3.105 \(\int \sin ^4(a+b x) \tan (a+b x) \, dx\)

Optimal. Leaf size=40 \[ -\frac{\cos ^4(a+b x)}{4 b}+\frac{\cos ^2(a+b x)}{b}-\frac{\log (\cos (a+b x))}{b} \]

[Out]

Cos[a + b*x]^2/b - Cos[a + b*x]^4/(4*b) - Log[Cos[a + b*x]]/b

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Rubi [A]  time = 0.0256355, antiderivative size = 40, 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 = {2590, 266, 43} \[ -\frac{\cos ^4(a+b x)}{4 b}+\frac{\cos ^2(a+b x)}{b}-\frac{\log (\cos (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^4*Tan[a + b*x],x]

[Out]

Cos[a + b*x]^2/b - Cos[a + b*x]^4/(4*b) - Log[Cos[a + b*x]]/b

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0330667, size = 35, normalized size = 0.88 \[ -\frac{\frac{1}{4} \cos ^4(a+b x)-\cos ^2(a+b x)+\log (\cos (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((-Cos[a + b*x]^2 + Cos[a + b*x]^4/4 + Log[Cos[a + b*x]])/b)

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Maple [A]  time = 0.014, size = 40, normalized size = 1. \begin{align*} -{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{4}}{4\,b}}-{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{2}}{2\,b}}-{\frac{\ln \left ( \cos \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)^5,x)

[Out]

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

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Maxima [A]  time = 0.956812, size = 50, normalized size = 1.25 \begin{align*} -\frac{\sin \left (b x + a\right )^{4} + 2 \, \sin \left (b x + a\right )^{2} + 2 \, \log \left (\sin \left (b x + a\right )^{2} - 1\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.70497, size = 90, normalized size = 2.25 \begin{align*} -\frac{\cos \left (b x + a\right )^{4} - 4 \, \cos \left (b x + a\right )^{2} + 4 \, \log \left (-\cos \left (b x + a\right )\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(cos(b*x + a)^4 - 4*cos(b*x + a)^2 + 4*log(-cos(b*x + a)))/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)**5,x)

[Out]

Timed out

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Giac [B]  time = 1.18879, size = 305, normalized size = 7.62 \begin{align*} -\frac{\frac{3 \,{\left (\frac{\cos \left (b x + a\right ) + 1}{\cos \left (b x + a\right ) - 1} + \frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1}\right )}^{2} - \frac{20 \,{\left (\cos \left (b x + a\right ) + 1\right )}}{\cos \left (b x + a\right ) - 1} - \frac{20 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + 44}{{\left (\frac{\cos \left (b x + a\right ) + 1}{\cos \left (b x + a\right ) - 1} + \frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 2\right )}^{2}} - 2 \, \log \left ({\left | -\frac{\cos \left (b x + a\right ) + 1}{\cos \left (b x + a\right ) - 1} - \frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 2 \right |}\right ) + 2 \, \log \left ({\left | -\frac{\cos \left (b x + a\right ) + 1}{\cos \left (b x + a\right ) - 1} - \frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 2 \right |}\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((3*((cos(b*x + a) + 1)/(cos(b*x + a) - 1) + (cos(b*x + a) - 1)/(cos(b*x + a) + 1))^2 - 20*(cos(b*x + a)
+ 1)/(cos(b*x + a) - 1) - 20*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 44)/((cos(b*x + a) + 1)/(cos(b*x + a) - 1
) + (cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 2)^2 - 2*log(abs(-(cos(b*x + a) + 1)/(cos(b*x + a) - 1) - (cos(b*x
+ a) - 1)/(cos(b*x + a) + 1) + 2)) + 2*log(abs(-(cos(b*x + a) + 1)/(cos(b*x + a) - 1) - (cos(b*x + a) - 1)/(c
os(b*x + a) + 1) - 2)))/b