\(\int \csc ^2(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\) [57]

Optimal result

Integrand size = 23, antiderivative size = 94 \[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {5 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}-\frac {a^3 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}-\frac {a^2 \cot (c+d x) \sqrt {a+a \sin (c+d x)}}{d} \] Output:

-5*a^(5/2)*arctanh(a^(1/2)*cos(d*x+c)/(a+a*sin(d*x+c))^(1/2))/d-a^3*cos(d* 
x+c)/d/(a+a*sin(d*x+c))^(1/2)-a^2*cot(d*x+c)*(a+a*sin(d*x+c))^(1/2)/d
 

Mathematica [A] (verified)

Time = 6.29 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.94 \[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {a^2 \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sin (c+d x))} \left (2 \cos \left (\frac {3}{2} (c+d x)\right )+5 \left (\log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sin (c+d x)+2 \sin \left (\frac {3}{2} (c+d x)\right )\right )}{d \left (1+\cot \left (\frac {1}{2} (c+d x)\right )\right ) \left (\csc \left (\frac {1}{4} (c+d x)\right )-\sec \left (\frac {1}{4} (c+d x)\right )\right ) \left (\csc \left (\frac {1}{4} (c+d x)\right )+\sec \left (\frac {1}{4} (c+d x)\right )\right )} \] Input:

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

Output:

-((a^2*Csc[(c + d*x)/2]^4*Sqrt[a*(1 + Sin[c + d*x])]*(2*Cos[(3*(c + d*x))/ 
2] + 5*(Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[1 - Cos[(c + d* 
x)/2] + Sin[(c + d*x)/2]])*Sin[c + d*x] + 2*Sin[(3*(c + d*x))/2]))/(d*(1 + 
 Cot[(c + d*x)/2])*(Csc[(c + d*x)/4] - Sec[(c + d*x)/4])*(Csc[(c + d*x)/4] 
 + Sec[(c + d*x)/4])))
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3241, 27, 3042, 3460, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3241

\(\displaystyle -a \int -\frac {1}{2} \csc (c+d x) \sqrt {\sin (c+d x) a+a} (\sin (c+d x) a+5 a)dx-\frac {a^2 \cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} a \int \csc (c+d x) \sqrt {\sin (c+d x) a+a} (\sin (c+d x) a+5 a)dx-\frac {a^2 \cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} a \int \frac {\sqrt {\sin (c+d x) a+a} (\sin (c+d x) a+5 a)}{\sin (c+d x)}dx-\frac {a^2 \cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}\)

\(\Big \downarrow \) 3460

\(\displaystyle \frac {1}{2} a \left (5 a \int \csc (c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )-\frac {a^2 \cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} a \left (5 a \int \frac {\sqrt {\sin (c+d x) a+a}}{\sin (c+d x)}dx-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )-\frac {a^2 \cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {1}{2} a \left (-\frac {10 a^2 \int \frac {1}{a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )-\frac {a^2 \cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} a \left (-\frac {10 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )-\frac {a^2 \cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}\)

Input:

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

Output:

-((a^2*Cot[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/d) + (a*((-10*a^(3/2)*ArcTan 
h[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/d - (2*a^2*Cos[c + d*x 
])/(d*Sqrt[a + a*Sin[c + d*x]])))/2
 

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 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 3460

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( 
f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + 
b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b 
*d*(2*n + 3))   Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] 
 /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 = 2.13 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.35

method	result	size
default	\(-\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {3}{2}} \left (2 \sqrt {a -a \sin \left (d x +c \right )}\, \sin \left (d x +c \right ) \sqrt {a}+5 \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right ) \sin \left (d x +c \right ) a +\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {a}\right )}{\sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(127\)

Input:

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

Output:

-(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*a^(3/2)*(2*(a-a*sin(d*x+c))^(1/2 
)*sin(d*x+c)*a^(1/2)+5*arctanh((a-a*sin(d*x+c))^(1/2)/a^(1/2))*sin(d*x+c)* 
a+(a-a*sin(d*x+c))^(1/2)*a^(1/2))/sin(d*x+c)/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. 308 vs. \(2 (84) = 168\).

Time = 0.15 (sec) , antiderivative size = 308, normalized size of antiderivative = 3.28 \[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\frac {5 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2} - {\left (a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) + 4 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2} \cos \left (d x + c\right ) - a^{2} + {\left (2 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{4 \, {\left (d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right ) + d\right )} \sin \left (d x + c\right ) - d\right )}} \] Input:

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

Output:

1/4*(5*(a^2*cos(d*x + c)^2 - a^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)*sqrt(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*(2*a^2*cos(d*x + c)^2 + a^2*cos(d*x 
 + c) - a^2 + (2*a^2*cos(d*x + c) + a^2)*sin(d*x + c))*sqrt(a*sin(d*x + c) 
 + a))/(d*cos(d*x + c)^2 - (d*cos(d*x + c) + d)*sin(d*x + c) - d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate(csc(d*x+c)^2*(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)^2, x)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.69 \[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {\sqrt {2} {\left (5 \, \sqrt {2} a^{2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 8 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {4 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}\right )} \sqrt {a}}{4 \, d} \] Input:

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

Output:

-1/4*sqrt(2)*(5*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*p 
i + 1/2*d*x + 1/2*c)) - 8*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) + 4*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))*sqrt(a)/ 
d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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