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

Optimal. Leaf size=38 \[ -\frac{\sin ^3(a+b x)}{3 b}-\frac{\sin (a+b x)}{b}+\frac{\tanh ^{-1}(\sin (a+b x))}{b} \]

[Out]

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

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Rubi [A]  time = 0.0266243, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2592, 302, 206} \[ -\frac{\sin ^3(a+b x)}{3 b}-\frac{\sin (a+b x)}{b}+\frac{\tanh ^{-1}(\sin (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^3*Tan[a + b*x],x]

[Out]

ArcTanh[Sin[a + b*x]]/b - Sin[a + b*x]/b - Sin[a + b*x]^3/(3*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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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

Mathematica [A]  time = 0.0146207, size = 38, normalized size = 1. \[ -\frac{\sin ^3(a+b x)}{3 b}-\frac{\sin (a+b x)}{b}+\frac{\tanh ^{-1}(\sin (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.016, size = 44, normalized size = 1.2 \begin{align*} -{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{3}}{3\,b}}-{\frac{\sin \left ( bx+a \right ) }{b}}+{\frac{\ln \left ( \sec \left ( bx+a \right ) +\tan \left ( bx+a \right ) \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.03131, size = 62, normalized size = 1.63 \begin{align*} -\frac{2 \, \sin \left (b x + a\right )^{3} - 3 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 3 \, \log \left (\sin \left (b x + a\right ) - 1\right ) + 6 \, \sin \left (b x + a\right )}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*sin(b*x + a)^3 - 3*log(sin(b*x + a) + 1) + 3*log(sin(b*x + a) - 1) + 6*sin(b*x + a))/b

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Fricas [A]  time = 1.70495, size = 132, normalized size = 3.47 \begin{align*} \frac{2 \,{\left (\cos \left (b x + a\right )^{2} - 4\right )} \sin \left (b x + a\right ) + 3 \, \log \left (\sin \left (b x + a\right ) + 1\right ) - 3 \, \log \left (-\sin \left (b x + a\right ) + 1\right )}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.17344, size = 65, normalized size = 1.71 \begin{align*} -\frac{2 \, \sin \left (b x + a\right )^{3} - 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 ) + 6 \, \sin \left (b x + a\right )}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*sin(b*x + a)^3 - 3*log(abs(sin(b*x + a) + 1)) + 3*log(abs(sin(b*x + a) - 1)) + 6*sin(b*x + a))/b