Integrand size = 19, antiderivative size = 58 \[ \int \csc (e+f x) \sqrt {b \sec (e+f x)} \, dx=\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{f}-\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{f} \] Output:
b^(1/2)*arctan((b*sec(f*x+e))^(1/2)/b^(1/2))/f-b^(1/2)*arctanh((b*sec(f*x+ e))^(1/2)/b^(1/2))/f
Time = 0.47 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.26 \[ \int \csc (e+f x) \sqrt {b \sec (e+f x)} \, dx=\frac {\left (2 \arctan \left (\sqrt {\sec (e+f x)}\right )+\log \left (1-\sqrt {\sec (e+f x)}\right )-\log \left (1+\sqrt {\sec (e+f x)}\right )\right ) \sqrt {b \sec (e+f x)}}{2 f \sqrt {\sec (e+f x)}} \] Input:
Integrate[Csc[e + f*x]*Sqrt[b*Sec[e + f*x]],x]
Output:
((2*ArcTan[Sqrt[Sec[e + f*x]]] + Log[1 - Sqrt[Sec[e + f*x]]] - Log[1 + Sqr t[Sec[e + f*x]]])*Sqrt[b*Sec[e + f*x]])/(2*f*Sqrt[Sec[e + f*x]])
Time = 0.23 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.88, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3042, 3102, 25, 27, 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.
\(\displaystyle \int \csc (e+f x) \sqrt {b \sec (e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc (e+f x) \sqrt {b \sec (e+f x)}dx\) |
\(\Big \downarrow \) 3102 |
\(\displaystyle \frac {\int -\frac {b^2 \sqrt {b \sec (e+f x)}}{b^2-b^2 \sec ^2(e+f x)}d(b \sec (e+f x))}{b f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {b^2 \sqrt {b \sec (e+f x)}}{b^2-b^2 \sec ^2(e+f x)}d(b \sec (e+f x))}{b f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {b \int \frac {\sqrt {b \sec (e+f x)}}{b^2-b^2 \sec ^2(e+f x)}d(b \sec (e+f x))}{f}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle -\frac {2 b \int \frac {b^2 \sec ^2(e+f x)}{b^2-b^4 \sec ^4(e+f x)}d\sqrt {b \sec (e+f x)}}{f}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle -\frac {2 b \left (\frac {1}{2} \int \frac {1}{b-b^2 \sec ^2(e+f x)}d\sqrt {b \sec (e+f x)}-\frac {1}{2} \int \frac {1}{b^2 \sec ^2(e+f x)+b}d\sqrt {b \sec (e+f x)}\right )}{f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {2 b \left (\frac {1}{2} \int \frac {1}{b-b^2 \sec ^2(e+f x)}d\sqrt {b \sec (e+f x)}-\frac {\arctan \left (\sqrt {b} \sec (e+f x)\right )}{2 \sqrt {b}}\right )}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {2 b \left (\frac {\text {arctanh}\left (\sqrt {b} \sec (e+f x)\right )}{2 \sqrt {b}}-\frac {\arctan \left (\sqrt {b} \sec (e+f x)\right )}{2 \sqrt {b}}\right )}{f}\) |
Input:
Int[Csc[e + f*x]*Sqrt[b*Sec[e + f*x]],x]
Output:
(-2*b*(-1/2*ArcTan[Sqrt[b]*Sec[e + f*x]]/Sqrt[b] + ArcTanh[Sqrt[b]*Sec[e + f*x]]/(2*Sqrt[b])))/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] :> 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. \(150\) vs. \(2(46)=92\).
Time = 2.84 (sec) , antiderivative size = 151, normalized size of antiderivative = 2.60
method | result | size |
default | \(-\frac {\left (-\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )+\ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-2 \cos \left (f x +e \right )+2}{\cos \left (f x +e \right )+1}\right )\right ) \sqrt {b \sec \left (f x +e \right )}\, \cos \left (f x +e \right )}{2 f \left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\) | \(151\) |
Input:
int(csc(f*x+e)*(b*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/2/f*(-arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))+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)))*(b*sec(f*x+e))^(1/2)*cos(f*x+e)/(cos(f *x+e)+1)/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (46) = 92\).
