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

Optimal. Leaf size=23 \[ -\frac{\log \left (a \cos ^3(e+f x)+b\right )}{3 a f} \]

[Out]

-Log[b + a*Cos[e + f*x]^3]/(3*a*f)

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Rubi [A]  time = 0.0307096, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {4138, 260} \[ -\frac{\log \left (a \cos ^3(e+f x)+b\right )}{3 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Log[b + a*Cos[e + f*x]^3]/(3*a*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 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

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

Mathematica [A]  time = 0.0186448, size = 23, normalized size = 1. \[ -\frac{\log \left (a \cos ^3(e+f x)+b\right )}{3 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Log[b + a*Cos[e + f*x]^3]/(3*a*f)

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Maple [A]  time = 0.026, size = 37, normalized size = 1.6 \begin{align*} -{\frac{\ln \left ( a+b \left ( \sec \left ( fx+e \right ) \right ) ^{3} \right ) }{3\,fa}}+{\frac{\ln \left ( \sec \left ( fx+e \right ) \right ) }{fa}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/f/a*ln(a+b*sec(f*x+e)^3)+1/f/a*ln(sec(f*x+e))

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Maxima [A]  time = 0.998469, size = 28, normalized size = 1.22 \begin{align*} -\frac{\log \left (a \cos \left (f x + e\right )^{3} + b\right )}{3 \, a f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*log(a*cos(f*x + e)^3 + b)/(a*f)

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Fricas [A]  time = 0.552398, size = 51, normalized size = 2.22 \begin{align*} -\frac{\log \left (a \cos \left (f x + e\right )^{3} + b\right )}{3 \, a f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*log(a*cos(f*x + e)^3 + b)/(a*f)

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Sympy [A]  time = 82.6005, size = 170, normalized size = 7.39 \begin{align*} \begin{cases} \frac{\tilde{\infty } x \tan{\left (e \right )}}{\sec ^{3}{\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 )}}{2 a f} & \text{for}\: b = 0 \\\frac{x \tan{\left (e \right )}}{a + b \sec ^{3}{\left (e \right )}} & \text{for}\: f = 0 \\- \frac{1}{3 b f \sec ^{3}{\left (e + f x \right )}} & \text{for}\: a = 0 \\- \frac{\log{\left (- \sqrt [3]{-1} \sqrt [3]{a} \sqrt [3]{\frac{1}{b}} + \sec{\left (e + f x \right )} \right )}}{3 a f} + \frac{\log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 a f} - \frac{\log{\left (4 \left (-1\right )^{\frac{2}{3}} a^{\frac{2}{3}} \left (\frac{1}{b}\right )^{\frac{2}{3}} + 4 \sqrt [3]{-1} \sqrt [3]{a} \sqrt [3]{\frac{1}{b}} \sec{\left (e + f x \right )} + 4 \sec ^{2}{\left (e + f x \right )} \right )}}{3 a f} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x*tan(e)/sec(e)**3, Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), (log(tan(e + f*x)**2 + 1)/(2*a*f), Eq(b, 0
)), (x*tan(e)/(a + b*sec(e)**3), Eq(f, 0)), (-1/(3*b*f*sec(e + f*x)**3), Eq(a, 0)), (-log(-(-1)**(1/3)*a**(1/3
)*(1/b)**(1/3) + sec(e + f*x))/(3*a*f) + log(tan(e + f*x)**2 + 1)/(2*a*f) - log(4*(-1)**(2/3)*a**(2/3)*(1/b)**
(2/3) + 4*(-1)**(1/3)*a**(1/3)*(1/b)**(1/3)*sec(e + f*x) + 4*sec(e + f*x)**2)/(3*a*f), True))

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Giac [B]  time = 1.22856, size = 258, normalized size = 11.22 \begin{align*} \frac{\frac{3 \, \log \left (-\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1\right )}{a} - \frac{\log \left ({\left | a + b + \frac{3 \, a{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac{3 \, b{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{3 \, a{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{3 \, b{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{a{\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac{b{\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} \right |}\right )}{a}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*log(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 1)/a - log(abs(a + b + 3*a*(cos(f*x + e) - 1)/(cos(f*x + e
) + 1) - 3*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 3*a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 3*b*(cos(
f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + a*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - b*(cos(f*x + e) - 1)^3/(c
os(f*x + e) + 1)^3))/a)/f