\(\int \csc ^4(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\) [59]

Optimal result

Integrand size = 23, antiderivative size = 144 \[ \int \csc ^4(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {25 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{8 d}-\frac {25 a^3 \cot (c+d x)}{8 d \sqrt {a+a \sin (c+d x)}}-\frac {13 a^3 \cot (c+d x) \csc (c+d x)}{12 d \sqrt {a+a \sin (c+d x)}}-\frac {a^2 \cot (c+d x) \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{3 d} \] Output:

-25/8*a^(5/2)*arctanh(a^(1/2)*cos(d*x+c)/(a+a*sin(d*x+c))^(1/2))/d-25/8*a^ 
3*cot(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-13/12*a^3*cot(d*x+c)*csc(d*x+c)/d/(a 
+a*sin(d*x+c))^(1/2)-1/3*a^2*cot(d*x+c)*csc(d*x+c)^2*(a+a*sin(d*x+c))^(1/2 
)/d
 

Mathematica [A] (verified)

Time = 6.55 (sec) , antiderivative size = 288, normalized size of antiderivative = 2.00 \[ \int \csc ^4(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\frac {a^2 \csc ^{10}\left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sin (c+d x))} \left (-228 \cos \left (\frac {1}{2} (c+d x)\right )+14 \cos \left (\frac {3}{2} (c+d x)\right )+150 \cos \left (\frac {5}{2} (c+d x)\right )+228 \sin \left (\frac {1}{2} (c+d x)\right )-225 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+225 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+14 \sin \left (\frac {3}{2} (c+d x)\right )-150 \sin \left (\frac {5}{2} (c+d x)\right )+75 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))-75 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))\right )}{24 d \left (1+\cot \left (\frac {1}{2} (c+d x)\right )\right ) \left (\csc ^2\left (\frac {1}{4} (c+d x)\right )-\sec ^2\left (\frac {1}{4} (c+d x)\right )\right )^3} \] Input:

Integrate[Csc[c + d*x]^4*(a + a*Sin[c + d*x])^(5/2),x]
 

Output:

(a^2*Csc[(c + d*x)/2]^10*Sqrt[a*(1 + Sin[c + d*x])]*(-228*Cos[(c + d*x)/2] 
 + 14*Cos[(3*(c + d*x))/2] + 150*Cos[(5*(c + d*x))/2] + 228*Sin[(c + d*x)/ 
2] - 225*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[c + d*x] + 225*L 
og[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*x] + 14*Sin[(3*(c + 
d*x))/2] - 150*Sin[(5*(c + d*x))/2] + 75*Log[1 + Cos[(c + d*x)/2] - Sin[(c 
 + d*x)/2]]*Sin[3*(c + d*x)] - 75*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x) 
/2]]*Sin[3*(c + d*x)]))/(24*d*(1 + Cot[(c + d*x)/2])*(Csc[(c + d*x)/4]^2 - 
 Sec[(c + d*x)/4]^2)^3)
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 3241, 27, 3042, 3459, 3042, 3251, 3042, 3252, 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.

\(\displaystyle \int \csc ^4(c+d x) (a \sin (c+d x)+a)^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (c+d x)+a)^{5/2}}{\sin (c+d x)^4}dx\)

\(\Big \downarrow \) 3241

\(\displaystyle -\frac {1}{3} a \int -\frac {1}{2} \csc ^3(c+d x) \sqrt {\sin (c+d x) a+a} (9 \sin (c+d x) a+13 a)dx-\frac {a^2 \cot (c+d x) \csc ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} a \int \csc ^3(c+d x) \sqrt {\sin (c+d x) a+a} (9 \sin (c+d x) a+13 a)dx-\frac {a^2 \cot (c+d x) \csc ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} a \int \frac {\sqrt {\sin (c+d x) a+a} (9 \sin (c+d x) a+13 a)}{\sin (c+d x)^3}dx-\frac {a^2 \cot (c+d x) \csc ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3459

\(\displaystyle \frac {1}{6} a \left (\frac {75}{4} a \int \csc ^2(c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {13 a^2 \cot (c+d x) \csc (c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {a^2 \cot (c+d x) \csc ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} a \left (\frac {75}{4} a \int \frac {\sqrt {\sin (c+d x) a+a}}{\sin (c+d x)^2}dx-\frac {13 a^2 \cot (c+d x) \csc (c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {a^2 \cot (c+d x) \csc ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3251

