\(\int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx\) [250]

Optimal result

Integrand size = 21, antiderivative size = 115 \[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=-\frac {7 \arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b d^{5/2}}-\frac {7 \text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b d^{5/2}}+\frac {7}{6 b d (d \cos (a+b x))^{3/2}}-\frac {\csc ^2(a+b x)}{2 b d (d \cos (a+b x))^{3/2}} \] Output:

-7/4*arctan((d*cos(b*x+a))^(1/2)/d^(1/2))/b/d^(5/2)-7/4*arctanh((d*cos(b*x 
+a))^(1/2)/d^(1/2))/b/d^(5/2)+7/6/b/d/(d*cos(b*x+a))^(3/2)-1/2*csc(b*x+a)^ 
2/b/d/(d*cos(b*x+a))^(3/2)
 

Mathematica [C] (verified)

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

Time = 0.71 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.80 \[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\frac {\sqrt [4]{-\cot ^2(a+b x)} \left (4-3 \cot ^2(a+b x)\right )+7 \cot ^2(a+b x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{4},\frac {7}{4},\csc ^2(a+b x)\right )}{6 b d (d \cos (a+b x))^{3/2} \sqrt [4]{-\cot ^2(a+b x)}} \] Input:

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

Output:

((-Cot[a + b*x]^2)^(1/4)*(4 - 3*Cot[a + b*x]^2) + 7*Cot[a + b*x]^2*Hyperge 
ometric2F1[3/4, 3/4, 7/4, Csc[a + b*x]^2])/(6*b*d*(d*Cos[a + b*x])^(3/2)*( 
-Cot[a + b*x]^2)^(1/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.47 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3045, 27, 253, 264, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (a+b x)^3 (d \cos (a+b x))^{5/2}}dx\)

\(\Big \downarrow \) 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

\(\displaystyle -\frac {d^3 \int \frac {1}{(d \cos (a+b x))^{5/2} \left (d^2-d^2 \cos ^2(a+b x)\right )^2}d(d \cos (a+b x))}{b}\)

\(\Big \downarrow \) 253

\(\displaystyle -\frac {d^3 \left (\frac {7 \int \frac {1}{(d \cos (a+b x))^{5/2} \left (d^2-d^2 \cos ^2(a+b x)\right )}d(d \cos (a+b x))}{4 d^2}+\frac {1}{2 d^2 (d \cos (a+b x))^{3/2} \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\)

\(\Big \downarrow \) 264

\(\displaystyle -\frac {d^3 \left (\frac {7 \left (\frac {\int \frac {1}{\sqrt {d \cos (a+b x)} \left (d^2-d^2 \cos ^2(a+b x)\right )}d(d \cos (a+b x))}{d^2}-\frac {2}{3 d^2 (d \cos (a+b x))^{3/2}}\right )}{4 d^2}+\frac {1}{2 d^2 (d \cos (a+b x))^{3/2} \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\)

\(\Big \downarrow \) 266

\(\displaystyle -\frac {d^3 \left (\frac {7 \left (\frac {2 \int \frac {1}{d^2-d^4 \cos ^4(a+b x)}d\sqrt {d \cos (a+b x)}}{d^2}-\frac {2}{3 d^2 (d \cos (a+b x))^{3/2}}\right )}{4 d^2}+\frac {1}{2 d^2 (d \cos (a+b x))^{3/2} \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\)

\(\Big \downarrow \) 756

\(\displaystyle -\frac {d^3 \left (\frac {7 \left (\frac {2 \left (\frac {\int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}}{2 d}+\frac {\int \frac {1}{d^2 \cos ^2(a+b x)+d}d\sqrt {d \cos (a+b x)}}{2 d}\right )}{d^2}-\frac {2}{3 d^2 (d \cos (a+b x))^{3/2}}\right )}{4 d^2}+\frac {1}{2 d^2 (d \cos (a+b x))^{3/2} \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {d^3 \left (\frac {7 \left (\frac {2 \left (\frac {\int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}}{2 d}+\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 d^{3/2}}\right )}{d^2}-\frac {2}{3 d^2 (d \cos (a+b x))^{3/2}}\right )}{4 d^2}+\frac {1}{2 d^2 (d \cos (a+b x))^{3/2} \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {d^3 \left (\frac {7 \left (\frac {2 \left (\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 d^{3/2}}+\frac {\text {arctanh}\left (\sqrt {d} \cos (a+b x)\right )}{2 d^{3/2}}\right )}{d^2}-\frac {2}{3 d^2 (d \cos (a+b x))^{3/2}}\right )}{4 d^2}+\frac {1}{2 d^2 (d \cos (a+b x))^{3/2} \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\)

Input:

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

Output:

-((d^3*(1/(2*d^2*(d*Cos[a + b*x])^(3/2)*(d^2 - d^2*Cos[a + b*x]^2)) + (7*( 
(2*(ArcTan[Sqrt[d]*Cos[a + b*x]]/(2*d^(3/2)) + ArcTanh[Sqrt[d]*Cos[a + b*x 
]]/(2*d^(3/2))))/d^2 - 2/(3*d^2*(d*Cos[a + b*x])^(3/2))))/(4*d^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 253

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x 
)^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 
2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m 
}, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && 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 756

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]
 

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. \(884\) vs. \(2(91)=182\).

