3.1.67 \(\int \frac {\csc ^6(e+f x)}{a+b \tan ^2(e+f x)} \, dx\) [67]

Optimal. Leaf size=105 \[ -\frac {(a-b)^2 \sqrt {b} \text {ArcTan}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{a^{7/2} f}-\frac {(a-b)^2 \cot (e+f x)}{a^3 f}-\frac {(2 a-b) \cot ^3(e+f x)}{3 a^2 f}-\frac {\cot ^5(e+f x)}{5 a f} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3744, 472, 211} \begin {gather*} -\frac {\sqrt {b} (a-b)^2 \text {ArcTan}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{a^{7/2} f}-\frac {(a-b)^2 \cot (e+f x)}{a^3 f}-\frac {(2 a-b) \cot ^3(e+f x)}{3 a^2 f}-\frac {\cot ^5(e+f x)}{5 a f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^6/(a + b*Tan[e + f*x]^2),x]

[Out]

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

Rule 211

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

Rule 472

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[(e*x)^m*((a + b*x^n)^p/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

Rule 3744

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 ^6(e+f x)}{a+b \tan ^2(e+f x)} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^2}{x^6 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\text {Subst}\left (\int \left (\frac {1}{a x^6}+\frac {2 a-b}{a^2 x^4}+\frac {(a-b)^2}{a^3 x^2}-\frac {(a-b)^2 b}{a^3 \left (a+b x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(a-b)^2 \cot (e+f x)}{a^3 f}-\frac {(2 a-b) \cot ^3(e+f x)}{3 a^2 f}-\frac {\cot ^5(e+f x)}{5 a f}-\frac {\left ((a-b)^2 b\right ) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{a^3 f}\\ &=-\frac {(a-b)^2 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{a^{7/2} f}-\frac {(a-b)^2 \cot (e+f x)}{a^3 f}-\frac {(2 a-b) \cot ^3(e+f x)}{3 a^2 f}-\frac {\cot ^5(e+f x)}{5 a f}\\ \end {align*}

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Mathematica [A]
time = 0.55, size = 103, normalized size = 0.98 \begin {gather*} \frac {-15 (a-b)^2 \sqrt {b} \text {ArcTan}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )-\sqrt {a} \cot (e+f x) \left (8 a^2-25 a b+15 b^2+a (4 a-5 b) \csc ^2(e+f x)+3 a^2 \csc ^4(e+f x)\right )}{15 a^{7/2} f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^6/(a + b*Tan[e + f*x]^2),x]

[Out]

(-15*(a - b)^2*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]] - Sqrt[a]*Cot[e + f*x]*(8*a^2 - 25*a*b + 15*b^2
+ a*(4*a - 5*b)*Csc[e + f*x]^2 + 3*a^2*Csc[e + f*x]^4))/(15*a^(7/2)*f)

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Maple [A]
time = 0.30, size = 99, normalized size = 0.94

method	result	size
derivativedivides	\(\frac {-\frac {1}{5 a \tan \left (f x +e \right )^{5}}-\frac {2 a -b}{3 a^{2} \tan \left (f x +e \right )^{3}}-\frac {a^{2}-2 a b +b^{2}}{a^{3} \tan \left (f x +e \right )}-\frac {b \left (a^{2}-2 a b +b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{a^{3} \sqrt {a b}}}{f}\)	\(99\)
default	\(\frac {-\frac {1}{5 a \tan \left (f x +e \right )^{5}}-\frac {2 a -b}{3 a^{2} \tan \left (f x +e \right )^{3}}-\frac {a^{2}-2 a b +b^{2}}{a^{3} \tan \left (f x +e \right )}-\frac {b \left (a^{2}-2 a b +b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{a^{3} \sqrt {a b}}}{f}\)	\(99\)
risch	\(\frac {2 i \left (15 a b \,{\mathrm e}^{8 i \left (f x +e \right )}-15 b^{2} {\mathrm e}^{8 i \left (f x +e \right )}-90 a b \,{\mathrm e}^{6 i \left (f x +e \right )}+60 b^{2} {\mathrm e}^{6 i \left (f x +e \right )}-80 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}+160 a b \,{\mathrm e}^{4 i \left (f x +e \right )}-90 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+40 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}-110 a b \,{\mathrm e}^{2 i \left (f x +e \right )}+60 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}-8 a^{2}+25 a b -15 b^{2}\right )}{15 f \,a^{3} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{5}}-\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right )}{2 a^{2} f}+\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right ) b}{a^{3} f}-\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right ) b^{2}}{2 a^{4} f}+\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right )}{2 a^{2} f}-\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right ) b}{a^{3} f}+\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right ) b^{2}}{2 a^{4} f}\)	\(467\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^6/(a+b*tan(f*x+e)^2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.50, size = 108, normalized size = 1.03 \begin {gather*} -\frac {\frac {15 \, {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} + \frac {15 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{4} + 5 \, {\left (2 \, a^{2} - a b\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2}}{a^{3} \tan \left (f x + e\right )^{5}}}{15 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

