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]

-(a-2*b)*cot(f*x+e)/a^3/f-1/3*cot(f*x+e)^3/a^2/f-1/2*(3*a-5*b)*arctan(b^(1/2)*tan(f*x+e)/a^(1/2))*b^(1/2)/a^(7
/2)/f-1/2*(a-b)*b*tan(f*x+e)/a^3/f/(a+b*tan(f*x+e)^2)

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Rubi [A]  time = 0.15, antiderivative size = 116, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.99, 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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fricas [B]  time = 0.65, size = 587, normalized size = 5.06 \[ \left [-\frac {4 \, {\left (4 \, a^{2} - 19 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - 8 \, {\left (3 \, a^{2} - 14 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left ({\left (3 \, a^{2} - 8 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (3 \, a^{2} - 11 \, a b + 10 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 3 \, a b + 5 \, b^{2}\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 (3 \, a b - 5 \, b^{2}\right )} \cos \left (f x + e\right )}{24 \, {\left ({\left (a^{4} - a^{3} b\right )} f \cos \left (f x + e\right )^{4} - a^{3} b f - {\left (a^{4} - 2 \, a^{3} b\right )} f \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}, -\frac {2 \, {\left (4 \, a^{2} - 19 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - 4 \, {\left (3 \, a^{2} - 14 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left ({\left (3 \, a^{2} - 8 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (3 \, a^{2} - 11 \, a b + 10 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 3 \, a b + 5 \, b^{2}\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 (3 \, a b - 5 \, b^{2}\right )} \cos \left (f x + e\right )}{12 \, {\left ({\left (a^{4} - a^{3} b\right )} f \cos \left (f x + e\right )^{4} - a^{3} b f - {\left (a^{4} - 2 \, a^{3} b\right )} f \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ] \]

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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giac [A]  time = 3.34, size = 142, normalized size = 1.22 \[ -\frac {\frac {3 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )} {\left (3 \, a b - 5 \, b^{2}\right )}}{\sqrt {a b} a^{3}} + \frac {3 \, {\left (a b \tan \left (f x + e\right ) - b^{2} \tan \left (f x + e\right )\right )}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )} a^{3}} + \frac {2 \, {\left (3 \, a \tan \left (f x + e\right )^{2} - 6 \, b \tan \left (f x + e\right )^{2} + a\right )}}{a^{3} \tan \left (f x + e\right )^{3}}}{6 \, f} \]

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

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maple [A]  time = 0.62, size = 169, normalized size = 1.46 \[ -\frac {b \tan \left (f x +e \right )}{2 f \,a^{2} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}+\frac {b^{2} \tan \left (f x +e \right )}{2 f \,a^{3} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}-\frac {3 b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right )}{2 f \,a^{2} \sqrt {a b}}+\frac {5 b^{2} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right )}{2 f \,a^{3} \sqrt {a b}}-\frac {1}{3 f \,a^{2} \tan \left (f x +e \right )^{3}}-\frac {1}{f \,a^{2} \tan \left (f x +e \right )}+\frac {2 b}{f \,a^{3} \tan \left (f x +e \right )} \]

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]

-1/2/f*b/a^2*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*b/a^2/(a*b)^(1/2)
*arctan(tan(f*x+e)*b/(a*b)^(1/2))+5/2/f/a^3*b^2/(a*b)^(1/2)*arctan(tan(f*x+e)*b/(a*b)^(1/2))-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

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maxima [A]  time = 0.62, size = 115, normalized size = 0.99 \[ -\frac {\frac {3 \, {\left (3 \, a b - 5 \, b^{2}\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (3 \, a^{2} - 5 \, a b\right )} \tan \left (f x + e\right )^{2} + 2 \, a^{2}}{a^{3} b \tan \left (f x + e\right )^{5} + a^{4} \tan \left (f x + e\right )^{3}} + \frac {3 \, {\left (3 \, a b - 5 \, b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}}}{6 \, f} \]

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]

-1/6*((3*(3*a*b - 5*b^2)*tan(f*x + e)^4 + 2*(3*a^2 - 5*a*b)*tan(f*x + e)^2 + 2*a^2)/(a^3*b*tan(f*x + e)^5 + a^
4*tan(f*x + e)^3) + 3*(3*a*b - 5*b^2)*arctan(b*tan(f*x + e)/sqrt(a*b))/(sqrt(a*b)*a^3))/f

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mupad [B]  time = 11.37, size = 108, normalized size = 0.93 \[ -\frac {\frac {1}{3\,a}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (3\,a-5\,b\right )}{3\,a^2}+\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (3\,a-5\,b\right )}{2\,a^3}}{f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^5+a\,{\mathrm {tan}\left (e+f\,x\right )}^3\right )}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )}{\sqrt {a}}\right )\,\left (3\,a-5\,b\right )}{2\,a^{7/2}\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(e + f*x)^4*(a + b*tan(e + f*x)^2)^2),x)

[Out]

- (1/(3*a) + (tan(e + f*x)^2*(3*a - 5*b))/(3*a^2) + (b*tan(e + f*x)^4*(3*a - 5*b))/(2*a^3))/(f*(a*tan(e + f*x)
^3 + b*tan(e + f*x)^5)) - (b^(1/2)*atan((b^(1/2)*tan(e + f*x))/a^(1/2))*(3*a - 5*b))/(2*a^(7/2)*f)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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