\(\int (d \cos (a+b x))^{11/2} \csc ^3(a+b x) \, dx\) [242]

Optimal result

Integrand size = 21, antiderivative size = 135 \[ \int (d \cos (a+b x))^{11/2} \csc ^3(a+b x) \, dx=\frac {9 d^{11/2} \arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b}+\frac {9 d^{11/2} \text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b}-\frac {9 d^5 \sqrt {d \cos (a+b x)}}{2 b}-\frac {9 d^3 (d \cos (a+b x))^{5/2}}{10 b}-\frac {d (d \cos (a+b x))^{9/2} \csc ^2(a+b x)}{2 b} \] Output:

9/4*d^(11/2)*arctan((d*cos(b*x+a))^(1/2)/d^(1/2))/b+9/4*d^(11/2)*arctanh(( 
d*cos(b*x+a))^(1/2)/d^(1/2))/b-9/2*d^5*(d*cos(b*x+a))^(1/2)/b-9/10*d^3*(d* 
cos(b*x+a))^(5/2)/b-1/2*d*(d*cos(b*x+a))^(9/2)*csc(b*x+a)^2/b
 

Mathematica [A] (verified)

Time = 3.09 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.01 \[ \int (d \cos (a+b x))^{11/2} \csc ^3(a+b x) \, dx=\frac {d (d \cos (a+b x))^{9/2} \left (45 \arctan \left (\sqrt {\cos (a+b x)}\right )+24 \text {arctanh}\left (\sqrt {\cos (a+b x)}\right )-2 \sqrt {\cos (a+b x)} \left (2 \cos (2 (a+b x))+5 \csc ^2(a+b x)\right )-\frac {21}{2} \left (8 \sqrt {\cos (a+b x)}+\log \left (1-\sqrt {\cos (a+b x)}\right )-\log \left (1+\sqrt {\cos (a+b x)}\right )\right )\right )}{20 b \cos ^{\frac {9}{2}}(a+b x)} \] Input:

Integrate[(d*Cos[a + b*x])^(11/2)*Csc[a + b*x]^3,x]
 

Output:

(d*(d*Cos[a + b*x])^(9/2)*(45*ArcTan[Sqrt[Cos[a + b*x]]] + 24*ArcTanh[Sqrt 
[Cos[a + b*x]]] - 2*Sqrt[Cos[a + b*x]]*(2*Cos[2*(a + b*x)] + 5*Csc[a + b*x 
]^2) - (21*(8*Sqrt[Cos[a + b*x]] + Log[1 - Sqrt[Cos[a + b*x]]] - Log[1 + S 
qrt[Cos[a + b*x]]]))/2))/(20*b*Cos[a + b*x]^(9/2))
 

Rubi [A] (warning: unable to verify)

Time = 0.51 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.99, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3045, 27, 252, 262, 262, 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.

\(\displaystyle \int \csc ^3(a+b x) (d \cos (a+b x))^{11/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(d \cos (a+b x))^{11/2}}{\sin (a+b x)^3}dx\)

\(\Big \downarrow \) 3045

\(\displaystyle -\frac {\int \frac {d^4 (d \cos (a+b x))^{11/2}}{\left (d^2-d^2 \cos ^2(a+b x)\right )^2}d(d \cos (a+b x))}{b d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {d^3 \int \frac {(d \cos (a+b x))^{11/2}}{\left (d^2-d^2 \cos ^2(a+b x)\right )^2}d(d \cos (a+b x))}{b}\)

\(\Big \downarrow \) 252

\(\displaystyle -\frac {d^3 \left (\frac {(d \cos (a+b x))^{9/2}}{2 \left (d^2-d^2 \cos ^2(a+b x)\right )}-\frac {9}{4} \int \frac {(d \cos (a+b x))^{7/2}}{d^2-d^2 \cos ^2(a+b x)}d(d \cos (a+b x))\right )}{b}\)