-1/15*(15*(a^2*b - 2*a*b^2 + b^3)*arctan(b*tan(f*x + e)/sqrt(a*b))/(sqrt(a*b)*a^3) + (15*(a^2 - 2*a*b + b^2)*t
an(f*x + e)^4 + 5*(2*a^2 - a*b)*tan(f*x + e)^2 + 3*a^2)/(a^3*tan(f*x + e)^5))/f

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (97) = 194\).
time = 4.45, size = 571, normalized size = 5.44 \begin {gather*} \left [-\frac {4 \, {\left (8 \, a^{2} - 25 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - 20 \, {\left (4 \, a^{2} - 11 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} - 2 \, a b + 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 ) + 60 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )}{60 \, {\left (a^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f\right )} \sin \left (f x + e\right )}, -\frac {2 \, {\left (8 \, a^{2} - 25 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - 10 \, {\left (4 \, a^{2} - 11 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} - 2 \, a b + 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 ) + 30 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )}{30 \, {\left (a^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f\right )} \sin \left (f x + e\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*tan(f*x+e)^2),x, algorithm="fricas")

[Out]

[-1/60*(4*(8*a^2 - 25*a*b + 15*b^2)*cos(f*x + e)^5 - 20*(4*a^2 - 11*a*b + 6*b^2)*cos(f*x + e)^3 - 15*((a^2 - 2
*a*b + b^2)*cos(f*x + e)^4 - 2*(a^2 - 2*a*b + b^2)*cos(f*x + e)^2 + a^2 - 2*a*b + 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))*s
in(f*x + e) + 60*(a^2 - 2*a*b + b^2)*cos(f*x + e))/((a^3*f*cos(f*x + e)^4 - 2*a^3*f*cos(f*x + e)^2 + a^3*f)*si
n(f*x + e)), -1/30*(2*(8*a^2 - 25*a*b + 15*b^2)*cos(f*x + e)^5 - 10*(4*a^2 - 11*a*b + 6*b^2)*cos(f*x + e)^3 -
15*((a^2 - 2*a*b + b^2)*cos(f*x + e)^4 - 2*(a^2 - 2*a*b + b^2)*cos(f*x + e)^2 + a^2 - 2*a*b + b^2)*sqrt(b/a)*a
rctan(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) + 30*(a^2 - 2*a*b
+ b^2)*cos(f*x + e))/((a^3*f*cos(f*x + e)^4 - 2*a^3*f*cos(f*x + e)^2 + a^3*f)*sin(f*x + e))]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**6/(a+b*tan(f*x+e)**2),x)

[Out]

Timed out

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Giac [A]
time = 0.72, size = 151, normalized size = 1.44 \begin {gather*} -\frac {\frac {15 \, {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} {\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 )}}{\sqrt {a b} a^{3}} + \frac {15 \, a^{2} \tan \left (f x + e\right )^{4} - 30 \, a b \tan \left (f x + e\right )^{4} + 15 \, b^{2} \tan \left (f x + e\right )^{4} + 10 \, a^{2} \tan \left (f x + e\right )^{2} - 5 \, a b \tan \left (f x + e\right )^{2} + 3 \, a^{2}}{a^{3} \tan \left (f x + e\right )^{5}}}{15 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*tan(f*x+e)^2),x, algorithm="giac")

[Out]

-1/15*(15*(a^2*b - 2*a*b^2 + b^3)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))/(sq
rt(a*b)*a^3) + (15*a^2*tan(f*x + e)^4 - 30*a*b*tan(f*x + e)^4 + 15*b^2*tan(f*x + e)^4 + 10*a^2*tan(f*x + e)^2
- 5*a*b*tan(f*x + e)^2 + 3*a^2)/(a^3*tan(f*x + e)^5))/f

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Mupad [B]
time = 11.39, size = 115, normalized size = 1.10 \begin {gather*} -\frac {\frac {1}{5\,a}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,a-b\right )}{3\,a^2}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (a^2-2\,a\,b+b^2\right )}{a^3}}{f\,{\mathrm {tan}\left (e+f\,x\right )}^5}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )\,{\left (a-b\right )}^2}{\sqrt {a}\,\left (a^2-2\,a\,b+b^2\right )}\right )\,{\left (a-b\right )}^2}{a^{7/2}\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(e + f*x)^6*(a + b*tan(e + f*x)^2)),x)

[Out]

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

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