3.2.15 \(\int \sec ^6(a+b x) \tan ^5(a+b x) \, dx\) [115]

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

[Out]

1/6*sec(b*x+a)^6/b-1/4*sec(b*x+a)^8/b+1/10*sec(b*x+a)^10/b

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Rubi [A]
time = 0.03, 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 = {2686, 272, 45} \begin {gather*} \frac {\sec ^{10}(a+b x)}{10 b}-\frac {\sec ^8(a+b x)}{4 b}+\frac {\sec ^6(a+b x)}{6 b} \end {gather*}

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 45

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 272

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 2686

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 ^6(a+b x) \tan ^5(a+b x) \, dx &=\frac {\text {Subst}\left (\int x^5 \left (-1+x^2\right )^2 \, dx,x,\sec (a+b x)\right )}{b}\\ &=\frac {\text {Subst}\left (\int (-1+x)^2 x^2 \, dx,x,\sec ^2(a+b x)\right )}{2 b}\\ &=\frac {\text {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*}

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Mathematica [A]
time = 0.04, size = 38, normalized size = 0.83 \begin {gather*} \frac {10 \sec ^6(a+b x)-15 \sec ^8(a+b x)+6 \sec ^{10}(a+b x)}{60 b} \end {gather*}

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 = 0.06, size = 60, normalized size = 1.30

method	result	size
derivativedivides	\(\frac {\frac {\sin ^{6}\left (b x +a \right )}{10 \cos \left (b x +a \right )^{10}}+\frac {\sin ^{6}\left (b x +a \right )}{20 \cos \left (b x +a \right )^{8}}+\frac {\sin ^{6}\left (b x +a \right )}{60 \cos \left (b x +a \right )^{6}}}{b}\)	\(60\)
default	\(\frac {\frac {\sin ^{6}\left (b x +a \right )}{10 \cos \left (b x +a \right )^{10}}+\frac {\sin ^{6}\left (b x +a \right )}{20 \cos \left (b x +a \right )^{8}}+\frac {\sin ^{6}\left (b x +a \right )}{60 \cos \left (b x +a \right )^{6}}}{b}\)	\(60\)
risch	\(\frac {\frac {32 \,{\mathrm e}^{14 i \left (b x +a \right )}}{3}-\frac {64 \,{\mathrm e}^{12 i \left (b x +a \right )}}{3}+\frac {192 \,{\mathrm e}^{10 i \left (b x +a \right )}}{5}-\frac {64 \,{\mathrm e}^{8 i \left (b x +a \right )}}{3}+\frac {32 \,{\mathrm e}^{6 i \left (b x +a \right )}}{3}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{10}}\)	\(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.30, size = 79, normalized size = 1.72 \begin {gather*} -\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 {gather*}

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 = 0.35, size = 35, normalized size = 0.76 \begin {gather*} \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 {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (40) = 80\).
time = 5.27, size = 139, normalized size = 3.02 \begin {gather*} -\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 {gather*}

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)

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Mupad [B]
time = 0.44, size = 35, normalized size = 0.76 \begin {gather*} \frac {{\mathrm {tan}\left (a+b\,x\right )}^6\,\left (6\,{\mathrm {tan}\left (a+b\,x\right )}^4+15\,{\mathrm {tan}\left (a+b\,x\right )}^2+10\right )}{60\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^5/cos(a + b*x)^11,x)

[Out]

(tan(a + b*x)^6*(15*tan(a + b*x)^2 + 6*tan(a + b*x)^4 + 10))/(60*b)

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