\(\int \sec ^5(a+b x) \tan ^2(a+b x) \, dx\) [65]

Optimal result

Integrand size = 17, antiderivative size = 76 \[ \int \sec ^5(a+b x) \tan ^2(a+b x) \, dx=-\frac {\text {arctanh}(\sin (a+b x))}{16 b}-\frac {\sec (a+b x) \tan (a+b x)}{16 b}-\frac {\sec ^3(a+b x) \tan (a+b x)}{24 b}+\frac {\sec ^5(a+b x) \tan (a+b x)}{6 b} \]

[Out]

-1/16*arctanh(sin(b*x+a))/b-1/16*sec(b*x+a)*tan(b*x+a)/b-1/24*sec(b*x+a)^3*tan(b*x+a)/b+1/6*sec(b*x+a)^5*tan(b
*x+a)/b

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2691, 3853, 3855} \[ \int \sec ^5(a+b x) \tan ^2(a+b x) \, dx=-\frac {\text {arctanh}(\sin (a+b x))}{16 b}+\frac {\tan (a+b x) \sec ^5(a+b x)}{6 b}-\frac {\tan (a+b x) \sec ^3(a+b x)}{24 b}-\frac {\tan (a+b x) \sec (a+b x)}{16 b} \]

[In]

Int[Sec[a + b*x]^5*Tan[a + b*x]^2,x]

[Out]

-1/16*ArcTanh[Sin[a + b*x]]/b - (Sec[a + b*x]*Tan[a + b*x])/(16*b) - (Sec[a + b*x]^3*Tan[a + b*x])/(24*b) + (S
ec[a + b*x]^5*Tan[a + b*x])/(6*b)

Rule 2691

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[b^2*((n - 1)/(m + n - 1)), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\sec ^5(a+b x) \tan (a+b x)}{6 b}-\frac {1}{6} \int \sec ^5(a+b x) \, dx \\ & = -\frac {\sec ^3(a+b x) \tan (a+b x)}{24 b}+\frac {\sec ^5(a+b x) \tan (a+b x)}{6 b}-\frac {1}{8} \int \sec ^3(a+b x) \, dx \\ & = -\frac {\sec (a+b x) \tan (a+b x)}{16 b}-\frac {\sec ^3(a+b x) \tan (a+b x)}{24 b}+\frac {\sec ^5(a+b x) \tan (a+b x)}{6 b}-\frac {1}{16} \int \sec (a+b x) \, dx \\ & = -\frac {\text {arctanh}(\sin (a+b x))}{16 b}-\frac {\sec (a+b x) \tan (a+b x)}{16 b}-\frac {\sec ^3(a+b x) \tan (a+b x)}{24 b}+\frac {\sec ^5(a+b x) \tan (a+b x)}{6 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00 \[ \int \sec ^5(a+b x) \tan ^2(a+b x) \, dx=-\frac {\text {arctanh}(\sin (a+b x))}{16 b}-\frac {\sec (a+b x) \tan (a+b x)}{16 b}-\frac {\sec ^3(a+b x) \tan (a+b x)}{24 b}+\frac {\sec ^5(a+b x) \tan (a+b x)}{6 b} \]

[In]

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

[Out]

-1/16*ArcTanh[Sin[a + b*x]]/b - (Sec[a + b*x]*Tan[a + b*x])/(16*b) - (Sec[a + b*x]^3*Tan[a + b*x])/(24*b) + (S
ec[a + b*x]^5*Tan[a + b*x])/(6*b)

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.11

method	result	size
derivativedivides	\(\frac {\frac {\sin ^{3}\left (b x +a \right )}{6 \cos \left (b x +a \right )^{6}}+\frac {\sin ^{3}\left (b x +a \right )}{8 \cos \left (b x +a \right )^{4}}+\frac {\sin ^{3}\left (b x +a \right )}{16 \cos \left (b x +a \right )^{2}}+\frac {\sin \left (b x +a \right )}{16}-\frac {\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{16}}{b}\)	\(84\)
default	\(\frac {\frac {\sin ^{3}\left (b x +a \right )}{6 \cos \left (b x +a \right )^{6}}+\frac {\sin ^{3}\left (b x +a \right )}{8 \cos \left (b x +a \right )^{4}}+\frac {\sin ^{3}\left (b x +a \right )}{16 \cos \left (b x +a \right )^{2}}+\frac {\sin \left (b x +a \right )}{16}-\frac {\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{16}}{b}\)	\(84\)
risch	\(\frac {i \left (3 \,{\mathrm e}^{11 i \left (b x +a \right )}+17 \,{\mathrm e}^{9 i \left (b x +a \right )}-114 \,{\mathrm e}^{7 i \left (b x +a \right )}+114 \,{\mathrm e}^{5 i \left (b x +a \right )}-17 \,{\mathrm e}^{3 i \left (b x +a \right )}-3 \,{\mathrm e}^{i \left (b x +a \right )}\right )}{24 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{6}}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{16 b}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{16 b}\)	\(124\)
norman	\(\frac {\frac {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{8 b}+\frac {47 \left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{24 b}+\frac {13 \left (\tan ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}+\frac {13 \left (\tan ^{7}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}+\frac {47 \left (\tan ^{9}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{24 b}+\frac {\tan ^{11}\left (\frac {b x}{2}+\frac {a}{2}\right )}{8 b}}{\left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{6}}+\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{16 b}-\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{16 b}\)	\(147\)
parallelrisch	\(\frac {\left (3 \cos \left (6 b x +6 a \right )+18 \cos \left (4 b x +4 a \right )+45 \cos \left (2 b x +2 a \right )+30\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )+\left (-45 \cos \left (2 b x +2 a \right )-18 \cos \left (4 b x +4 a \right )-3 \cos \left (6 b x +6 a \right )-30\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )+228 \sin \left (b x +a \right )-34 \sin \left (3 b x +3 a \right )-6 \sin \left (5 b x +5 a \right )}{48 b \left (\cos \left (6 b x +6 a \right )+6 \cos \left (4 b x +4 a \right )+15 \cos \left (2 b x +2 a \right )+10\right )}\)	\(168\)

