\(\int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx\) [224]

Optimal result

Integrand size = 19, antiderivative size = 78 \[ \int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx=\frac {d^{5/2} \arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b}-\frac {d^{5/2} \text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b}+\frac {2 d (d \cos (a+b x))^{3/2}}{3 b} \] Output:

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
 

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.87 \[ \int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx=\frac {(d \cos (a+b x))^{5/2} \left (3 \arctan \left (\sqrt {\cos (a+b x)}\right )-3 \text {arctanh}\left (\sqrt {\cos (a+b x)}\right )+2 \cos ^{\frac {3}{2}}(a+b x)\right )}{3 b \cos ^{\frac {5}{2}}(a+b x)} \] Input:

Integrate[(d*Cos[a + b*x])^(5/2)*Csc[a + b*x],x]
 

Output:

((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))
 

Rubi [A] (warning: unable to verify)

Time = 0.46 (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

\(\displaystyle -\frac {d \left (2 d^2 \left (\frac {\text {arctanh}\left (\sqrt {d} \cos (a+b x)\right )}{2 \sqrt {d}}-\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 \sqrt {d}}\right )-\frac {2}{3} (d \cos (a+b x))^{3/2}\right )}{b}\)

Input:

Int[(d*Cos[a + b*x])^(5/2)*Csc[a + b*x],x]
 

Output:

-((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)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

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])
 

rule 219

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])
 

rule 262

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]
 

rule 266

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]
 

rule 827

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3045

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])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(252\) vs. \(2(62)=124\).

Time = 3.37 (sec) , antiderivative size = 253, normalized size of antiderivative = 3.24

method	result	size
default	\(-\frac {3 d^{\frac {5}{2}} \ln \left (\frac {4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right ) \sqrt {-d}+3 d^{\frac {5}{2}} \ln \left (-\frac {2 \left (2 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-\sqrt {d}\, \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}+d \right )}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right ) \sqrt {-d}+8 \sqrt {-d}\, \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d^{2}-4 d^{2} \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}\, \sqrt {-d}+6 d^{3} \ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right )}{6 \sqrt {-d}\, b}\)	\(253\)

Input:

int((d*cos(b*x+a))^(5/2)*csc(b*x+a),x,method=_RETURNVERBOSE)
 

Output:

-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*sin(1/2*b*x+1/2*a)^2*d+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*sin(1/2*b 
*x+1/2*a)^2*d+d)^(1/2)+d))*(-d)^(1/2)+8*(-d)^(1/2)*(-2*sin(1/2*b*x+1/2*a)^ 
2*d+d)^(1/2)*sin(1/2*b*x+1/2*a)^2*d^2-4*d^2*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^ 
(1/2)*(-d)^(1/2)+6*d^3*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*sin(1/2*b*x 
+1/2*a)^2*d+d)^(1/2)-d)))/b
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (62) = 124\).

Time = 0.19 (sec) , antiderivative size = 290, normalized size of antiderivative = 3.72 \[ \int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx=\left [\frac {6 \, \sqrt {-d} d^{2} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) + 1\right )}}{2 \, d \cos \left (b x + a\right )}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )} d^{2} \cos \left (b x + a\right ) + 3 \, \sqrt {-d} d^{2} \log \left (-\frac {d \cos \left (b x + a\right )^{2} + 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right )}{12 \, b}, \frac {6 \, d^{\frac {5}{2}} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - 1\right )}}{2 \, \sqrt {d} \cos \left (b x + a\right )}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )} d^{2} \cos \left (b x + a\right ) + 3 \, d^{\frac {5}{2}} \log \left (-\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right )}{12 \, b}\right ] \] Input:

integrate((d*cos(b*x+a))^(5/2)*csc(b*x+a),x, algorithm="fricas")
 

Output:

[1/12*(6*sqrt(-d)*d^2*arctan(1/2*sqrt(d*cos(b*x + a))*sqrt(-d)*(cos(b*x + 
a) + 1)/(d*cos(b*x + a))) + 8*sqrt(d*cos(b*x + a))*d^2*cos(b*x + a) + 3*sq 
rt(-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(1/2*sqrt(d*cos(b*x + a))*(cos(b*x + a) - 1)/( 
sqrt(d)*cos(b*x + a))) + 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*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]
 

Sympy [F(-1)]

Timed out. \[ \int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx=\text {Timed out} \] Input:

integrate((d*cos(b*x+a))**(5/2)*csc(b*x+a),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.06 \[ \int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx=\frac {6 \, d^{\frac {7}{2}} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right ) + 3 \, d^{\frac {7}{2}} \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right ) + 4 \, \left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}} d^{2}}{6 \, b d} \] Input:

integrate((d*cos(b*x+a))^(5/2)*csc(b*x+a),x, algorithm="maxima")
 

Output:

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)
 

Giac [F]

\[ \int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} \csc \left (b x + a\right ) \,d x } \] Input:

integrate((d*cos(b*x+a))^(5/2)*csc(b*x+a),x, algorithm="giac")
 

Output:

integrate((d*cos(b*x + a))^(5/2)*csc(b*x + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx=\int \frac {{\left (d\,\cos \left (a+b\,x\right )\right )}^{5/2}}{\sin \left (a+b\,x\right )} \,d x \] Input:

int((d*cos(a + b*x))^(5/2)/sin(a + b*x),x)
 

Output:

int((d*cos(a + b*x))^(5/2)/sin(a + b*x), x)
 

Reduce [F]

\[ \int (d \cos (a+b x))^{5/2} \csc (a+b x) \, dx=\sqrt {d}\, \left (\int \sqrt {\cos \left (b x +a \right )}\, \cos \left (b x +a \right )^{2} \csc \left (b x +a \right )d x \right ) d^{2} \] Input:

int((d*cos(b*x+a))^(5/2)*csc(b*x+a),x)
 

Output:

sqrt(d)*int(sqrt(cos(a + b*x))*cos(a + b*x)**2*csc(a + b*x),x)*d**2