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

Optimal. Leaf size=55 \[ \frac{3 \tanh ^{-1}(\sin (a+b x))}{8 b}+\frac{\tan ^3(a+b x) \sec (a+b x)}{4 b}-\frac{3 \tan (a+b x) \sec (a+b x)}{8 b} \]

[Out]

(3*ArcTanh[Sin[a + b*x]])/(8*b) - (3*Sec[a + b*x]*Tan[a + b*x])/(8*b) + (Sec[a + b*x]*Tan[a + b*x]^3)/(4*b)

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Rubi [A]  time = 0.0423837, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2611, 3770} \[ \frac{3 \tanh ^{-1}(\sin (a+b x))}{8 b}+\frac{\tan ^3(a+b x) \sec (a+b x)}{4 b}-\frac{3 \tan (a+b x) \sec (a+b x)}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]*Tan[a + b*x]^4,x]

[Out]

(3*ArcTanh[Sin[a + b*x]])/(8*b) - (3*Sec[a + b*x]*Tan[a + b*x])/(8*b) + (Sec[a + b*x]*Tan[a + b*x]^3)/(4*b)

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3770

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

Rubi steps

\begin{align*} \int \sec (a+b x) \tan ^4(a+b x) \, dx &=\frac{\sec (a+b x) \tan ^3(a+b x)}{4 b}-\frac{3}{4} \int \sec (a+b x) \tan ^2(a+b x) \, dx\\ &=-\frac{3 \sec (a+b x) \tan (a+b x)}{8 b}+\frac{\sec (a+b x) \tan ^3(a+b x)}{4 b}+\frac{3}{8} \int \sec (a+b x) \, dx\\ &=\frac{3 \tanh ^{-1}(\sin (a+b x))}{8 b}-\frac{3 \sec (a+b x) \tan (a+b x)}{8 b}+\frac{\sec (a+b x) \tan ^3(a+b x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.120879, size = 45, normalized size = 0.82 \[ \frac{6 \tanh ^{-1}(\sin (a+b x))-(5 \cos (2 (a+b x))+1) \tan (a+b x) \sec ^3(a+b x)}{16 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]*Tan[a + b*x]^4,x]

[Out]

(6*ArcTanh[Sin[a + b*x]] - (1 + 5*Cos[2*(a + b*x)])*Sec[a + b*x]^3*Tan[a + b*x])/(16*b)

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Maple [A]  time = 0.019, size = 87, normalized size = 1.6 \begin{align*}{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{5}}{4\,b \left ( \cos \left ( bx+a \right ) \right ) ^{4}}}-{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{5}}{8\,b \left ( \cos \left ( bx+a \right ) \right ) ^{2}}}-{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{3}}{8\,b}}-{\frac{3\,\sin \left ( bx+a \right ) }{8\,b}}+{\frac{3\,\ln \left ( \sec \left ( bx+a \right ) +\tan \left ( bx+a \right ) \right ) }{8\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^5*sin(b*x+a)^4,x)

[Out]

1/4/b*sin(b*x+a)^5/cos(b*x+a)^4-1/8/b*sin(b*x+a)^5/cos(b*x+a)^2-1/8*sin(b*x+a)^3/b-3/8*sin(b*x+a)/b+3/8/b*ln(s
ec(b*x+a)+tan(b*x+a))

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Maxima [A]  time = 1.06134, size = 96, normalized size = 1.75 \begin{align*} \frac{\frac{2 \,{\left (5 \, \sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )\right )}}{\sin \left (b x + a\right )^{4} - 2 \, \sin \left (b x + a\right )^{2} + 1} + 3 \, \log \left (\sin \left (b x + a\right ) + 1\right ) - 3 \, \log \left (\sin \left (b x + a\right ) - 1\right )}{16 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^5*sin(b*x+a)^4,x, algorithm="maxima")

[Out]

1/16*(2*(5*sin(b*x + a)^3 - 3*sin(b*x + a))/(sin(b*x + a)^4 - 2*sin(b*x + a)^2 + 1) + 3*log(sin(b*x + a) + 1)
- 3*log(sin(b*x + a) - 1))/b

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Fricas [A]  time = 1.68203, size = 200, normalized size = 3.64 \begin{align*} \frac{3 \, \cos \left (b x + a\right )^{4} \log \left (\sin \left (b x + a\right ) + 1\right ) - 3 \, \cos \left (b x + a\right )^{4} \log \left (-\sin \left (b x + a\right ) + 1\right ) - 2 \,{\left (5 \, \cos \left (b x + a\right )^{2} - 2\right )} \sin \left (b x + a\right )}{16 \, b \cos \left (b x + a\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^5*sin(b*x+a)^4,x, algorithm="fricas")

[Out]

1/16*(3*cos(b*x + a)^4*log(sin(b*x + a) + 1) - 3*cos(b*x + a)^4*log(-sin(b*x + a) + 1) - 2*(5*cos(b*x + a)^2 -
2)*sin(b*x + a))/(b*cos(b*x + a)^4)

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

[Out]

Timed out

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Giac [A]  time = 1.21393, size = 85, normalized size = 1.55 \begin{align*} \frac{\frac{2 \,{\left (5 \, \sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )\right )}}{{\left (\sin \left (b x + a\right )^{2} - 1\right )}^{2}} + 3 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) - 3 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{16 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^5*sin(b*x+a)^4,x, algorithm="giac")

[Out]

1/16*(2*(5*sin(b*x + a)^3 - 3*sin(b*x + a))/(sin(b*x + a)^2 - 1)^2 + 3*log(abs(sin(b*x + a) + 1)) - 3*log(abs(
sin(b*x + a) - 1)))/b