\(\int \csc (c+d x) (a+a \sin (c+d x))^{3/2} \, dx\) [48]

Optimal result

Integrand size = 21, antiderivative size = 66 \[ \int \csc (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 1.37 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.79 \[ \int \csc (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {\left (-2 \cos \left (\frac {1}{2} (c+d x)\right )-\log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sin \left (\frac {1}{2} (c+d x)\right )\right ) (a (1+\sin (c+d x)))^{3/2}}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3} \] Input:

Integrate[Csc[c + d*x]*(a + a*Sin[c + d*x])^(3/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3242, 27, 2011, 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.

\(\displaystyle \int \csc (c+d x) (a \sin (c+d x)+a)^{3/2} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3242

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3252

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

\(\Big \downarrow \) 219

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

Input:

Int[Csc[c + d*x]*(a + a*Sin[c + d*x])^(3/2),x]
 

Output:

(-2*a^(3/2)*ArcTanh[(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]])
 

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 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x 
, a + b*x])
 

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]
 

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.27

method	result	size
default	\(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a \left (\sqrt {a}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right )+\sqrt {a -a \sin \left (d x +c \right )}\right )}{\cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(84\)

Input:

int(csc(d*x+c)*(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-2*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*a*(a^(1/2)*arctanh((a-a*sin(d* 
x+c))^(1/2)/a^(1/2))+(a-a*sin(d*x+c))^(1/2))/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. 239 vs. \(2 (58) = 116\).

Time = 0.09 (sec) , antiderivative size = 239, normalized size of antiderivative = 3.62 \[ \int \csc (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {{\left (a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) - 4 \, {\left (a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{2 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \] Input:

integrate(csc(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \csc (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \csc {\left (c + d x \right )}\, dx \] Input:

integrate(csc(d*x+c)*(a+a*sin(d*x+c))**(3/2),x)
 

Output:

Integral((a*(sin(c + d*x) + 1))**(3/2)*csc(c + d*x), x)
 

Maxima [F]

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

integrate(csc(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

integrate((a*sin(d*x + c) + a)^(3/2)*csc(d*x + c), x)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.58 \[ \int \csc (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {\sqrt {2} {\left (\sqrt {2} a \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 4 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{2 \, d} \] Input:

integrate(csc(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

-1/2*sqrt(2)*(sqrt(2)*a*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*a*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. \[ \int \csc (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}}{\sin \left (c+d\,x\right )} \,d x \] Input:

int((a + a*sin(c + d*x))^(3/2)/sin(c + d*x),x)
 

Output:

int((a + a*sin(c + d*x))^(3/2)/sin(c + d*x), x)
 

Reduce [F]

\[ \int \csc (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\sqrt {a}\, a \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \csc \left (d x +c \right ) \sin \left (d x +c \right )d x +\int \sqrt {\sin \left (d x +c \right )+1}\, \csc \left (d x +c \right )d x \right ) \] Input:

int(csc(d*x+c)*(a+a*sin(d*x+c))^(3/2),x)
 

Output:

sqrt(a)*a*(int(sqrt(sin(c + d*x) + 1)*csc(c + d*x)*sin(c + d*x),x) + int(s 
qrt(sin(c + d*x) + 1)*csc(c + d*x),x))