3.403 \(\int \csc ^5(e+f x) (b \sec (e+f x))^{5/2} \, dx\)
Optimal. Leaf size=143 \[ -\frac{\cot ^4(e+f x) (b \sec (e+f x))^{11/2}}{4 b^3 f}+\frac{77 b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{32 f}-\frac{77 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{32 f}+\frac{77 b (b \sec (e+f x))^{3/2}}{48 f}-\frac{11 \cot ^2(e+f x) (b \sec (e+f x))^{7/2}}{16 b f} \]
[Out]
(77*b^(5/2)*ArcTan[Sqrt[b*Sec[e + f*x]]/Sqrt[b]])/(32*f) - (77*b^(5/2)*ArcTanh[Sqrt[b*Sec[e + f*x]]/Sqrt[b]])/
(32*f) + (77*b*(b*Sec[e + f*x])^(3/2))/(48*f) - (11*Cot[e + f*x]^2*(b*Sec[e + f*x])^(7/2))/(16*b*f) - (Cot[e +
f*x]^4*(b*Sec[e + f*x])^(11/2))/(4*b^3*f)
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Rubi [A] time = 0.097547, antiderivative size = 143, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used =
{2622, 288, 321, 329, 298, 203, 206} \[ -\frac{\cot ^4(e+f x) (b \sec (e+f x))^{11/2}}{4 b^3 f}+\frac{77 b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{32 f}-\frac{77 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{32 f}+\frac{77 b (b \sec (e+f x))^{3/2}}{48 f}-\frac{11 \cot ^2(e+f x) (b \sec (e+f x))^{7/2}}{16 b f} \]
Antiderivative was successfully verified.
[In]
Int[Csc[e + f*x]^5*(b*Sec[e + f*x])^(5/2),x]
[Out]
(77*b^(5/2)*ArcTan[Sqrt[b*Sec[e + f*x]]/Sqrt[b]])/(32*f) - (77*b^(5/2)*ArcTanh[Sqrt[b*Sec[e + f*x]]/Sqrt[b]])/
(32*f) + (77*b*(b*Sec[e + f*x])^(3/2))/(48*f) - (11*Cot[e + f*x]^2*(b*Sec[e + f*x])^(7/2))/(16*b*f) - (Cot[e +
f*x]^4*(b*Sec[e + f*x])^(11/2))/(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 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 ^5(e+f x) (b \sec (e+f x))^{5/2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^{13/2}}{\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))^{11/2}}{4 b^3 f}+\frac{11 \operatorname{Subst}\left (\int \frac{x^{9/2}}{\left (-1+\frac{x^2}{b^2}\right )^2} \, dx,x,b \sec (e+f x)\right )}{8 b^3 f}\\ &=-\frac{11 \cot ^2(e+f x) (b \sec (e+f x))^{7/2}}{16 b f}-\frac{\cot ^4(e+f x) (b \sec (e+f x))^{11/2}}{4 b^3 f}+\frac{77 \operatorname{Subst}\left (\int \frac{x^{5/2}}{-1+\frac{x^2}{b^2}} \, dx,x,b \sec (e+f x)\right )}{32 b f}\\ &=\frac{77 b (b \sec (e+f x))^{3/2}}{48 f}-\frac{11 \cot ^2(e+f x) (b \sec (e+f x))^{7/2}}{16 b f}-\frac{\cot ^4(e+f x) (b \sec (e+f x))^{11/2}}{4 b^3 f}+\frac{(77 b) \operatorname{Subst}\left (\int \frac{\sqrt{x}}{-1+\frac{x^2}{b^2}} \, dx,x,b \sec (e+f x)\right )}{32 f}\\ &=\frac{77 b (b \sec (e+f x))^{3/2}}{48 f}-\frac{11 \cot ^2(e+f x) (b \sec (e+f x))^{7/2}}{16 b f}-\frac{\cot ^4(e+f x) (b \sec (e+f x))^{11/2}}{4 b^3 f}+\frac{(77 b) \operatorname{Subst}\left (\int \frac{x^2}{-1+\frac{x^4}{b^2}} \, dx,x,\sqrt{b \sec (e+f x)}\right )}{16 f}\\ &=\frac{77 b (b \sec (e+f x))^{3/2}}{48 f}-\frac{11 \cot ^2(e+f x) (b \sec (e+f x))^{7/2}}{16 b f}-\frac{\cot ^4(e+f x) (b \sec (e+f x))^{11/2}}{4 b^3 f}-\frac{\left (77 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{b-x^2} \, dx,x,\sqrt{b \sec (e+f x)}\right )}{32 f}+\frac{\left (77 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{b+x^2} \, dx,x,\sqrt{b \sec (e+f x)}\right )}{32 f}\\ &=\frac{77 b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{32 f}-\frac{77 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{32 f}+\frac{77 b (b \sec (e+f x))^{3/2}}{48 f}-\frac{11 \cot ^2(e+f x) (b \sec (e+f x))^{7/2}}{16 b f}-\frac{\cot ^4(e+f x) (b \sec (e+f x))^{11/2}}{4 b^3 f}\\ \end{align*}
Mathematica [A] time = 1.19865, size = 119, normalized size = 0.83 \[ \frac{b^3 \left (-48 \csc ^4(e+f x)-180 \csc ^2(e+f x)+128 \sec ^2(e+f x)+231 \sqrt{\sec (e+f x)} \left (\log \left (1-\sqrt{\sec (e+f x)}\right )-\log \left (\sqrt{\sec (e+f x)}+1\right )\right )+462 \sqrt{\sec (e+f x)} \tan ^{-1}\left (\sqrt{\sec (e+f x)}\right )\right )}{192 f \sqrt{b \sec (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[Csc[e + f*x]^5*(b*Sec[e + f*x])^(5/2),x]
[Out]
(b^3*(-180*Csc[e + f*x]^2 - 48*Csc[e + f*x]^4 + 462*ArcTan[Sqrt[Sec[e + f*x]]]*Sqrt[Sec[e + f*x]] + 231*(Log[1
