\(\int \sec ^2(a+b x) \tan ^3(a+b x) \, dx\) [75]

Optimal result

Integrand size = 17, antiderivative size = 15 \[ \int \sec ^2(a+b x) \tan ^3(a+b x) \, dx=\frac {\tan ^4(a+b x)}{4 b} \]

[Out]

1/4*tan(b*x+a)^4/b

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2687, 30} \[ \int \sec ^2(a+b x) \tan ^3(a+b x) \, dx=\frac {\tan ^4(a+b x)}{4 b} \]

[In]

Int[Sec[a + b*x]^2*Tan[a + b*x]^3,x]

[Out]

Tan[a + b*x]^4/(4*b)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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^3 \, dx,x,\tan (a+b x)\right )}{b} \\ & = \frac {\tan ^4(a+b x)}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \sec ^2(a+b x) \tan ^3(a+b x) \, dx=\frac {\tan ^4(a+b x)}{4 b} \]

[In]

Integrate[Sec[a + b*x]^2*Tan[a + b*x]^3,x]

[Out]

Tan[a + b*x]^4/(4*b)

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.73

method	result	size
derivativedivides	\(\frac {\frac {\left (\sec ^{4}\left (b x +a \right )\right )}{4}-\frac {\left (\sec ^{2}\left (b x +a \right )\right )}{2}}{b}\)	\(26\)
default	\(\frac {\frac {\left (\sec ^{4}\left (b x +a \right )\right )}{4}-\frac {\left (\sec ^{2}\left (b x +a \right )\right )}{2}}{b}\)	\(26\)
norman	\(\frac {4 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{4}}\)	\(32\)
risch	\(-\frac {2 \left ({\mathrm e}^{6 i \left (b x +a \right )}+{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{4}}\)	\(38\)
parallelrisch	\(\frac {4 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{4} \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )^{4}}\)	\(43\)

[In]

int(sec(b*x+a)^5*sin(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

1/b*(1/4*sec(b*x+a)^4-1/2*sec(b*x+a)^2)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.67 \[ \int \sec ^2(a+b x) \tan ^3(a+b x) \, dx=-\frac {2 \, \cos \left (b x + a\right )^{2} - 1}{4 \, b \cos \left (b x + a\right )^{4}} \]

[In]

integrate(sec(b*x+a)^5*sin(b*x+a)^3,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \sec ^2(a+b x) \tan ^3(a+b x) \, dx=\text {Timed out} \]

[In]

integrate(sec(b*x+a)**5*sin(b*x+a)**3,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (13) = 26\).

Time = 0.18 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.60 \[ \int \sec ^2(a+b x) \tan ^3(a+b x) \, dx=\frac {2 \, \sin \left (b x + a\right )^{2} - 1}{4 \, {\left (\sin \left (b x + a\right )^{4} - 2 \, \sin \left (b x + a\right )^{2} + 1\right )} b} \]

[In]

integrate(sec(b*x+a)^5*sin(b*x+a)^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.67 \[ \int \sec ^2(a+b x) \tan ^3(a+b x) \, dx=-\frac {2 \, \cos \left (b x + a\right )^{2} - 1}{4 \, b \cos \left (b x + a\right )^{4}} \]

[In]

integrate(sec(b*x+a)^5*sin(b*x+a)^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \sec ^2(a+b x) \tan ^3(a+b x) \, dx=\frac {{\mathrm {tan}\left (a+b\,x\right )}^4}{4\,b} \]

[In]

int(sin(a + b*x)^3/cos(a + b*x)^5,x)

[Out]

tan(a + b*x)^4/(4*b)