\(\displaystyle \frac {1}{6} a \left (\frac {75}{4} a \left (\frac {1}{2} \int \csc (c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {a \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )-\frac {13 a^2 \cot (c+d x) \csc (c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {a^2 \cot (c+d x) \csc ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} a \left (\frac {75}{4} a \left (\frac {1}{2} \int \frac {\sqrt {\sin (c+d x) a+a}}{\sin (c+d x)}dx-\frac {a \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )-\frac {13 a^2 \cot (c+d x) \csc (c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {a^2 \cot (c+d x) \csc ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {1}{6} a \left (\frac {75}{4} a \left (-\frac {a \int \frac {1}{a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}-\frac {a \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )-\frac {13 a^2 \cot (c+d x) \csc (c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {a^2 \cot (c+d x) \csc ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{6} a \left (\frac {75}{4} a \left (-\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}-\frac {a \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )-\frac {13 a^2 \cot (c+d x) \csc (c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {a^2 \cot (c+d x) \csc ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\)

Input:

Int[Csc[c + d*x]^4*(a + a*Sin[c + d*x])^(5/2),x]
 

Output:

-1/3*(a^2*Cot[c + d*x]*Csc[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]])/d + (a*((- 
13*a^2*Cot[c + d*x]*Csc[c + d*x])/(2*d*Sqrt[a + a*Sin[c + d*x]]) + (75*a*( 
-((Sqrt[a]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/d) - 
(a*Cot[c + d*x])/(d*Sqrt[a + a*Sin[c + d*x]])))/4))/6
 

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 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 3042

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

rule 3241

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(b*c - a*d)*Cos[e + f*x]*(a + b 
*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a* 
d))), x] + Simp[b^2/(d*(n + 1)*(b*c + a*d))   Int[(a + b*Sin[e + f*x])^(m - 
 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b* 
c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, 
d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] 
 && GtQ[m, 1] && LtQ[n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || 
 (IntegerQ[m] && EqQ[c, 0]))
 

rule 3251

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((c + d*Sin[e 
+ f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] + Sim 
p[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2)))   Int[Sqrt[a + b*Sin[e 
+ f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x 
] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, 
-1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]
 

rule 3252

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f)   Subst[Int[1/(b*c + a*d - d*x^2), 
x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, 
 e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3459

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( 
f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1) 
*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b* 
c - 2*a*d*(n + 1)))/(2*d*(n + 1)*(b*c + a*d))   Int[Sqrt[a + b*Sin[e + f*x] 
]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x 
] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, 
-1]
 

Maple [A] (verified)

Time = 25.99 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00

method	result	size
default	\(-\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (75 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {5}{2}} a^{\frac {3}{2}}+75 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) a^{4} \sin \left (d x +c \right )^{3}-184 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} a^{\frac {5}{2}}+117 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {7}{2}}\right )}{24 \sin \left (d x +c \right )^{3} a^{\frac {3}{2}} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(144\)

Input:

int(csc(d*x+c)^4*(a+a*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/24*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(75*(-a*(sin(d*x+c)-1))^(5/ 
2)*a^(3/2)+75*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*a^4*sin(d*x+c)^3- 
184*(-a*(sin(d*x+c)-1))^(3/2)*a^(5/2)+117*(-a*(sin(d*x+c)-1))^(1/2)*a^(7/2 
))/sin(d*x+c)^3/a^(3/2)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 408 vs. \(2 (124) = 248\).

