\(\int \sec ^3(a+b x) \tan ^3(a+b x) \, dx\) [76]

Optimal result

Integrand size = 17, antiderivative size = 31 \[ \int \sec ^3(a+b x) \tan ^3(a+b x) \, dx=-\frac {\sec ^3(a+b x)}{3 b}+\frac {\sec ^5(a+b x)}{5 b} \]

[Out]

-1/3*sec(b*x+a)^3/b+1/5*sec(b*x+a)^5/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 = {2686, 14} \[ \int \sec ^3(a+b x) \tan ^3(a+b x) \, dx=\frac {\sec ^5(a+b x)}{5 b}-\frac {\sec ^3(a+b x)}{3 b} \]

[In]

Int[Sec[a + b*x]^3*Tan[a + b*x]^3,x]

[Out]

-1/3*Sec[a + b*x]^3/b + Sec[a + b*x]^5/(5*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*} \text {integral}& = \frac {\text {Subst}\left (\int x^2 \left (-1+x^2\right ) \, dx,x,\sec (a+b x)\right )}{b} \\ & = \frac {\text {Subst}\left (\int \left (-x^2+x^4\right ) \, dx,x,\sec (a+b x)\right )}{b} \\ & = -\frac {\sec ^3(a+b x)}{3 b}+\frac {\sec ^5(a+b x)}{5 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \sec ^3(a+b x) \tan ^3(a+b x) \, dx=-\frac {\sec ^3(a+b x)}{3 b}+\frac {\sec ^5(a+b x)}{5 b} \]

[In]

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

[Out]

-1/3*Sec[a + b*x]^3/b + Sec[a + b*x]^5/(5*b)

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(\frac {\frac {\left (\sec ^{5}\left (b x +a \right )\right )}{5}-\frac {\left (\sec ^{3}\left (b x +a \right )\right )}{3}}{b}\)	\(26\)
default	\(\frac {\frac {\left (\sec ^{5}\left (b x +a \right )\right )}{5}-\frac {\left (\sec ^{3}\left (b x +a \right )\right )}{3}}{b}\)	\(26\)
risch	\(-\frac {8 \left (5 \,{\mathrm e}^{7 i \left (b x +a \right )}-2 \,{\mathrm e}^{5 i \left (b x +a \right )}+5 \,{\mathrm e}^{3 i \left (b x +a \right )}\right )}{15 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{5}}\)	\(53\)
norman	\(\frac {\frac {4}{15 b}-\frac {4 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}-\frac {4 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}-\frac {4 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}}{\left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{5}}\)	\(71\)
parallelrisch	\(\frac {\frac {4}{15}-4 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\frac {4 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3}-\frac {4 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3}}{b \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{5} \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )^{5}}\)	\(73\)

[In]

int(sec(b*x+a)^6*sin(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

1/b*(1/5*sec(b*x+a)^5-1/3*sec(b*x+a)^3)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \sec ^3(a+b x) \tan ^3(a+b x) \, dx=-\frac {5 \, \cos \left (b x + a\right )^{2} - 3}{15 \, b \cos \left (b x + a\right )^{5}} \]

[In]

integrate(sec(b*x+a)^6*sin(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/15*(5*cos(b*x + a)^2 - 3)/(b*cos(b*x + a)^5)

Sympy [F(-1)]

Timed out. \[ \int \sec ^3(a+b x) \tan ^3(a+b x) \, dx=\text {Timed out} \]

[In]

integrate(sec(b*x+a)**6*sin(b*x+a)**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \sec ^3(a+b x) \tan ^3(a+b x) \, dx=-\frac {5 \, \cos \left (b x + a\right )^{2} - 3}{15 \, b \cos \left (b x + a\right )^{5}} \]

[In]

integrate(sec(b*x+a)^6*sin(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/15*(5*cos(b*x + a)^2 - 3)/(b*cos(b*x + a)^5)

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \sec ^3(a+b x) \tan ^3(a+b x) \, dx=-\frac {5 \, \cos \left (b x + a\right )^{2} - 3}{15 \, b \cos \left (b x + a\right )^{5}} \]

[In]

integrate(sec(b*x+a)^6*sin(b*x+a)^3,x, algorithm="giac")

[Out]

-1/15*(5*cos(b*x + a)^2 - 3)/(b*cos(b*x + a)^5)

Mupad [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \sec ^3(a+b x) \tan ^3(a+b x) \, dx=-\frac {\frac {{\cos \left (a+b\,x\right )}^2}{3}-\frac {1}{5}}{b\,{\cos \left (a+b\,x\right )}^5} \]

[In]

int(sin(a + b*x)^3/cos(a + b*x)^6,x)

[Out]

-(cos(a + b*x)^2/3 - 1/5)/(b*cos(a + b*x)^5)