3.66 \(\int \frac{\csc ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx\)

Optimal. Leaf size=76 \[ -\frac{\sqrt{b} (a-b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{a^{5/2} f}-\frac{(a-b) \cot (e+f x)}{a^2 f}-\frac{\cot ^3(e+f x)}{3 a f} \]

[Out]

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

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Rubi [A]  time = 0.0902362, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3663, 453, 325, 205} \[ -\frac{\sqrt{b} (a-b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{a^{5/2} f}-\frac{(a-b) \cot (e+f x)}{a^2 f}-\frac{\cot ^3(e+f x)}{3 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a - b)*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(a^(5/2)*f)) - ((a - b)*Cot[e + f*x])/(a^2*f) - Cot
[e + f*x]^3/(3*a*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 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.291959, size = 73, normalized size = 0.96 \[ \frac{3 \sqrt{b} (b-a) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )-\sqrt{a} \cot (e+f x) \left (a \csc ^2(e+f x)+2 a-3 b\right )}{3 a^{5/2} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.075, size = 107, normalized size = 1.4 \begin{align*} -{\frac{1}{3\,fa \left ( \tan \left ( fx+e \right ) \right ) ^{3}}}-{\frac{1}{fa\tan \left ( fx+e \right ) }}+{\frac{b}{f{a}^{2}\tan \left ( fx+e \right ) }}-{\frac{b}{fa}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{{b}^{2}}{f{a}^{2}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \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),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.82012, size = 884, normalized size = 11.63 \begin{align*} \left [-\frac{4 \,{\left (2 \, a - 3 \, b\right )} \cos \left (f x + e\right )^{3} + 3 \,{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - a + b\right )} \sqrt{-\frac{b}{a}} \log \left (\frac{{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (3 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \,{\left ({\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - a b \cos \left (f x + e\right )\right )} \sqrt{-\frac{b}{a}} \sin \left (f x + e\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \,{\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}}\right ) \sin \left (f x + e\right ) - 12 \,{\left (a - b\right )} \cos \left (f x + e\right )}{12 \,{\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f\right )} \sin \left (f x + e\right )}, -\frac{2 \,{\left (2 \, a - 3 \, b\right )} \cos \left (f x + e\right )^{3} - 3 \,{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - a + b\right )} \sqrt{\frac{b}{a}} \arctan \left (\frac{{\left ({\left (a + b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt{\frac{b}{a}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 6 \,{\left (a - b\right )} \cos \left (f x + e\right )}{6 \,{\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f\right )} \sin \left (f x + e\right )}\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),x, algorithm="fricas")

[Out]

[-1/12*(4*(2*a - 3*b)*cos(f*x + e)^3 + 3*((a - b)*cos(f*x + e)^2 - a + b)*sqrt(-b/a)*log(((a^2 + 6*a*b + b^2)*
cos(f*x + e)^4 - 2*(3*a*b + b^2)*cos(f*x + e)^2 - 4*((a^2 + a*b)*cos(f*x + e)^3 - a*b*cos(f*x + e))*sqrt(-b/a)
*sin(f*x + e) + b^2)/((a^2 - 2*a*b + b^2)*cos(f*x + e)^4 + 2*(a*b - b^2)*cos(f*x + e)^2 + b^2))*sin(f*x + e) -
12*(a - b)*cos(f*x + e))/((a^2*f*cos(f*x + e)^2 - a^2*f)*sin(f*x + e)), -1/6*(2*(2*a - 3*b)*cos(f*x + e)^3 -
3*((a - b)*cos(f*x + e)^2 - a + b)*sqrt(b/a)*arctan(1/2*((a + b)*cos(f*x + e)^2 - b)*sqrt(b/a)/(b*cos(f*x + e)
*sin(f*x + e)))*sin(f*x + e) - 6*(a - b)*cos(f*x + e))/((a^2*f*cos(f*x + e)^2 - a^2*f)*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),x)

[Out]

Timed out

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Giac [A]  time = 1.44278, size = 131, normalized size = 1.72 \begin{align*} -\frac{\frac{3 \,{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b}}\right )\right )}{\left (a b - b^{2}\right )}}{\sqrt{a b} a^{2}} + \frac{3 \, a \tan \left (f x + e\right )^{2} - 3 \, b \tan \left (f x + e\right )^{2} + a}{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),x, algorithm="giac")

[Out]

-1/3*(3*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))*(a*b - b^2)/(sqrt(a*b)*a^2) +
(3*a*tan(f*x + e)^2 - 3*b*tan(f*x + e)^2 + a)/(a^2*tan(f*x + e)^3))/f