3.1.79 \(\int \sec ^6(a+b x) \tan ^3(a+b x) \, dx\) [79]

Optimal. Leaf size=31 \[ -\frac {\sec ^6(a+b x)}{6 b}+\frac {\sec ^8(a+b x)}{8 b} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2686, 14} \begin {gather*} \frac {\sec ^8(a+b x)}{8 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]^3,x]

[Out]

-1/6*Sec[a + b*x]^6/b + Sec[a + b*x]^8/(8*b)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 ^3(a+b x) \, dx &=\frac {\text {Subst}\left (\int x^5 \left (-1+x^2\right ) \, dx,x,\sec (a+b x)\right )}{b}\\ &=\frac {\text {Subst}\left (\int \left (-x^5+x^7\right ) \, dx,x,\sec (a+b x)\right )}{b}\\ &=-\frac {\sec ^6(a+b x)}{6 b}+\frac {\sec ^8(a+b x)}{8 b}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 28, normalized size = 0.90 \begin {gather*} -\frac {4 \sec ^6(a+b x)-3 \sec ^8(a+b x)}{24 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^6*Tan[a + b*x]^3,x]

[Out]

-1/24*(4*Sec[a + b*x]^6 - 3*Sec[a + b*x]^8)/b

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(59\) vs. \(2(27)=54\).
time = 0.09, size = 60, normalized size = 1.94

method	result	size
risch	\(-\frac {32 \left ({\mathrm e}^{10 i \left (b x +a \right )}-{\mathrm e}^{8 i \left (b x +a \right )}+{\mathrm e}^{6 i \left (b x +a \right )}\right )}{3 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{8}}\)	\(49\)
derivativedivides	\(\frac {\frac {\sin ^{4}\left (b x +a \right )}{8 \cos \left (b x +a \right )^{8}}+\frac {\sin ^{4}\left (b x +a \right )}{12 \cos \left (b x +a \right )^{6}}+\frac {\sin ^{4}\left (b x +a \right )}{24 \cos \left (b x +a \right )^{4}}}{b}\)	\(60\)
default	\(\frac {\frac {\sin ^{4}\left (b x +a \right )}{8 \cos \left (b x +a \right )^{8}}+\frac {\sin ^{4}\left (b x +a \right )}{12 \cos \left (b x +a \right )^{6}}+\frac {\sin ^{4}\left (b x +a \right )}{24 \cos \left (b x +a \right )^{4}}}{b}\)	\(60\)
norman	\(\frac {\frac {4 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}+\frac {4 \left (\tan ^{12}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}+\frac {16 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}+\frac {16 \left (\tan ^{10}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}+\frac {40 \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}}{\left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{8}}\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^9*sin(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

1/b*(1/8*sin(b*x+a)^4/cos(b*x+a)^8+1/12*sin(b*x+a)^4/cos(b*x+a)^6+1/24*sin(b*x+a)^4/cos(b*x+a)^4)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (27) = 54\).
time = 0.29, size = 59, normalized size = 1.90 \begin {gather*} \frac {4 \, \sin \left (b x + a\right )^{2} - 1}{24 \, {\left (\sin \left (b x + a\right )^{8} - 4 \, \sin \left (b x + a\right )^{6} + 6 \, \sin \left (b x + a\right )^{4} - 4 \, \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)^9*sin(b*x+a)^3,x, algorithm="maxima")

[Out]

1/24*(4*sin(b*x + a)^2 - 1)/((sin(b*x + a)^8 - 4*sin(b*x + a)^6 + 6*sin(b*x + a)^4 - 4*sin(b*x + a)^2 + 1)*b)

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Fricas [A]
time = 0.35, size = 25, normalized size = 0.81 \begin {gather*} -\frac {4 \, \cos \left (b x + a\right )^{2} - 3}{24 \, b \cos \left (b x + a\right )^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^9*sin(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/24*(4*cos(b*x + a)^2 - 3)/(b*cos(b*x + a)^8)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**9*sin(b*x+a)**3,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 8569 deep

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Giac [A]
time = 3.92, size = 25, normalized size = 0.81 \begin {gather*} -\frac {4 \, \cos \left (b x + a\right )^{2} - 3}{24 \, b \cos \left (b x + a\right )^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^9*sin(b*x+a)^3,x, algorithm="giac")

[Out]

-1/24*(4*cos(b*x + a)^2 - 3)/(b*cos(b*x + a)^8)

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Mupad [B]
time = 0.42, size = 35, normalized size = 1.13 \begin {gather*} \frac {{\mathrm {tan}\left (a+b\,x\right )}^4\,\left (3\,{\mathrm {tan}\left (a+b\,x\right )}^4+8\,{\mathrm {tan}\left (a+b\,x\right )}^2+6\right )}{24\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^3/cos(a + b*x)^9,x)

[Out]

(tan(a + b*x)^4*(8*tan(a + b*x)^2 + 3*tan(a + b*x)^4 + 6))/(24*b)

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