\(\int (d \cos (a+b x))^{5/2} \csc ^3(a+b x) \, dx\) [245]

Optimal result

Integrand size = 21, antiderivative size = 91 \[ \int (d \cos (a+b x))^{5/2} \csc ^3(a+b x) \, dx=-\frac {3 d^{5/2} \arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b}+\frac {3 d^{5/2} \text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b}-\frac {d (d \cos (a+b x))^{3/2} \csc ^2(a+b x)}{2 b} \] Output:

-3/4*d^(5/2)*arctan((d*cos(b*x+a))^(1/2)/d^(1/2))/b+3/4*d^(5/2)*arctanh((d 
*cos(b*x+a))^(1/2)/d^(1/2))/b-1/2*d*(d*cos(b*x+a))^(3/2)*csc(b*x+a)^2/b
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.61 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.71 \[ \int (d \cos (a+b x))^{5/2} \csc ^3(a+b x) \, dx=-\frac {d^3 \left (\cot ^2(a+b x)-3 \sqrt [4]{-\cot ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{4},\frac {5}{4},\csc ^2(a+b x)\right )\right )}{2 b \sqrt {d \cos (a+b x)}} \] Input:

Integrate[(d*Cos[a + b*x])^(5/2)*Csc[a + b*x]^3,x]
 

Output:

-1/2*(d^3*(Cot[a + b*x]^2 - 3*(-Cot[a + b*x]^2)^(1/4)*Hypergeometric2F1[1/ 
4, 1/4, 5/4, Csc[a + b*x]^2]))/(b*Sqrt[d*Cos[a + b*x]])
 

Rubi [A] (warning: unable to verify)

Time = 0.48 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3045, 27, 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.

\(\displaystyle \int \csc ^3(a+b x) (d \cos (a+b x))^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(d \cos (a+b x))^{5/2}}{\sin (a+b x)^3}dx\)

\(\Big \downarrow \) 3045

\(\displaystyle -\frac {\int \frac {d^4 (d \cos (a+b x))^{5/2}}{\left (d^2-d^2 \cos ^2(a+b x)\right )^2}d(d \cos (a+b x))}{b d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {d^3 \int \frac {(d \cos (a+b x))^{5/2}}{\left (d^2-d^2 \cos ^2(a+b x)\right )^2}d(d \cos (a+b x))}{b}\)

\(\Big \downarrow \) 252

\(\displaystyle -\frac {d^3 \left (\frac {(d \cos (a+b x))^{3/2}}{2 \left (d^2-d^2 \cos ^2(a+b x)\right )}-\frac {3}{4} \int \frac {\sqrt {d \cos (a+b x)}}{d^2-d^2 \cos ^2(a+b x)}d(d \cos (a+b x))\right )}{b}\)

\(\Big \downarrow \) 266

\(\displaystyle -\frac {d^3 \left (\frac {(d \cos (a+b x))^{3/2}}{2 \left (d^2-d^2 \cos ^2(a+b x)\right )}-\frac {3}{2} \int \frac {d^2 \cos ^2(a+b x)}{d^2-d^4 \cos ^4(a+b x)}d\sqrt {d \cos (a+b x)}\right )}{b}\)

\(\Big \downarrow \) 827

\(\displaystyle -\frac {d^3 \left (\frac {(d \cos (a+b x))^{3/2}}{2 \left (d^2-d^2 \cos ^2(a+b x)\right )}-\frac {3}{2} \left (\frac {1}{2} \int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}-\frac {1}{2} \int \frac {1}{d^2 \cos ^2(a+b x)+d}d\sqrt {d \cos (a+b x)}\right )\right )}{b}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {d^3 \left (\frac {(d \cos (a+b x))^{3/2}}{2 \left (d^2-d^2 \cos ^2(a+b x)\right )}-\frac {3}{2} \left (\frac {1}{2} \int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}-\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 \sqrt {d}}\right )\right )}{b}\)

\(\Big \downarrow \) 219

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

Input:

Int[(d*Cos[a + b*x])^(5/2)*Csc[a + b*x]^3,x]
 

Output:

-((d^3*((-3*(-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)/(2*(d^2 - d^2*Cos[a 
+ b*x]^2))))/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 252

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]
 

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. \(285\) vs. \(2(71)=142\).

Time = 9.77 (sec) , antiderivative size = 286, normalized size of antiderivative = 3.14

method	result	size
default	\(\frac {\frac {d^{2} \sqrt {2 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-d}}{8 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}+\frac {3 d^{3} \ln \left (\frac {-2 d +2 \sqrt {-d}\, \sqrt {2 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-d}}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right )}{4 \sqrt {-d}}-\frac {d^{2} \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}}{16 \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}+\frac {3 d^{\frac {5}{2}} \ln \left (\frac {-4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right )}{8}+\frac {d^{2} \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}}{16 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-16}+\frac {3 d^{\frac {5}{2}} \ln \left (\frac {4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right )}{8}}{b}\)	\(286\)

Input:

int((d*cos(b*x+a))^(5/2)*csc(b*x+a)^3,x,method=_RETURNVERBOSE)
 

Output:

(1/8*d^2/cos(1/2*b*x+1/2*a)^2*(2*d*cos(1/2*b*x+1/2*a)^2-d)^(1/2)+3/4*d^3/( 
-d)^(1/2)*ln((-2*d+2*(-d)^(1/2)*(2*d*cos(1/2*b*x+1/2*a)^2-d)^(1/2))/cos(1/ 
2*b*x+1/2*a))-1/16*d^2/(cos(1/2*b*x+1/2*a)+1)*(-2*sin(1/2*b*x+1/2*a)^2*d+d 
)^(1/2)+3/8*d^(5/2)*ln((-4*d*cos(1/2*b*x+1/2*a)+2*d^(1/2)*(-2*sin(1/2*b*x+ 
1/2*a)^2*d+d)^(1/2)-2*d)/(cos(1/2*b*x+1/2*a)+1))+1/16*d^2/(cos(1/2*b*x+1/2 
*a)-1)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)+3/8*d^(5/2)*ln((4*d*cos(1/2*b*x 
+1/2*a)+2*d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-2*d)/(cos(1/2*b*x+1/ 
2*a)-1)))/b
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (71) = 142\).

