3.94 \(\int \sin (a+b x) \tan ^3(a+b x) \, dx\)

Optimal. Leaf size=49 \[ \frac{3 \sin (a+b x)}{2 b}+\frac{\sin (a+b x) \tan ^2(a+b x)}{2 b}-\frac{3 \tanh ^{-1}(\sin (a+b x))}{2 b} \]

[Out]

(-3*ArcTanh[Sin[a + b*x]])/(2*b) + (3*Sin[a + b*x])/(2*b) + (Sin[a + b*x]*Tan[a + b*x]^2)/(2*b)

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Rubi [A]  time = 0.0292958, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {2592, 288, 321, 206} \[ \frac{3 \sin (a+b x)}{2 b}+\frac{\sin (a+b x) \tan ^2(a+b x)}{2 b}-\frac{3 \tanh ^{-1}(\sin (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]*Tan[a + b*x]^3,x]

[Out]

(-3*ArcTanh[Sin[a + b*x]])/(2*b) + (3*Sin[a + b*x])/(2*b) + (Sin[a + b*x]*Tan[a + b*x]^2)/(2*b)

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0907542, size = 40, normalized size = 0.82 \[ \frac{(\cos (2 (a+b x))+2) \tan (a+b x) \sec (a+b x)-3 \tanh ^{-1}(\sin (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*ArcTanh[Sin[a + b*x]] + (2 + Cos[2*(a + b*x)])*Sec[a + b*x]*Tan[a + b*x])/(2*b)

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Maple [A]  time = 0.02, size = 66, normalized size = 1.4 \begin{align*}{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{5}}{2\,b \left ( \cos \left ( bx+a \right ) \right ) ^{2}}}+{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{3}}{2\,b}}+{\frac{3\,\sin \left ( bx+a \right ) }{2\,b}}-{\frac{3\,\ln \left ( \sec \left ( bx+a \right ) +\tan \left ( bx+a \right ) \right ) }{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^3*sin(b*x+a)^4,x)

[Out]

1/2/b*sin(b*x+a)^5/cos(b*x+a)^2+1/2*sin(b*x+a)^3/b+3/2*sin(b*x+a)/b-3/2/b*ln(sec(b*x+a)+tan(b*x+a))

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Maxima [A]  time = 1.01236, size = 76, normalized size = 1.55 \begin{align*} -\frac{\frac{2 \, \sin \left (b x + a\right )}{\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 ) - 4 \, \sin \left (b x + a\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3*sin(b*x+a)^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.72964, size = 200, normalized size = 4.08 \begin{align*} -\frac{3 \, \cos \left (b x + a\right )^{2} \log \left (\sin \left (b x + a\right ) + 1\right ) - 3 \, \cos \left (b x + a\right )^{2} \log \left (-\sin \left (b x + a\right ) + 1\right ) - 2 \,{\left (2 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right )}{4 \, b \cos \left (b x + a\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3*sin(b*x+a)^4,x, algorithm="fricas")

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.19916, size = 78, normalized size = 1.59 \begin{align*} -\frac{\frac{2 \, \sin \left (b x + a\right )}{\sin \left (b x + a\right )^{2} - 1} + 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 ) - 4 \, \sin \left (b x + a\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3*sin(b*x+a)^4,x, algorithm="giac")

[Out]

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