3.85 \(\int \tan ^4(a+b x) \, dx\)

Optimal. Leaf size=28 \[ \frac{\tan ^3(a+b x)}{3 b}-\frac{\tan (a+b x)}{b}+x \]

[Out]

x - Tan[a + b*x]/b + Tan[a + b*x]^3/(3*b)

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Rubi [A]  time = 0.0160507, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3473, 8} \[ \frac{\tan ^3(a+b x)}{3 b}-\frac{\tan (a+b x)}{b}+x \]

Antiderivative was successfully verified.

[In]

Int[Tan[a + b*x]^4,x]

[Out]

x - Tan[a + b*x]/b + Tan[a + b*x]^3/(3*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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \tan ^4(a+b x) \, dx &=\frac{\tan ^3(a+b x)}{3 b}-\int \tan ^2(a+b x) \, dx\\ &=-\frac{\tan (a+b x)}{b}+\frac{\tan ^3(a+b x)}{3 b}+\int 1 \, dx\\ &=x-\frac{\tan (a+b x)}{b}+\frac{\tan ^3(a+b x)}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.0094322, size = 38, normalized size = 1.36 \[ \frac{\tan ^3(a+b x)}{3 b}+\frac{\tan ^{-1}(\tan (a+b x))}{b}-\frac{\tan (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[a + b*x]^4,x]

[Out]

ArcTan[Tan[a + b*x]]/b - Tan[a + b*x]/b + Tan[a + b*x]^3/(3*b)

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Maple [A]  time = 0.02, size = 28, normalized size = 1. \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \tan \left ( bx+a \right ) \right ) ^{3}}{3}}-\tan \left ( bx+a \right ) +bx+a \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^4*sin(b*x+a)^4,x)

[Out]

1/b*(1/3*tan(b*x+a)^3-tan(b*x+a)+b*x+a)

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Maxima [A]  time = 1.50201, size = 39, normalized size = 1.39 \begin{align*} \frac{\tan \left (b x + a\right )^{3} + 3 \, b x + 3 \, a - 3 \, \tan \left (b x + a\right )}{3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4*sin(b*x+a)^4,x, algorithm="maxima")

[Out]

1/3*(tan(b*x + a)^3 + 3*b*x + 3*a - 3*tan(b*x + a))/b

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Fricas [A]  time = 1.6484, size = 115, normalized size = 4.11 \begin{align*} \frac{3 \, b x \cos \left (b x + a\right )^{3} -{\left (4 \, \cos \left (b x + a\right )^{2} - 1\right )} \sin \left (b x + a\right )}{3 \, b \cos \left (b x + a\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4*sin(b*x+a)^4,x, algorithm="fricas")

[Out]

1/3*(3*b*x*cos(b*x + a)^3 - (4*cos(b*x + a)^2 - 1)*sin(b*x + a))/(b*cos(b*x + a)^3)

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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)**4*sin(b*x+a)**4,x)

[Out]

Timed out

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Giac [A]  time = 1.69654, size = 39, normalized size = 1.39 \begin{align*} \frac{\tan \left (b x + a\right )^{3} + 3 \, b x + 3 \, a - 3 \, \tan \left (b x + a\right )}{3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4*sin(b*x+a)^4,x, algorithm="giac")

[Out]

1/3*(tan(b*x + a)^3 + 3*b*x + 3*a - 3*tan(b*x + a))/b