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

Optimal. Leaf size=78 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{b^{3/2} f}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{b^{3/2} f}+\frac {2}{b f \sqrt {b \sec (e+f x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2622, 325, 329, 298, 203, 206} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{b^{3/2} f}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{b^{3/2} f}+\frac {2}{b f \sqrt {b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 203

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

Rule 206

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

Rule 298

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 325

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

Rule 329

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 2622

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*} \int \frac {\csc (e+f x)}{(b \sec (e+f x))^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^{3/2} \left (-1+\frac {x^2}{b^2}\right )} \, dx,x,b \sec (e+f x)\right )}{b f}\\ &=\frac {2}{b f \sqrt {b \sec (e+f x)}}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {x}}{-1+\frac {x^2}{b^2}} \, dx,x,b \sec (e+f x)\right )}{b^3 f}\\ &=\frac {2}{b f \sqrt {b \sec (e+f x)}}+\frac {2 \operatorname {Subst}\left (\int \frac {x^2}{-1+\frac {x^4}{b^2}} \, dx,x,\sqrt {b \sec (e+f x)}\right )}{b^3 f}\\ &=\frac {2}{b f \sqrt {b \sec (e+f x)}}-\frac {\operatorname {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt {b \sec (e+f x)}\right )}{b f}+\frac {\operatorname {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt {b \sec (e+f x)}\right )}{b f}\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{b^{3/2} f}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{b^{3/2} f}+\frac {2}{b f \sqrt {b \sec (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 2.00, size = 89, normalized size = 1.14 \[ \frac {\sqrt {\sec (e+f x)} \left (\log \left (1-\sqrt {\sec (e+f x)}\right )-\log \left (\sqrt {\sec (e+f x)}+1\right )\right )+2 \sqrt {\sec (e+f x)} \tan ^{-1}\left (\sqrt {\sec (e+f x)}\right )+4}{2 b f \sqrt {b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4 + 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]])/(2*b*f*Sqrt[b*Sec[e + f*x]])

________________________________________________________________________________________

fricas [B]  time = 1.02, size = 314, normalized size = 4.03 \[ \left [\frac {2 \, \sqrt {-b} \arctan \left (\frac {2 \, \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{b \cos \left (f x + e\right ) + b}\right ) + 8 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) - \sqrt {-b} \log \left (-\frac {b \cos \left (f x + e\right )^{2} + 4 \, {\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right )}{4 \, b^{2} f}, \frac {2 \, \sqrt {b} \arctan \left (\frac {2 \, \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{b \cos \left (f x + e\right ) - b}\right ) + 8 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) + \sqrt {b} \log \left (-\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right )}{4 \, b^{2} f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*sqrt(-b)*arctan(2*sqrt(-b)*sqrt(b/cos(f*x + e))*cos(f*x + e)/(b*cos(f*x + e) + b)) + 8*sqrt(b/cos(f*x
+ e))*cos(f*x + e) - 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)))/(b^2*f), 1/4*(2*sqrt(b)*arctan(2*sqrt
(b)*sqrt(b/cos(f*x + e))*cos(f*x + e)/(b*cos(f*x + e) - b)) + 8*sqrt(b/cos(f*x + e))*cos(f*x + e) + sqrt(b)*lo
g(-(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)))/(b^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)/(b*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)2/b/f*2/(sqrt(-b)*tan((f*x+exp(1))/2)^2-sqrt(-b*tan((f*x+exp(1))/2)^4+b)+sqrt(-b))/sign(tan((f*x+exp(1))/2
)^2-1)

________________________________________________________________________________________

maple [B]  time = 0.17, size = 221, normalized size = 2.83 \[ -\frac {\left (-1+\cos \left (f x +e \right )\right ) \left (4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\ln \left (-\frac {2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1}{\sin \left (f x +e \right )^{2}}\right )+\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\right )}{2 f \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} \left (\frac {b}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.62, size = 89, normalized size = 1.14 \[ \frac {b {\left (\frac {2 \, \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b}}\right )}{b^{\frac {5}{2}}} + \frac {\log \left (-\frac {\sqrt {b} - \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b} + \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right )}{b^{\frac {5}{2}}} + \frac {4}{b^{2} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right )}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*(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) + 4/(b^2*sqrt(b/cos(f*x + e))))/f

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sin \left (e+f\,x\right )\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________