\(\int \frac {\csc ^3(e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\) [106]

Optimal result

Integrand size = 25, antiderivative size = 134 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {(a-3 b) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{2 a^{5/2} f}-\frac {b (a+3 b) \cos (e+f x)}{2 a^2 (a+b) f \sqrt {a+b-b \cos ^2(e+f x)}}-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \sqrt {a+b-b \cos ^2(e+f x)}} \] Output:

-1/2*(a-3*b)*arctanh(a^(1/2)*cos(f*x+e)/(a+b-b*cos(f*x+e)^2)^(1/2))/a^(5/2 
)/f-1/2*b*(a+3*b)*cos(f*x+e)/a^2/(a+b)/f/(a+b-b*cos(f*x+e)^2)^(1/2)-1/2*co 
t(f*x+e)*csc(f*x+e)/a/f/(a+b-b*cos(f*x+e)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 1.09 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {-\frac {(a-3 b) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \cos (e+f x)}{\sqrt {2 a+b-b \cos (2 (e+f x))}}\right )}{a^{5/2}}+\frac {\left (-2 a^2-3 a b-3 b^2+b (a+3 b) \cos (2 (e+f x))\right ) \cot (e+f x) \csc (e+f x)}{\sqrt {2} a^2 (a+b) \sqrt {2 a+b-b \cos (2 (e+f x))}}}{2 f} \] Input:

Integrate[Csc[e + f*x]^3/(a + b*Sin[e + f*x]^2)^(3/2),x]
 

Output:

(-(((a - 3*b)*ArcTanh[(Sqrt[2]*Sqrt[a]*Cos[e + f*x])/Sqrt[2*a + b - b*Cos[ 
2*(e + f*x)]]])/a^(5/2)) + ((-2*a^2 - 3*a*b - 3*b^2 + b*(a + 3*b)*Cos[2*(e 
 + f*x)])*Cot[e + f*x]*Csc[e + f*x])/(Sqrt[2]*a^2*(a + b)*Sqrt[2*a + b - b 
*Cos[2*(e + f*x)]]))/(2*f)
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3665, 316, 402, 25, 27, 291, 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.

\(\displaystyle \int \frac {\csc ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (e+f x)^3 \left (a+b \sin (e+f x)^2\right )^{3/2}}dx\)

\(\Big \downarrow \) 3665

\(\displaystyle -\frac {\int \frac {1}{\left (1-\cos ^2(e+f x)\right )^2 \left (-b \cos ^2(e+f x)+a+b\right )^{3/2}}d\cos (e+f x)}{f}\)

