3.416 \(\int \frac{\csc ^5(e+f x)}{\sqrt{b \sec (e+f x)}} \, dx\)

Optimal. Leaf size=123 \[ -\frac{\cot ^4(e+f x) (b \sec (e+f x))^{5/2}}{4 b^3 f}-\frac{5 \cot ^2(e+f x) \sqrt{b \sec (e+f x)}}{16 b f}-\frac{5 \tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{32 \sqrt{b} f}-\frac{5 \tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{32 \sqrt{b} f} \]

[Out]

(-5*ArcTan[Sqrt[b*Sec[e + f*x]]/Sqrt[b]])/(32*Sqrt[b]*f) - (5*ArcTanh[Sqrt[b*Sec[e + f*x]]/Sqrt[b]])/(32*Sqrt[
b]*f) - (5*Cot[e + f*x]^2*Sqrt[b*Sec[e + f*x]])/(16*b*f) - (Cot[e + f*x]^4*(b*Sec[e + f*x])^(5/2))/(4*b^3*f)

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Rubi [A]  time = 0.0785193, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2622, 288, 329, 212, 206, 203} \[ -\frac{\cot ^4(e+f x) (b \sec (e+f x))^{5/2}}{4 b^3 f}-\frac{5 \cot ^2(e+f x) \sqrt{b \sec (e+f x)}}{16 b f}-\frac{5 \tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{32 \sqrt{b} f}-\frac{5 \tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{32 \sqrt{b} f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^5/Sqrt[b*Sec[e + f*x]],x]

[Out]

(-5*ArcTan[Sqrt[b*Sec[e + f*x]]/Sqrt[b]])/(32*Sqrt[b]*f) - (5*ArcTanh[Sqrt[b*Sec[e + f*x]]/Sqrt[b]])/(32*Sqrt[
b]*f) - (5*Cot[e + f*x]^2*Sqrt[b*Sec[e + f*x]])/(16*b*f) - (Cot[e + f*x]^4*(b*Sec[e + f*x])^(5/2))/(4*b^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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 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 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 ^5(e+f x)}{\sqrt{b \sec (e+f x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^{7/2}}{\left (-1+\frac{x^2}{b^2}\right )^3} \, dx,x,b \sec (e+f x)\right )}{b^5 f}\\ &=-\frac{\cot ^4(e+f x) (b \sec (e+f x))^{5/2}}{4 b^3 f}+\frac{5 \operatorname{Subst}\left (\int \frac{x^{3/2}}{\left (-1+\frac{x^2}{b^2}\right )^2} \, dx,x,b \sec (e+f x)\right )}{8 b^3 f}\\ &=-\frac{5 \cot ^2(e+f x) \sqrt{b \sec (e+f x)}}{16 b f}-\frac{\cot ^4(e+f x) (b \sec (e+f x))^{5/2}}{4 b^3 f}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (-1+\frac{x^2}{b^2}\right )} \, dx,x,b \sec (e+f x)\right )}{32 b f}\\ &=-\frac{5 \cot ^2(e+f x) \sqrt{b \sec (e+f x)}}{16 b f}-\frac{\cot ^4(e+f x) (b \sec (e+f x))^{5/2}}{4 b^3 f}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{-1+\frac{x^4}{b^2}} \, dx,x,\sqrt{b \sec (e+f x)}\right )}{16 b f}\\ &=-\frac{5 \cot ^2(e+f x) \sqrt{b \sec (e+f x)}}{16 b f}-\frac{\cot ^4(e+f x) (b \sec (e+f x))^{5/2}}{4 b^3 f}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{b-x^2} \, dx,x,\sqrt{b \sec (e+f x)}\right )}{32 f}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{b+x^2} \, dx,x,\sqrt{b \sec (e+f x)}\right )}{32 f}\\ &=-\frac{5 \tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{32 \sqrt{b} f}-\frac{5 \tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{32 \sqrt{b} f}-\frac{5 \cot ^2(e+f x) \sqrt{b \sec (e+f x)}}{16 b f}-\frac{\cot ^4(e+f x) (b \sec (e+f x))^{5/2}}{4 b^3 f}\\ \end{align*}

