Integrand size = 19, antiderivative size = 62 \[ \int \sqrt {\csc (a+b x)} \sec ^3(a+b x) \, dx=-\frac {3 \arctan \left (\sqrt {\csc (a+b x)}\right )}{4 b}+\frac {3 \text {arctanh}\left (\sqrt {\csc (a+b x)}\right )}{4 b}+\frac {\sec ^2(a+b x)}{2 b \sqrt {\csc (a+b x)}} \]
-3/4*arctan(csc(b*x+a)^(1/2))/b+3/4*arctanh(csc(b*x+a)^(1/2))/b+1/2*sec(b* x+a)^2/b/csc(b*x+a)^(1/2)
Time = 0.09 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.19 \[ \int \sqrt {\csc (a+b x)} \sec ^3(a+b x) \, dx=\frac {\sqrt {\csc (a+b x)} \left (3 \arctan \left (\sqrt {\sin (a+b x)}\right )+3 \text {arctanh}\left (\sqrt {\sin (a+b x)}\right )+2 \sec ^2(a+b x) \sqrt {\sin (a+b x)}\right ) \sqrt {\sin (a+b x)}}{4 b} \]
(Sqrt[Csc[a + b*x]]*(3*ArcTan[Sqrt[Sin[a + b*x]]] + 3*ArcTanh[Sqrt[Sin[a + b*x]]] + 2*Sec[a + b*x]^2*Sqrt[Sin[a + b*x]])*Sqrt[Sin[a + b*x]])/(4*b)
Time = 0.23 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3042, 3101, 252, 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.
\(\displaystyle \int \sqrt {\csc (a+b x)} \sec ^3(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {\csc (a+b x)} \sec (a+b x)^3dx\) |
\(\Big \downarrow \) 3101 |
\(\displaystyle -\frac {\int \frac {\csc ^{\frac {5}{2}}(a+b x)}{\left (1-\csc ^2(a+b x)\right )^2}d\csc (a+b x)}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {\frac {\csc ^{\frac {3}{2}}(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {3}{4} \int \frac {\sqrt {\csc (a+b x)}}{1-\csc ^2(a+b x)}d\csc (a+b x)}{b}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle -\frac {\frac {\csc ^{\frac {3}{2}}(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {3}{2} \int \frac {\csc (a+b x)}{1-\csc ^2(a+b x)}d\sqrt {\csc (a+b x)}}{b}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle -\frac {\frac {\csc ^{\frac {3}{2}}(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {3}{2} \left (\frac {1}{2} \int \frac {1}{1-\csc (a+b x)}d\sqrt {\csc (a+b x)}-\frac {1}{2} \int \frac {1}{\csc (a+b x)+1}d\sqrt {\csc (a+b x)}\right )}{b}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {\frac {\csc ^{\frac {3}{2}}(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {3}{2} \left (\frac {1}{2} \int \frac {1}{1-\csc (a+b x)}d\sqrt {\csc (a+b x)}-\frac {1}{2} \arctan \left (\sqrt {\csc (a+b x)}\right )\right )}{b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\frac {\csc ^{\frac {3}{2}}(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {3}{2} \left (\frac {1}{2} \text {arctanh}\left (\sqrt {\csc (a+b x)}\right )-\frac {1}{2} \arctan \left (\sqrt {\csc (a+b x)}\right )\right )}{b}\) |
-(((-3*(-1/2*ArcTan[Sqrt[Csc[a + b*x]]] + ArcTanh[Sqrt[Csc[a + b*x]]]/2))/ 2 + Csc[a + b*x]^(3/2)/(2*(1 - Csc[a + b*x]^2)))/b)
3.3.75.3.1 Defintions of rubi rules used
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)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[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_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_S ymbol] :> Simp[-(f*a^n)^(-1) Subst[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
Time = 1.14 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.18
method | result | size |
default | \(\frac {-\left (-3 \ln \left (1+\sqrt {\sin }\left (b x +a \right )\right )+3 \ln \left (\sqrt {\sin }\left (b x +a \right )-1\right )-6 \arctan \left (\sqrt {\sin }\left (b x +a \right )\right )\right ) \left (\cos ^{2}\left (b x +a \right )\right )+4 \left (\sqrt {\sin }\left (b x +a \right )\right )}{8 \cos \left (b x +a \right )^{2} b}\) | \(73\) |
1/8*(-(-3*ln(1+sin(b*x+a)^(1/2))+3*ln(sin(b*x+a)^(1/2)-1)-6*arctan(sin(b*x +a)^(1/2)))*cos(b*x+a)^2+4*sin(b*x+a)^(1/2))/cos(b*x+a)^2/b
Leaf count of result is larger than twice the leaf count of optimal. 131 vs. \(2 (50) = 100\).
