\(\int \sec ^2(a+b x) \tan ^5(a+b x) \, dx\) [111]

Optimal result

Integrand size = 17, antiderivative size = 15 \[ \int \sec ^2(a+b x) \tan ^5(a+b x) \, dx=\frac {\tan ^6(a+b x)}{6 b} \]

[Out]

1/6*tan(b*x+a)^6/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} \[ \int \sec ^2(a+b x) \tan ^5(a+b x) \, dx=\frac {\tan ^6(a+b x)}{6 b} \]

[In]

Int[Sec[a + b*x]^2*Tan[a + b*x]^5,x]

[Out]

Tan[a + b*x]^6/(6*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^5 \, dx,x,\tan (a+b x)\right )}{b} \\ & = \frac {\tan ^6(a+b x)}{6 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \sec ^2(a+b x) \tan ^5(a+b x) \, dx=\frac {\tan ^6(a+b x)}{6 b} \]

[In]

Integrate[Sec[a + b*x]^2*Tan[a + b*x]^5,x]

[Out]

Tan[a + b*x]^6/(6*b)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(31\) vs. \(2(13)=26\).

Time = 0.14 (sec) , antiderivative size = 32, normalized size of antiderivative = 2.13

method	result	size
norman	\(\frac {32 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{6}}\)	\(32\)
parallelrisch	\(\frac {32 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{6}}\)	\(32\)
derivativedivides	\(\frac {\frac {\left (\sec ^{6}\left (b x +a \right )\right )}{6}-\frac {\left (\sec ^{4}\left (b x +a \right )\right )}{2}+\frac {\left (\sec ^{2}\left (b x +a \right )\right )}{2}}{b}\)	\(36\)
default	\(\frac {\frac {\left (\sec ^{6}\left (b x +a \right )\right )}{6}-\frac {\left (\sec ^{4}\left (b x +a \right )\right )}{2}+\frac {\left (\sec ^{2}\left (b x +a \right )\right )}{2}}{b}\)	\(36\)
risch	\(\frac {2 \,{\mathrm e}^{10 i \left (b x +a \right )}+\frac {20 \,{\mathrm e}^{6 i \left (b x +a \right )}}{3}+2 \,{\mathrm e}^{2 i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{6}}\)	\(53\)

[In]

int(sec(b*x+a)^7*sin(b*x+a)^5,x,method=_RETURNVERBOSE)

[Out]

32/3/b*tan(1/2*b*x+1/2*a)^6/(tan(1/2*b*x+1/2*a)^2-1)^6

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (13) = 26\).

Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 2.33 \[ \int \sec ^2(a+b x) \tan ^5(a+b x) \, dx=\frac {3 \, \cos \left (b x + a\right )^{4} - 3 \, \cos \left (b x + a\right )^{2} + 1}{6 \, b \cos \left (b x + a\right )^{6}} \]

[In]

integrate(sec(b*x+a)^7*sin(b*x+a)^5,x, algorithm="fricas")

[Out]

1/6*(3*cos(b*x + a)^4 - 3*cos(b*x + a)^2 + 1)/(b*cos(b*x + a)^6)

Sympy [F(-1)]

Timed out. \[ \int \sec ^2(a+b x) \tan ^5(a+b x) \, dx=\text {Timed out} \]

[In]

integrate(sec(b*x+a)**7*sin(b*x+a)**5,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (13) = 26\).

Time = 0.19 (sec) , antiderivative size = 59, normalized size of antiderivative = 3.93 \[ \int \sec ^2(a+b x) \tan ^5(a+b x) \, dx=-\frac {3 \, \sin \left (b x + a\right )^{4} - 3 \, \sin \left (b x + a\right )^{2} + 1}{6 \, {\left (\sin \left (b x + a\right )^{6} - 3 \, \sin \left (b x + a\right )^{4} + 3 \, \sin \left (b x + a\right )^{2} - 1\right )} b} \]

[In]

integrate(sec(b*x+a)^7*sin(b*x+a)^5,x, algorithm="maxima")

[Out]

-1/6*(3*sin(b*x + a)^4 - 3*sin(b*x + a)^2 + 1)/((sin(b*x + a)^6 - 3*sin(b*x + a)^4 + 3*sin(b*x + a)^2 - 1)*b)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (13) = 26\).

Time = 0.42 (sec) , antiderivative size = 48, normalized size of antiderivative = 3.20 \[ \int \sec ^2(a+b x) \tan ^5(a+b x) \, dx=-\frac {32 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{3 \, b {\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}^{6} {\left (\cos \left (b x + a\right ) + 1\right )}^{3}} \]

[In]

integrate(sec(b*x+a)^7*sin(b*x+a)^5,x, algorithm="giac")

[Out]

-32/3*(cos(b*x + a) - 1)^3/(b*((cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 1)^6*(cos(b*x + a) + 1)^3)

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \sec ^2(a+b x) \tan ^5(a+b x) \, dx=\frac {{\mathrm {tan}\left (a+b\,x\right )}^6}{6\,b} \]

[In]

int(sin(a + b*x)^5/cos(a + b*x)^7,x)

[Out]

tan(a + b*x)^6/(6*b)