Integrand size = 19, antiderivative size = 77 \[ \int (d \cos (a+b x))^{3/2} \csc (a+b x) \, dx=-\frac {d^{3/2} \arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b}-\frac {d^{3/2} \text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b}+\frac {2 d \sqrt {d \cos (a+b x)}}{b} \]
-d^(3/2)*arctan((d*cos(b*x+a))^(1/2)/d^(1/2))/b-d^(3/2)*arctanh((d*cos(b*x +a))^(1/2)/d^(1/2))/b+2*d*(d*cos(b*x+a))^(1/2)/b
Time = 0.16 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.84 \[ \int (d \cos (a+b x))^{3/2} \csc (a+b x) \, dx=\frac {\left (-\arctan \left (\sqrt {\cos (a+b x)}\right )-\text {arctanh}\left (\sqrt {\cos (a+b x)}\right )+2 \sqrt {\cos (a+b x)}\right ) (d \cos (a+b x))^{3/2}}{b \cos ^{\frac {3}{2}}(a+b x)} \]
((-ArcTan[Sqrt[Cos[a + b*x]]] - ArcTanh[Sqrt[Cos[a + b*x]]] + 2*Sqrt[Cos[a + b*x]])*(d*Cos[a + b*x])^(3/2))/(b*Cos[a + b*x]^(3/2))
Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.92, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3042, 3045, 27, 262, 266, 756, 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.
\(\displaystyle \int \csc (a+b x) (d \cos (a+b x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(d \cos (a+b x))^{3/2}}{\sin (a+b x)}dx\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle -\frac {\int \frac {d^2 (d \cos (a+b x))^{3/2}}{d^2-d^2 \cos ^2(a+b x)}d(d \cos (a+b x))}{b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {d \int \frac {(d \cos (a+b x))^{3/2}}{d^2-d^2 \cos ^2(a+b x)}d(d \cos (a+b x))}{b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -\frac {d \left (d^2 \int \frac {1}{\sqrt {d \cos (a+b x)} \left (d^2-d^2 \cos ^2(a+b x)\right )}d(d \cos (a+b x))-2 \sqrt {d \cos (a+b x)}\right )}{b}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle -\frac {d \left (2 d^2 \int \frac {1}{d^2-d^4 \cos ^4(a+b x)}d\sqrt {d \cos (a+b x)}-2 \sqrt {d \cos (a+b x)}\right )}{b}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle -\frac {d \left (2 d^2 \left (\frac {\int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}}{2 d}+\frac {\int \frac {1}{d^2 \cos ^2(a+b x)+d}d\sqrt {d \cos (a+b x)}}{2 d}\right )-2 \sqrt {d \cos (a+b x)}\right )}{b}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {d \left (2 d^2 \left (\frac {\int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}}{2 d}+\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 d^{3/2}}\right )-2 \sqrt {d \cos (a+b x)}\right )}{b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {d \left (2 d^2 \left (\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 d^{3/2}}+\frac {\text {arctanh}\left (\sqrt {d} \cos (a+b x)\right )}{2 d^{3/2}}\right )-2 \sqrt {d \cos (a+b x)}\right )}{b}\) |
-((d*(2*d^2*(ArcTan[Sqrt[d]*Cos[a + b*x]]/(2*d^(3/2)) + ArcTanh[Sqrt[d]*Co s[a + b*x]]/(2*d^(3/2))) - 2*Sqrt[d*Cos[a + b*x]]))/b)
3.3.25.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> Simp[-(a*f)^(-1) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
Leaf count of result is larger than twice the leaf count of optimal. \(211\) vs. \(2(63)=126\).
Time = 0.11 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.75
method | result | size |
default | \(\frac {-d^{\frac {3}{2}} \ln \left (\frac {4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right ) \sqrt {-d}-d^{\frac {3}{2}} \ln \left (-\frac {2 \left (2 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-\sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}+d \right )}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right ) \sqrt {-d}+4 d \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}\, \sqrt {-d}+2 d^{2} \ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right )}{2 \sqrt {-d}\, b}\) | \(212\) |
1/2*(-d^(3/2)*ln(2/(cos(1/2*b*x+1/2*a)-1)*(2*d*cos(1/2*b*x+1/2*a)+d^(1/2)* (-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)-d))*(-d)^(1/2)-d^(3/2)*ln(-2/(cos(1/2* b*x+1/2*a)+1)*(2*d*cos(1/2*b*x+1/2*a)-d^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d )^(1/2)+d))*(-d)^(1/2)+4*d*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)*(-d)^(1/2)+ 2*d^2*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1 /2)-d)))/(-d)^(1/2)/b
Time = 0.37 (sec) , antiderivative size = 259, normalized size of antiderivative = 3.36 \[ \int (d \cos (a+b x))^{3/2} \csc (a+b x) \, dx=\left [\frac {2 \, \sqrt {-d} d \arctan \left (\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d}}{d \cos \left (b x + a\right ) + d}\right ) + \sqrt {-d} d \log \left (-\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )} d}{4 \, b}, \frac {2 \, d^{\frac {3}{2}} \arctan \left (\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d}}{d \cos \left (b x + a\right ) - d}\right ) + d^{\frac {3}{2}} \log \left (-\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )} d}{4 \, b}\right ] \]
[1/4*(2*sqrt(-d)*d*arctan(2*sqrt(d*cos(b*x + a))*sqrt(-d)/(d*cos(b*x + a) + d)) + sqrt(-d)*d*log(-(d*cos(b*x + a)^2 - 4*sqrt(d*cos(b*x + a))*sqrt(-d )*(cos(b*x + a) - 1) - 6*d*cos(b*x + a) + d)/(cos(b*x + a)^2 + 2*cos(b*x + a) + 1)) + 8*sqrt(d*cos(b*x + a))*d)/b, 1/4*(2*d^(3/2)*arctan(2*sqrt(d*co s(b*x + a))*sqrt(d)/(d*cos(b*x + a) - d)) + d^(3/2)*log(-(d*cos(b*x + a)^2 - 4*sqrt(d*cos(b*x + a))*sqrt(d)*(cos(b*x + a) + 1) + 6*d*cos(b*x + a) + d)/(cos(b*x + a)^2 - 2*cos(b*x + a) + 1)) + 8*sqrt(d*cos(b*x + a))*d)/b]
Timed out. \[ \int (d \cos (a+b x))^{3/2} \csc (a+b x) \, dx=\text {Timed out} \]
Time = 0.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.08 \[ \int (d \cos (a+b x))^{3/2} \csc (a+b x) \, dx=-\frac {2 \, d^{\frac {5}{2}} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right ) - d^{\frac {5}{2}} \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right ) - 4 \, \sqrt {d \cos \left (b x + a\right )} d^{2}}{2 \, b d} \]
-1/2*(2*d^(5/2)*arctan(sqrt(d*cos(b*x + a))/sqrt(d)) - d^(5/2)*log((sqrt(d *cos(b*x + a)) - sqrt(d))/(sqrt(d*cos(b*x + a)) + sqrt(d))) - 4*sqrt(d*cos (b*x + a))*d^2)/(b*d)
\[ \int (d \cos (a+b x))^{3/2} \csc (a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}} \csc \left (b x + a\right ) \,d x } \]
Timed out. \[ \int (d \cos (a+b x))^{3/2} \csc (a+b x) \, dx=\int \frac {{\left (d\,\cos \left (a+b\,x\right )\right )}^{3/2}}{\sin \left (a+b\,x\right )} \,d x \]