3.14 \(\int \csc ^6(e+f x) (a+b \sec ^2(e+f x)) \, dx\)

Optimal. Leaf size=68 \[ -\frac{(a+b) \cot ^5(e+f x)}{5 f}-\frac{(2 a+3 b) \cot ^3(e+f x)}{3 f}-\frac{(a+3 b) \cot (e+f x)}{f}+\frac{b \tan (e+f x)}{f} \]

[Out]

-(((a + 3*b)*Cot[e + f*x])/f) - ((2*a + 3*b)*Cot[e + f*x]^3)/(3*f) - ((a + b)*Cot[e + f*x]^5)/(5*f) + (b*Tan[e
+ f*x])/f

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Rubi [A]  time = 0.0560181, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {4132, 448} \[ -\frac{(a+b) \cot ^5(e+f x)}{5 f}-\frac{(2 a+3 b) \cot ^3(e+f x)}{3 f}-\frac{(a+3 b) \cot (e+f x)}{f}+\frac{b \tan (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^6*(a + b*Sec[e + f*x]^2),x]

[Out]

-(((a + 3*b)*Cot[e + f*x])/f) - ((2*a + 3*b)*Cot[e + f*x]^3)/(3*f) - ((a + b)*Cot[e + f*x]^5)/(5*f) + (b*Tan[e
+ f*x])/f

Rule 4132

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/2]

Rule 448

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

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

Mathematica [A]  time = 0.0459084, size = 128, normalized size = 1.88 \[ -\frac{8 a \cot (e+f x)}{15 f}-\frac{a \cot (e+f x) \csc ^4(e+f x)}{5 f}-\frac{4 a \cot (e+f x) \csc ^2(e+f x)}{15 f}+\frac{b \tan (e+f x)}{f}-\frac{11 b \cot (e+f x)}{5 f}-\frac{b \cot (e+f x) \csc ^4(e+f x)}{5 f}-\frac{3 b \cot (e+f x) \csc ^2(e+f x)}{5 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^6*(a + b*Sec[e + f*x]^2),x]

[Out]

(-8*a*Cot[e + f*x])/(15*f) - (11*b*Cot[e + f*x])/(5*f) - (4*a*Cot[e + f*x]*Csc[e + f*x]^2)/(15*f) - (3*b*Cot[e
+ f*x]*Csc[e + f*x]^2)/(5*f) - (a*Cot[e + f*x]*Csc[e + f*x]^4)/(5*f) - (b*Cot[e + f*x]*Csc[e + f*x]^4)/(5*f)
+ (b*Tan[e + f*x])/f

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Maple [A]  time = 0.053, size = 101, normalized size = 1.5 \begin{align*}{\frac{1}{f} \left ( a \left ( -{\frac{8}{15}}-{\frac{ \left ( \csc \left ( fx+e \right ) \right ) ^{4}}{5}}-{\frac{4\, \left ( \csc \left ( fx+e \right ) \right ) ^{2}}{15}} \right ) \cot \left ( fx+e \right ) +b \left ( -{\frac{1}{5\, \left ( \sin \left ( fx+e \right ) \right ) ^{5}\cos \left ( fx+e \right ) }}-{\frac{2}{5\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}\cos \left ( fx+e \right ) }}+{\frac{8}{5\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }}-{\frac{16\,\cot \left ( fx+e \right ) }{5}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^6*(a+b*sec(f*x+e)^2),x)

[Out]

1/f*(a*(-8/15-1/5*csc(f*x+e)^4-4/15*csc(f*x+e)^2)*cot(f*x+e)+b*(-1/5/sin(f*x+e)^5/cos(f*x+e)-2/5/sin(f*x+e)^3/
cos(f*x+e)+8/5/sin(f*x+e)/cos(f*x+e)-16/5*cot(f*x+e)))

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Maxima [A]  time = 1.01311, size = 86, normalized size = 1.26 \begin{align*} \frac{15 \, b \tan \left (f x + e\right ) - \frac{15 \,{\left (a + 3 \, b\right )} \tan \left (f x + e\right )^{4} + 5 \,{\left (2 \, a + 3 \, b\right )} \tan \left (f x + e\right )^{2} + 3 \, a + 3 \, b}{\tan \left (f x + e\right )^{5}}}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6*(a+b*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

1/15*(15*b*tan(f*x + e) - (15*(a + 3*b)*tan(f*x + e)^4 + 5*(2*a + 3*b)*tan(f*x + e)^2 + 3*a + 3*b)/tan(f*x + e
)^5)/f

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Fricas [A]  time = 0.469611, size = 236, normalized size = 3.47 \begin{align*} -\frac{8 \,{\left (a + 6 \, b\right )} \cos \left (f x + e\right )^{6} - 20 \,{\left (a + 6 \, b\right )} \cos \left (f x + e\right )^{4} + 15 \,{\left (a + 6 \, b\right )} \cos \left (f x + e\right )^{2} - 15 \, b}{15 \,{\left (f \cos \left (f x + e\right )^{5} - 2 \, f \cos \left (f x + e\right )^{3} + f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6*(a+b*sec(f*x+e)^2),x, algorithm="fricas")

[Out]

-1/15*(8*(a + 6*b)*cos(f*x + e)^6 - 20*(a + 6*b)*cos(f*x + e)^4 + 15*(a + 6*b)*cos(f*x + e)^2 - 15*b)/((f*cos(
f*x + e)^5 - 2*f*cos(f*x + e)^3 + f*cos(f*x + e))*sin(f*x + e))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**6*(a+b*sec(f*x+e)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.3392, size = 111, normalized size = 1.63 \begin{align*} \frac{15 \, b \tan \left (f x + e\right ) - \frac{15 \, a \tan \left (f x + e\right )^{4} + 45 \, b \tan \left (f x + e\right )^{4} + 10 \, a \tan \left (f x + e\right )^{2} + 15 \, b \tan \left (f x + e\right )^{2} + 3 \, a + 3 \, b}{\tan \left (f x + e\right )^{5}}}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6*(a+b*sec(f*x+e)^2),x, algorithm="giac")

[Out]

1/15*(15*b*tan(f*x + e) - (15*a*tan(f*x + e)^4 + 45*b*tan(f*x + e)^4 + 10*a*tan(f*x + e)^2 + 15*b*tan(f*x + e)
^2 + 3*a + 3*b)/tan(f*x + e)^5)/f