Time = 0.14 (sec) , antiderivative size = 265, normalized size of antiderivative = 4.57 \[ \int \csc (e+f x) \sqrt {b \sec (e+f x)} \, dx=\left [\frac {2 \, \sqrt {-b} \arctan \left (\frac {2 \, \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{b \cos \left (f x + e\right ) + b}\right ) + \sqrt {-b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right )}{4 \, f}, \frac {2 \, \sqrt {b} \arctan \left (\frac {2 \, \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{b \cos \left (f x + e\right ) - b}\right ) + \sqrt {b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right )}{4 \, f}\right ] \] Input:
integrate(csc(f*x+e)*(b*sec(f*x+e))^(1/2),x, algorithm="fricas")
Output:
[1/4*(2*sqrt(-b)*arctan(2*sqrt(-b)*sqrt(b/cos(f*x + e))*cos(f*x + e)/(b*co s(f*x + e) + b)) + sqrt(-b)*log((b*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 - co s(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)))/f, 1/4*(2*sqrt(b)*arctan(2*sqrt(b)*sqrt(b/ cos(f*x + e))*cos(f*x + e)/(b*cos(f*x + e) - b)) + 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)))/f]
\[ \int \csc (e+f x) \sqrt {b \sec (e+f x)} \, dx=\int \sqrt {b \sec {\left (e + f x \right )}} \csc {\left (e + f x \right )}\, dx \] Input:
integrate(csc(f*x+e)*(b*sec(f*x+e))**(1/2),x)
Output:
Integral(sqrt(b*sec(e + f*x))*csc(e + f*x), x)
Time = 0.11 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.24 \[ \int \csc (e+f x) \sqrt {b \sec (e+f x)} \, dx=\frac {b {\left (\frac {2 \, \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b}}\right )}{\sqrt {b}} + \frac {\log \left (-\frac {\sqrt {b} - \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b} + \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right )}{\sqrt {b}}\right )}}{2 \, f} \] Input:
integrate(csc(f*x+e)*(b*sec(f*x+e))^(1/2),x, algorithm="maxima")
Output:
1/2*b*(2*arctan(sqrt(b/cos(f*x + e))/sqrt(b))/sqrt(b) + log(-(sqrt(b) - sq rt(b/cos(f*x + e)))/(sqrt(b) + sqrt(b/cos(f*x + e))))/sqrt(b))/f
Time = 0.13 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.05 \[ \int \csc (e+f x) \sqrt {b \sec (e+f x)} \, dx=\frac {b^{2} {\left (\frac {\arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {-b}}\right )}{\sqrt {-b} b} - \frac {\arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {b}}\right )}{b^{\frac {3}{2}}}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{f} \] Input:
integrate(csc(f*x+e)*(b*sec(f*x+e))^(1/2),x, algorithm="giac")
Output:
b^2*(arctan(sqrt(b*cos(f*x + e))/sqrt(-b))/(sqrt(-b)*b) - arctan(sqrt(b*co s(f*x + e))/sqrt(b))/b^(3/2))*sgn(cos(f*x + e))/f
Timed out. \[ \int \csc (e+f x) \sqrt {b \sec (e+f x)} \, dx=\int \frac {\sqrt {\frac {b}{\cos \left (e+f\,x\right )}}}{\sin \left (e+f\,x\right )} \,d x \] Input:
int((b/cos(e + f*x))^(1/2)/sin(e + f*x),x)
Output:
int((b/cos(e + f*x))^(1/2)/sin(e + f*x), x)
\[ \int \csc (e+f x) \sqrt {b \sec (e+f x)} \, dx=\sqrt {b}\, \left (\int \sqrt {\sec \left (f x +e \right )}\, \csc \left (f x +e \right )d x \right ) \] Input:
int(csc(f*x+e)*(b*sec(f*x+e))^(1/2),x)
Output:
sqrt(b)*int(sqrt(sec(e + f*x))*csc(e + f*x),x)