3.88 \(\int \sec ^6(a+b x) \tan ^4(a+b x) \, dx\)

Optimal. Leaf size=46 \[ \frac {\tan ^9(a+b x)}{9 b}+\frac {2 \tan ^7(a+b x)}{7 b}+\frac {\tan ^5(a+b x)}{5 b} \]

[Out]

1/5*tan(b*x+a)^5/b+2/7*tan(b*x+a)^7/b+1/9*tan(b*x+a)^9/b

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Rubi [A]  time = 0.04, antiderivative size = 46, 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 = {2607, 270} \[ \frac {\tan ^9(a+b x)}{9 b}+\frac {2 \tan ^7(a+b x)}{7 b}+\frac {\tan ^5(a+b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^6*Tan[a + b*x]^4,x]

[Out]

Tan[a + b*x]^5/(5*b) + (2*Tan[a + b*x]^7)/(7*b) + Tan[a + b*x]^9/(9*b)

Rule 270

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

Rule 2607

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

Rubi steps

\begin {align*} \int \sec ^6(a+b x) \tan ^4(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int x^4 \left (1+x^2\right )^2 \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \left (x^4+2 x^6+x^8\right ) \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac {\tan ^5(a+b x)}{5 b}+\frac {2 \tan ^7(a+b x)}{7 b}+\frac {\tan ^9(a+b x)}{9 b}\\ \end {align*}

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Mathematica [B]  time = 0.03, size = 98, normalized size = 2.13 \[ \frac {8 \tan (a+b x)}{315 b}+\frac {\tan (a+b x) \sec ^8(a+b x)}{9 b}-\frac {10 \tan (a+b x) \sec ^6(a+b x)}{63 b}+\frac {\tan (a+b x) \sec ^4(a+b x)}{105 b}+\frac {4 \tan (a+b x) \sec ^2(a+b x)}{315 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^6*Tan[a + b*x]^4,x]

[Out]

(8*Tan[a + b*x])/(315*b) + (4*Sec[a + b*x]^2*Tan[a + b*x])/(315*b) + (Sec[a + b*x]^4*Tan[a + b*x])/(105*b) - (
10*Sec[a + b*x]^6*Tan[a + b*x])/(63*b) + (Sec[a + b*x]^8*Tan[a + b*x])/(9*b)

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fricas [A]  time = 0.44, size = 61, normalized size = 1.33 \[ \frac {{\left (8 \, \cos \left (b x + a\right )^{8} + 4 \, \cos \left (b x + a\right )^{6} + 3 \, \cos \left (b x + a\right )^{4} - 50 \, \cos \left (b x + a\right )^{2} + 35\right )} \sin \left (b x + a\right )}{315 \, b \cos \left (b x + a\right )^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^10*sin(b*x+a)^4,x, algorithm="fricas")

[Out]

1/315*(8*cos(b*x + a)^8 + 4*cos(b*x + a)^6 + 3*cos(b*x + a)^4 - 50*cos(b*x + a)^2 + 35)*sin(b*x + a)/(b*cos(b*
x + a)^9)

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giac [A]  time = 0.25, size = 36, normalized size = 0.78 \[ \frac {35 \, \tan \left (b x + a\right )^{9} + 90 \, \tan \left (b x + a\right )^{7} + 63 \, \tan \left (b x + a\right )^{5}}{315 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^10*sin(b*x+a)^4,x, algorithm="giac")

[Out]

1/315*(35*tan(b*x + a)^9 + 90*tan(b*x + a)^7 + 63*tan(b*x + a)^5)/b

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maple [A]  time = 0.04, size = 60, normalized size = 1.30 \[ \frac {\frac {\sin ^{5}\left (b x +a \right )}{9 \cos \left (b x +a \right )^{9}}+\frac {4 \left (\sin ^{5}\left (b x +a \right )\right )}{63 \cos \left (b x +a \right )^{7}}+\frac {8 \left (\sin ^{5}\left (b x +a \right )\right )}{315 \cos \left (b x +a \right )^{5}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^10*sin(b*x+a)^4,x)

[Out]

1/b*(1/9*sin(b*x+a)^5/cos(b*x+a)^9+4/63*sin(b*x+a)^5/cos(b*x+a)^7+8/315*sin(b*x+a)^5/cos(b*x+a)^5)

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maxima [A]  time = 0.45, size = 36, normalized size = 0.78 \[ \frac {35 \, \tan \left (b x + a\right )^{9} + 90 \, \tan \left (b x + a\right )^{7} + 63 \, \tan \left (b x + a\right )^{5}}{315 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^10*sin(b*x+a)^4,x, algorithm="maxima")

[Out]

1/315*(35*tan(b*x + a)^9 + 90*tan(b*x + a)^7 + 63*tan(b*x + a)^5)/b

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mupad [B]  time = 0.43, size = 35, normalized size = 0.76 \[ \frac {{\mathrm {tan}\left (a+b\,x\right )}^5\,\left (35\,{\mathrm {tan}\left (a+b\,x\right )}^4+90\,{\mathrm {tan}\left (a+b\,x\right )}^2+63\right )}{315\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^4/cos(a + b*x)^10,x)

[Out]

(tan(a + b*x)^5*(90*tan(a + b*x)^2 + 35*tan(a + b*x)^4 + 63))/(315*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)**10*sin(b*x+a)**4,x)

[Out]

Timed out

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