\(\Big \downarrow \) 262

\(\displaystyle -\frac {d^3 \left (\frac {(d \cos (a+b x))^{9/2}}{2 \left (d^2-d^2 \cos ^2(a+b x)\right )}-\frac {9}{4} \left (d^2 \int \frac {(d \cos (a+b x))^{3/2}}{d^2-d^2 \cos ^2(a+b x)}d(d \cos (a+b x))-\frac {2}{5} (d \cos (a+b x))^{5/2}\right )\right )}{b}\)

\(\Big \downarrow \) 262

\(\displaystyle -\frac {d^3 \left (\frac {(d \cos (a+b x))^{9/2}}{2 \left (d^2-d^2 \cos ^2(a+b x)\right )}-\frac {9}{4} \left (d^2 \left (d^2 \int \frac {1}{\sqrt {d \cos (a+b x)} \left (d^2-d^2 \cos ^2(a+b x)\right )}d(d \cos (a+b x))-2 \sqrt {d \cos (a+b x)}\right )-\frac {2}{5} (d \cos (a+b x))^{5/2}\right )\right )}{b}\)

\(\Big \downarrow \) 266

\(\displaystyle -\frac {d^3 \left (\frac {(d \cos (a+b x))^{9/2}}{2 \left (d^2-d^2 \cos ^2(a+b x)\right )}-\frac {9}{4} \left (d^2 \left (2 d^2 \int \frac {1}{d^2-d^4 \cos ^4(a+b x)}d\sqrt {d \cos (a+b x)}-2 \sqrt {d \cos (a+b x)}\right )-\frac {2}{5} (d \cos (a+b x))^{5/2}\right )\right )}{b}\)

\(\Big \downarrow \) 756

\(\displaystyle -\frac {d^3 \left (\frac {(d \cos (a+b x))^{9/2}}{2 \left (d^2-d^2 \cos ^2(a+b x)\right )}-\frac {9}{4} \left (d^2 \left (2 d^2 \left (\frac {\int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}}{2 d}+\frac {\int \frac {1}{d^2 \cos ^2(a+b x)+d}d\sqrt {d \cos (a+b x)}}{2 d}\right )-2 \sqrt {d \cos (a+b x)}\right )-\frac {2}{5} (d \cos (a+b x))^{5/2}\right )\right )}{b}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {d^3 \left (\frac {(d \cos (a+b x))^{9/2}}{2 \left (d^2-d^2 \cos ^2(a+b x)\right )}-\frac {9}{4} \left (d^2 \left (2 d^2 \left (\frac {\int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}}{2 d}+\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 d^{3/2}}\right )-2 \sqrt {d \cos (a+b x)}\right )-\frac {2}{5} (d \cos (a+b x))^{5/2}\right )\right )}{b}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {d^3 \left (\frac {(d \cos (a+b x))^{9/2}}{2 \left (d^2-d^2 \cos ^2(a+b x)\right )}-\frac {9}{4} \left (d^2 \left (2 d^2 \left (\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 d^{3/2}}+\frac {\text {arctanh}\left (\sqrt {d} \cos (a+b x)\right )}{2 d^{3/2}}\right )-2 \sqrt {d \cos (a+b x)}\right )-\frac {2}{5} (d \cos (a+b x))^{5/2}\right )\right )}{b}\)

Input:

Int[(d*Cos[a + b*x])^(11/2)*Csc[a + b*x]^3,x]
 

Output:

-((d^3*((d*Cos[a + b*x])^(9/2)/(2*(d^2 - d^2*Cos[a + b*x]^2)) - (9*((-2*(d 
*Cos[a + b*x])^(5/2))/5 + d^2*(2*d^2*(ArcTan[Sqrt[d]*Cos[a + b*x]]/(2*d^(3 
/2)) + ArcTanh[Sqrt[d]*Cos[a + b*x]]/(2*d^(3/2))) - 2*Sqrt[d*Cos[a + b*x]] 
)))/4))/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 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 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. \(406\) vs. \(2(107)=214\).

