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

Optimal. Leaf size=31 \[ \frac {\sec ^7(a+b x)}{7 b}-\frac {\sec ^5(a+b x)}{5 b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 31, 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, 14} \[ \frac {\sec ^7(a+b x)}{7 b}-\frac {\sec ^5(a+b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^5*Tan[a + b*x]^3,x]

[Out]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 31, normalized size = 1.00 \[ \frac {\sec ^7(a+b x)}{7 b}-\frac {\sec ^5(a+b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*Sec[a + b*x]^5/b + Sec[a + b*x]^7/(7*b)

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 25, normalized size = 0.81 \[ -\frac {7 \, \cos \left (b x + a\right )^{2} - 5}{35 \, b \cos \left (b x + a\right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^8*sin(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/35*(7*cos(b*x + a)^2 - 5)/(b*cos(b*x + a)^7)

________________________________________________________________________________________

giac [A]  time = 0.23, size = 25, normalized size = 0.81 \[ -\frac {7 \, \cos \left (b x + a\right )^{2} - 5}{35 \, b \cos \left (b x + a\right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^8*sin(b*x+a)^3,x, algorithm="giac")

[Out]

-1/35*(7*cos(b*x + a)^2 - 5)/(b*cos(b*x + a)^7)

________________________________________________________________________________________

maple [B]  time = 0.04, size = 96, normalized size = 3.10 \[ \frac {\frac {\sin ^{4}\left (b x +a \right )}{7 \cos \left (b x +a \right )^{7}}+\frac {3 \left (\sin ^{4}\left (b x +a \right )\right )}{35 \cos \left (b x +a \right )^{5}}+\frac {\sin ^{4}\left (b x +a \right )}{35 \cos \left (b x +a \right )^{3}}-\frac {\sin ^{4}\left (b x +a \right )}{35 \cos \left (b x +a \right )}-\frac {\left (2+\sin ^{2}\left (b x +a \right )\right ) \cos \left (b x +a \right )}{35}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^8*sin(b*x+a)^3,x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.32, size = 25, normalized size = 0.81 \[ -\frac {7 \, \cos \left (b x + a\right )^{2} - 5}{35 \, b \cos \left (b x + a\right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^8*sin(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/35*(7*cos(b*x + a)^2 - 5)/(b*cos(b*x + a)^7)

________________________________________________________________________________________

mupad [B]  time = 0.63, size = 25, normalized size = 0.81 \[ -\frac {7\,{\cos \left (a+b\,x\right )}^2-5}{35\,b\,{\cos \left (a+b\,x\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^3/cos(a + b*x)^8,x)

[Out]

-(7*cos(a + b*x)^2 - 5)/(35*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)**3,x)

[Out]

Timed out

________________________________________________________________________________________