3.59 \(\int \frac {\csc ^3(e+f x)}{a+b \tan ^2(e+f x)} \, dx\)

Optimal. Leaf size=89 \[ -\frac {(a-2 b) \tanh ^{-1}(\cos (e+f x))}{2 a^2 f}-\frac {\sqrt {b} \sqrt {a-b} \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{a^2 f}-\frac {\cot (e+f x) \csc (e+f x)}{2 a f} \]

[Out]

-1/2*(a-2*b)*arctanh(cos(f*x+e))/a^2/f-1/2*cot(f*x+e)*csc(f*x+e)/a/f-arctan(sec(f*x+e)*b^(1/2)/(a-b)^(1/2))*(a
-b)^(1/2)*b^(1/2)/a^2/f

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3664, 471, 522, 207, 205} \[ -\frac {(a-2 b) \tanh ^{-1}(\cos (e+f x))}{2 a^2 f}-\frac {\sqrt {b} \sqrt {a-b} \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{a^2 f}-\frac {\cot (e+f x) \csc (e+f x)}{2 a f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^3/(a + b*Tan[e + f*x]^2),x]

[Out]

-((Sqrt[a - b]*Sqrt[b]*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(a^2*f)) - ((a - 2*b)*ArcTanh[Cos[e + f*x]]
)/(2*a^2*f) - (Cot[e + f*x]*Csc[e + f*x])/(2*a*f)

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 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 471

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

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 3664

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a - b + b*ff^2*x^2)^p)/x^(
m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

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

________________________________________________________________________________________

Mathematica [B]  time = 0.65, size = 195, normalized size = 2.19 \[ \frac {8 \sqrt {b} \sqrt {a-b} \tan ^{-1}\left (\frac {\sqrt {a-b}-\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )+8 \sqrt {b} \sqrt {a-b} \tan ^{-1}\left (\frac {\sqrt {a-b}+\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )-a \csc ^2\left (\frac {1}{2} (e+f x)\right )+a \sec ^2\left (\frac {1}{2} (e+f x)\right )+4 a \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )-4 a \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )-8 b \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )+8 b \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )}{8 a^2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^3/(a + b*Tan[e + f*x]^2),x]

[Out]

(8*Sqrt[a - b]*Sqrt[b]*ArcTan[(Sqrt[a - b] - Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]] + 8*Sqrt[a - b]*Sqrt[b]*ArcTan
[(Sqrt[a - b] + Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]] - a*Csc[(e + f*x)/2]^2 - 4*a*Log[Cos[(e + f*x)/2]] + 8*b*Lo
g[Cos[(e + f*x)/2]] + 4*a*Log[Sin[(e + f*x)/2]] - 8*b*Log[Sin[(e + f*x)/2]] + a*Sec[(e + f*x)/2]^2)/(8*a^2*f)

________________________________________________________________________________________

fricas [A]  time = 0.58, size = 327, normalized size = 3.67 \[ \left [\frac {2 \, \sqrt {-a b + b^{2}} {\left (\cos \left (f x + e\right )^{2} - 1\right )} \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-a b + b^{2}} \cos \left (f x + e\right ) - b}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right ) + 2 \, a \cos \left (f x + e\right ) - {\left ({\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{2} - a + 2 \, b\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + {\left ({\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{2} - a + 2 \, b\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{4 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f\right )}}, \frac {4 \, \sqrt {a b - b^{2}} {\left (\cos \left (f x + e\right )^{2} - 1\right )} \arctan \left (\frac {\sqrt {a b - b^{2}} \cos \left (f x + e\right )}{b}\right ) + 2 \, a \cos \left (f x + e\right ) - {\left ({\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{2} - a + 2 \, b\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + {\left ({\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{2} - a + 2 \, b\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{4 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3/(a+b*tan(f*x+e)^2),x, algorithm="fricas")

[Out]

[1/4*(2*sqrt(-a*b + b^2)*(cos(f*x + e)^2 - 1)*log(-((a - b)*cos(f*x + e)^2 + 2*sqrt(-a*b + b^2)*cos(f*x + e) -
b)/((a - b)*cos(f*x + e)^2 + b)) + 2*a*cos(f*x + e) - ((a - 2*b)*cos(f*x + e)^2 - a + 2*b)*log(1/2*cos(f*x +
e) + 1/2) + ((a - 2*b)*cos(f*x + e)^2 - a + 2*b)*log(-1/2*cos(f*x + e) + 1/2))/(a^2*f*cos(f*x + e)^2 - a^2*f),
1/4*(4*sqrt(a*b - b^2)*(cos(f*x + e)^2 - 1)*arctan(sqrt(a*b - b^2)*cos(f*x + e)/b) + 2*a*cos(f*x + e) - ((a -
2*b)*cos(f*x + e)^2 - a + 2*b)*log(1/2*cos(f*x + e) + 1/2) + ((a - 2*b)*cos(f*x + e)^2 - a + 2*b)*log(-1/2*co
s(f*x + e) + 1/2))/(a^2*f*cos(f*x + e)^2 - a^2*f)]

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3/(a+b*tan(f*x+e)^2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)2/f*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*1/16/a+(-2*(1-cos
(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a+4*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b-a)*1/16/a^2/(1-cos(f*x+exp(1))
)*(1+cos(f*x+exp(1)))+(-2*a*b+2*b^2)*1/4/a^2/sqrt(-b^2+a*b)*atan((-a*cos(f*x+exp(1))+b*cos(f*x+exp(1))+b)/(sqr
t(-b^2+a*b)*cos(f*x+exp(1))+sqrt(-b^2+a*b)))+(a-2*b)*1/8/a^2*ln(abs(1-cos(f*x+exp(1)))/abs(1+cos(f*x+exp(1))))
)

