3.351 \(\int (b \sec (e+f x))^m \tan ^5(e+f x) \, dx\)

Optimal. Leaf size=67 \[ -\frac{2 (b \sec (e+f x))^{m+2}}{b^2 f (m+2)}+\frac{(b \sec (e+f x))^{m+4}}{b^4 f (m+4)}+\frac{(b \sec (e+f x))^m}{f m} \]

[Out]

(b*Sec[e + f*x])^m/(f*m) - (2*(b*Sec[e + f*x])^(2 + m))/(b^2*f*(2 + m)) + (b*Sec[e + f*x])^(4 + m)/(b^4*f*(4 +
m))

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Rubi [A]  time = 0.0623153, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2606, 270} \[ -\frac{2 (b \sec (e+f x))^{m+2}}{b^2 f (m+2)}+\frac{(b \sec (e+f x))^{m+4}}{b^4 f (m+4)}+\frac{(b \sec (e+f x))^m}{f m} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sec[e + f*x])^m*Tan[e + f*x]^5,x]

[Out]

(b*Sec[e + f*x])^m/(f*m) - (2*(b*Sec[e + f*x])^(2 + m))/(b^2*f*(2 + m)) + (b*Sec[e + f*x])^(4 + m)/(b^4*f*(4 +
m))

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])

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]

Rubi steps

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

Mathematica [A]  time = 0.351632, size = 47, normalized size = 0.7 \[ \frac{\left (\frac{\sec ^4(e+f x)}{m+4}-\frac{2 \sec ^2(e+f x)}{m+2}+\frac{1}{m}\right ) (b \sec (e+f x))^m}{f} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[e + f*x])^m*Tan[e + f*x]^5,x]

[Out]

((b*Sec[e + f*x])^m*(m^(-1) - (2*Sec[e + f*x]^2)/(2 + m) + Sec[e + f*x]^4/(4 + m)))/f

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Maple [C]  time = 0.603, size = 6797, normalized size = 101.5 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(f*x+e))^m*tan(f*x+e)^5,x)

[Out]

result too large to display

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Maxima [A]  time = 1.0213, size = 104, normalized size = 1.55 \begin{align*} \frac{\frac{b^{m} \cos \left (f x + e\right )^{-m}}{m} - \frac{2 \, b^{m} \cos \left (f x + e\right )^{-m}}{{\left (m + 2\right )} \cos \left (f x + e\right )^{2}} + \frac{b^{m} \cos \left (f x + e\right )^{-m}}{{\left (m + 4\right )} \cos \left (f x + e\right )^{4}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^m*tan(f*x+e)^5,x, algorithm="maxima")

[Out]

(b^m*cos(f*x + e)^(-m)/m - 2*b^m*cos(f*x + e)^(-m)/((m + 2)*cos(f*x + e)^2) + b^m*cos(f*x + e)^(-m)/((m + 4)*c
os(f*x + e)^4))/f

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Fricas [A]  time = 1.72904, size = 188, normalized size = 2.81 \begin{align*} \frac{{\left ({\left (m^{2} + 6 \, m + 8\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (m^{2} + 4 \, m\right )} \cos \left (f x + e\right )^{2} + m^{2} + 2 \, m\right )} \left (\frac{b}{\cos \left (f x + e\right )}\right )^{m}}{{\left (f m^{3} + 6 \, f m^{2} + 8 \, f m\right )} \cos \left (f x + e\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^m*tan(f*x+e)^5,x, algorithm="fricas")

[Out]

((m^2 + 6*m + 8)*cos(f*x + e)^4 - 2*(m^2 + 4*m)*cos(f*x + e)^2 + m^2 + 2*m)*(b/cos(f*x + e))^m/((f*m^3 + 6*f*m
^2 + 8*f*m)*cos(f*x + e)^4)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))**m*tan(f*x+e)**5,x)

[Out]

Exception raised: TypeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \sec \left (f x + e\right )\right )^{m} \tan \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^m*tan(f*x+e)^5,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e))^m*tan(f*x + e)^5, x)