\(\int \frac {\csc (a+b x)}{(d \cos (a+b x))^{3/2}} \, dx\) [228]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 19, antiderivative size = 78 \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{3/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{3/2}}+\frac {2}{b d \sqrt {d \cos (a+b x)}} \]

[Out]

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/d^(3/2)+2/b/d/(d*cos(b*
x+a))^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2645, 331, 335, 304, 209, 212} \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{3/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{3/2}}+\frac {2}{b d \sqrt {d \cos (a+b x)}} \]

[In]

Int[Csc[a + b*x]/(d*Cos[a + b*x])^(3/2),x]

[Out]

ArcTan[Sqrt[d*Cos[a + b*x]]/Sqrt[d]]/(b*d^(3/2)) - ArcTanh[Sqrt[d*Cos[a + b*x]]/Sqrt[d]]/(b*d^(3/2)) + 2/(b*d*
Sqrt[d*Cos[a + b*x]])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
 FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
 && FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(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])

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {1}{x^{3/2} \left (1-\frac {x^2}{d^2}\right )} \, dx,x,d \cos (a+b x)\right )}{b d} \\ & = \frac {2}{b d \sqrt {d \cos (a+b x)}}-\frac {\text {Subst}\left (\int \frac {\sqrt {x}}{1-\frac {x^2}{d^2}} \, dx,x,d \cos (a+b x)\right )}{b d^3} \\ & = \frac {2}{b d \sqrt {d \cos (a+b x)}}-\frac {2 \text {Subst}\left (\int \frac {x^2}{1-\frac {x^4}{d^2}} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{b d^3} \\ & = \frac {2}{b d \sqrt {d \cos (a+b x)}}-\frac {\text {Subst}\left (\int \frac {1}{d-x^2} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{b d}+\frac {\text {Subst}\left (\int \frac {1}{d+x^2} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{b d} \\ & = \frac {\arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{3/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{3/2}}+\frac {2}{b d \sqrt {d \cos (a+b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.85 \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\frac {2+\arctan \left (\sqrt {\cos (a+b x)}\right ) \sqrt {\cos (a+b x)}-\text {arctanh}\left (\sqrt {\cos (a+b x)}\right ) \sqrt {\cos (a+b x)}}{b d \sqrt {d \cos (a+b x)}} \]

[In]

Integrate[Csc[a + b*x]/(d*Cos[a + b*x])^(3/2),x]

[Out]

(2 + ArcTan[Sqrt[Cos[a + b*x]]]*Sqrt[Cos[a + b*x]] - ArcTanh[Sqrt[Cos[a + b*x]]]*Sqrt[Cos[a + b*x]])/(b*d*Sqrt
[d*Cos[a + b*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(440\) vs. \(2(64)=128\).

Time = 0.10 (sec) , antiderivative size = 441, normalized size of antiderivative = 5.65

method result size
default \(-\frac {4 d^{\frac {5}{2}} \ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+2 \sqrt {-d}\, \ln \left (-\frac {2 \left (2 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-\sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}+d \right )}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d^{2}+2 \sqrt {-d}\, \ln \left (\frac {4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d^{2}-2 \ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right ) d^{\frac {5}{2}}+4 \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}\, \sqrt {-d}\, d^{\frac {3}{2}}-\ln \left (-\frac {2 \left (2 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-\sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}+d \right )}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right ) \sqrt {-d}\, d^{2}-\ln \left (\frac {4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right ) \sqrt {-d}\, d^{2}}{2 \sqrt {-d}\, d^{\frac {7}{2}} \left (2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) b}\) \(441\)

[In]

int(csc(b*x+a)/(d*cos(b*x+a))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/2/(-d)^(1/2)/d^(7/2)/(2*sin(1/2*b*x+1/2*a)^2-1)*(4*d^(5/2)*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))*sin(1/2*b*x+1/2*a)^2+2*(-d)^(1/2)*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))*sin(1/2*b*x+1/2*a)^2*d^2+2*(-d)^(1/2)*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))*sin(1/2*b*x+1/2*a)^2*d^2-2*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^(5/2)+4*(-2*d*sin(1/2*b*x+1/2*a)^2
+d)^(1/2)*(-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)*d^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^2)/b

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 148 vs. \(2 (64) = 128\).

Time = 0.32 (sec) , antiderivative size = 309, normalized size of antiderivative = 3.96 \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\left [\frac {2 \, \sqrt {-d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) + 1\right )}}{2 \, d \cos \left (b x + a\right )}\right ) \cos \left (b x + a\right ) - \sqrt {-d} \cos \left (b x + a\right ) \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )}}{4 \, b d^{2} \cos \left (b x + a\right )}, \frac {2 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - 1\right )}}{2 \, \sqrt {d} \cos \left (b x + a\right )}\right ) \cos \left (b x + a\right ) + \sqrt {d} \cos \left (b x + a\right ) \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )}}{4 \, b d^{2} \cos \left (b x + a\right )}\right ] \]

[In]

integrate(csc(b*x+a)/(d*cos(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

[1/4*(2*sqrt(-d)*arctan(1/2*sqrt(d*cos(b*x + a))*sqrt(-d)*(cos(b*x + a) + 1)/(d*cos(b*x + a)))*cos(b*x + a) -
sqrt(-d)*cos(b*x + a)*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)))/(b*d^2*cos(b*x + a)), 1/4*(2*sqrt(
d)*arctan(1/2*sqrt(d*cos(b*x + a))*(cos(b*x + a) - 1)/(sqrt(d)*cos(b*x + a)))*cos(b*x + a) + sqrt(d)*cos(b*x +
 a)*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)))/(b*d^2*cos(b*x + a))]

Sympy [F]

\[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\int \frac {\csc {\left (a + b x \right )}}{\left (d \cos {\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(csc(b*x+a)/(d*cos(b*x+a))**(3/2),x)

[Out]

Integral(csc(a + b*x)/(d*cos(a + b*x))**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.01 \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\frac {\frac {2 \, \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right )}{\sqrt {d}} + \frac {\log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right )}{\sqrt {d}} + \frac {4}{\sqrt {d \cos \left (b x + a\right )}}}{2 \, b d} \]

[In]

integrate(csc(b*x+a)/(d*cos(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

1/2*(2*arctan(sqrt(d*cos(b*x + a))/sqrt(d))/sqrt(d) + log((sqrt(d*cos(b*x + a)) - sqrt(d))/(sqrt(d*cos(b*x + a
)) + sqrt(d)))/sqrt(d) + 4/sqrt(d*cos(b*x + a)))/(b*d)

Giac [F]

\[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\int { \frac {\csc \left (b x + a\right )}{\left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate(csc(b*x+a)/(d*cos(b*x+a))^(3/2),x, algorithm="giac")

[Out]

integrate(csc(b*x + a)/(d*cos(b*x + a))^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\int \frac {1}{\sin \left (a+b\,x\right )\,{\left (d\,\cos \left (a+b\,x\right )\right )}^{3/2}} \,d x \]

[In]

int(1/(sin(a + b*x)*(d*cos(a + b*x))^(3/2)),x)

[Out]

int(1/(sin(a + b*x)*(d*cos(a + b*x))^(3/2)), x)