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

Optimal. Leaf size=80 \[ -\frac{b \sec (e+f x)}{a f (a+b) \sqrt{a+b \sec ^2(e+f x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{f (a+b)^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0921376, antiderivative size = 80, 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 = {4134, 382, 377, 207} \[ -\frac{b \sec (e+f x)}{a f (a+b) \sqrt{a+b \sec ^2(e+f x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{f (a+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4134

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Cos[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^
n)^p)/x^(m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (Gt
Q[m, 0] || EqQ[n, 2] || EqQ[n, 4])

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

Mathematica [A]  time = 0.837625, size = 113, normalized size = 1.41 \[ -\frac{\sec ^3(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (a \sqrt{-a \sin ^2(e+f x)+a+b} \tanh ^{-1}\left (\frac{\sqrt{-a \sin ^2(e+f x)+a+b}}{\sqrt{a+b}}\right )+b \sqrt{a+b}\right )}{2 a f (a+b)^{3/2} \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.355, size = 1094, normalized size = 13.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/f/(a+b)^(5/2)/a*(b+a*cos(f*x+e)^2)*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-2/(a+b)^(1
/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos
(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)*a^2+cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+
e))^2)^(1/2)*ln(-2/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(
1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)*a*b+cos(f*x+e)*((b+
a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1
/2)+a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/(-1+cos(f*x+e)))*a^2+cos(f*x+e)*((
b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^
(1/2)+a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/(-1+cos(f*x+e)))*a*b+((b+a*cos(f
*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-2/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+
e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)
*a^2+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-2/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e
)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+
b)/sin(f*x+e)^2)*a*b+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(
f*x+e))^2)^(1/2)*(a+b)^(1/2)+a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/(-1+cos(f
*x+e)))*a^2+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2
)^(1/2)*(a+b)^(1/2)+a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/(-1+cos(f*x+e)))*a
*b+2*(a+b)^(3/2)*b)/cos(f*x+e)^3/((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(3/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (f x + e\right )}{{\left (b \sec \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*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 0.847261, size = 821, normalized size = 10.26 \begin{align*} \left [-\frac{2 \,{\left (a b + b^{2}\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) -{\left (a^{2} \cos \left (f x + e\right )^{2} + a b\right )} \sqrt{a + b} \log \left (\frac{2 \,{\left (a \cos \left (f x + e\right )^{2} - 2 \, \sqrt{a + b} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + 2 \, 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 + 2 \, a^{2} b^{2} + a b^{3}\right )} f\right )}}, \frac{{\left (a^{2} \cos \left (f x + e\right )^{2} + a b\right )} \sqrt{-a - b} \arctan \left (\frac{\sqrt{-a - b} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a + b}\right ) -{\left (a b + b^{2}\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\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 + 2 \, a^{2} b^{2} + a b^{3}\right )} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (f x + e\right )}{{\left (b \sec \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*sec(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

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