Time = 0.13 (sec) , antiderivative size = 371, normalized size of antiderivative = 4.08 \[ \int (d \cos (a+b x))^{5/2} \csc ^3(a+b x) \, dx=\left [\frac {8 \, \sqrt {d \cos \left (b x + a\right )} d^{2} \cos \left (b x + a\right ) - 6 \, {\left (d^{2} \cos \left (b x + a\right )^{2} - d^{2}\right )} \sqrt {-d} \arctan \left (\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d}}{d \cos \left (b x + a\right ) + d}\right ) + 3 \, {\left (d^{2} \cos \left (b x + a\right )^{2} - d^{2}\right )} \sqrt {-d} \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right )}{16 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}}, \frac {8 \, \sqrt {d \cos \left (b x + a\right )} d^{2} \cos \left (b x + a\right ) + 6 \, {\left (d^{2} \cos \left (b x + a\right )^{2} - d^{2}\right )} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d}}{d \cos \left (b x + a\right ) - d}\right ) + 3 \, {\left (d^{2} \cos \left (b x + a\right )^{2} - d^{2}\right )} \sqrt {d} \log \left (\frac {d \cos \left (b x + a\right )^{2} + 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right )}{16 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}}\right ] \] Input:

integrate((d*cos(b*x+a))^(5/2)*csc(b*x+a)^3,x, algorithm="fricas")
 

Output:

[1/16*(8*sqrt(d*cos(b*x + a))*d^2*cos(b*x + a) - 6*(d^2*cos(b*x + a)^2 - d 
^2)*sqrt(-d)*arctan(2*sqrt(d*cos(b*x + a))*sqrt(-d)/(d*cos(b*x + a) + d)) 
+ 3*(d^2*cos(b*x + a)^2 - d^2)*sqrt(-d)*log((d*cos(b*x + a)^2 - 4*sqrt(d*c 
os(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*cos(b*x + a)^2 - b), 1/16*(8*sqrt(d*cos( 
b*x + a))*d^2*cos(b*x + a) + 6*(d^2*cos(b*x + a)^2 - d^2)*sqrt(d)*arctan(2 
*sqrt(d*cos(b*x + a))*sqrt(d)/(d*cos(b*x + a) - d)) + 3*(d^2*cos(b*x + a)^ 
2 - d^2)*sqrt(d)*log((d*cos(b*x + a)^2 + 4*sqrt(d*cos(b*x + a))*sqrt(d)*(c 
os(b*x + a) + 1) + 6*d*cos(b*x + a) + d)/(cos(b*x + a)^2 - 2*cos(b*x + a) 
+ 1)))/(b*cos(b*x + a)^2 - b)]
 

Sympy [F(-1)]

Timed out. \[ \int (d \cos (a+b x))^{5/2} \csc ^3(a+b x) \, dx=\text {Timed out} \] Input:

integrate((d*cos(b*x+a))**(5/2)*csc(b*x+a)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.13 \[ \int (d \cos (a+b x))^{5/2} \csc ^3(a+b x) \, dx=\frac {\frac {4 \, \left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}} d^{4}}{d^{2} \cos \left (b x + a\right )^{2} - d^{2}} - 6 \, d^{\frac {7}{2}} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right ) - 3 \, d^{\frac {7}{2}} \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right )}{8 \, b d} \] Input:

integrate((d*cos(b*x+a))^(5/2)*csc(b*x+a)^3,x, algorithm="maxima")
 

Output:

1/8*(4*(d*cos(b*x + a))^(3/2)*d^4/(d^2*cos(b*x + a)^2 - d^2) - 6*d^(7/2)*a 
rctan(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))))/(b*d)
 

Giac [F]

\[ \int (d \cos (a+b x))^{5/2} \csc ^3(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} \csc \left (b x + a\right )^{3} \,d x } \] Input:

integrate((d*cos(b*x+a))^(5/2)*csc(b*x+a)^3,x, algorithm="giac")
 

Output:

integrate((d*cos(b*x + a))^(5/2)*csc(b*x + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int (d \cos (a+b x))^{5/2} \csc ^3(a+b x) \, dx=\int \frac {{\left (d\,\cos \left (a+b\,x\right )\right )}^{5/2}}{{\sin \left (a+b\,x\right )}^3} \,d x \] Input:

int((d*cos(a + b*x))^(5/2)/sin(a + b*x)^3,x)
 

Output:

int((d*cos(a + b*x))^(5/2)/sin(a + b*x)^3, x)
 

Reduce [F]

\[ \int (d \cos (a+b x))^{5/2} \csc ^3(a+b x) \, dx=\sqrt {d}\, \left (\int \sqrt {\cos \left (b x +a \right )}\, \cos \left (b x +a \right )^{2} \csc \left (b x +a \right )^{3}d x \right ) d^{2} \] Input:

int((d*cos(b*x+a))^(5/2)*csc(b*x+a)^3,x)
 

Output:

sqrt(d)*int(sqrt(cos(a + b*x))*cos(a + b*x)**2*csc(a + b*x)**3,x)*d**2