3.110 \(\int \sec (a+b x) \tan ^5(a+b x) \, dx\)

Optimal. Leaf size=41 \[ \frac {\sec ^5(a+b x)}{5 b}-\frac {2 \sec ^3(a+b x)}{3 b}+\frac {\sec (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 41, 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 = {2606, 194} \[ \frac {\sec ^5(a+b x)}{5 b}-\frac {2 \sec ^3(a+b x)}{3 b}+\frac {\sec (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]*Tan[a + b*x]^5,x]

[Out]

Sec[a + b*x]/b - (2*Sec[a + b*x]^3)/(3*b) + Sec[a + b*x]^5/(5*b)

Rule 194

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

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 41, normalized size = 1.00 \[ \frac {\sec ^5(a+b x)}{5 b}-\frac {2 \sec ^3(a+b x)}{3 b}+\frac {\sec (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]*Tan[a + b*x]^5,x]

[Out]

Sec[a + b*x]/b - (2*Sec[a + b*x]^3)/(3*b) + Sec[a + b*x]^5/(5*b)

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fricas [A]  time = 0.42, size = 35, normalized size = 0.85 \[ \frac {15 \, \cos \left (b x + a\right )^{4} - 10 \, \cos \left (b x + a\right )^{2} + 3}{15 \, b \cos \left (b x + a\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^6*sin(b*x+a)^5,x, algorithm="fricas")

[Out]

1/15*(15*cos(b*x + a)^4 - 10*cos(b*x + a)^2 + 3)/(b*cos(b*x + a)^5)

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giac [A]  time = 0.23, size = 72, normalized size = 1.76 \[ \frac {16 \, {\left (\frac {5 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {10 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 1\right )}}{15 \, b {\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^6*sin(b*x+a)^5,x, algorithm="giac")

[Out]

16/15*(5*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 10*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 1)/(b*((cos(b*
x + a) - 1)/(cos(b*x + a) + 1) + 1)^5)

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maple [B]  time = 0.03, size = 88, normalized size = 2.15 \[ \frac {\frac {\sin ^{6}\left (b x +a \right )}{5 \cos \left (b x +a \right )^{5}}-\frac {\sin ^{6}\left (b x +a \right )}{15 \cos \left (b x +a \right )^{3}}+\frac {\sin ^{6}\left (b x +a \right )}{5 \cos \left (b x +a \right )}+\frac {\left (\frac {8}{3}+\sin ^{4}\left (b x +a \right )+\frac {4 \left (\sin ^{2}\left (b x +a \right )\right )}{3}\right ) \cos \left (b x +a \right )}{5}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^6*sin(b*x+a)^5,x)

[Out]

1/b*(1/5*sin(b*x+a)^6/cos(b*x+a)^5-1/15*sin(b*x+a)^6/cos(b*x+a)^3+1/5*sin(b*x+a)^6/cos(b*x+a)+1/5*(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 = 0.30, size = 35, normalized size = 0.85 \[ \frac {15 \, \cos \left (b x + a\right )^{4} - 10 \, \cos \left (b x + a\right )^{2} + 3}{15 \, b \cos \left (b x + a\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^6*sin(b*x+a)^5,x, algorithm="maxima")

[Out]

1/15*(15*cos(b*x + a)^4 - 10*cos(b*x + a)^2 + 3)/(b*cos(b*x + a)^5)

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mupad [B]  time = 0.54, size = 35, normalized size = 0.85 \[ \frac {15\,{\cos \left (a+b\,x\right )}^4-10\,{\cos \left (a+b\,x\right )}^2+3}{15\,b\,{\cos \left (a+b\,x\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^5/cos(a + b*x)^6,x)

[Out]

(15*cos(a + b*x)^4 - 10*cos(a + b*x)^2 + 3)/(15*b*cos(a + b*x)^5)

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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)**6*sin(b*x+a)**5,x)

[Out]

Timed out

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