3.441 \(\int \frac{\csc (e+f x)}{(b \sec (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=81 \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{b^{5/2} f}-\frac{\tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{b^{5/2} f}+\frac{2}{3 b f (b \sec (e+f x))^{3/2}} \]

[Out]

-(ArcTan[Sqrt[b*Sec[e + f*x]]/Sqrt[b]]/(b^(5/2)*f)) - ArcTanh[Sqrt[b*Sec[e + f*x]]/Sqrt[b]]/(b^(5/2)*f) + 2/(3
*b*f*(b*Sec[e + f*x])^(3/2))

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Rubi [A]  time = 0.055331, antiderivative size = 81, normalized size of antiderivative = 1., 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, 212, 206, 203} \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{b^{5/2} f}-\frac{\tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{b^{5/2} f}+\frac{2}{3 b f (b \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]/(b*Sec[e + f*x])^(5/2),x]

[Out]

-(ArcTan[Sqrt[b*Sec[e + f*x]]/Sqrt[b]]/(b^(5/2)*f)) - ArcTanh[Sqrt[b*Sec[e + f*x]]/Sqrt[b]]/(b^(5/2)*f) + 2/(3
*b*f*(b*Sec[e + f*x])^(3/2))

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])

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 212

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

Rubi steps

\begin{align*} \int \frac{\csc (e+f x)}{(b \sec (e+f x))^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^{5/2} \left (-1+\frac{x^2}{b^2}\right )} \, dx,x,b \sec (e+f x)\right )}{b f}\\ &=\frac{2}{3 b f (b \sec (e+f x))^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (-1+\frac{x^2}{b^2}\right )} \, dx,x,b \sec (e+f x)\right )}{b^3 f}\\ &=\frac{2}{3 b f (b \sec (e+f x))^{3/2}}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{-1+\frac{x^4}{b^2}} \, dx,x,\sqrt{b \sec (e+f x)}\right )}{b^3 f}\\ &=\frac{2}{3 b f (b \sec (e+f x))^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{1}{b-x^2} \, dx,x,\sqrt{b \sec (e+f x)}\right )}{b^2 f}-\frac{\operatorname{Subst}\left (\int \frac{1}{b+x^2} \, dx,x,\sqrt{b \sec (e+f x)}\right )}{b^2 f}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{b^{5/2} f}-\frac{\tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{b^{5/2} f}+\frac{2}{3 b f (b \sec (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.205727, size = 90, normalized size = 1.11 \[ \frac{\sqrt{\sec (e+f x)} \left (\frac{4}{\sec ^{\frac{3}{2}}(e+f x)}+3 \log \left (1-\sqrt{\sec (e+f x)}\right )-3 \log \left (\sqrt{\sec (e+f x)}+1\right )-6 \tan ^{-1}\left (\sqrt{\sec (e+f x)}\right )\right )}{6 b^2 f \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]/(b*Sec[e + f*x])^(5/2),x]

[Out]

((-6*ArcTan[Sqrt[Sec[e + f*x]]] + 3*Log[1 - Sqrt[Sec[e + f*x]]] - 3*Log[1 + Sqrt[Sec[e + f*x]]] + 4/Sec[e + f*
x]^(3/2))*Sqrt[Sec[e + f*x]])/(6*b^2*f*Sqrt[b*Sec[e + f*x]])

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Maple [B]  time = 0.121, size = 377, normalized size = 4.7 \begin{align*}{\frac{ \left ( -1+\cos \left ( fx+e \right ) \right ) ^{2} \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}{6\,f \left ( \cos \left ( fx+e \right ) \right ) ^{3} \left ( \sin \left ( fx+e \right ) \right ) ^{4}} \left ( -3\,\cos \left ( fx+e \right ) \sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}\ln \left ( -{\frac{1}{ \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \left ( 2\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,\cos \left ( fx+e \right ) -2\,\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}-1 \right ) } \right ) -3\,\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}\arctan \left ( 1/2\,{\frac{1}{\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}}} \right ) \cos \left ( fx+e \right ) +4\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-3\,\ln \left ( -{\frac{1}{ \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \left ( 2\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,\cos \left ( fx+e \right ) -2\,\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}-1 \right ) } \right ) \sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}-3\,\arctan \left ( 1/2\,{\frac{1}{\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}}} \right ) \sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}} \right ) \left ({\frac{b}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)/(b*sec(f*x+e))^(5/2),x)

[Out]

1/6/f*(-1+cos(f*x+e))^2*(-3*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)/(
cos(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)-3*(-c
os(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))*cos(f*x+e)+4*cos(f*x+e)^2-3
*ln(-(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(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)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-3*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))*(cos(f*x+e)+1)^2/cos(f*x+e)^3/sin(f*x+e)^4/(b/cos(f*x+e))^(5/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)/(b*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.73854, size = 826, normalized size = 10.2 \begin{align*} \left [\frac{8 \, \sqrt{\frac{b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} + 6 \, \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 ) - 3 \, \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 )}{12 \, b^{3} f}, \frac{8 \, \sqrt{\frac{b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} - 6 \, \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 ) + 3 \, \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 )}{12 \, b^{3} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)/(b*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

[1/12*(8*sqrt(b/cos(f*x + e))*cos(f*x + e)^2 + 6*sqrt(-b)*arctan(2*sqrt(-b)*sqrt(b/cos(f*x + e))*cos(f*x + e)/
(b*cos(f*x + e) + b)) - 3*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^3*f), 1/12*(8*sqrt(b/cos(f*x
+ e))*cos(f*x + e)^2 - 6*sqrt(b)*arctan(2*sqrt(b)*sqrt(b/cos(f*x + e))*cos(f*x + e)/(b*cos(f*x + e) - b)) + 3*
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^3*f)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)/(b*sec(f*x+e))**(5/2),x)

[Out]

Timed out

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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 \sec \left (f x + e\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)/(b*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate(csc(f*x + e)/(b*sec(f*x + e))^(5/2), x)