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]

3/8*arctanh(sin(b*x+a))/b-3/8*sec(b*x+a)*tan(b*x+a)/b+1/4*sec(b*x+a)*tan(b*x+a)^3/b

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Rubi [A]  time = 0.04, antiderivative size = 55, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.11, 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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fricas [A]  time = 0.47, size = 74, normalized size = 1.35 \[ \frac {3 \, \cos \left (b x + a\right )^{4} \log \left (\sin \left (b x + a\right ) + 1\right ) - 3 \, \cos \left (b x + a\right )^{4} \log \left (-\sin \left (b x + a\right ) + 1\right ) - 2 \, {\left (5 \, \cos \left (b x + a\right )^{2} - 2\right )} \sin \left (b x + a\right )}{16 \, b \cos \left (b x + a\right )^{4}} \]

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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giac [A]  time = 0.39, size = 63, normalized size = 1.15 \[ \frac {\frac {2 \, {\left (5 \, \sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )\right )}}{{\left (\sin \left (b x + a\right )^{2} - 1\right )}^{2}} + 3 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) - 3 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{16 \, b} \]

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

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maple [A]  time = 0.04, size = 87, normalized size = 1.58 \[ \frac {\sin ^{5}\left (b x +a \right )}{4 b \cos \left (b x +a \right )^{4}}-\frac {\sin ^{5}\left (b x +a \right )}{8 b \cos \left (b x +a \right )^{2}}-\frac {\sin ^{3}\left (b x +a \right )}{8 b}-\frac {3 \sin \left (b x +a \right )}{8 b}+\frac {3 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{8 b} \]

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 = 0.51, size = 71, normalized size = 1.29 \[ \frac {\frac {2 \, {\left (5 \, \sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )\right )}}{\sin \left (b x + a\right )^{4} - 2 \, \sin \left (b x + a\right )^{2} + 1} + 3 \, \log \left (\sin \left (b x + a\right ) + 1\right ) - 3 \, \log \left (\sin \left (b x + a\right ) - 1\right )}{16 \, b} \]

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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mupad [B]  time = 6.59, size = 126, normalized size = 2.29 \[ \frac {3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )\right )}{4\,b}-\frac {\frac {3\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^7}{4}-\frac {11\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^5}{4}-\frac {11\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^3}{4}+\frac {3\,\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}{4}}{b\,\left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^4/cos(a + b*x)^5,x)

[Out]

(3*atanh(tan(a/2 + (b*x)/2)))/(4*b) - ((3*tan(a/2 + (b*x)/2))/4 - (11*tan(a/2 + (b*x)/2)^3)/4 - (11*tan(a/2 +
(b*x)/2)^5)/4 + (3*tan(a/2 + (b*x)/2)^7)/4)/(b*(6*tan(a/2 + (b*x)/2)^4 - 4*tan(a/2 + (b*x)/2)^2 - 4*tan(a/2 +
(b*x)/2)^6 + tan(a/2 + (b*x)/2)^8 + 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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