3.119 \(\int \sin (a+b x) \tan ^6(a+b x) \, dx\)

Optimal. Leaf size=50 \[ \frac{\cos (a+b x)}{b}+\frac{\sec ^5(a+b x)}{5 b}-\frac{\sec ^3(a+b x)}{b}+\frac{3 \sec (a+b x)}{b} \]

[Out]

Cos[a + b*x]/b + (3*Sec[a + b*x])/b - Sec[a + b*x]^3/b + Sec[a + b*x]^5/(5*b)

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Rubi [A]  time = 0.0279641, antiderivative size = 50, normalized size of antiderivative = 1., 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 = {2590, 270} \[ \frac{\cos (a+b x)}{b}+\frac{\sec ^5(a+b x)}{5 b}-\frac{\sec ^3(a+b x)}{b}+\frac{3 \sec (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]*Tan[a + b*x]^6,x]

[Out]

Cos[a + b*x]/b + (3*Sec[a + b*x])/b - Sec[a + b*x]^3/b + Sec[a + b*x]^5/(5*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 270

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

Rubi steps

\begin{align*} \int \sin (a+b x) \tan ^6(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^3}{x^6} \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int \left (-1+\frac{1}{x^6}-\frac{3}{x^4}+\frac{3}{x^2}\right ) \, dx,x,\cos (a+b x)\right )}{b}\\ &=\frac{\cos (a+b x)}{b}+\frac{3 \sec (a+b x)}{b}-\frac{\sec ^3(a+b x)}{b}+\frac{\sec ^5(a+b x)}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.0400032, size = 50, normalized size = 1. \[ \frac{\cos (a+b x)}{b}+\frac{\sec ^5(a+b x)}{5 b}-\frac{\sec ^3(a+b x)}{b}+\frac{3 \sec (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]*Tan[a + b*x]^6,x]

[Out]

Cos[a + b*x]/b + (3*Sec[a + b*x])/b - Sec[a + b*x]^3/b + Sec[a + b*x]^5/(5*b)

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Maple [A]  time = 0.026, size = 96, normalized size = 1.9 \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{8}}{5\, \left ( \cos \left ( bx+a \right ) \right ) ^{5}}}-{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{8}}{5\, \left ( \cos \left ( bx+a \right ) \right ) ^{3}}}+{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{8}}{\cos \left ( bx+a \right ) }}+ \left ({\frac{16}{5}}+ \left ( \sin \left ( bx+a \right ) \right ) ^{6}+{\frac{6\, \left ( \sin \left ( bx+a \right ) \right ) ^{4}}{5}}+{\frac{8\, \left ( \sin \left ( bx+a \right ) \right ) ^{2}}{5}} \right ) \cos \left ( bx+a \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^6*sin(b*x+a)^7,x)

[Out]

1/b*(1/5*sin(b*x+a)^8/cos(b*x+a)^5-1/5*sin(b*x+a)^8/cos(b*x+a)^3+sin(b*x+a)^8/cos(b*x+a)+(16/5+sin(b*x+a)^6+6/
5*sin(b*x+a)^4+8/5*sin(b*x+a)^2)*cos(b*x+a))

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Maxima [A]  time = 0.975018, size = 61, normalized size = 1.22 \begin{align*} \frac{\frac{15 \, \cos \left (b x + a\right )^{4} - 5 \, \cos \left (b x + a\right )^{2} + 1}{\cos \left (b x + a\right )^{5}} + 5 \, \cos \left (b x + a\right )}{5 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^6*sin(b*x+a)^7,x, algorithm="maxima")

[Out]

1/5*((15*cos(b*x + a)^4 - 5*cos(b*x + a)^2 + 1)/cos(b*x + a)^5 + 5*cos(b*x + a))/b

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Fricas [A]  time = 1.58501, size = 116, normalized size = 2.32 \begin{align*} \frac{5 \, \cos \left (b x + a\right )^{6} + 15 \, \cos \left (b x + a\right )^{4} - 5 \, \cos \left (b x + a\right )^{2} + 1}{5 \, b \cos \left (b x + a\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^6*sin(b*x+a)^7,x, algorithm="fricas")

[Out]

1/5*(5*cos(b*x + a)^6 + 15*cos(b*x + a)^4 - 5*cos(b*x + a)^2 + 1)/(b*cos(b*x + a)^5)

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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)**6*sin(b*x+a)**7,x)

[Out]

Timed out

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Giac [B]  time = 1.21745, size = 194, normalized size = 3.88 \begin{align*} -\frac{2 \,{\left (\frac{5}{\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1} - \frac{\frac{50 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac{80 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac{30 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} + \frac{5 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} + 11}{{\left (\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}^{5}}\right )}}{5 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^6*sin(b*x+a)^7,x, algorithm="giac")

[Out]

-2/5*(5/((cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 1) - (50*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 80*(cos(b*x
+ a) - 1)^2/(cos(b*x + a) + 1)^2 + 30*(cos(b*x + a) - 1)^3/(cos(b*x + a) + 1)^3 + 5*(cos(b*x + a) - 1)^4/(cos(
b*x + a) + 1)^4 + 11)/((cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 1)^5)/b