3.64 \(\int \sec ^3(a+b x) \tan ^2(a+b x) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0448754, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {2611, 3768, 3770} \[ -\frac{\tanh ^{-1}(\sin (a+b x))}{8 b}+\frac{\tan (a+b x) \sec ^3(a+b x)}{4 b}-\frac{\tan (a+b x) \sec (a+b x)}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^3*Tan[a + b*x]^2,x]

[Out]

-ArcTanh[Sin[a + b*x]]/(8*b) - (Sec[a + b*x]*Tan[a + b*x])/(8*b) + (Sec[a + b*x]^3*Tan[a + b*x])/(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 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 ^3(a+b x) \tan ^2(a+b x) \, dx &=\frac{\sec ^3(a+b x) \tan (a+b x)}{4 b}-\frac{1}{4} \int \sec ^3(a+b x) \, dx\\ &=-\frac{\sec (a+b x) \tan (a+b x)}{8 b}+\frac{\sec ^3(a+b x) \tan (a+b x)}{4 b}-\frac{1}{8} \int \sec (a+b x) \, dx\\ &=-\frac{\tanh ^{-1}(\sin (a+b x))}{8 b}-\frac{\sec (a+b x) \tan (a+b x)}{8 b}+\frac{\sec ^3(a+b x) \tan (a+b x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.0447106, size = 55, normalized size = 1. \[ -\frac{\tanh ^{-1}(\sin (a+b x))}{8 b}+\frac{\tan (a+b x) \sec ^3(a+b x)}{4 b}-\frac{\tan (a+b x) \sec (a+b x)}{8 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^3*Tan[a + b*x]^2,x]

[Out]

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

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Maple [A]  time = 0.021, size = 74, normalized size = 1.4 \begin{align*}{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{3}}{4\,b \left ( \cos \left ( bx+a \right ) \right ) ^{4}}}+{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{3}}{8\,b \left ( \cos \left ( bx+a \right ) \right ) ^{2}}}+{\frac{\sin \left ( bx+a \right ) }{8\,b}}-{\frac{\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)^2,x)

[Out]

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

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Maxima [A]  time = 1.04326, size = 88, normalized size = 1.6 \begin{align*} \frac{\frac{2 \,{\left (\sin \left (b x + a\right )^{3} + \sin \left (b x + a\right )\right )}}{\sin \left (b x + a\right )^{4} - 2 \, \sin \left (b x + a\right )^{2} + 1} - \log \left (\sin \left (b x + a\right ) + 1\right ) + \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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.70726, size = 193, normalized size = 3.51 \begin{align*} -\frac{\cos \left (b x + a\right )^{4} \log \left (\sin \left (b x + a\right ) + 1\right ) - \cos \left (b x + a\right )^{4} \log \left (-\sin \left (b x + a\right ) + 1\right ) + 2 \,{\left (\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)^2,x, algorithm="fricas")

[Out]

-1/16*(cos(b*x + a)^4*log(sin(b*x + a) + 1) - cos(b*x + a)^4*log(-sin(b*x + a) + 1) + 2*(cos(b*x + a)^2 - 2)*s
in(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)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.25782, size = 111, normalized size = 2.02 \begin{align*} \frac{\frac{4 \,{\left (\frac{1}{\sin \left (b x + a\right )} + \sin \left (b x + a\right )\right )}}{{\left (\frac{1}{\sin \left (b x + a\right )} + \sin \left (b x + a\right )\right )}^{2} - 4} - \log \left ({\left | \frac{1}{\sin \left (b x + a\right )} + \sin \left (b x + a\right ) + 2 \right |}\right ) + \log \left ({\left | \frac{1}{\sin \left (b x + a\right )} + \sin \left (b x + a\right ) - 2 \right |}\right )}{32 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^5*sin(b*x+a)^2,x, algorithm="giac")

[Out]

1/32*(4*(1/sin(b*x + a) + sin(b*x + a))/((1/sin(b*x + a) + sin(b*x + a))^2 - 4) - log(abs(1/sin(b*x + a) + sin
(b*x + a) + 2)) + log(abs(1/sin(b*x + a) + sin(b*x + a) - 2)))/b