[In]

int(sec(b*x+a)^7*sin(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

1/b*(1/6*sin(b*x+a)^3/cos(b*x+a)^6+1/8*sin(b*x+a)^3/cos(b*x+a)^4+1/16*sin(b*x+a)^3/cos(b*x+a)^2+1/16*sin(b*x+a
)-1/16*ln(sec(b*x+a)+tan(b*x+a)))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.11 \[ \int \sec ^5(a+b x) \tan ^2(a+b x) \, dx=-\frac {3 \, \cos \left (b x + a\right )^{6} \log \left (\sin \left (b x + a\right ) + 1\right ) - 3 \, \cos \left (b x + a\right )^{6} \log \left (-\sin \left (b x + a\right ) + 1\right ) + 2 \, {\left (3 \, \cos \left (b x + a\right )^{4} + 2 \, \cos \left (b x + a\right )^{2} - 8\right )} \sin \left (b x + a\right )}{96 \, b \cos \left (b x + a\right )^{6}} \]

[In]

integrate(sec(b*x+a)^7*sin(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/96*(3*cos(b*x + a)^6*log(sin(b*x + a) + 1) - 3*cos(b*x + a)^6*log(-sin(b*x + a) + 1) + 2*(3*cos(b*x + a)^4
+ 2*cos(b*x + a)^2 - 8)*sin(b*x + a))/(b*cos(b*x + a)^6)

Sympy [F(-1)]

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

[In]

integrate(sec(b*x+a)**7*sin(b*x+a)**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.20 \[ \int \sec ^5(a+b x) \tan ^2(a+b x) \, dx=\frac {\frac {2 \, {\left (3 \, \sin \left (b x + a\right )^{5} - 8 \, \sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )\right )}}{\sin \left (b x + a\right )^{6} - 3 \, \sin \left (b x + a\right )^{4} + 3 \, \sin \left (b x + a\right )^{2} - 1} - 3 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 3 \, \log \left (\sin \left (b x + a\right ) - 1\right )}{96 \, b} \]

[In]

integrate(sec(b*x+a)^7*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

1/96*(2*(3*sin(b*x + a)^5 - 8*sin(b*x + a)^3 - 3*sin(b*x + a))/(sin(b*x + a)^6 - 3*sin(b*x + a)^4 + 3*sin(b*x
+ a)^2 - 1) - 3*log(sin(b*x + a) + 1) + 3*log(sin(b*x + a) - 1))/b

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.96 \[ \int \sec ^5(a+b x) \tan ^2(a+b x) \, dx=\frac {\frac {2 \, {\left (3 \, \sin \left (b x + a\right )^{5} - 8 \, \sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )\right )}}{{\left (\sin \left (b x + a\right )^{2} - 1\right )}^{3}} - 3 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) + 3 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{96 \, b} \]

[In]

integrate(sec(b*x+a)^7*sin(b*x+a)^2,x, algorithm="giac")

[Out]

1/96*(2*(3*sin(b*x + a)^5 - 8*sin(b*x + a)^3 - 3*sin(b*x + a))/(sin(b*x + a)^2 - 1)^3 - 3*log(abs(sin(b*x + a)
+ 1)) + 3*log(abs(sin(b*x + a) - 1)))/b

Mupad [B] (verification not implemented)

Time = 7.17 (sec) , antiderivative size = 177, normalized size of antiderivative = 2.33 \[ \int \sec ^5(a+b x) \tan ^2(a+b x) \, dx=\frac {\frac {{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^{11}}{8}+\frac {47\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^9}{24}+\frac {13\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^7}{4}+\frac {13\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^5}{4}+\frac {47\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^3}{24}+\frac {\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}{8}}{b\,\left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2+1\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )\right )}{8\,b} \]

[In]

int(sin(a + b*x)^2/cos(a + b*x)^7,x)

[Out]

(tan(a/2 + (b*x)/2)/8 + (47*tan(a/2 + (b*x)/2)^3)/24 + (13*tan(a/2 + (b*x)/2)^5)/4 + (13*tan(a/2 + (b*x)/2)^7)
/4 + (47*tan(a/2 + (b*x)/2)^9)/24 + tan(a/2 + (b*x)/2)^11/8)/(b*(15*tan(a/2 + (b*x)/2)^4 - 6*tan(a/2 + (b*x)/2
)^2 - 20*tan(a/2 + (b*x)/2)^6 + 15*tan(a/2 + (b*x)/2)^8 - 6*tan(a/2 + (b*x)/2)^10 + tan(a/2 + (b*x)/2)^12 + 1)
) - atanh(tan(a/2 + (b*x)/2))/(8*b)