3.112 \(\int \sec ^3(a+b x) \tan ^5(a+b x) \, dx\)

Optimal. Leaf size=46 \[ \frac {\sec ^7(a+b x)}{7 b}-\frac {2 \sec ^5(a+b x)}{5 b}+\frac {\sec ^3(a+b x)}{3 b} \]

[Out]

1/3*sec(b*x+a)^3/b-2/5*sec(b*x+a)^5/b+1/7*sec(b*x+a)^7/b

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2606, 270} \[ \frac {\sec ^7(a+b x)}{7 b}-\frac {2 \sec ^5(a+b x)}{5 b}+\frac {\sec ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^3*Tan[a + b*x]^5,x]

[Out]

Sec[a + b*x]^3/(3*b) - (2*Sec[a + b*x]^5)/(5*b) + Sec[a + b*x]^7/(7*b)

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2606

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*} \int \sec ^3(a+b x) \tan ^5(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int x^2 \left (-1+x^2\right )^2 \, dx,x,\sec (a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\sec (a+b x)\right )}{b}\\ &=\frac {\sec ^3(a+b x)}{3 b}-\frac {2 \sec ^5(a+b x)}{5 b}+\frac {\sec ^7(a+b x)}{7 b}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 46, normalized size = 1.00 \[ \frac {\sec ^7(a+b x)}{7 b}-\frac {2 \sec ^5(a+b x)}{5 b}+\frac {\sec ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^3*Tan[a + b*x]^5,x]

[Out]

Sec[a + b*x]^3/(3*b) - (2*Sec[a + b*x]^5)/(5*b) + Sec[a + b*x]^7/(7*b)

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 35, normalized size = 0.76 \[ \frac {35 \, \cos \left (b x + a\right )^{4} - 42 \, \cos \left (b x + a\right )^{2} + 15}{105 \, b \cos \left (b x + a\right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^8*sin(b*x+a)^5,x, algorithm="fricas")

[Out]

1/105*(35*cos(b*x + a)^4 - 42*cos(b*x + a)^2 + 15)/(b*cos(b*x + a)^7)

________________________________________________________________________________________

giac [B]  time = 0.26, size = 116, normalized size = 2.52 \[ \frac {16 \, {\left (\frac {7 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {21 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - \frac {35 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} + \frac {70 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} + 1\right )}}{105 \, b {\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^8*sin(b*x+a)^5,x, algorithm="giac")

[Out]

16/105*(7*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 21*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 - 35*(cos(b*x +
a) - 1)^3/(cos(b*x + a) + 1)^3 + 70*(cos(b*x + a) - 1)^4/(cos(b*x + a) + 1)^4 + 1)/(b*((cos(b*x + a) - 1)/(co
s(b*x + a) + 1) + 1)^7)

________________________________________________________________________________________

maple [B]  time = 0.04, size = 106, normalized size = 2.30 \[ \frac {\frac {\sin ^{6}\left (b x +a \right )}{7 \cos \left (b x +a \right )^{7}}+\frac {\sin ^{6}\left (b x +a \right )}{35 \cos \left (b x +a \right )^{5}}-\frac {\sin ^{6}\left (b x +a \right )}{105 \cos \left (b x +a \right )^{3}}+\frac {\sin ^{6}\left (b x +a \right )}{35 \cos \left (b x +a \right )}+\frac {\left (\frac {8}{3}+\sin ^{4}\left (b x +a \right )+\frac {4 \left (\sin ^{2}\left (b x +a \right )\right )}{3}\right ) \cos \left (b x +a \right )}{35}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^8*sin(b*x+a)^5,x)

[Out]

1/b*(1/7*sin(b*x+a)^6/cos(b*x+a)^7+1/35*sin(b*x+a)^6/cos(b*x+a)^5-1/105*sin(b*x+a)^6/cos(b*x+a)^3+1/35*sin(b*x
+a)^6/cos(b*x+a)+1/35*(8/3+sin(b*x+a)^4+4/3*sin(b*x+a)^2)*cos(b*x+a))

________________________________________________________________________________________

maxima [A]  time = 0.75, size = 35, normalized size = 0.76 \[ \frac {35 \, \cos \left (b x + a\right )^{4} - 42 \, \cos \left (b x + a\right )^{2} + 15}{105 \, b \cos \left (b x + a\right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^8*sin(b*x+a)^5,x, algorithm="maxima")

[Out]

1/105*(35*cos(b*x + a)^4 - 42*cos(b*x + a)^2 + 15)/(b*cos(b*x + a)^7)

________________________________________________________________________________________

mupad [B]  time = 0.59, size = 35, normalized size = 0.76 \[ \frac {35\,{\cos \left (a+b\,x\right )}^4-42\,{\cos \left (a+b\,x\right )}^2+15}{105\,b\,{\cos \left (a+b\,x\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^5/cos(a + b*x)^8,x)

[Out]

(35*cos(a + b*x)^4 - 42*cos(a + b*x)^2 + 15)/(105*b*cos(a + b*x)^7)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**8*sin(b*x+a)**5,x)

[Out]

Timed out

________________________________________________________________________________________