3.73 \(\int \tan ^3(a+b x) \, dx\)

Optimal. Leaf size=27 \[ \frac{\tan ^2(a+b x)}{2 b}+\frac{\log (\cos (a+b x))}{b} \]

[Out]

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

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Rubi [A]  time = 0.0111813, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3473, 3475} \[ \frac{\tan ^2(a+b x)}{2 b}+\frac{\log (\cos (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Tan[a + b*x]^3,x]

[Out]

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

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \tan ^3(a+b x) \, dx &=\frac{\tan ^2(a+b x)}{2 b}-\int \tan (a+b x) \, dx\\ &=\frac{\log (\cos (a+b x))}{b}+\frac{\tan ^2(a+b x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.024339, size = 25, normalized size = 0.93 \[ \frac{\tan ^2(a+b x)+2 \log (\cos (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[a + b*x]^3,x]

[Out]

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

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Maple [A]  time = 0.019, size = 26, normalized size = 1. \begin{align*}{\frac{\ln \left ( \cos \left ( bx+a \right ) \right ) }{b}}+{\frac{ \left ( \tan \left ( bx+a \right ) \right ) ^{2}}{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^3*sin(b*x+a)^3,x)

[Out]

ln(cos(b*x+a))/b+1/2*tan(b*x+a)^2/b

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Maxima [A]  time = 1.00059, size = 42, normalized size = 1.56 \begin{align*} -\frac{\frac{1}{\sin \left (b x + a\right )^{2} - 1} - \log \left (\sin \left (b x + a\right )^{2} - 1\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3*sin(b*x+a)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.63733, size = 89, normalized size = 3.3 \begin{align*} \frac{2 \, \cos \left (b x + a\right )^{2} \log \left (-\cos \left (b x + a\right )\right ) + 1}{2 \, b \cos \left (b x + a\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3*sin(b*x+a)^3,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**3*sin(b*x+a)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.18609, size = 57, normalized size = 2.11 \begin{align*} \frac{\log \left (\frac{\cos \left (b x + a\right )^{2}}{b^{2}}\right )}{2 \, b} - \frac{\cos \left (b x + a\right )^{2} - 1}{2 \, b \cos \left (b x + a\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3*sin(b*x+a)^3,x, algorithm="giac")

[Out]

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