3.26 \(\int \csc ^4(e+f x) (a+b \sec ^2(e+f x))^2 \, dx\)

Optimal. Leaf size=76 \[ \frac{b (2 a+3 b) \tan (e+f x)}{f}-\frac{(a+b)^2 \cot ^3(e+f x)}{3 f}-\frac{(a+b) (a+3 b) \cot (e+f x)}{f}+\frac{b^2 \tan ^3(e+f x)}{3 f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0756403, antiderivative size = 76, 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 = {4132, 448} \[ \frac{b (2 a+3 b) \tan (e+f x)}{f}-\frac{(a+b)^2 \cot ^3(e+f x)}{3 f}-\frac{(a+b) (a+3 b) \cot (e+f x)}{f}+\frac{b^2 \tan ^3(e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^4*(a + b*Sec[e + f*x]^2)^2,x]

[Out]

-(((a + b)*(a + 3*b)*Cot[e + f*x])/f) - ((a + b)^2*Cot[e + f*x]^3)/(3*f) + (b*(2*a + 3*b)*Tan[e + f*x])/f + (b
^2*Tan[e + f*x]^3)/(3*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 ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right ) \left (a+b+b x^2\right )^2}{x^4} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left (b (2 a+3 b)+\frac{(a+b)^2}{x^4}+\frac{(a+b) (a+3 b)}{x^2}+b^2 x^2\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{(a+b) (a+3 b) \cot (e+f x)}{f}-\frac{(a+b)^2 \cot ^3(e+f x)}{3 f}+\frac{b (2 a+3 b) \tan (e+f x)}{f}+\frac{b^2 \tan ^3(e+f x)}{3 f}\\ \end{align*}

Mathematica [A]  time = 1.30718, size = 151, normalized size = 1.99 \[ -\frac{\csc (2 e) \csc ^3(2 (e+f x)) \left (-3 a^2 \sin (2 (e+f x))+a^2 \sin (6 (e+f x))+3 a^2 \sin (4 e+2 f x)+a^2 \sin (4 e+6 f x)-6 a b \sin (2 (e+f x))+2 a b \sin (6 (e+f x))+8 a b \sin (4 e+6 f x)+8 a (a+2 b) \sin (2 e)-6 (a+2 b)^2 \sin (2 f x)+8 b^2 \sin (4 e+6 f x)\right )}{6 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^4*(a + b*Sec[e + f*x]^2)^2,x]

[Out]

-(Csc[2*e]*Csc[2*(e + f*x)]^3*(8*a*(a + 2*b)*Sin[2*e] - 6*(a + 2*b)^2*Sin[2*f*x] - 3*a^2*Sin[2*(e + f*x)] - 6*
a*b*Sin[2*(e + f*x)] + a^2*Sin[6*(e + f*x)] + 2*a*b*Sin[6*(e + f*x)] + 3*a^2*Sin[4*e + 2*f*x] + a^2*Sin[4*e +
6*f*x] + 8*a*b*Sin[4*e + 6*f*x] + 8*b^2*Sin[4*e + 6*f*x]))/(6*f)

________________________________________________________________________________________

Maple [A]  time = 0.061, size = 144, normalized size = 1.9 \begin{align*}{\frac{1}{f} \left ({a}^{2} \left ( -{\frac{2}{3}}-{\frac{ \left ( \csc \left ( fx+e \right ) \right ) ^{2}}{3}} \right ) \cot \left ( fx+e \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}{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }}-8/3\,\cot \left ( fx+e \right ) \right ) +{b}^{2} \left ({\frac{1}{3\, \left ( \sin \left ( fx+e \right ) \right ) ^{3} \left ( \cos \left ( fx+e \right ) \right ) ^{3}}}-{\frac{2}{3\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}\cos \left ( fx+e \right ) }}+{\frac{8}{3\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }}-{\frac{16\,\cot \left ( fx+e \right ) }{3}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^4*(a+b*sec(f*x+e)^2)^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.987961, size = 108, normalized size = 1.42 \begin{align*} \frac{b^{2} \tan \left (f x + e\right )^{3} + 3 \,{\left (2 \, a b + 3 \, b^{2}\right )} \tan \left (f x + e\right ) - \frac{3 \,{\left (a^{2} + 4 \, a b + 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}}{\tan \left (f x + e\right )^{3}}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 0.477801, size = 240, normalized size = 3.16 \begin{align*} -\frac{2 \,{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{6} - 3 \,{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 6 \,{\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}}{3 \,{\left (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)^4*(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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)**4*(a+b*sec(f*x+e)**2)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.2773, size = 142, normalized size = 1.87 \begin{align*} \frac{b^{2} \tan \left (f x + e\right )^{3} + 6 \, a b \tan \left (f x + e\right ) + 9 \, b^{2} \tan \left (f x + e\right ) - \frac{3 \, a^{2} \tan \left (f x + e\right )^{2} + 12 \, a b \tan \left (f x + e\right )^{2} + 9 \, b^{2} \tan \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}}{\tan \left (f x + e\right )^{3}}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")

[Out]

1/3*(b^2*tan(f*x + e)^3 + 6*a*b*tan(f*x + e) + 9*b^2*tan(f*x + e) - (3*a^2*tan(f*x + e)^2 + 12*a*b*tan(f*x + e
)^2 + 9*b^2*tan(f*x + e)^2 + a^2 + 2*a*b + b^2)/tan(f*x + e)^3)/f