________________________________________________________________________________________

maple [B]  time = 0.69, size = 189, normalized size = 2.12 \[ \frac {b \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right )}{f a \sqrt {\left (a -b \right ) b}}-\frac {b^{2} \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right )}{f \,a^{2} \sqrt {\left (a -b \right ) b}}+\frac {1}{4 f a \left (-1+\cos \left (f x +e \right )\right )}+\frac {\ln \left (-1+\cos \left (f x +e \right )\right )}{4 f a}-\frac {\ln \left (-1+\cos \left (f x +e \right )\right ) b}{2 f \,a^{2}}+\frac {1}{4 f a \left (1+\cos \left (f x +e \right )\right )}-\frac {\ln \left (1+\cos \left (f x +e \right )\right )}{4 f a}+\frac {\ln \left (1+\cos \left (f x +e \right )\right ) b}{2 f \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^3/(a+b*tan(f*x+e)^2),x)

[Out]

1/f/a*b/((a-b)*b)^(1/2)*arctan((a-b)*cos(f*x+e)/((a-b)*b)^(1/2))-1/f*b^2/a^2/((a-b)*b)^(1/2)*arctan((a-b)*cos(
f*x+e)/((a-b)*b)^(1/2))+1/4/f/a/(-1+cos(f*x+e))+1/4/f/a*ln(-1+cos(f*x+e))-1/2/f/a^2*ln(-1+cos(f*x+e))*b+1/4/f/
a/(1+cos(f*x+e))-1/4/f/a*ln(1+cos(f*x+e))+1/2/f/a^2*ln(1+cos(f*x+e))*b

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3/(a+b*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is b-a positive or negative?

________________________________________________________________________________________

mupad [B]  time = 13.39, size = 591, normalized size = 6.64 \[ -\frac {a\,\left ({\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )+2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )-\frac {1}{2}\right )+4\,\mathrm {atan}\left (\frac {2\,a^4\,b\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-a^4\,b-10\,a^3\,b^2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+4\,a^3\,b^2+20\,a^2\,b^3\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-6\,a^2\,b^3-18\,a\,b^4\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+3\,a\,b^4+6\,b^5\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{6\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\left (a\,b-b^2\right )}^{5/2}-2\,a^2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\left (a\,b-b^2\right )}^{3/2}}\right )\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\sqrt {a\,b-b^2}-4\,\mathrm {atan}\left (\frac {2\,a^4\,b\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-a^4\,b-10\,a^3\,b^2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+4\,a^3\,b^2+20\,a^2\,b^3\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-6\,a^2\,b^3-18\,a\,b^4\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+3\,a\,b^4+6\,b^5\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{6\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\left (a\,b-b^2\right )}^{5/2}-2\,a^2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\left (a\,b-b^2\right )}^{3/2}}\right )\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\sqrt {a\,b-b^2}+4\,b\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )-4\,b\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{4\,a^2\,f\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-4\,a^2\,f\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(e + f*x)^3*(a + b*tan(e + f*x)^2)),x)

[Out]

-(a*(cos(e/2 + (f*x)/2)^2 - 2*cos(e/2 + (f*x)/2)^2*log(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)) + 2*cos(e/2 + (f
*x)/2)^4*log(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)) - 1/2) + 4*atan((6*b^5*cos(e/2 + (f*x)/2)^2 + 3*a*b^4 - a^
4*b - 6*a^2*b^3 + 4*a^3*b^2 + 20*a^2*b^3*cos(e/2 + (f*x)/2)^2 - 10*a^3*b^2*cos(e/2 + (f*x)/2)^2 - 18*a*b^4*cos
(e/2 + (f*x)/2)^2 + 2*a^4*b*cos(e/2 + (f*x)/2)^2)/(6*cos(e/2 + (f*x)/2)^2*(a*b - b^2)^(5/2) - 2*a^2*cos(e/2 +
(f*x)/2)^2*(a*b - b^2)^(3/2)))*cos(e/2 + (f*x)/2)^2*(a*b - b^2)^(1/2) - 4*atan((6*b^5*cos(e/2 + (f*x)/2)^2 + 3
*a*b^4 - a^4*b - 6*a^2*b^3 + 4*a^3*b^2 + 20*a^2*b^3*cos(e/2 + (f*x)/2)^2 - 10*a^3*b^2*cos(e/2 + (f*x)/2)^2 - 1
8*a*b^4*cos(e/2 + (f*x)/2)^2 + 2*a^4*b*cos(e/2 + (f*x)/2)^2)/(6*cos(e/2 + (f*x)/2)^2*(a*b - b^2)^(5/2) - 2*a^2
*cos(e/2 + (f*x)/2)^2*(a*b - b^2)^(3/2)))*cos(e/2 + (f*x)/2)^4*(a*b - b^2)^(1/2) + 4*b*cos(e/2 + (f*x)/2)^2*lo
g(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)) - 4*b*cos(e/2 + (f*x)/2)^4*log(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2))
)/(4*a^2*f*cos(e/2 + (f*x)/2)^2 - 4*a^2*f*cos(e/2 + (f*x)/2)^4)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{3}{\left (e + f x \right )}}{a + b \tan ^{2}{\left (e + f x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**3/(a+b*tan(f*x+e)**2),x)

[Out]

Integral(csc(e + f*x)**3/(a + b*tan(e + f*x)**2), x)

________________________________________________________________________________________