Integrand size = 19, antiderivative size = 78 \[ \int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx=\frac {d^{5/2} \arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b}-\frac {d^{5/2} \text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b}+\frac {2 d (d \cos (a+b x))^{3/2}}{3 b} \]
d^(5/2)*arctan((d*cos(b*x+a))^(1/2)/d^(1/2))/b-d^(5/2)*arctanh((d*cos(b*x+ a))^(1/2)/d^(1/2))/b+2/3*d*(d*cos(b*x+a))^(3/2)/b
Time = 0.22 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.87 \[ \int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx=\frac {(d \cos (a+b x))^{5/2} \left (3 \arctan \left (\sqrt {\cos (a+b x)}\right )-3 \text {arctanh}\left (\sqrt {\cos (a+b x)}\right )+2 \cos ^{\frac {3}{2}}(a+b x)\right )}{3 b \cos ^{\frac {5}{2}}(a+b x)} \]
((d*Cos[a + b*x])^(5/2)*(3*ArcTan[Sqrt[Cos[a + b*x]]] - 3*ArcTanh[Sqrt[Cos [a + b*x]]] + 2*Cos[a + b*x]^(3/2)))/(3*b*Cos[a + b*x]^(5/2))
Time = 0.26 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.94, 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, 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.
\(\displaystyle \int \csc (a+b x) (d \cos (a+b x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(d \cos (a+b x))^{5/2}}{\sin (a+b x)}dx\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle -\frac {\int \frac {d^2 (d \cos (a+b x))^{5/2}}{d^2-d^2 \cos ^2(a+b x)}d(d \cos (a+b x))}{b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {d \int \frac {(d \cos (a+b x))^{5/2}}{d^2-d^2 \cos ^2(a+b x)}d(d \cos (a+b x))}{b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -\frac {d \left (d^2 \int \frac {\sqrt {d \cos (a+b x)}}{d^2-d^2 \cos ^2(a+b x)}d(d \cos (a+b x))-\frac {2}{3} (d \cos (a+b x))^{3/2}\right )}{b}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle -\frac {d \left (2 d^2 \int \frac {d^2 \cos ^2(a+b x)}{d^2-d^4 \cos ^4(a+b x)}d\sqrt {d \cos (a+b x)}-\frac {2}{3} (d \cos (a+b x))^{3/2}\right )}{b}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle -\frac {d \left (2 d^2 \left (\frac {1}{2} \int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}-\frac {1}{2} \int \frac {1}{d^2 \cos ^2(a+b x)+d}d\sqrt {d \cos (a+b x)}\right )-\frac {2}{3} (d \cos (a+b x))^{3/2}\right )}{b}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {d \left (2 d^2 \left (\frac {1}{2} \int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}-\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 \sqrt {d}}\right )-\frac {2}{3} (d \cos (a+b x))^{3/2}\right )}{b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {d \left (2 d^2 \left (\frac {\text {arctanh}\left (\sqrt {d} \cos (a+b x)\right )}{2 \sqrt {d}}-\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 \sqrt {d}}\right )-\frac {2}{3} (d \cos (a+b x))^{3/2}\right )}{b}\) |
-((d*(2*d^2*(-1/2*ArcTan[Sqrt[d]*Cos[a + b*x]]/Sqrt[d] + ArcTanh[Sqrt[d]*C os[a + b*x]]/(2*Sqrt[d])) - (2*(d*Cos[a + b*x])^(3/2))/3))/b)
3.3.24.3.1 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[(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[(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. \(252\) vs. \(2(62)=124\).
Time = 0.36 (sec) , antiderivative size = 253, normalized size of antiderivative = 3.24
method | result | size |
default | \(-\frac {3 d^{\frac {5}{2}} \ln \left (\frac {4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right ) \sqrt {-d}+3 d^{\frac {5}{2}} \ln \left (-\frac {2 \left (2 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-\sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}+d \right )}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right ) \sqrt {-d}+8 d^{2} \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}\, \sqrt {-d}\, \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-4 d^{2} \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}\, \sqrt {-d}+6 d^{3} \ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right )}{6 \sqrt {-d}\, b}\) | \(253\) |
-1/6/(-d)^(1/2)*(3*d^(5/2)*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)+3*d^(5/2)* 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)+8*d^2*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^ (1/2)*(-d)^(1/2)*sin(1/2*b*x+1/2*a)^2-4*d^2*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^ (1/2)*(-d)^(1/2)+6*d^3*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)))/b
Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (62) = 124\).
Time = 0.42 (sec) , antiderivative size = 281, normalized size of antiderivative = 3.60 \[ \int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx=\left [\frac {6 \, \sqrt {-d} d^{2} \arctan \left (\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d}}{d \cos \left (b x + a\right ) + d}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )} d^{2} \cos \left (b x + a\right ) + 3 \, \sqrt {-d} d^{2} \log \left (-\frac {d \cos \left (b x + a\right )^{2} + 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right )}{12 \, b}, -\frac {6 \, d^{\frac {5}{2}} \arctan \left (\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d}}{d \cos \left (b x + a\right ) - d}\right ) - 8 \, \sqrt {d \cos \left (b x + a\right )} d^{2} \cos \left (b x + a\right ) - 3 \, d^{\frac {5}{2}} \log \left (-\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right )}{12 \, b}\right ] \]
[1/12*(6*sqrt(-d)*d^2*arctan(2*sqrt(d*cos(b*x + a))*sqrt(-d)/(d*cos(b*x + a) + d)) + 8*sqrt(d*cos(b*x + a))*d^2*cos(b*x + a) + 3*sqrt(-d)*d^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)))/b, -1/12*(6*d^ (5/2)*arctan(2*sqrt(d*cos(b*x + a))*sqrt(d)/(d*cos(b*x + a) - d)) - 8*sqrt (d*cos(b*x + a))*d^2*cos(b*x + a) - 3*d^(5/2)*log(-(d*cos(b*x + a)^2 - 4*s qrt(d*cos(b*x + a))*sqrt(d)*(cos(b*x + a) + 1) + 6*d*cos(b*x + a) + d)/(co s(b*x + a)^2 - 2*cos(b*x + a) + 1)))/b]
Timed out. \[ \int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx=\text {Timed out} \]
Time = 0.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.06 \[ \int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx=\frac {6 \, d^{\frac {7}{2}} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right ) + 3 \, d^{\frac {7}{2}} \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right ) + 4 \, \left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}} d^{2}}{6 \, b d} \]
1/6*(6*d^(7/2)*arctan(sqrt(d*cos(b*x + a))/sqrt(d)) + 3*d^(7/2)*log((sqrt( d*cos(b*x + a)) - sqrt(d))/(sqrt(d*cos(b*x + a)) + sqrt(d))) + 4*(d*cos(b* x + a))^(3/2)*d^2)/(b*d)
\[ \int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} \csc \left (b x + a\right ) \,d x } \]
Timed out. \[ \int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx=\int \frac {{\left (d\,\cos \left (a+b\,x\right )\right )}^{5/2}}{\sin \left (a+b\,x\right )} \,d x \]