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

Optimal. Leaf size=93 \[ -\frac{\left (a^2+4 a b+b^2\right ) \cot (e+f x)}{f}-\frac{a^2 \cot ^5(e+f x)}{5 f}+\frac{2 b (a+b) \tan (e+f x)}{f}-\frac{2 a (a+b) \cot ^3(e+f x)}{3 f}+\frac{b^2 \tan ^3(e+f x)}{3 f} \]

[Out]

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

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Rubi [A]  time = 0.0898972, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {3663, 448} \[ -\frac{\left (a^2+4 a b+b^2\right ) \cot (e+f x)}{f}-\frac{a^2 \cot ^5(e+f x)}{5 f}+\frac{2 b (a+b) \tan (e+f x)}{f}-\frac{2 a (a+b) \cot ^3(e+f x)}{3 f}+\frac{b^2 \tan ^3(e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3663

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff^(m + 1))/f, Subst[Int[(x^m*(a + b*(ff*x)^n)^p)/(c^2 + ff^2*x^2
)^(m/2 + 1), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/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 \tan ^2(e+f x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2 \left (a+b x^2\right )^2}{x^6} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left (2 b (a+b)+\frac{a^2}{x^6}+\frac{2 a (a+b)}{x^4}+\frac{a^2+4 a b+b^2}{x^2}+b^2 x^2\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\left (a^2+4 a b+b^2\right ) \cot (e+f x)}{f}-\frac{2 a (a+b) \cot ^3(e+f x)}{3 f}-\frac{a^2 \cot ^5(e+f x)}{5 f}+\frac{2 b (a+b) \tan (e+f x)}{f}+\frac{b^2 \tan ^3(e+f x)}{3 f}\\ \end{align*}

Mathematica [A]  time = 0.774006, size = 88, normalized size = 0.95 \[ \frac{5 b \tan (e+f x) \left (6 a+b \sec ^2(e+f x)+5 b\right )-\cot (e+f x) \left (3 a^2 \csc ^4(e+f x)+8 a^2+2 a (2 a+5 b) \csc ^2(e+f x)+50 a b+15 b^2\right )}{15 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Cot[e + f*x]*(8*a^2 + 50*a*b + 15*b^2 + 2*a*(2*a + 5*b)*Csc[e + f*x]^2 + 3*a^2*Csc[e + f*x]^4)) + 5*b*(6*a
+ 5*b + b*Sec[e + f*x]^2)*Tan[e + f*x])/(15*f)

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Maple [A]  time = 0.059, size = 136, normalized size = 1.5 \begin{align*}{\frac{1}{f} \left ({b}^{2} \left ({\frac{1}{3\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}}}+{\frac{4}{3\,\cos \left ( fx+e \right ) \sin \left ( fx+e \right ) }}-{\frac{8\,\cot \left ( fx+e \right ) }{3}} \right ) +2\,ab \left ( -1/3\,{\frac{1}{ \left ( \sin \left ( fx+e \right ) \right ) ^{3}\cos \left ( fx+e \right ) }}+4/3\,{\frac{1}{\cos \left ( fx+e \right ) \sin \left ( fx+e \right ) }}-8/3\,\cot \left ( fx+e \right ) \right ) +{a}^{2} \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 ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.966183, size = 119, normalized size = 1.28 \begin{align*} \frac{5 \, b^{2} \tan \left (f x + e\right )^{3} + 30 \,{\left (a b + b^{2}\right )} \tan \left (f x + e\right ) - \frac{15 \,{\left (a^{2} + 4 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{4} + 10 \,{\left (a^{2} + a b\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2}}{\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*tan(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

1/15*(5*b^2*tan(f*x + e)^3 + 30*(a*b + b^2)*tan(f*x + e) - (15*(a^2 + 4*a*b + b^2)*tan(f*x + e)^4 + 10*(a^2 +
a*b)*tan(f*x + e)^2 + 3*a^2)/tan(f*x + e)^5)/f

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Fricas [A]  time = 2.0376, size = 339, normalized size = 3.65 \begin{align*} -\frac{8 \,{\left (a^{2} + 10 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{8} - 20 \,{\left (a^{2} + 10 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{6} + 15 \,{\left (a^{2} + 10 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 10 \,{\left (3 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - 5 \, b^{2}}{15 \,{\left (f \cos \left (f x + e\right )^{7} - 2 \, f \cos \left (f x + e\right )^{5} + f \cos \left (f x + e\right )^{3}\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*tan(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

-1/15*(8*(a^2 + 10*a*b + 5*b^2)*cos(f*x + e)^8 - 20*(a^2 + 10*a*b + 5*b^2)*cos(f*x + e)^6 + 15*(a^2 + 10*a*b +
5*b^2)*cos(f*x + e)^4 - 10*(3*a*b + b^2)*cos(f*x + e)^2 - 5*b^2)/((f*cos(f*x + e)^7 - 2*f*cos(f*x + e)^5 + f*
cos(f*x + e)^3)*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*tan(f*x+e)**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.67698, size = 173, normalized size = 1.86 \begin{align*} \frac{5 \, b^{2} \tan \left (f x + e\right )^{3} + 30 \, a b \tan \left (f x + e\right ) + 30 \, b^{2} \tan \left (f x + e\right ) - \frac{15 \, a^{2} \tan \left (f x + e\right )^{4} + 60 \, a b \tan \left (f x + e\right )^{4} + 15 \, b^{2} \tan \left (f x + e\right )^{4} + 10 \, a^{2} \tan \left (f x + e\right )^{2} + 10 \, a b \tan \left (f x + e\right )^{2} + 3 \, a^{2}}{\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*tan(f*x+e)^2)^2,x, algorithm="giac")

[Out]

1/15*(5*b^2*tan(f*x + e)^3 + 30*a*b*tan(f*x + e) + 30*b^2*tan(f*x + e) - (15*a^2*tan(f*x + e)^4 + 60*a*b*tan(f
*x + e)^4 + 15*b^2*tan(f*x + e)^4 + 10*a^2*tan(f*x + e)^2 + 10*a*b*tan(f*x + e)^2 + 3*a^2)/tan(f*x + e)^5)/f