Time = 3.51 (sec) , antiderivative size = 885, normalized size of antiderivative = 7.70

method	result	size
default	\(\text {Expression too large to display}\)	\(885\)

Input:

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

Output:

-1/24/d^(7/2)/(-d)^(1/2)/sin(1/2*b*x+1/2*a)^2/(4*sin(1/2*b*x+1/2*a)^6-8*si 
n(1/2*b*x+1/2*a)^4+5*sin(1/2*b*x+1/2*a)^2-1)*(-3*(-2*sin(1/2*b*x+1/2*a)^2* 
d+d)^(1/2)*(-d)^(1/2)*d^(1/2)+84*(-2*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))*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*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)+d))* 
(-d)^(1/2)*d+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)*d)*sin(1/2*b*x+1/2*a)^8- 
168*(-2*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))*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*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)+d))*(-d)^(1/2)*d+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)*d)*sin(1/2*b*x+1/2*a)^6-7*(-6*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))*d^(3/2)+3*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)*d-4*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*(-d)^(1 
/2)*d^(1/2)+3*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)*d)*sin(1/2*b*x+1/2*a)^2 
+7*(-30*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))*d^(3/2)+15*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)*d-4*(-2*sin(...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (91) = 182\).

Time = 0.14 (sec) , antiderivative size = 409, normalized size of antiderivative = 3.56 \[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\left [\frac {42 \, {\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \sqrt {-d} \arctan \left (\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d}}{d \cos \left (b x + a\right ) + d}\right ) - 21 \, {\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \sqrt {-d} \log \left (\frac {d \cos \left (b x + a\right )^{2} + 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )} {\left (7 \, \cos \left (b x + a\right )^{2} - 4\right )}}{48 \, {\left (b d^{3} \cos \left (b x + a\right )^{4} - b d^{3} \cos \left (b x + a\right )^{2}\right )}}, \frac {42 \, {\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d}}{d \cos \left (b x + a\right ) - d}\right ) + 21 \, {\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \sqrt {d} \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )} {\left (7 \, \cos \left (b x + a\right )^{2} - 4\right )}}{48 \, {\left (b d^{3} \cos \left (b x + a\right )^{4} - b d^{3} \cos \left (b x + a\right )^{2}\right )}}\right ] \] Input:

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

Output:

[1/48*(42*(cos(b*x + a)^4 - cos(b*x + a)^2)*sqrt(-d)*arctan(2*sqrt(d*cos(b 
*x + a))*sqrt(-d)/(d*cos(b*x + a) + d)) - 21*(cos(b*x + a)^4 - cos(b*x + a 
)^2)*sqrt(-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))*(7*cos(b*x + a)^2 - 4))/(b*d^3*cos(b*x + a)^4 
 - b*d^3*cos(b*x + a)^2), 1/48*(42*(cos(b*x + a)^4 - cos(b*x + a)^2)*sqrt( 
d)*arctan(2*sqrt(d*cos(b*x + a))*sqrt(d)/(d*cos(b*x + a) - d)) + 21*(cos(b 
*x + a)^4 - cos(b*x + a)^2)*sqrt(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))*(7*cos(b*x + a)^2 - 4))/ 
(b*d^3*cos(b*x + a)^4 - b*d^3*cos(b*x + a)^2)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.02 \[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\frac {\frac {4 \, {\left (7 \, d^{2} \cos \left (b x + a\right )^{2} - 4 \, d^{2}\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {7}{2}} - \left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}} d^{2}} - \frac {42 \, \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right )}{d^{\frac {3}{2}}} + \frac {21 \, \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right )}{d^{\frac {3}{2}}}}{24 \, b d} \] Input:

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

Output:

1/24*(4*(7*d^2*cos(b*x + a)^2 - 4*d^2)/((d*cos(b*x + a))^(7/2) - (d*cos(b* 
x + a))^(3/2)*d^2) - 42*arctan(sqrt(d*cos(b*x + a))/sqrt(d))/d^(3/2) + 21* 
log((sqrt(d*cos(b*x + a)) - sqrt(d))/(sqrt(d*cos(b*x + a)) + sqrt(d)))/d^( 
3/2))/(b*d)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx=\int \frac {1}{{\sin \left (a+b\,x\right )}^3\,{\left (d\,\cos \left (a+b\,x\right )\right )}^{5/2}} \,d x \] Input:

int(1/(sin(a + b*x)^3*(d*cos(a + b*x))^(5/2)),x)
 

Output:

int(1/(sin(a + b*x)^3*(d*cos(a + b*x))^(5/2)), x)
 

Reduce [F]

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

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

Output:

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