2\,a-2\,b \right ) bf}}-{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }{2\,f \left ( a-b \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.06747, size = 72, normalized size = 1.44 \begin{align*} \frac{\frac{a \log \left (-{\left (a - b\right )} \sin \left (f x + e\right )^{2} + a\right )}{a b - b^{2}} - \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*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.17745, size = 151, normalized size = 3.02 \begin{align*} \frac{a \log \left (\frac{b \tan \left (f x + e\right )^{2} + a}{\tan \left (f x + e\right )^{2} + 1}\right ) -{\left (a - b\right )} \log \left (\frac{1}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \,{\left (a b - b^{2}\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 4.97766, size = 240, normalized size = 4.8 \begin{align*} \begin{cases} \tilde{\infty } x \tan{\left (e \right )} & \text{for}\: a = 0 \wedge b = 0 \wedge f = 0 \\\frac{\log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )} \tan ^{2}{\left (e + f x \right )}}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} + \frac{\log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} + \frac{1}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} & \text{for}\: a = b \\\frac{- \frac{\log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac{\tan ^{2}{\left (e + f x \right )}}{2 f}}{a} & \text{for}\: b = 0 \\\frac{x \tan ^{3}{\left (e \right )}}{a + b \tan ^{2}{\left (e \right )}} & \text{for}\: f = 0 \\\frac{a \log{\left (- i \sqrt{a} \sqrt{\frac{1}{b}} + \tan{\left (e + f x \right )} \right )}}{2 a b f - 2 b^{2} f} + \frac{a \log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + \tan{\left (e + f x \right )} \right )}}{2 a b f - 2 b^{2} f} - \frac{b \log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 a b f - 2 b^{2} f} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x*tan(e), Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), (log(tan(e + f*x)**2 + 1)*tan(e + f*x)**2/(2*b*f*tan
(e + f*x)**2 + 2*b*f) + log(tan(e + f*x)**2 + 1)/(2*b*f*tan(e + f*x)**2 + 2*b*f) + 1/(2*b*f*tan(e + f*x)**2 +
2*b*f), Eq(a, b)), ((-log(tan(e + f*x)**2 + 1)/(2*f) + tan(e + f*x)**2/(2*f))/a, Eq(b, 0)), (x*tan(e)**3/(a +
b*tan(e)**2), Eq(f, 0)), (a*log(-I*sqrt(a)*sqrt(1/b) + tan(e + f*x))/(2*a*b*f - 2*b**2*f) + a*log(I*sqrt(a)*sq
rt(1/b) + tan(e + f*x))/(2*a*b*f - 2*b**2*f) - b*log(tan(e + f*x)**2 + 1)/(2*a*b*f - 2*b**2*f), True))

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Giac [B]  time = 1.97193, size = 252, normalized size = 5.04 \begin{align*} \frac{\frac{a^{2} \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 )} - 2 \, a + 4 \, 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 - b} - \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*tan(f*x+e)^2),x, algorithm="giac")

[Out]

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