\(\int \csc (c+d x) (a+a \sin (c+d x))^{5/2} \, dx\) [56]

Optimal result

Integrand size = 21, antiderivative size = 98 \[ \int \csc (c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {2 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}-\frac {14 a^3 \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d} \] Output:

-2*a^(5/2)*arctanh(a^(1/2)*cos(d*x+c)/(a+a*sin(d*x+c))^(1/2))/d-14/3*a^3*c 
os(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-2/3*a^2*cos(d*x+c)*(a+a*sin(d*x+c))^(1/ 
2)/d
 

Mathematica [A] (verified)

Time = 5.79 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.46 \[ \int \csc (c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {(a (1+\sin (c+d x)))^{5/2} \left (15 \cos \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {3}{2} (c+d x)\right )+3 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-3 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-15 \sin \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {3}{2} (c+d x)\right )\right )}{3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^5} \] Input:

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

Output:

-1/3*((a*(1 + Sin[c + d*x]))^(5/2)*(15*Cos[(c + d*x)/2] + Cos[(3*(c + d*x) 
)/2] + 3*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 3*Log[1 - Cos[(c + 
 d*x)/2] + Sin[(c + d*x)/2]] - 15*Sin[(c + d*x)/2] + Sin[(3*(c + d*x))/2]) 
)/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5)
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3242, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3242

\(\displaystyle \frac {2}{3} \int \frac {1}{2} \csc (c+d x) \sqrt {\sin (c+d x) a+a} \left (7 \sin (c+d x) a^2+3 a^2\right )dx-\frac {2 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \int \csc (c+d x) \sqrt {\sin (c+d x) a+a} \left (7 \sin (c+d x) a^2+3 a^2\right )dx-\frac {2 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \int \frac {\sqrt {\sin (c+d x) a+a} \left (7 \sin (c+d x) a^2+3 a^2\right )}{\sin (c+d x)}dx-\frac {2 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3460

\(\displaystyle \frac {1}{3} \left (3 a^2 \int \csc (c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {14 a^3 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (3 a^2 \int \frac {\sqrt {\sin (c+d x) a+a}}{\sin (c+d x)}dx-\frac {14 a^3 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {1}{3} \left (-\frac {6 a^3 \int \frac {1}{a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}-\frac {14 a^3 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{3} \left (-\frac {6 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}-\frac {14 a^3 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\)

Input:

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

Output:

(-2*a^2*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(3*d) + ((-6*a^(5/2)*ArcTan 
h[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/d - (14*a^3*Cos[c + d* 
x])/(d*Sqrt[a + a*Sin[c + d*x]]))/3
 

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 3242

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x 
])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m 
+ n))   Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c* 
(m - 2) + b^2*d*(n + 1) + a^2*d*(m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 
 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, 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 = 0.70 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.05

method	result	size
default	\(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a \left (3 a^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right )-\left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}}+9 \sqrt {a -a \sin \left (d x +c \right )}\, a \right )}{3 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(103\)

Input:

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

Output:

-2/3*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*a*(3*a^(3/2)*arctanh((a-a*si 
n(d*x+c))^(1/2)/a^(1/2))-(a-a*sin(d*x+c))^(3/2)+9*(a-a*sin(d*x+c))^(1/2)*a 
)/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. 279 vs. \(2 (84) = 168\).

Time = 0.09 (sec) , antiderivative size = 279, normalized size of antiderivative = 2.85 \[ \int \csc (c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\frac {3 \, {\left (a^{2} \cos \left (d x + c\right ) + a^{2} \sin \left (d x + c\right ) + a^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) - 4 \, {\left (a^{2} \cos \left (d x + c\right )^{2} + 8 \, a^{2} \cos \left (d x + c\right ) + 7 \, a^{2} + {\left (a^{2} \cos \left (d x + c\right ) - 7 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{6 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \] Input:

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

Output:

1/6*(3*(a^2*cos(d*x + c) + a^2*sin(d*x + c) + a^2)*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*(a^2*cos(d*x + c)^2 + 8*a^2*cos(d*x + c) + 7*a^2 + (a^2*co 
s(d*x + c) - 7*a^2)*sin(d*x + c))*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c 
) + d*sin(d*x + c) + d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.44 \[ \int \csc (c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {\sqrt {2} {\left (8 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, \sqrt {2} a^{2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 36 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{6 \, d} \] Input:

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

Output:

-1/6*sqrt(2)*(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)^3 + 3*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)) - 36*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))*sqrt(a)/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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