\(\int \sec ^4(a+b x) \tan ^4(a+b x) \, dx\) [87]

Optimal result

Integrand size = 17, antiderivative size = 31 \[ \int \sec ^4(a+b x) \tan ^4(a+b x) \, dx=\frac {\tan ^5(a+b x)}{5 b}+\frac {\tan ^7(a+b x)}{7 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2687, 14} \[ \int \sec ^4(a+b x) \tan ^4(a+b x) \, dx=\frac {\tan ^7(a+b x)}{7 b}+\frac {\tan ^5(a+b x)}{5 b} \]

[In]

Int[Sec[a + b*x]^4*Tan[a + b*x]^4,x]

[Out]

Tan[a + b*x]^5/(5*b) + Tan[a + b*x]^7/(7*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 2687

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(77\) vs. \(2(31)=62\).

Time = 0.11 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.48 \[ \int \sec ^4(a+b x) \tan ^4(a+b x) \, dx=\frac {2 \tan (a+b x)}{35 b}+\frac {\sec ^2(a+b x) \tan (a+b x)}{35 b}-\frac {8 \sec ^4(a+b x) \tan (a+b x)}{35 b}+\frac {\sec ^6(a+b x) \tan (a+b x)}{7 b} \]

[In]

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

[Out]

(2*Tan[a + b*x])/(35*b) + (Sec[a + b*x]^2*Tan[a + b*x])/(35*b) - (8*Sec[a + b*x]^4*Tan[a + b*x])/(35*b) + (Sec
[a + b*x]^6*Tan[a + b*x])/(7*b)

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35

method	result	size
derivativedivides	\(\frac {\frac {\sin ^{5}\left (b x +a \right )}{7 \cos \left (b x +a \right )^{7}}+\frac {2 \left (\sin ^{5}\left (b x +a \right )\right )}{35 \cos \left (b x +a \right )^{5}}}{b}\)	\(42\)
default	\(\frac {\frac {\sin ^{5}\left (b x +a \right )}{7 \cos \left (b x +a \right )^{7}}+\frac {2 \left (\sin ^{5}\left (b x +a \right )\right )}{35 \cos \left (b x +a \right )^{5}}}{b}\)	\(42\)
parallelrisch	\(-\frac {32 \left (\tan ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (7 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+6 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+7\right )}{35 b \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{7}}\)	\(60\)
norman	\(\frac {-\frac {32 \left (\tan ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{5 b}-\frac {192 \left (\tan ^{7}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{35 b}-\frac {32 \left (\tan ^{9}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{5 b}}{\left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{7}}\)	\(66\)
risch	\(\frac {4 i \left (35 \,{\mathrm e}^{10 i \left (b x +a \right )}-35 \,{\mathrm e}^{8 i \left (b x +a \right )}+70 \,{\mathrm e}^{6 i \left (b x +a \right )}-14 \,{\mathrm e}^{4 i \left (b x +a \right )}+7 \,{\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{35 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{7}}\)	\(77\)

[In]

int(sec(b*x+a)^8*sin(b*x+a)^4,x,method=_RETURNVERBOSE)

[Out]

1/b*(1/7*sin(b*x+a)^5/cos(b*x+a)^7+2/35*sin(b*x+a)^5/cos(b*x+a)^5)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.58 \[ \int \sec ^4(a+b x) \tan ^4(a+b x) \, dx=\frac {{\left (2 \, \cos \left (b x + a\right )^{6} + \cos \left (b x + a\right )^{4} - 8 \, \cos \left (b x + a\right )^{2} + 5\right )} \sin \left (b x + a\right )}{35 \, b \cos \left (b x + a\right )^{7}} \]

[In]

integrate(sec(b*x+a)^8*sin(b*x+a)^4,x, algorithm="fricas")

[Out]

1/35*(2*cos(b*x + a)^6 + cos(b*x + a)^4 - 8*cos(b*x + a)^2 + 5)*sin(b*x + a)/(b*cos(b*x + a)^7)

Sympy [F(-1)]

Timed out. \[ \int \sec ^4(a+b x) \tan ^4(a+b x) \, dx=\text {Timed out} \]

[In]

integrate(sec(b*x+a)**8*sin(b*x+a)**4,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \sec ^4(a+b x) \tan ^4(a+b x) \, dx=\frac {5 \, \tan \left (b x + a\right )^{7} + 7 \, \tan \left (b x + a\right )^{5}}{35 \, b} \]

[In]

integrate(sec(b*x+a)^8*sin(b*x+a)^4,x, algorithm="maxima")

[Out]

1/35*(5*tan(b*x + a)^7 + 7*tan(b*x + a)^5)/b

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \sec ^4(a+b x) \tan ^4(a+b x) \, dx=\frac {5 \, \tan \left (b x + a\right )^{7} + 7 \, \tan \left (b x + a\right )^{5}}{35 \, b} \]

[In]

integrate(sec(b*x+a)^8*sin(b*x+a)^4,x, algorithm="giac")

[Out]

1/35*(5*tan(b*x + a)^7 + 7*tan(b*x + a)^5)/b

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \sec ^4(a+b x) \tan ^4(a+b x) \, dx=\frac {{\mathrm {tan}\left (a+b\,x\right )}^5\,\left (5\,{\mathrm {tan}\left (a+b\,x\right )}^2+7\right )}{35\,b} \]

[In]

int(sin(a + b*x)^4/cos(a + b*x)^8,x)

[Out]

(tan(a + b*x)^5*(5*tan(a + b*x)^2 + 7))/(35*b)