- Sqrt[Sec[e + f*x]]] - Log[1 + Sqrt[Sec[e + f*x]]])*Sqrt[Sec[e + f*x]] + 128*Sec[e + f*x]^2))/(192*f*Sqrt[b*
Sec[e + f*x]])
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Maple [B] time = 0.126, size = 1161, normalized size = 8.1 \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))^(5/2),x)
[Out]
-1/192/f*(408*cos(f*x+e)^5*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)+360*cos(f*x+e)^4*(-cos(f*x+e)/(cos(f*x+e)+1)^2
)^(3/2)-57*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)^5-231*arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))*cos(
f*x+e)^5+288*cos(f*x+e)^5*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)-504*cos(f*x+e)^3*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)
+100*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)^4+57*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)^4+231*arctan
(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))*cos(f*x+e)^4-288*cos(f*x+e)^4*ln(-2*(2*cos(f*x+e)^2*(-cos(f*x+e)/(c
os(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)-456*co
s(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)-456*cos(f*x+e)^3*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+57*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)+231*cos(f*x+e)^3*arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))-288*cos
(f*x+e)^3*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)+484*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-57*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)-231*cos(f*x+e)^2*arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))+288*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)-128*(-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)^4
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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))^(5/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 3.0086, size = 1395, normalized size = 9.76 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)^5*(b*sec(f*x+e))^(5/2),x, algorithm="fricas")
[Out]
[1/384*(462*(b^2*cos(f*x + e)^5 - 2*b^2*cos(f*x + e)^3 + b^2*cos(f*x + e))*sqrt(-b)*arctan(1/2*sqrt(-b)*sqrt(b
/cos(f*x + e))*(cos(f*x + e) + 1)/b) + 231*(b^2*cos(f*x + e)^5 - 2*b^2*cos(f*x + e)^3 + b^2*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)) + 8*(77*b^2*cos(f*x + e)^4 - 121*b^2*cos(f*x + e)^2 + 32*b^2)*sq
rt(b/cos(f*x + e)))/(f*cos(f*x + e)^5 - 2*f*cos(f*x + e)^3 + f*cos(f*x + e)), -1/384*(462*(b^2*cos(f*x + e)^5
- 2*b^2*cos(f*x + e)^3 + b^2*cos(f*x + e))*sqrt(b)*arctan(1/2*sqrt(b/cos(f*x + e))*(cos(f*x + e) - 1)/sqrt(b))
- 231*(b^2*cos(f*x + e)^5 - 2*b^2*cos(f*x + e)^3 + b^2*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)) - 8*(77*b^2*cos(f*x + e)^4 - 121*b^2*cos(f*x + e)^2 + 32*b^2)*sqrt(b/cos(f*x + e)))/(f*cos(f*x + e)^5
- 2*f*cos(f*x + e)^3 + 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)**5*(b*sec(f*x+e))**(5/2),x)
[Out]
Timed out
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Giac [A] time = 1.15855, size = 215, normalized size = 1.5 \begin{align*} \frac{b^{10}{\left (\frac{231 \, \arctan \left (\frac{\sqrt{b \cos \left (f x + e\right )}}{\sqrt{-b}}\right )}{\sqrt{-b} b^{7}} - \frac{231 \, \arctan \left (\frac{\sqrt{b \cos \left (f x + e\right )}}{\sqrt{b}}\right )}{b^{\frac{15}{2}}} + \frac{6 \,{\left (15 \, \sqrt{b \cos \left (f x + e\right )} b^{2} \cos \left (f x + e\right )^{2} - 19 \, \sqrt{b \cos \left (f x + e\right )} b^{2}\right )}}{{\left (b^{2} \cos \left (f x + e\right )^{2} - b^{2}\right )}^{2} b^{6}} + \frac{64}{\sqrt{b \cos \left (f x + e\right )} b^{7} \cos \left (f x + e\right )}\right )} \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{96 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)^5*(b*sec(f*x+e))^(5/2),x, algorithm="giac")
[Out]
1/96*b^10*(231*arctan(sqrt(b*cos(f*x + e))/sqrt(-b))/(sqrt(-b)*b^7) - 231*arctan(sqrt(b*cos(f*x + e))/sqrt(b))
/b^(15/2) + 6*(15*sqrt(b*cos(f*x + e))*b^2*cos(f*x + e)^2 - 19*sqrt(b*cos(f*x + e))*b^2)/((b^2*cos(f*x + e)^2
- b^2)^2*b^6) + 64/(sqrt(b*cos(f*x + e))*b^7*cos(f*x + e)))*sgn(cos(f*x + e))/f