Time = 0.10 (sec) , antiderivative size = 408, normalized size of antiderivative = 2.83 \[ \int \csc ^4(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\frac {75 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{3} + a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right ) - a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) + 4 \, {\left (75 \, a^{2} \cos \left (d x + c\right )^{3} + 41 \, a^{2} \cos \left (d x + c\right )^{2} - 83 \, a^{2} \cos \left (d x + c\right ) - 49 \, a^{2} - {\left (75 \, a^{2} \cos \left (d x + c\right )^{2} + 34 \, a^{2} \cos \left (d x + c\right ) - 49 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{96 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) - d\right )} \sin \left (d x + c\right ) + d\right )}} \] Input:

integrate(csc(d*x+c)^4*(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

1/96*(75*(a^2*cos(d*x + c)^4 - 2*a^2*cos(d*x + c)^2 + a^2 - (a^2*cos(d*x + 
 c)^3 + a^2*cos(d*x + c)^2 - a^2*cos(d*x + c) - a^2)*sin(d*x + c))*sqrt(a) 
*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 + (cos(d*x 
 + c) + 3)*sin(d*x + c) - 2*cos(d*x + c) - 3)*sqrt(a*sin(d*x + c) + a)*sqr 
t(a) - 9*a*cos(d*x + c) + (a*cos(d*x + c)^2 + 8*a*cos(d*x + c) - a)*sin(d* 
x + c) - a)/(cos(d*x + c)^3 + cos(d*x + c)^2 + (cos(d*x + c)^2 - 1)*sin(d* 
x + c) - cos(d*x + c) - 1)) + 4*(75*a^2*cos(d*x + c)^3 + 41*a^2*cos(d*x + 
c)^2 - 83*a^2*cos(d*x + c) - 49*a^2 - (75*a^2*cos(d*x + c)^2 + 34*a^2*cos( 
d*x + c) - 49*a^2)*sin(d*x + c))*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c) 
^4 - 2*d*cos(d*x + c)^2 - (d*cos(d*x + c)^3 + d*cos(d*x + c)^2 - d*cos(d*x 
 + c) - d)*sin(d*x + c) + d)
 

Sympy [F(-1)]

Timed out. \[ \int \csc ^4(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:

integrate(csc(d*x+c)**4*(a+a*sin(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \csc ^4(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \csc \left (d x + c\right )^{4} \,d x } \] Input:

integrate(csc(d*x+c)^4*(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

integrate((a*sin(d*x + c) + a)^(5/2)*csc(d*x + c)^4, x)
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.36 \[ \int \csc ^4(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {\sqrt {2} {\left (75 \, \sqrt {2} a^{2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + \frac {4 \, {\left (300 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 368 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 117 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}\right )} \sqrt {a}}{96 \, d} \] Input:

integrate(csc(d*x+c)^4*(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

-1/96*sqrt(2)*(75*sqrt(2)*a^2*log(abs(-2*sqrt(2) + 4*sin(-1/4*pi + 1/2*d*x 
 + 1/2*c))/abs(2*sqrt(2) + 4*sin(-1/4*pi + 1/2*d*x + 1/2*c)))*sgn(cos(-1/4 
*pi + 1/2*d*x + 1/2*c)) + 4*(300*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*s 
in(-1/4*pi + 1/2*d*x + 1/2*c)^5 - 368*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2* 
c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 + 117*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 
 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c))/(2*sin(-1/4*pi + 1/2*d*x + 1/2*c) 
^2 - 1)^3)*sqrt(a)/d
 

Mupad [F(-1)]

Timed out. \[ \int \csc ^4(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}}{{\sin \left (c+d\,x\right )}^4} \,d x \] Input:

int((a + a*sin(c + d*x))^(5/2)/sin(c + d*x)^4,x)
 

Output:

int((a + a*sin(c + d*x))^(5/2)/sin(c + d*x)^4, x)
 

Reduce [F]

\[ \int \csc ^4(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\sqrt {a}\, a^{2} \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \csc \left (d x +c \right )^{4} \sin \left (d x +c \right )^{2}d x +2 \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \csc \left (d x +c \right )^{4} \sin \left (d x +c \right )d x \right )+\int \sqrt {\sin \left (d x +c \right )+1}\, \csc \left (d x +c \right )^{4}d x \right ) \] Input:

int(csc(d*x+c)^4*(a+a*sin(d*x+c))^(5/2),x)
 

Output:

sqrt(a)*a**2*(int(sqrt(sin(c + d*x) + 1)*csc(c + d*x)**4*sin(c + d*x)**2,x 
) + 2*int(sqrt(sin(c + d*x) + 1)*csc(c + d*x)**4*sin(c + d*x),x) + int(sqr 
t(sin(c + d*x) + 1)*csc(c + d*x)**4,x))