\(\int \frac {\csc ^3(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx\) [429]

Optimal result

Integrand size = 21, antiderivative size = 93 \[ \int \frac {\csc ^3(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{4 b^{3/2} f}+\frac {\text {arctanh}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{4 b^{3/2} f}-\frac {\cot ^2(e+f x) (b \sec (e+f x))^{3/2}}{2 b^3 f} \]

[Out]

-1/4*arctan((b*sec(f*x+e))^(1/2)/b^(1/2))/b^(3/2)/f+1/4*arctanh((b*sec(f*x+e))^(1/2)/b^(1/2))/b^(3/2)/f-1/2*co
t(f*x+e)^2*(b*sec(f*x+e))^(3/2)/b^3/f

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2702, 296, 335, 304, 209, 212} \[ \int \frac {\csc ^3(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{4 b^{3/2} f}+\frac {\text {arctanh}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{4 b^{3/2} f}-\frac {\cot ^2(e+f x) (b \sec (e+f x))^{3/2}}{2 b^3 f} \]

[In]

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

[Out]

-1/4*ArcTan[Sqrt[b*Sec[e + f*x]]/Sqrt[b]]/(b^(3/2)*f) + ArcTanh[Sqrt[b*Sec[e + f*x]]/Sqrt[b]]/(4*b^(3/2)*f) -
(Cot[e + f*x]^2*(b*Sec[e + f*x])^(3/2))/(2*b^3*f)

Rule 209

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

Rule 212

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

Rule 296

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2702

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\sqrt {x}}{\left (-1+\frac {x^2}{b^2}\right )^2} \, dx,x,b \sec (e+f x)\right )}{b^3 f} \\ & = -\frac {\cot ^2(e+f x) (b \sec (e+f x))^{3/2}}{2 b^3 f}-\frac {\text {Subst}\left (\int \frac {\sqrt {x}}{-1+\frac {x^2}{b^2}} \, dx,x,b \sec (e+f x)\right )}{4 b^3 f} \\ & = -\frac {\cot ^2(e+f x) (b \sec (e+f x))^{3/2}}{2 b^3 f}-\frac {\text {Subst}\left (\int \frac {x^2}{-1+\frac {x^4}{b^2}} \, dx,x,\sqrt {b \sec (e+f x)}\right )}{2 b^3 f} \\ & = -\frac {\cot ^2(e+f x) (b \sec (e+f x))^{3/2}}{2 b^3 f}+\frac {\text {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt {b \sec (e+f x)}\right )}{4 b f}-\frac {\text {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt {b \sec (e+f x)}\right )}{4 b f} \\ & = -\frac {\arctan \left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{4 b^{3/2} f}+\frac {\text {arctanh}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{4 b^{3/2} f}-\frac {\cot ^2(e+f x) (b \sec (e+f x))^{3/2}}{2 b^3 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.05 \[ \int \frac {\csc ^3(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {-4 \csc ^2(e+f x)-2 \arctan \left (\sqrt {\sec (e+f x)}\right ) \sqrt {\sec (e+f x)}+\left (-\log \left (1-\sqrt {\sec (e+f x)}\right )+\log \left (1+\sqrt {\sec (e+f x)}\right )\right ) \sqrt {\sec (e+f x)}}{8 b f \sqrt {b \sec (e+f x)}} \]

[In]

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

[Out]

(-4*Csc[e + f*x]^2 - 2*ArcTan[Sqrt[Sec[e + f*x]]]*Sqrt[Sec[e + f*x]] + (-Log[1 - Sqrt[Sec[e + f*x]]] + Log[1 +
Sqrt[Sec[e + f*x]]])*Sqrt[Sec[e + f*x]])/(8*b*f*Sqrt[b*Sec[e + f*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(280\) vs. \(2(73)=146\).

Time = 0.19 (sec) , antiderivative size = 281, normalized size of antiderivative = 3.02

method	result	size
default	\(\frac {\ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right ) \cos \left (f x +e \right )-\cos \left (f x +e \right ) \arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right )+\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )}{8 f \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sqrt {b \sec \left (f x +e \right )}\, b \left (\cos ^{2}\left (f x +e \right )-1\right )}\)	\(281\)

[In]

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

[Out]

1/8/f*(ln((2*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)
+1)/(cos(f*x+e)+1))*cos(f*x+e)-cos(f*x+e)*arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))+4*(-cos(f*x+e)/(cos
(f*x+e)+1)^2)^(1/2)-ln((2*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/
2)-cos(f*x+e)+1)/(cos(f*x+e)+1))+arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)))/(-cos(f*x+e)/(cos(f*x+e)+1)
^2)^(1/2)/(b*sec(f*x+e))^(1/2)/b/(cos(f*x+e)^2-1)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (73) = 146\).

