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

Optimal. Leaf size=84 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{a^{3/2} f}-\frac{b \sec (e+f x)}{a f (a-b) \sqrt{a+b \sec ^2(e+f x)-b}} \]

[Out]

-(ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]]/(a^(3/2)*f)) - (b*Sec[e + f*x])/(a*(a - b)*f*
Sqrt[a - b + b*Sec[e + f*x]^2])

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Rubi [A]  time = 0.0965029, antiderivative size = 84, 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 = {3664, 382, 377, 207} \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{a^{3/2} f}-\frac{b \sec (e+f x)}{a f (a-b) \sqrt{a+b \sec ^2(e+f x)-b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]]/(a^(3/2)*f)) - (b*Sec[e + f*x])/(a*(a - b)*f*
Sqrt[a - b + b*Sec[e + f*x]^2])

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]

Rule 382

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

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])

Rubi steps

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

Mathematica [B]  time = 4.8542, size = 333, normalized size = 3.96 \[ -\frac{\cos (e+f x) \sec ^6\left (\frac{1}{2} (e+f x)\right ) \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (\sqrt{2} \sqrt{a} b (\cos (e+f x)+1) \sqrt{\sec ^4\left (\frac{1}{2} (e+f x)\right ) ((a-b) \cos (2 (e+f x))+a+b)}+(a-b) ((a-b) \cos (2 (e+f x))+a+b) \tanh ^{-1}\left (\frac{a-(a-2 b) \tan ^2\left (\frac{1}{2} (e+f x)\right )}{\sqrt{a} \sqrt{a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac{1}{2} (e+f x)\right )}}\right )+(a-b) ((a-b) \cos (2 (e+f x))+a+b) \tanh ^{-1}\left (\frac{a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )+2 b}{\sqrt{a} \sqrt{a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac{1}{2} (e+f x)\right )}}\right )\right )}{2 a^{3/2} f (a-b) \left (\sec ^4\left (\frac{1}{2} (e+f x)\right ) ((a-b) \cos (2 (e+f x))+a+b)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[e + f*x]*Sec[(e + f*x)/2]^6*((a - b)*ArcTanh[(a - (a - 2*b)*Tan[(e + f*x)/2]^2)/(Sqrt[a]*Sqrt[4*b*Tan[(e
+ f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2])]*(a + b + (a - b)*Cos[2*(e + f*x)]) + (a - b)*ArcTanh[(2*b + a*
(-1 + Tan[(e + f*x)/2]^2))/(Sqrt[a]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2])]*(a + b + (a
- b)*Cos[2*(e + f*x)]) + Sqrt[2]*Sqrt[a]*b*(1 + Cos[e + f*x])*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[(e
+ f*x)/2]^4])*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(2*a^(3/2)*(a - b)*f*((a + b + (a - b)*
Cos[2*(e + f*x)])*Sec[(e + f*x)/2]^4)^(3/2))

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Maple [B]  time = 0.211, size = 3491, normalized size = 41.6 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/f/a^(5/2)/(a-b)*(ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+
1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/
2)+b)/sin(f*x+e)^2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)^3*a^3-2*ln(-2/a^(1/2
)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b
*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*((a*cos(f*x+e)
^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)^3*a^2*b+ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*c
os(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f
*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1
)^2)^(1/2)*cos(f*x+e)^3*a*b^2+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*ln(-4*(cos(f*x+e)*((a
*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+((a*cos(f*x+e)^2-cos
(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/(cos(f*x+e)-1))*cos(f*x+e)^3*a^3-2*((a*cos(f*x+e)^2-cos(f*x+
e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*ln(-4*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)
*a^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/(cos(
f*x+e)-1))*cos(f*x+e)^3*a^2*b+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*ln(-4*(cos(f*x+e)*((a
*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+((a*cos(f*x+e)^2-cos
(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/(cos(f*x+e)-1))*cos(f*x+e)^3*a*b^2+2*cos(f*x+e)^2*a^(5/2)*b+
ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-co
s(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*((
a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)^2*a^3-2*ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(
f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*
x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(
cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)^2*a^2*b+ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)
^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+
e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)^
2*a*b^2+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*ln(-4*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+
e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*
x+e)+1)^2)^(1/2)*a^(1/2)+b)/(cos(f*x+e)-1))*cos(f*x+e)^2*a^3-2*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+
1)^2)^(1/2)*ln(-4*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+cos(f*x+e)*a-
b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/(cos(f*x+e)-1))*cos(f*x+e)^
2*a^2*b+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*ln(-4*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+
e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*
x+e)+1)^2)^(1/2)*a^(1/2)+b)/(cos(f*x+e)-1))*cos(f*x+e)^2*a*b^2-2*cos(f*x+e)^2*a^(3/2)*b^2+ln(-2/a^(1/2)*(cos(f
*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x
+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*((a*cos(f*x+e)^2-cos(f
*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)*a^2*b-ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^
2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+
b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)
*cos(f*x+e)*a*b^2+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*ln(-4*(cos(f*x+e)*((a*cos(f*x+e)^
2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+
b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/(cos(f*x+e)-1))*cos(f*x+e)*a^2*b-((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos
(f*x+e)+1)^2)^(1/2)*ln(-4*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+cos(f
*x+e)*a-b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/(cos(f*x+e)-1))*cos
(f*x+e)*a*b^2+ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1
/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/si
n(f*x+e)^2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^2*b-ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos
(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f
*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/
(cos(f*x+e)+1)^2)^(1/2)*a*b^2+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*ln(-4*(cos(f*x+e)*((a
*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+((a*cos(f*x+e)^2-cos
(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/(cos(f*x+e)-1))*a^2*b-((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(co
s(f*x+e)+1)^2)^(1/2)*ln(-4*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+cos(
f*x+e)*a-b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/(cos(f*x+e)-1))*a*
b^2+2*b^2*a^(3/2))/cos(f*x+e)^3/((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 3.41413, size = 842, normalized size = 10.02 \begin{align*} \left [-\frac{2 \, a b \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) -{\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a b - b^{2}\right )} \sqrt{a} \log \left (-\frac{2 \,{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt{a} \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right )}{2 \,{\left ({\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{3} b - a^{2} b^{2}\right )} f\right )}}, -\frac{a b \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) -{\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a b - b^{2}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a}\right )}{{\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{3} b - a^{2} b^{2}\right )} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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