3.97 \(\int \sec ^5(a+b x) \tan ^4(a+b x) \, dx\)

Optimal. Leaf size=99 \[ \frac{3 \tanh ^{-1}(\sin (a+b x))}{128 b}+\frac{\tan ^3(a+b x) \sec ^5(a+b x)}{8 b}-\frac{\tan (a+b x) \sec ^5(a+b x)}{16 b}+\frac{\tan (a+b x) \sec ^3(a+b x)}{64 b}+\frac{3 \tan (a+b x) \sec (a+b x)}{128 b} \]

[Out]

(3*ArcTanh[Sin[a + b*x]])/(128*b) + (3*Sec[a + b*x]*Tan[a + b*x])/(128*b) + (Sec[a + b*x]^3*Tan[a + b*x])/(64*
b) - (Sec[a + b*x]^5*Tan[a + b*x])/(16*b) + (Sec[a + b*x]^5*Tan[a + b*x]^3)/(8*b)

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Rubi [A]  time = 0.0893559, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {2611, 3768, 3770} \[ \frac{3 \tanh ^{-1}(\sin (a+b x))}{128 b}+\frac{\tan ^3(a+b x) \sec ^5(a+b x)}{8 b}-\frac{\tan (a+b x) \sec ^5(a+b x)}{16 b}+\frac{\tan (a+b x) \sec ^3(a+b x)}{64 b}+\frac{3 \tan (a+b x) \sec (a+b x)}{128 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^5*Tan[a + b*x]^4,x]

[Out]

(3*ArcTanh[Sin[a + b*x]])/(128*b) + (3*Sec[a + b*x]*Tan[a + b*x])/(128*b) + (Sec[a + b*x]^3*Tan[a + b*x])/(64*
b) - (Sec[a + b*x]^5*Tan[a + b*x])/(16*b) + (Sec[a + b*x]^5*Tan[a + b*x]^3)/(8*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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[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 ^5(a+b x) \tan ^4(a+b x) \, dx &=\frac{\sec ^5(a+b x) \tan ^3(a+b x)}{8 b}-\frac{3}{8} \int \sec ^5(a+b x) \tan ^2(a+b x) \, dx\\ &=-\frac{\sec ^5(a+b x) \tan (a+b x)}{16 b}+\frac{\sec ^5(a+b x) \tan ^3(a+b x)}{8 b}+\frac{1}{16} \int \sec ^5(a+b x) \, dx\\ &=\frac{\sec ^3(a+b x) \tan (a+b x)}{64 b}-\frac{\sec ^5(a+b x) \tan (a+b x)}{16 b}+\frac{\sec ^5(a+b x) \tan ^3(a+b x)}{8 b}+\frac{3}{64} \int \sec ^3(a+b x) \, dx\\ &=\frac{3 \sec (a+b x) \tan (a+b x)}{128 b}+\frac{\sec ^3(a+b x) \tan (a+b x)}{64 b}-\frac{\sec ^5(a+b x) \tan (a+b x)}{16 b}+\frac{\sec ^5(a+b x) \tan ^3(a+b x)}{8 b}+\frac{3}{128} \int \sec (a+b x) \, dx\\ &=\frac{3 \tanh ^{-1}(\sin (a+b x))}{128 b}+\frac{3 \sec (a+b x) \tan (a+b x)}{128 b}+\frac{\sec ^3(a+b x) \tan (a+b x)}{64 b}-\frac{\sec ^5(a+b x) \tan (a+b x)}{16 b}+\frac{\sec ^5(a+b x) \tan ^3(a+b x)}{8 b}\\ \end{align*}

Mathematica [A]  time = 0.312647, size = 64, normalized size = 0.65 \[ \frac{96 \tanh ^{-1}(\sin (a+b x))+(-307 \cos (2 (a+b x))+26 \cos (4 (a+b x))+3 \cos (6 (a+b x))+182) \tan (a+b x) \sec ^7(a+b x)}{4096 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^5*Tan[a + b*x]^4,x]

[Out]

(96*ArcTanh[Sin[a + b*x]] + (182 - 307*Cos[2*(a + b*x)] + 26*Cos[4*(a + b*x)] + 3*Cos[6*(a + b*x)])*Sec[a + b*
x]^7*Tan[a + b*x])/(4096*b)

