3.115 \(\int \sec ^6(a+b x) \tan ^5(a+b x) \, dx\)

Optimal. Leaf size=46 \[ \frac{\sec ^{10}(a+b x)}{10 b}-\frac{\sec ^8(a+b x)}{4 b}+\frac{\sec ^6(a+b x)}{6 b} \]

[Out]

Sec[a + b*x]^6/(6*b) - Sec[a + b*x]^8/(4*b) + Sec[a + b*x]^10/(10*b)

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Rubi [A]  time = 0.0391812, antiderivative size = 46, normalized size of antiderivative = 1., 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} \[ \frac{\sec ^{10}(a+b x)}{10 b}-\frac{\sec ^8(a+b x)}{4 b}+\frac{\sec ^6(a+b x)}{6 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^6*Tan[a + b*x]^5,x]

[Out]

Sec[a + b*x]^6/(6*b) - Sec[a + b*x]^8/(4*b) + Sec[a + b*x]^10/(10*b)

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])

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 \sec ^6(a+b x) \tan ^5(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int x^5 \left (-1+x^2\right )^2 \, dx,x,\sec (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int (-1+x)^2 x^2 \, dx,x,\sec ^2(a+b x)\right )}{2 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (x^2-2 x^3+x^4\right ) \, dx,x,\sec ^2(a+b x)\right )}{2 b}\\ &=\frac{\sec ^6(a+b x)}{6 b}-\frac{\sec ^8(a+b x)}{4 b}+\frac{\sec ^{10}(a+b x)}{10 b}\\ \end{align*}

Mathematica [A]  time = 0.0586623, size = 38, normalized size = 0.83 \[ \frac{6 \sec ^{10}(a+b x)-15 \sec ^8(a+b x)+10 \sec ^6(a+b x)}{60 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^6*Tan[a + b*x]^5,x]

[Out]

(10*Sec[a + b*x]^6 - 15*Sec[a + b*x]^8 + 6*Sec[a + b*x]^10)/(60*b)

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Maple [A]  time = 0.023, size = 60, normalized size = 1.3 \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{6}}{10\, \left ( \cos \left ( bx+a \right ) \right ) ^{10}}}+{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{6}}{20\, \left ( \cos \left ( bx+a \right ) \right ) ^{8}}}+{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{6}}{60\, \left ( \cos \left ( bx+a \right ) \right ) ^{6}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^11*sin(b*x+a)^5,x)

[Out]

1/b*(1/10*sin(b*x+a)^6/cos(b*x+a)^10+1/20*sin(b*x+a)^6/cos(b*x+a)^8+1/60*sin(b*x+a)^6/cos(b*x+a)^6)

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Maxima [A]  time = 0.972727, size = 107, normalized size = 2.33 \begin{align*} -\frac{10 \, \sin \left (b x + a\right )^{4} - 5 \, \sin \left (b x + a\right )^{2} + 1}{60 \,{\left (\sin \left (b x + a\right )^{10} - 5 \, \sin \left (b x + a\right )^{8} + 10 \, \sin \left (b x + a\right )^{6} - 10 \, \sin \left (b x + a\right )^{4} + 5 \, \sin \left (b x + a\right )^{2} - 1\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^11*sin(b*x+a)^5,x, algorithm="maxima")

[Out]

-1/60*(10*sin(b*x + a)^4 - 5*sin(b*x + a)^2 + 1)/((sin(b*x + a)^10 - 5*sin(b*x + a)^8 + 10*sin(b*x + a)^6 - 10
*sin(b*x + a)^4 + 5*sin(b*x + a)^2 - 1)*b)

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Fricas [A]  time = 1.59344, size = 95, normalized size = 2.07 \begin{align*} \frac{10 \, \cos \left (b x + a\right )^{4} - 15 \, \cos \left (b x + a\right )^{2} + 6}{60 \, b \cos \left (b x + a\right )^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^11*sin(b*x+a)^5,x, algorithm="fricas")

[Out]

1/60*(10*cos(b*x + a)^4 - 15*cos(b*x + a)^2 + 6)/(b*cos(b*x + a)^10)

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

[Out]

Timed out

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Giac [B]  time = 1.19066, size = 188, normalized size = 4.09 \begin{align*} -\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{10 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} + \frac{18 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{5}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{5}} - \frac{10 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{6}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{6}} + \frac{5 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{7}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{7}}\right )}}{15 \, b{\left (\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^11*sin(b*x+a)^5,x, algorithm="giac")

[Out]

-32/15*(5*(cos(b*x + a) - 1)^3/(cos(b*x + a) + 1)^3 - 10*(cos(b*x + a) - 1)^4/(cos(b*x + a) + 1)^4 + 18*(cos(b
*x + a) - 1)^5/(cos(b*x + a) + 1)^5 - 10*(cos(b*x + a) - 1)^6/(cos(b*x + a) + 1)^6 + 5*(cos(b*x + a) - 1)^7/(c
os(b*x + a) + 1)^7)/(b*((cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 1)^10)