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

Optimal. Leaf size=70 \[ -\frac{a^2 \cot ^3(e+f x)}{3 f}+\frac{b (2 a+b) \tan (e+f x)}{f}-\frac{a (a+2 b) \cot (e+f x)}{f}+\frac{b^2 \tan ^3(e+f x)}{3 f} \]

[Out]

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

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Rubi [A]  time = 0.0724582, antiderivative size = 70, 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} \[ -\frac{a^2 \cot ^3(e+f x)}{3 f}+\frac{b (2 a+b) \tan (e+f x)}{f}-\frac{a (a+2 b) \cot (e+f x)}{f}+\frac{b^2 \tan ^3(e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.441543, size = 59, normalized size = 0.84 \[ \frac{b \tan (e+f x) \left (6 a+b \sec ^2(e+f x)+2 b\right )-a \cot (e+f x) \left (a \csc ^2(e+f x)+2 a+6 b\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.082, size = 81, normalized size = 1.2 \begin{align*}{\frac{1}{f} \left ( -{b}^{2} \left ( -{\frac{2}{3}}-{\frac{ \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{3}} \right ) \tan \left ( fx+e \right ) +2\,ab \left ({\frac{1}{\cos \left ( fx+e \right ) \sin \left ( fx+e \right ) }}-2\,\cot \left ( fx+e \right ) \right ) +{a}^{2} \left ( -{\frac{2}{3}}-{\frac{ \left ( \csc \left ( fx+e \right ) \right ) ^{2}}{3}} \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)^4*(a+b*tan(f*x+e)^2)^2,x)

[Out]

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

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Maxima [A]  time = 0.979427, size = 89, normalized size = 1.27 \begin{align*} \frac{b^{2} \tan \left (f x + e\right )^{3} + 3 \,{\left (2 \, a b + b^{2}\right )} \tan \left (f x + e\right ) - \frac{3 \,{\left (a^{2} + 2 \, a b\right )} \tan \left (f x + e\right )^{2} + a^{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*tan(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.95461, size = 224, normalized size = 3.2 \begin{align*} -\frac{2 \,{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{6} - 3 \,{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 6 \, a b \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*tan(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

-1/3*(2*(a^2 + 6*a*b + b^2)*cos(f*x + e)^6 - 3*(a^2 + 6*a*b + b^2)*cos(f*x + e)^4 + 6*a*b*cos(f*x + e)^2 + b^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)**4*(a+b*tan(f*x+e)**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.65381, size = 113, normalized size = 1.61 \begin{align*} \frac{b^{2} \tan \left (f x + e\right )^{3} + 6 \, a b \tan \left (f x + e\right ) + 3 \, b^{2} \tan \left (f x + e\right ) - \frac{3 \, a^{2} \tan \left (f x + e\right )^{2} + 6 \, a b \tan \left (f x + e\right )^{2} + a^{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*tan(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) + 3*b^2*tan(f*x + e) - (3*a^2*tan(f*x + e)^2 + 6*a*b*tan(f*x + e)
^2 + a^2)/tan(f*x + e)^3)/f