3.117 \(\int \sec ^8(a+b x) \tan ^5(a+b x) \, dx\)

Optimal. Leaf size=46 \[ \frac {\sec ^{12}(a+b x)}{12 b}-\frac {\sec ^{10}(a+b x)}{5 b}+\frac {\sec ^8(a+b x)}{8 b} \]

[Out]

1/8*sec(b*x+a)^8/b-1/5*sec(b*x+a)^10/b+1/12*sec(b*x+a)^12/b

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Rubi [A]  time = 0.04, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2606, 266, 43} \[ \frac {\sec ^{12}(a+b x)}{12 b}-\frac {\sec ^{10}(a+b x)}{5 b}+\frac {\sec ^8(a+b x)}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^8*Tan[a + b*x]^5,x]

[Out]

Sec[a + b*x]^8/(8*b) - Sec[a + b*x]^10/(5*b) + Sec[a + b*x]^12/(12*b)

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])

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 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 ^8(a+b x) \tan ^5(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int x^7 \left (-1+x^2\right )^2 \, dx,x,\sec (a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int (-1+x)^2 x^3 \, dx,x,\sec ^2(a+b x)\right )}{2 b}\\ &=\frac {\operatorname {Subst}\left (\int \left (x^3-2 x^4+x^5\right ) \, dx,x,\sec ^2(a+b x)\right )}{2 b}\\ &=\frac {\sec ^8(a+b x)}{8 b}-\frac {\sec ^{10}(a+b x)}{5 b}+\frac {\sec ^{12}(a+b x)}{12 b}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 38, normalized size = 0.83 \[ \frac {10 \sec ^{12}(a+b x)-24 \sec ^{10}(a+b x)+15 \sec ^8(a+b x)}{120 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^8*Tan[a + b*x]^5,x]

[Out]

(15*Sec[a + b*x]^8 - 24*Sec[a + b*x]^10 + 10*Sec[a + b*x]^12)/(120*b)

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fricas [A]  time = 0.43, size = 35, normalized size = 0.76 \[ \frac {15 \, \cos \left (b x + a\right )^{4} - 24 \, \cos \left (b x + a\right )^{2} + 10}{120 \, b \cos \left (b x + a\right )^{12}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^13*sin(b*x+a)^5,x, algorithm="fricas")

[Out]

1/120*(15*cos(b*x + a)^4 - 24*cos(b*x + a)^2 + 10)/(b*cos(b*x + a)^12)

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giac [B]  time = 0.38, size = 183, normalized size = 3.98 \[ -\frac {32 \, {\left (\frac {5 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} - \frac {15 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} + \frac {39 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{5}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{5}} - \frac {42 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{6}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{6}} + \frac {39 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{7}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{7}} - \frac {15 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{8}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{8}} + \frac {5 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{9}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{9}}\right )}}{15 \, b {\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}^{12}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^13*sin(b*x+a)^5,x, algorithm="giac")

[Out]

-32/15*(5*(cos(b*x + a) - 1)^3/(cos(b*x + a) + 1)^3 - 15*(cos(b*x + a) - 1)^4/(cos(b*x + a) + 1)^4 + 39*(cos(b
*x + a) - 1)^5/(cos(b*x + a) + 1)^5 - 42*(cos(b*x + a) - 1)^6/(cos(b*x + a) + 1)^6 + 39*(cos(b*x + a) - 1)^7/(
cos(b*x + a) + 1)^7 - 15*(cos(b*x + a) - 1)^8/(cos(b*x + a) + 1)^8 + 5*(cos(b*x + a) - 1)^9/(cos(b*x + a) + 1)
^9)/(b*((cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 1)^12)

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maple [A]  time = 0.04, size = 78, normalized size = 1.70 \[ \frac {\frac {\sin ^{6}\left (b x +a \right )}{12 \cos \left (b x +a \right )^{12}}+\frac {\sin ^{6}\left (b x +a \right )}{20 \cos \left (b x +a \right )^{10}}+\frac {\sin ^{6}\left (b x +a \right )}{40 \cos \left (b x +a \right )^{8}}+\frac {\sin ^{6}\left (b x +a \right )}{120 \cos \left (b x +a \right )^{6}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^13*sin(b*x+a)^5,x)

[Out]

1/b*(1/12*sin(b*x+a)^6/cos(b*x+a)^12+1/20*sin(b*x+a)^6/cos(b*x+a)^10+1/40*sin(b*x+a)^6/cos(b*x+a)^8+1/120*sin(
b*x+a)^6/cos(b*x+a)^6)

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maxima [B]  time = 0.39, size = 89, normalized size = 1.93 \[ \frac {15 \, \sin \left (b x + a\right )^{4} - 6 \, \sin \left (b x + a\right )^{2} + 1}{120 \, {\left (\sin \left (b x + a\right )^{12} - 6 \, \sin \left (b x + a\right )^{10} + 15 \, \sin \left (b x + a\right )^{8} - 20 \, \sin \left (b x + a\right )^{6} + 15 \, \sin \left (b x + a\right )^{4} - 6 \, \sin \left (b x + a\right )^{2} + 1\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^13*sin(b*x+a)^5,x, algorithm="maxima")

[Out]

1/120*(15*sin(b*x + a)^4 - 6*sin(b*x + a)^2 + 1)/((sin(b*x + a)^12 - 6*sin(b*x + a)^10 + 15*sin(b*x + a)^8 - 2
0*sin(b*x + a)^6 + 15*sin(b*x + a)^4 - 6*sin(b*x + a)^2 + 1)*b)

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mupad [B]  time = 0.42, size = 45, normalized size = 0.98 \[ \frac {\frac {{\mathrm {tan}\left (a+b\,x\right )}^{12}}{12}+\frac {3\,{\mathrm {tan}\left (a+b\,x\right )}^{10}}{10}+\frac {3\,{\mathrm {tan}\left (a+b\,x\right )}^8}{8}+\frac {{\mathrm {tan}\left (a+b\,x\right )}^6}{6}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^5/cos(a + b*x)^13,x)

[Out]

(tan(a + b*x)^6/6 + (3*tan(a + b*x)^8)/8 + (3*tan(a + b*x)^10)/10 + tan(a + b*x)^12/12)/b

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

[Out]

Timed out

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