Time = 0.32 (sec) , antiderivative size = 131, normalized size of antiderivative = 2.11 \[ \int \sqrt {\csc (a+b x)} \sec ^3(a+b x) \, dx=\frac {6 \, \arctan \left (\frac {\sin \left (b x + a\right ) - 1}{2 \, \sqrt {\sin \left (b x + a\right )}}\right ) \cos \left (b x + a\right )^{2} + 3 \, \cos \left (b x + a\right )^{2} \log \left (\frac {\cos \left (b x + a\right )^{2} + \frac {4 \, {\left (\cos \left (b x + a\right )^{2} - \sin \left (b x + a\right ) - 1\right )}}{\sqrt {\sin \left (b x + a\right )}} - 6 \, \sin \left (b x + a\right ) - 2}{\cos \left (b x + a\right )^{2} + 2 \, \sin \left (b x + a\right ) - 2}\right ) + 8 \, \sqrt {\sin \left (b x + a\right )}}{16 \, b \cos \left (b x + a\right )^{2}} \]
1/16*(6*arctan(1/2*(sin(b*x + a) - 1)/sqrt(sin(b*x + a)))*cos(b*x + a)^2 + 3*cos(b*x + a)^2*log((cos(b*x + a)^2 + 4*(cos(b*x + a)^2 - sin(b*x + a) - 1)/sqrt(sin(b*x + a)) - 6*sin(b*x + a) - 2)/(cos(b*x + a)^2 + 2*sin(b*x + a) - 2)) + 8*sqrt(sin(b*x + a)))/(b*cos(b*x + a)^2)
\[ \int \sqrt {\csc (a+b x)} \sec ^3(a+b x) \, dx=\int \sqrt {\csc {\left (a + b x \right )}} \sec ^{3}{\left (a + b x \right )}\, dx \]
Time = 0.35 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.05 \[ \int \sqrt {\csc (a+b x)} \sec ^3(a+b x) \, dx=\frac {\frac {4}{{\left (\frac {1}{\sin \left (b x + a\right )^{2}} - 1\right )} \sin \left (b x + a\right )^{\frac {3}{2}}} - 6 \, \arctan \left (\frac {1}{\sqrt {\sin \left (b x + a\right )}}\right ) + 3 \, \log \left (\frac {1}{\sqrt {\sin \left (b x + a\right )}} + 1\right ) - 3 \, \log \left (\frac {1}{\sqrt {\sin \left (b x + a\right )}} - 1\right )}{8 \, b} \]
1/8*(4/((1/sin(b*x + a)^2 - 1)*sin(b*x + a)^(3/2)) - 6*arctan(1/sqrt(sin(b *x + a))) + 3*log(1/sqrt(sin(b*x + a)) + 1) - 3*log(1/sqrt(sin(b*x + a)) - 1))/b
Time = 0.32 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.06 \[ \int \sqrt {\csc (a+b x)} \sec ^3(a+b x) \, dx=-\frac {\frac {4 \, \sqrt {\sin \left (b x + a\right )}}{\sin \left (b x + a\right )^{2} - 1} - 6 \, \arctan \left (\sqrt {\sin \left (b x + a\right )}\right ) - 3 \, \log \left (\sqrt {\sin \left (b x + a\right )} + 1\right ) + 3 \, \log \left ({\left | \sqrt {\sin \left (b x + a\right )} - 1 \right |}\right )}{8 \, b} \]
-1/8*(4*sqrt(sin(b*x + a))/(sin(b*x + a)^2 - 1) - 6*arctan(sqrt(sin(b*x + a))) - 3*log(sqrt(sin(b*x + a)) + 1) + 3*log(abs(sqrt(sin(b*x + a)) - 1))) /b
Timed out. \[ \int \sqrt {\csc (a+b x)} \sec ^3(a+b x) \, dx=\int \frac {\sqrt {\frac {1}{\sin \left (a+b\,x\right )}}}{{\cos \left (a+b\,x\right )}^3} \,d x \]