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

Optimal. Leaf size=78 \[ \frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{f}-\frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{f}+\frac{2 b (b \sec (e+f x))^{3/2}}{3 f} \]

[Out]

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

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Rubi [A]  time = 0.0543744, antiderivative size = 78, 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, 321, 329, 298, 203, 206} \[ \frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{f}-\frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{f}+\frac{2 b (b \sec (e+f x))^{3/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 321

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

Rubi steps

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

Mathematica [A]  time = 0.191296, size = 87, normalized size = 1.12 \[ \frac{(b \sec (e+f x))^{5/2} \left (4 \sec ^{\frac{3}{2}}(e+f x)+3 \log \left (1-\sqrt{\sec (e+f x)}\right )-3 \log \left (\sqrt{\sec (e+f x)}+1\right )+6 \tan ^{-1}\left (\sqrt{\sec (e+f x)}\right )\right )}{6 f \sec ^{\frac{5}{2}}(e+f x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Sec[e + f*x])^(5/2)*(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)))/(6*f*Sec[e + f*x]^(5/2))

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Maple [B]  time = 0.119, size = 237, normalized size = 3. \begin{align*} -{\frac{ \left ( -1+\cos \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) }{6\,f \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \left ( 3\, \left ( \cos \left ( fx+e \right ) \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 ) -3\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\ln \left ( -2\,{\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 ) +4\,\cos \left ( fx+e \right ) \sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}+4\,\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}}}{\frac{1}{\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{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))*(3*cos(f*x+e)^2*arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))-3*cos(f*x+e)^2*ln(-2*(
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)+4*cos(f*x+e)*(-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)*(b/cos(f*x+e))^(5/2)/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/sin(f*x+e)^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 = 2.69001, size = 868, normalized size = 11.13 \begin{align*} \left [\frac{6 \, \sqrt{-b} b^{2} \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 ) \cos \left (f x + e\right ) + 3 \, \sqrt{-b} b^{2} \cos \left (f x + e\right ) \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 \, b^{2} \sqrt{\frac{b}{\cos \left (f x + e\right )}}}{12 \, f \cos \left (f x + e\right )}, -\frac{6 \, b^{\frac{5}{2}} \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 ) \cos \left (f x + e\right ) - 3 \, b^{\frac{5}{2}} \cos \left (f x + e\right ) \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 \, b^{2} \sqrt{\frac{b}{\cos \left (f x + e\right )}}}{12 \, f \cos \left (f x + e\right )}\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*(6*sqrt(-b)*b^2*arctan(1/2*sqrt(-b)*sqrt(b/cos(f*x + e))*(cos(f*x + e) + 1)/b)*cos(f*x + e) + 3*sqrt(-b)
*b^2*cos(f*x + e)*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*b^2*sqrt(b/cos(f*x + e)))/(f*cos(f*x + e)), -1/
12*(6*b^(5/2)*arctan(1/2*sqrt(b/cos(f*x + e))*(cos(f*x + e) - 1)/sqrt(b))*cos(f*x + e) - 3*b^(5/2)*cos(f*x + e
)*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*b^2*sqrt(b/cos(f*x + e)))/(f*cos(f*x + e))]

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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 [A]  time = 1.1702, size = 123, normalized size = 1.58 \begin{align*} \frac{b^{6}{\left (\frac{3 \, \arctan \left (\frac{\sqrt{b \cos \left (f x + e\right )}}{\sqrt{-b}}\right )}{\sqrt{-b} b^{3}} - \frac{3 \, \arctan \left (\frac{\sqrt{b \cos \left (f x + e\right )}}{\sqrt{b}}\right )}{b^{\frac{7}{2}}} + \frac{2}{\sqrt{b \cos \left (f x + e\right )} b^{3} \cos \left (f x + e\right )}\right )} \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{3 \, f} \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]

1/3*b^6*(3*arctan(sqrt(b*cos(f*x + e))/sqrt(-b))/(sqrt(-b)*b^3) - 3*arctan(sqrt(b*cos(f*x + e))/sqrt(b))/b^(7/
2) + 2/(sqrt(b*cos(f*x + e))*b^3*cos(f*x + e)))*sgn(cos(f*x + e))/f