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

Optimal. Leaf size=38 \[ -\frac {\cos (a+b x)}{b}+\frac {\sec ^3(a+b x)}{3 b}-\frac {2 \sec (a+b x)}{b} \]

[Out]

-cos(b*x+a)/b-2*sec(b*x+a)/b+1/3*sec(b*x+a)^3/b

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Rubi [A]  time = 0.03, antiderivative size = 38, normalized size of antiderivative = 1.00, 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} \[ -\frac {\cos (a+b x)}{b}+\frac {\sec ^3(a+b x)}{3 b}-\frac {2 \sec (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]*Tan[a + b*x]^4,x]

[Out]

-(Cos[a + b*x]/b) - (2*Sec[a + b*x])/b + Sec[a + b*x]^3/(3*b)

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]

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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 38, normalized size = 1.00 \[ -\frac {\cos (a+b x)}{b}+\frac {\sec ^3(a+b x)}{3 b}-\frac {2 \sec (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[a + b*x]/b) - (2*Sec[a + b*x])/b + Sec[a + b*x]^3/(3*b)

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fricas [A]  time = 0.42, size = 35, normalized size = 0.92 \[ -\frac {3 \, \cos \left (b x + a\right )^{4} + 6 \, \cos \left (b x + a\right )^{2} - 1}{3 \, b \cos \left (b x + a\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4*sin(b*x+a)^5,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.24, size = 100, normalized size = 2.63 \[ \frac {2 \, {\left (\frac {3}{\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1} - \frac {\frac {12 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {3 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 5}{{\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}^{3}}\right )}}{3 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4*sin(b*x+a)^5,x, algorithm="giac")

[Out]

2/3*(3/((cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 1) - (12*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 3*(cos(b*x +
a) - 1)^2/(cos(b*x + a) + 1)^2 + 5)/((cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 1)^3)/b

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maple [A]  time = 0.03, size = 70, normalized size = 1.84 \[ \frac {\frac {\sin ^{6}\left (b x +a \right )}{3 \cos \left (b x +a \right )^{3}}-\frac {\sin ^{6}\left (b x +a \right )}{\cos \left (b x +a \right )}-\left (\frac {8}{3}+\sin ^{4}\left (b x +a \right )+\frac {4 \left (\sin ^{2}\left (b x +a \right )\right )}{3}\right ) \cos \left (b x +a \right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^4*sin(b*x+a)^5,x)

[Out]

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

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maxima [A]  time = 0.32, size = 35, normalized size = 0.92 \[ -\frac {\frac {6 \, \cos \left (b x + a\right )^{2} - 1}{\cos \left (b x + a\right )^{3}} + 3 \, \cos \left (b x + a\right )}{3 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4*sin(b*x+a)^5,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.53, size = 35, normalized size = 0.92 \[ -\frac {3\,{\cos \left (a+b\,x\right )}^4+6\,{\cos \left (a+b\,x\right )}^2-1}{3\,b\,{\cos \left (a+b\,x\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^5/cos(a + b*x)^4,x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**4*sin(b*x+a)**5,x)

[Out]

Timed out

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