Time = 9.64 (sec) , antiderivative size = 407, normalized size of antiderivative = 3.01

method	result	size
default	\(\frac {-\frac {d^{5} \sqrt {2 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-d}}{8 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}-\frac {9 d^{6} \ln \left (\frac {-2 d +2 \sqrt {-d}\, \sqrt {2 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-d}}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right )}{4 \sqrt {-d}}-6 d^{5} \sqrt {d \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )}+\frac {8 d^{5} \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} \sqrt {2 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-d}}{5}+\frac {8 d^{5} \sqrt {2 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-d}}{5}-\frac {8 d^{5} \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{4} \sqrt {2 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-d}}{5}-\frac {d^{5} \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}}{16 \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}+\frac {9 d^{\frac {11}{2}} \ln \left (\frac {-4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right )}{8}+\frac {d^{5} \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}}{16 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-16}+\frac {9 d^{\frac {11}{2}} \ln \left (\frac {4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right )}{8}}{b}\)	\(407\)

Input:

int((d*cos(b*x+a))^(11/2)*csc(b*x+a)^3,x,method=_RETURNVERBOSE)
 

Output:

(-1/8*d^5/cos(1/2*b*x+1/2*a)^2*(2*d*cos(1/2*b*x+1/2*a)^2-d)^(1/2)-9/4*d^6/ 
(-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))-6*d^5*(d*(2*cos(1/2*b*x+1/2*a)^2-1))^(1/2)+8/5*d^5*cos(1/2* 
b*x+1/2*a)^2*(2*d*cos(1/2*b*x+1/2*a)^2-d)^(1/2)+8/5*d^5*(2*d*cos(1/2*b*x+1 
/2*a)^2-d)^(1/2)-8/5*d^5*cos(1/2*b*x+1/2*a)^4*(2*d*cos(1/2*b*x+1/2*a)^2-d) 
^(1/2)-1/16*d^5/(cos(1/2*b*x+1/2*a)+1)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2) 
+9/8*d^(11/2)*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^5/(cos(1/2*b*x+1/2*a)-1) 
*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)+9/8*d^(11/2)*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 [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 428, normalized size of antiderivative = 3.17 \[ \int (d \cos (a+b x))^{11/2} \csc ^3(a+b x) \, dx=\left [-\frac {90 \, {\left (d^{5} \cos \left (b x + a\right )^{2} - d^{5}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) + 1\right )}}{2 \, d \cos \left (b x + a\right )}\right ) - 45 \, {\left (d^{5} \cos \left (b x + a\right )^{2} - d^{5}\right )} \sqrt {-d} \log \left (-\frac {d \cos \left (b x + a\right )^{2} + 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, {\left (4 \, d^{5} \cos \left (b x + a\right )^{4} + 36 \, d^{5} \cos \left (b x + a\right )^{2} - 45 \, d^{5}\right )} \sqrt {d \cos \left (b x + a\right )}}{80 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}}, \frac {90 \, {\left (d^{5} \cos \left (b x + a\right )^{2} - d^{5}\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - 1\right )}}{2 \, \sqrt {d} \cos \left (b x + a\right )}\right ) + 45 \, {\left (d^{5} \cos \left (b x + a\right )^{2} - d^{5}\right )} \sqrt {d} \log \left (-\frac {d \cos \left (b x + a\right )^{2} + 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right ) - 8 \, {\left (4 \, d^{5} \cos \left (b x + a\right )^{4} + 36 \, d^{5} \cos \left (b x + a\right )^{2} - 45 \, d^{5}\right )} \sqrt {d \cos \left (b x + a\right )}}{80 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}}\right ] \] Input:

integrate((d*cos(b*x+a))^(11/2)*csc(b*x+a)^3,x, algorithm="fricas")
 

Output:

