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

Optimal. Leaf size=116 \[ -\frac{\sqrt{b} (3 a-5 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{2 a^{7/2} f}-\frac{b (a-b) \tan (e+f x)}{2 a^3 f \left (a+b \tan ^2(e+f x)\right )}-\frac{(a-2 b) \cot (e+f x)}{a^3 f}-\frac{\cot ^3(e+f x)}{3 a^2 f} \]

[Out]

-((3*a - 5*b)*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(2*a^(7/2)*f) - ((a - 2*b)*Cot[e + f*x])/(a^3*f)
- Cot[e + f*x]^3/(3*a^2*f) - ((a - b)*b*Tan[e + f*x])/(2*a^3*f*(a + b*Tan[e + f*x]^2))

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Rubi [A]  time = 0.145864, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3663, 456, 1261, 205} \[ -\frac{\sqrt{b} (3 a-5 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{2 a^{7/2} f}-\frac{b (a-b) \tan (e+f x)}{2 a^3 f \left (a+b \tan ^2(e+f x)\right )}-\frac{(a-2 b) \cot (e+f x)}{a^3 f}-\frac{\cot ^3(e+f x)}{3 a^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*a - 5*b)*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(2*a^(7/2)*f) - ((a - 2*b)*Cot[e + f*x])/(a^3*f)
- Cot[e + f*x]^3/(3*a^2*f) - ((a - b)*b*Tan[e + f*x])/(2*a^3*f*(a + b*Tan[e + f*x]^2))

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 456

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

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)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1+x^2}{x^4 \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{(a-b) b \tan (e+f x)}{2 a^3 f \left (a+b \tan ^2(e+f x)\right )}-\frac{b \operatorname{Subst}\left (\int \frac{-\frac{2}{a b}-\frac{2 (a-b) x^2}{a^2 b}+\frac{(a-b) x^4}{a^3}}{x^4 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=-\frac{(a-b) b \tan (e+f x)}{2 a^3 f \left (a+b \tan ^2(e+f x)\right )}-\frac{b \operatorname{Subst}\left (\int \left (-\frac{2}{a^2 b x^4}-\frac{2 (a-2 b)}{a^3 b x^2}+\frac{3 a-5 b}{a^3 \left (a+b x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=-\frac{(a-2 b) \cot (e+f x)}{a^3 f}-\frac{\cot ^3(e+f x)}{3 a^2 f}-\frac{(a-b) b \tan (e+f x)}{2 a^3 f \left (a+b \tan ^2(e+f x)\right )}-\frac{((3 a-5 b) b) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{2 a^3 f}\\ &=-\frac{(3 a-5 b) \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{2 a^{7/2} f}-\frac{(a-2 b) \cot (e+f x)}{a^3 f}-\frac{\cot ^3(e+f x)}{3 a^2 f}-\frac{(a-b) b \tan (e+f x)}{2 a^3 f \left (a+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.762804, size = 112, normalized size = 0.97 \[ \frac{3 \sqrt{b} (5 b-3 a) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )+\sqrt{a} \left (\frac{3 b (b-a) \sin (2 (e+f x))}{(a-b) \cos (2 (e+f x))+a+b}-2 \cot (e+f x) \left (a \csc ^2(e+f x)+2 a-6 b\right )\right )}{6 a^{7/2} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sqrt[b]*(-3*a + 5*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]] + Sqrt[a]*(-2*Cot[e + f*x]*(2*a - 6*b + a*Csc[e
+ f*x]^2) + (3*b*(-a + b)*Sin[2*(e + f*x)])/(a + b + (a - b)*Cos[2*(e + f*x)])))/(6*a^(7/2)*f)

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Maple [A]  time = 0.095, size = 169, normalized size = 1.5 \begin{align*} -{\frac{1}{3\,f{a}^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}}-{\frac{1}{f{a}^{2}\tan \left ( fx+e \right ) }}+2\,{\frac{b}{f{a}^{3}\tan \left ( fx+e \right ) }}-{\frac{b\tan \left ( fx+e \right ) }{2\,f{a}^{2} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{{b}^{2}\tan \left ( fx+e \right ) }{2\,f{a}^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{3\,b}{2\,f{a}^{2}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{5\,{b}^{2}}{2\,f{a}^{3}}\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)^2,x)

[Out]

-1/3/f/a^2/tan(f*x+e)^3-1/f/a^2/tan(f*x+e)+2/f/a^3/tan(f*x+e)*b-1/2/f/a^2*b*tan(f*x+e)/(a+b*tan(f*x+e)^2)+1/2/
f/a^3*b^2*tan(f*x+e)/(a+b*tan(f*x+e)^2)-3/2/f/a^2*b/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2))+5/2/f/a^3*b^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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.20976, size = 1374, normalized size = 11.84 \begin{align*} \text{result too large to display} \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]

[-1/24*(4*(4*a^2 - 19*a*b + 15*b^2)*cos(f*x + e)^5 - 8*(3*a^2 - 14*a*b + 15*b^2)*cos(f*x + e)^3 + 3*((3*a^2 -
8*a*b + 5*b^2)*cos(f*x + e)^4 - (3*a^2 - 11*a*b + 10*b^2)*cos(f*x + e)^2 - 3*a*b + 5*b^2)*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*(3*a*b - 5*b^2)*cos(f*x + e))/(((a^4 - a^3*b)*f*cos(f*x + e)^4 - a^3*b*f - (a^4 - 2*a^3*b)
*f*cos(f*x + e)^2)*sin(f*x + e)), -1/12*(2*(4*a^2 - 19*a*b + 15*b^2)*cos(f*x + e)^5 - 4*(3*a^2 - 14*a*b + 15*b
^2)*cos(f*x + e)^3 - 3*((3*a^2 - 8*a*b + 5*b^2)*cos(f*x + e)^4 - (3*a^2 - 11*a*b + 10*b^2)*cos(f*x + e)^2 - 3*
a*b + 5*b^2)*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*(3*a*b - 5*b^2)*cos(f*x + e))/(((a^4 - a^3*b)*f*cos(f*x + e)^4 - a^3*b*f - (a^4 - 2*a^3*b)*f*cos(f*
x + e)^2)*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 = 1.38018, size = 192, normalized size = 1.66 \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 (3 \, a b - 5 \, b^{2}\right )}}{\sqrt{a b} a^{3}} + \frac{3 \,{\left (a b \tan \left (f x + e\right ) - b^{2} \tan \left (f x + e\right )\right )}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )} a^{3}} + \frac{2 \,{\left (3 \, a \tan \left (f x + e\right )^{2} - 6 \, b \tan \left (f x + e\right )^{2} + a\right )}}{a^{3} \tan \left (f x + e\right )^{3}}}{6 \, 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]

-1/6*(3*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))*(3*a*b - 5*b^2)/(sqrt(a*b)*a^
3) + 3*(a*b*tan(f*x + e) - b^2*tan(f*x + e))/((b*tan(f*x + e)^2 + a)*a^3) + 2*(3*a*tan(f*x + e)^2 - 6*b*tan(f*
x + e)^2 + a)/(a^3*tan(f*x + e)^3))/f