Integrand size = 19, antiderivative size = 78 \[ \int \csc (e+f x) (b \sec (e+f x))^{5/2} \, dx=\frac {b^{5/2} \arctan \left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{f}-\frac {b^{5/2} \text {arctanh}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{f}+\frac {2 b (b \sec (e+f x))^{3/2}}{3 f} \] Output:
b^(5/2)*arctan((b*sec(f*x+e))^(1/2)/b^(1/2))/f-b^(5/2)*arctanh((b*sec(f*x+ e))^(1/2)/b^(1/2))/f+2/3*b*(b*sec(f*x+e))^(3/2)/f
Time = 0.43 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.12 \[ \int \csc (e+f x) (b \sec (e+f x))^{5/2} \, dx=\frac {(b \sec (e+f x))^{5/2} \left (6 \arctan \left (\sqrt {\sec (e+f x)}\right )+3 \log \left (1-\sqrt {\sec (e+f x)}\right )-3 \log \left (1+\sqrt {\sec (e+f x)}\right )+4 \sec ^{\frac {3}{2}}(e+f x)\right )}{6 f \sec ^{\frac {5}{2}}(e+f x)} \] Input:
Integrate[Csc[e + f*x]*(b*Sec[e + f*x])^(5/2),x]
Output:
((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))
Time = 0.41 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.94, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {3042, 3102, 25, 27, 262, 266, 827, 216, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \csc (e+f x) (b \sec (e+f x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc (e+f x) (b \sec (e+f x))^{5/2}dx\) |
\(\Big \downarrow \) 3102 |
\(\displaystyle \frac {\int -\frac {b^2 (b \sec (e+f x))^{5/2}}{b^2-b^2 \sec ^2(e+f x)}d(b \sec (e+f x))}{b f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {b^2 (b \sec (e+f x))^{5/2}}{b^2-b^2 \sec ^2(e+f x)}d(b \sec (e+f x))}{b f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {b \int \frac {(b \sec (e+f x))^{5/2}}{b^2-b^2 \sec ^2(e+f x)}d(b \sec (e+f x))}{f}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -\frac {b \left (b^2 \int \frac {\sqrt {b \sec (e+f x)}}{b^2-b^2 \sec ^2(e+f x)}d(b \sec (e+f x))-\frac {2}{3} (b \sec (e+f x))^{3/2}\right )}{f}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle -\frac {b \left (2 b^2 \int \frac {b^2 \sec ^2(e+f x)}{b^2-b^4 \sec ^4(e+f x)}d\sqrt {b \sec (e+f x)}-\frac {2}{3} (b \sec (e+f x))^{3/2}\right )}{f}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle -\frac {b \left (2 b^2 \left (\frac {1}{2} \int \frac {1}{b-b^2 \sec ^2(e+f x)}d\sqrt {b \sec (e+f x)}-\frac {1}{2} \int \frac {1}{b^2 \sec ^2(e+f x)+b}d\sqrt {b \sec (e+f x)}\right )-\frac {2}{3} (b \sec (e+f x))^{3/2}\right )}{f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {b \left (2 b^2 \left (\frac {1}{2} \int \frac {1}{b-b^2 \sec ^2(e+f x)}d\sqrt {b \sec (e+f x)}-\frac {\arctan \left (\sqrt {b} \sec (e+f x)\right )}{2 \sqrt {b}}\right )-\frac {2}{3} (b \sec (e+f x))^{3/2}\right )}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {b \left (2 b^2 \left (\frac {\text {arctanh}\left (\sqrt {b} \sec (e+f x)\right )}{2 \sqrt {b}}-\frac {\arctan \left (\sqrt {b} \sec (e+f x)\right )}{2 \sqrt {b}}\right )-\frac {2}{3} (b \sec (e+f x))^{3/2}\right )}{f}\) |
Input:
Int[Csc[e + f*x]*(b*Sec[e + f*x])^(5/2),x]
Output:
-((b*(2*b^2*(-1/2*ArcTan[Sqrt[b]*Sec[e + f*x]]/Sqrt[b] + ArcTanh[Sqrt[b]*S ec[e + f*x]]/(2*Sqrt[b])) - (2*(b*Sec[e + f*x])^(3/2))/3))/f)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_S ymbol] :> Simp[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])
Leaf count of result is larger than twice the leaf count of optimal. \(192\) vs. \(2(62)=124\).