[-1/80*(90*(d^5*cos(b*x + a)^2 - d^5)*sqrt(-d)*arctan(1/2*sqrt(d*cos(b*x + 
 a))*sqrt(-d)*(cos(b*x + a) + 1)/(d*cos(b*x + a))) - 45*(d^5*cos(b*x + a)^ 
2 - d^5)*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*(4*d^5*cos(b*x + a)^4 + 36*d^5*cos(b*x + a)^2 - 45*d^5)*sqrt( 
d*cos(b*x + a)))/(b*cos(b*x + a)^2 - b), 1/80*(90*(d^5*cos(b*x + a)^2 - d^ 
5)*sqrt(d)*arctan(1/2*sqrt(d*cos(b*x + a))*(cos(b*x + a) - 1)/(sqrt(d)*cos 
(b*x + a))) + 45*(d^5*cos(b*x + a)^2 - d^5)*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*(4*d^5*cos(b*x + a)^4 + 36*d 
^5*cos(b*x + a)^2 - 45*d^5)*sqrt(d*cos(b*x + a)))/(b*cos(b*x + a)^2 - b)]
 

Sympy [F(-1)]

Timed out. \[ \int (d \cos (a+b x))^{11/2} \csc ^3(a+b x) \, dx=\text {Timed out} \] Input:

integrate((d*cos(b*x+a))**(11/2)*csc(b*x+a)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.99 \[ \int (d \cos (a+b x))^{11/2} \csc ^3(a+b x) \, dx=\frac {\frac {20 \, \sqrt {d \cos \left (b x + a\right )} d^{8}}{d^{2} \cos \left (b x + a\right )^{2} - d^{2}} + 90 \, d^{\frac {13}{2}} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right ) - 45 \, d^{\frac {13}{2}} \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right ) - 16 \, \left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} d^{4} - 160 \, \sqrt {d \cos \left (b x + a\right )} d^{6}}{40 \, b d} \] Input:

integrate((d*cos(b*x+a))^(11/2)*csc(b*x+a)^3,x, algorithm="maxima")
 

Output:

1/40*(20*sqrt(d*cos(b*x + a))*d^8/(d^2*cos(b*x + a)^2 - d^2) + 90*d^(13/2) 
*arctan(sqrt(d*cos(b*x + a))/sqrt(d)) - 45*d^(13/2)*log((sqrt(d*cos(b*x + 
a)) - sqrt(d))/(sqrt(d*cos(b*x + a)) + sqrt(d))) - 16*(d*cos(b*x + a))^(5/ 
2)*d^4 - 160*sqrt(d*cos(b*x + a))*d^6)/(b*d)
 

Giac [F]

\[ \int (d \cos (a+b x))^{11/2} \csc ^3(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {11}{2}} \csc \left (b x + a\right )^{3} \,d x } \] Input:

integrate((d*cos(b*x+a))^(11/2)*csc(b*x+a)^3,x, algorithm="giac")
 

Output:

integrate((d*cos(b*x + a))^(11/2)*csc(b*x + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int (d \cos (a+b x))^{11/2} \csc ^3(a+b x) \, dx=\int \frac {{\left (d\,\cos \left (a+b\,x\right )\right )}^{11/2}}{{\sin \left (a+b\,x\right )}^3} \,d x \] Input:

int((d*cos(a + b*x))^(11/2)/sin(a + b*x)^3,x)
 

Output:

int((d*cos(a + b*x))^(11/2)/sin(a + b*x)^3, x)
 

Reduce [F]

\[ \int (d \cos (a+b x))^{11/2} \csc ^3(a+b x) \, dx=\sqrt {d}\, \left (\int \sqrt {\cos \left (b x +a \right )}\, \cos \left (b x +a \right )^{5} \csc \left (b x +a \right )^{3}d x \right ) d^{5} \] Input:

int((d*cos(b*x+a))^(11/2)*csc(b*x+a)^3,x)
 

Output:

sqrt(d)*int(sqrt(cos(a + b*x))*cos(a + b*x)**5*csc(a + b*x)**3,x)*d**5