Mathematica [A]  time = 2.23631, size = 107, normalized size = 0.87 \[ -\frac{\sqrt{\sec (e+f x)} \left (-5 \log \left (1-\sqrt{\sec (e+f x)}\right )+5 \log \left (\sqrt{\sec (e+f x)}+1\right )+4 \left (4 \csc ^4(e+f x)+\csc ^2(e+f x)-5\right ) \sqrt{\sec (e+f x)}+10 \tan ^{-1}\left (\sqrt{\sec (e+f x)}\right )\right )}{64 f \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^5/Sqrt[b*Sec[e + f*x]],x]

[Out]

-((10*ArcTan[Sqrt[Sec[e + f*x]]] - 5*Log[1 - Sqrt[Sec[e + f*x]]] + 5*Log[1 + Sqrt[Sec[e + f*x]]] + 4*(-5 + Csc
[e + f*x]^2 + 4*Csc[e + f*x]^4)*Sqrt[Sec[e + f*x]])*Sqrt[Sec[e + f*x]])/(64*f*Sqrt[b*Sec[e + f*x]])

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Maple [B]  time = 0.146, size = 729, normalized size = 5.9 \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)^5/(b*sec(f*x+e))^(1/2),x)

[Out]

-1/64/f*(40*cos(f*x+e)^3*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)+24*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(
3/2)-20*cos(f*x+e)^3*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-5*cos(f*x+e)^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)-5*cos(f*
x+e)^3*arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))-72*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)+40*
cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+5*cos(f*x+e)^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)+5*cos(f*x+e)^2*a
rctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))-56*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)-20*cos(f*x+e)*(-cos(f*
x+e)/(cos(f*x+e)+1)^2)^(1/2)+5*cos(f*x+e)*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)+5*cos(f*x+e)*arctan(1/2/(-cos(f*x+e)/(c
os(f*x+e)+1)^2)^(1/2))-5*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)-5*arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)))/sin(
f*x+e)^4/(b/cos(f*x+e))^(1/2)/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)

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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)^5/(b*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.99466, size = 1188, normalized size = 9.66 \begin{align*} \left [\frac{10 \,{\left (\cos \left (f x + e\right )^{4} - 2 \, \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 ) - 5 \,{\left (\cos \left (f x + e\right )^{4} - 2 \, \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 \,{\left (5 \, \cos \left (f x + e\right )^{4} - 9 \, \cos \left (f x + e\right )^{2}\right )} \sqrt{\frac{b}{\cos \left (f x + e\right )}}}{128 \,{\left (b f \cos \left (f x + e\right )^{4} - 2 \, b f \cos \left (f x + e\right )^{2} + b f\right )}}, \frac{10 \,{\left (\cos \left (f x + e\right )^{4} - 2 \, \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 ) + 5 \,{\left (\cos \left (f x + e\right )^{4} - 2 \, \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 \,{\left (5 \, \cos \left (f x + e\right )^{4} - 9 \, \cos \left (f x + e\right )^{2}\right )} \sqrt{\frac{b}{\cos \left (f x + e\right )}}}{128 \,{\left (b f \cos \left (f x + e\right )^{4} - 2 \, b f \cos \left (f x + e\right )^{2} + b f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5/(b*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[1/128*(10*(cos(f*x + e)^4 - 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) - 5*(cos(f*x + e)^4 - 2*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
*(5*cos(f*x + e)^4 - 9*cos(f*x + e)^2)*sqrt(b/cos(f*x + e)))/(b*f*cos(f*x + e)^4 - 2*b*f*cos(f*x + e)^2 + b*f)
, 1/128*(10*(cos(f*x + e)^4 - 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)) + 5*(cos(f*x + e)^4 - 2*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*(5*c
os(f*x + e)^4 - 9*cos(f*x + e)^2)*sqrt(b/cos(f*x + e)))/(b*f*cos(f*x + e)^4 - 2*b*f*cos(f*x + e)^2 + b*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)**5/(b*sec(f*x+e))**(1/2),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 )^{5}}{\sqrt{b \sec \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5/(b*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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