Time = 0.37 (sec) , antiderivative size = 364, normalized size of antiderivative = 3.91 \[ \int \frac {\csc ^3(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\left [-\frac {2 \, {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} {\left (\cos \left (f x + e\right ) + 1\right )}}{2 \, b}\right ) + {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {-b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) - 8 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{16 \, {\left (b^{2} f \cos \left (f x + e\right )^{2} - b^{2} f\right )}}, \frac {2 \, {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {b} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}} {\left (\cos \left (f x + e\right ) - 1\right )}}{2 \, \sqrt {b}}\right ) + {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {b} \log \left (\frac {b \cos \left (f x + e\right )^{2} + 4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right ) + 8 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{16 \, {\left (b^{2} f \cos \left (f x + e\right )^{2} - b^{2} f\right )}}\right ] \]

[In]

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

[Out]

[-1/16*(2*(cos(f*x + e)^2 - 1)*sqrt(-b)*arctan(1/2*sqrt(-b)*sqrt(b/cos(f*x + e))*(cos(f*x + e) + 1)/b) + (cos(
f*x + e)^2 - 1)*sqrt(-b)*log((b*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 - cos(f*x + e))*sqrt(-b)*sqrt(b/cos(f*x + e
)) - 6*b*cos(f*x + e) + b)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)) - 8*sqrt(b/cos(f*x + e))*cos(f*x + e))/(b^2*
f*cos(f*x + e)^2 - b^2*f), 1/16*(2*(cos(f*x + e)^2 - 1)*sqrt(b)*arctan(1/2*sqrt(b/cos(f*x + e))*(cos(f*x + e)
- 1)/sqrt(b)) + (cos(f*x + e)^2 - 1)*sqrt(b)*log((b*cos(f*x + e)^2 + 4*(cos(f*x + e)^2 + cos(f*x + e))*sqrt(b)
*sqrt(b/cos(f*x + e)) + 6*b*cos(f*x + e) + b)/(cos(f*x + e)^2 - 2*cos(f*x + e) + 1)) + 8*sqrt(b/cos(f*x + e))*
cos(f*x + e))/(b^2*f*cos(f*x + e)^2 - b^2*f)]

Sympy [F]

\[ \int \frac {\csc ^3(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\int \frac {\csc ^{3}{\left (e + f x \right )}}{\left (b \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.14 \[ \int \frac {\csc ^3(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {b {\left (\frac {4 \, \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {3}{2}}}{b^{4} - \frac {b^{4}}{\cos \left (f x + e\right )^{2}}} - \frac {2 \, \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b}}\right )}{b^{\frac {5}{2}}} - \frac {\log \left (-\frac {\sqrt {b} - \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b} + \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right )}{b^{\frac {5}{2}}}\right )}}{8 \, f} \]

[In]

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

[Out]

1/8*b*(4*(b/cos(f*x + e))^(3/2)/(b^4 - b^4/cos(f*x + e)^2) - 2*arctan(sqrt(b/cos(f*x + e))/sqrt(b))/b^(5/2) -
log(-(sqrt(b) - sqrt(b/cos(f*x + e)))/(sqrt(b) + sqrt(b/cos(f*x + e))))/b^(5/2))/f

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00 \[ \int \frac {\csc ^3(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {\frac {2 \, \sqrt {b \cos \left (f x + e\right )}}{b^{2} \cos \left (f x + e\right )^{2} - b^{2}} - \frac {\arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {-b}}\right )}{\sqrt {-b} b} + \frac {\arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {b}}\right )}{b^{\frac {3}{2}}}}{4 \, f \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} \]

[In]

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

[Out]

1/4*(2*sqrt(b*cos(f*x + e))/(b^2*cos(f*x + e)^2 - b^2) - arctan(sqrt(b*cos(f*x + e))/sqrt(-b))/(sqrt(-b)*b) +
arctan(sqrt(b*cos(f*x + e))/sqrt(b))/b^(3/2))/(f*sgn(cos(f*x + e)))

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc ^3(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\int \frac {1}{{\sin \left (e+f\,x\right )}^3\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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