Time = 4.21 (sec) , antiderivative size = 193, normalized size of antiderivative = 2.47
method | result | size |
default | \(-\frac {\sqrt {b \sec \left (f x +e \right )}\, b^{2} \left (\sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (-4-4 \sec \left (f x +e \right )\right )+3 \ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-2 \cos \left (f x +e \right )+2}{\cos \left (f x +e \right )+1}\right ) \cos \left (f x +e \right )-3 \arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right ) \cos \left (f x +e \right )\right )}{6 f \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cos \left (f x +e \right )+1\right )}\) | \(193\) |
Input:
int(csc(f*x+e)*(b*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/6/f*(b*sec(f*x+e))^(1/2)*b^2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/(cos( f*x+e)+1)*((-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(-4-4*sec(f*x+e))+3*ln(2*( 2*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*(-cos(f*x+e)/(cos(f*x+ e)+1)^2)^(1/2)-cos(f*x+e)+1)/(cos(f*x+e)+1))*cos(f*x+e)-3*arctan(1/2/(-cos (f*x+e)/(cos(f*x+e)+1)^2)^(1/2))*cos(f*x+e))
Leaf count of result is larger than twice the leaf count of optimal. 166 vs. \(2 (62) = 124\).
Time = 0.14 (sec) , antiderivative size = 347, normalized size of antiderivative = 4.45 \[ \int \csc (e+f x) (b \sec (e+f x))^{5/2} \, dx=\left [\frac {6 \, \sqrt {-b} b^{2} \arctan \left (\frac {2 \, \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{b \cos \left (f x + e\right ) + b}\right ) \cos \left (f x + e\right ) + 3 \, \sqrt {-b} b^{2} \cos \left (f x + e\right ) \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) + 8 \, b^{2} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{12 \, f \cos \left (f x + e\right )}, \frac {6 \, b^{\frac {5}{2}} \arctan \left (\frac {2 \, \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{b \cos \left (f x + e\right ) - b}\right ) \cos \left (f x + e\right ) + 3 \, b^{\frac {5}{2}} \cos \left (f x + e\right ) \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right ) + 8 \, b^{2} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{12 \, f \cos \left (f x + e\right )}\right ] \] Input:
integrate(csc(f*x+e)*(b*sec(f*x+e))^(5/2),x, algorithm="fricas")
Output:
[1/12*(6*sqrt(-b)*b^2*arctan(2*sqrt(-b)*sqrt(b/cos(f*x + e))*cos(f*x + e)/ (b*cos(f*x + e) + b))*cos(f*x + e) + 3*sqrt(-b)*b^2*cos(f*x + e)*log((b*co s(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(2*sqrt(b) *sqrt(b/cos(f*x + e))*cos(f*x + e)/(b*cos(f*x + e) - 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))]
Timed out. \[ \int \csc (e+f x) (b \sec (e+f x))^{5/2} \, dx=\text {Timed out} \] Input:
integrate(csc(f*x+e)*(b*sec(f*x+e))**(5/2),x)
Output:
Timed out
Time = 0.13 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.12 \[ \int \csc (e+f x) (b \sec (e+f x))^{5/2} \, dx=\frac {{\left (6 \, b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b}}\right ) + 3 \, b^{\frac {3}{2}} \log \left (-\frac {\sqrt {b} - \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b} + \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right ) + 4 \, \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {3}{2}}\right )} b}{6 \, f} \] Input:
integrate(csc(f*x+e)*(b*sec(f*x+e))^(5/2),x, algorithm="maxima")
Output:
1/6*(6*b^(3/2)*arctan(sqrt(b/cos(f*x + e))/sqrt(b)) + 3*b^(3/2)*log(-(sqrt (b) - sqrt(b/cos(f*x + e)))/(sqrt(b) + sqrt(b/cos(f*x + e)))) + 4*(b/cos(f *x + e))^(3/2))*b/f
Time = 0.13 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.10 \[ \int \csc (e+f x) (b \sec (e+f x))^{5/2} \, dx=\frac {b^{6} {\left (\frac {3 \, \arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{3}} - \frac {3 \, \arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {b}}\right )}{b^{\frac {7}{2}}} + \frac {2}{\sqrt {b \cos \left (f x + e\right )} b^{3} \cos \left (f x + e\right )}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{3 \, f} \] Input:
integrate(csc(f*x+e)*(b*sec(f*x+e))^(5/2),x, algorithm="giac")
Output:
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
Timed out. \[ \int \csc (e+f x) (b \sec (e+f x))^{5/2} \, dx=\int \frac {{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{\sin \left (e+f\,x\right )} \,d x \] Input:
int((b/cos(e + f*x))^(5/2)/sin(e + f*x),x)
Output:
int((b/cos(e + f*x))^(5/2)/sin(e + f*x), x)
\[ \int \csc (e+f x) (b \sec (e+f x))^{5/2} \, dx=\sqrt {b}\, \left (\int \sqrt {\sec \left (f x +e \right )}\, \csc \left (f x +e \right ) \sec \left (f x +e \right )^{2}d x \right ) b^{2} \] Input:
int(csc(f*x+e)*(b*sec(f*x+e))^(5/2),x)
Output:
sqrt(b)*int(sqrt(sec(e + f*x))*csc(e + f*x)*sec(e + f*x)**2,x)*b**2