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Maple [A]  time = 0.022, size = 129, normalized size = 1.3 \begin{align*}{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{5}}{8\,b \left ( \cos \left ( bx+a \right ) \right ) ^{8}}}+{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{5}}{16\,b \left ( \cos \left ( bx+a \right ) \right ) ^{6}}}+{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{5}}{64\,b \left ( \cos \left ( bx+a \right ) \right ) ^{4}}}-{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{5}}{128\,b \left ( \cos \left ( bx+a \right ) \right ) ^{2}}}-{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{3}}{128\,b}}-{\frac{3\,\sin \left ( bx+a \right ) }{128\,b}}+{\frac{3\,\ln \left ( \sec \left ( bx+a \right ) +\tan \left ( bx+a \right ) \right ) }{128\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^9*sin(b*x+a)^4,x)

[Out]

1/8/b*sin(b*x+a)^5/cos(b*x+a)^8+1/16/b*sin(b*x+a)^5/cos(b*x+a)^6+1/64/b*sin(b*x+a)^5/cos(b*x+a)^4-1/128/b*sin(
b*x+a)^5/cos(b*x+a)^2-1/128*sin(b*x+a)^3/b-3/128*sin(b*x+a)/b+3/128/b*ln(sec(b*x+a)+tan(b*x+a))

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Maxima [A]  time = 1.08345, size = 150, normalized size = 1.52 \begin{align*} -\frac{\frac{2 \,{\left (3 \, \sin \left (b x + a\right )^{7} - 11 \, \sin \left (b x + a\right )^{5} - 11 \, \sin \left (b x + a\right )^{3} + 3 \, \sin \left (b x + a\right )\right )}}{\sin \left (b x + a\right )^{8} - 4 \, \sin \left (b x + a\right )^{6} + 6 \, \sin \left (b x + a\right )^{4} - 4 \, \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 )}{256 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^9*sin(b*x+a)^4,x, algorithm="maxima")

[Out]

-1/256*(2*(3*sin(b*x + a)^7 - 11*sin(b*x + a)^5 - 11*sin(b*x + a)^3 + 3*sin(b*x + a))/(sin(b*x + a)^8 - 4*sin(
b*x + a)^6 + 6*sin(b*x + a)^4 - 4*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 = 1.76918, size = 255, normalized size = 2.58 \begin{align*} \frac{3 \, \cos \left (b x + a\right )^{8} \log \left (\sin \left (b x + a\right ) + 1\right ) - 3 \, \cos \left (b x + a\right )^{8} \log \left (-\sin \left (b x + a\right ) + 1\right ) + 2 \,{\left (3 \, \cos \left (b x + a\right )^{6} + 2 \, \cos \left (b x + a\right )^{4} - 24 \, \cos \left (b x + a\right )^{2} + 16\right )} \sin \left (b x + a\right )}{256 \, b \cos \left (b x + a\right )^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^9*sin(b*x+a)^4,x, algorithm="fricas")

[Out]

1/256*(3*cos(b*x + a)^8*log(sin(b*x + a) + 1) - 3*cos(b*x + a)^8*log(-sin(b*x + a) + 1) + 2*(3*cos(b*x + a)^6
+ 2*cos(b*x + a)^4 - 24*cos(b*x + a)^2 + 16)*sin(b*x + a))/(b*cos(b*x + a)^8)

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

[Out]

Timed out

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Giac [A]  time = 1.21974, size = 144, normalized size = 1.45 \begin{align*} -\frac{\frac{4 \,{\left (3 \,{\left (\frac{1}{\sin \left (b x + a\right )} + \sin \left (b x + a\right )\right )}^{3} - \frac{20}{\sin \left (b x + a\right )} - 20 \, \sin \left (b x + a\right )\right )}}{{\left ({\left (\frac{1}{\sin \left (b x + a\right )} + \sin \left (b x + a\right )\right )}^{2} - 4\right )}^{2}} - 3 \, \log \left ({\left | \frac{1}{\sin \left (b x + a\right )} + \sin \left (b x + a\right ) + 2 \right |}\right ) + 3 \, \log \left ({\left | \frac{1}{\sin \left (b x + a\right )} + \sin \left (b x + a\right ) - 2 \right |}\right )}{512 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^9*sin(b*x+a)^4,x, algorithm="giac")

[Out]

-1/512*(4*(3*(1/sin(b*x + a) + sin(b*x + a))^3 - 20/sin(b*x + a) - 20*sin(b*x + a))/((1/sin(b*x + a) + sin(b*x
+ a))^2 - 4)^2 - 3*log(abs(1/sin(b*x + a) + sin(b*x + a) + 2)) + 3*log(abs(1/sin(b*x + a) + sin(b*x + a) - 2)
))/b