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 402

\(\displaystyle -\frac {\frac {\frac {b (a+3 b) \cos (e+f x)}{a (a+b) \sqrt {a-b \cos ^2(e+f x)+b}}-\frac {\int -\frac {(a-3 b) (a+b)}{\left (1-\cos ^2(e+f x)\right ) \sqrt {-b \cos ^2(e+f x)+a+b}}d\cos (e+f x)}{a (a+b)}}{2 a}+\frac {\cos (e+f x)}{2 a \left (1-\cos ^2(e+f x)\right ) \sqrt {a-b \cos ^2(e+f x)+b}}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\frac {\frac {\int \frac {(a-3 b) (a+b)}{\left (1-\cos ^2(e+f x)\right ) \sqrt {-b \cos ^2(e+f x)+a+b}}d\cos (e+f x)}{a (a+b)}+\frac {b (a+3 b) \cos (e+f x)}{a (a+b) \sqrt {a-b \cos ^2(e+f x)+b}}}{2 a}+\frac {\cos (e+f x)}{2 a \left (1-\cos ^2(e+f x)\right ) \sqrt {a-b \cos ^2(e+f x)+b}}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\frac {(a-3 b) \int \frac {1}{\left (1-\cos ^2(e+f x)\right ) \sqrt {-b \cos ^2(e+f x)+a+b}}d\cos (e+f x)}{a}+\frac {b (a+3 b) \cos (e+f x)}{a (a+b) \sqrt {a-b \cos ^2(e+f x)+b}}}{2 a}+\frac {\cos (e+f x)}{2 a \left (1-\cos ^2(e+f x)\right ) \sqrt {a-b \cos ^2(e+f x)+b}}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle -\frac {\frac {\frac {(a-3 b) \int \frac {1}{1-\frac {a \cos ^2(e+f x)}{-b \cos ^2(e+f x)+a+b}}d\frac {\cos (e+f x)}{\sqrt {-b \cos ^2(e+f x)+a+b}}}{a}+\frac {b (a+3 b) \cos (e+f x)}{a (a+b) \sqrt {a-b \cos ^2(e+f x)+b}}}{2 a}+\frac {\cos (e+f x)}{2 a \left (1-\cos ^2(e+f x)\right ) \sqrt {a-b \cos ^2(e+f x)+b}}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {\frac {\frac {(a-3 b) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{a^{3/2}}+\frac {b (a+3 b) \cos (e+f x)}{a (a+b) \sqrt {a-b \cos ^2(e+f x)+b}}}{2 a}+\frac {\cos (e+f x)}{2 a \left (1-\cos ^2(e+f x)\right ) \sqrt {a-b \cos ^2(e+f x)+b}}}{f}\)

Input:

Int[Csc[e + f*x]^3/(a + b*Sin[e + f*x]^2)^(3/2),x]
 

Output:

-((Cos[e + f*x]/(2*a*(1 - Cos[e + f*x]^2)*Sqrt[a + b - b*Cos[e + f*x]^2]) 
+ (((a - 3*b)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2 
]])/a^(3/2) + (b*(a + 3*b)*Cos[e + f*x])/(a*(a + b)*Sqrt[a + b - b*Cos[e + 
 f*x]^2]))/(2*a))/f)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 291

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

rule 316

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d))   Int[(a + b*x^2)^(p + 1)*(c + d*x 
^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x 
], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  ! 
( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, 
 p, q, x]
 

rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
(q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) 
*(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, q}, x] && LtQ[p, -1]
 

rule 3042

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

rule 3665

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ 
(p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f 
Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos[e + 
 f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(273\) vs. \(2(118)=236\).

Time = 0.40 (sec) , antiderivative size = 274, normalized size of antiderivative = 2.04

method	result	size
default	\(\frac {\sqrt {-\left (-a -b \sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )^{2}}\, \left (-\frac {\sqrt {-\left (-a -b \sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )^{2}}}{2 a^{2} \sin \left (f x +e \right )^{2}}-\frac {\ln \left (\frac {2 a +\left (-a +b \right ) \sin \left (f x +e \right )^{2}+2 \sqrt {a}\, \sqrt {-\left (-a -b \sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )^{2}}}{\sin \left (f x +e \right )^{2}}\right )}{4 a^{\frac {3}{2}}}+\frac {3 b \ln \left (\frac {2 a +\left (-a +b \right ) \sin \left (f x +e \right )^{2}+2 \sqrt {a}\, \sqrt {-\left (-a -b \sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )^{2}}}{\sin \left (f x +e \right )^{2}}\right )}{4 a^{\frac {5}{2}}}-\frac {b^{2} \cos \left (f x +e \right )^{2}}{a^{2} \left (a +b \right ) \sqrt {-\left (-a -b \sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )^{2}}}\right )}{\cos \left (f x +e \right ) \sqrt {a +b \sin \left (f x +e \right )^{2}}\, f}\)	\(274\)

Input:

int(csc(f*x+e)^3/(a+b*sin(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

(-(-a-b*sin(f*x+e)^2)*cos(f*x+e)^2)^(1/2)*(-1/2/a^2/sin(f*x+e)^2*(-(-a-b*s 
in(f*x+e)^2)*cos(f*x+e)^2)^(1/2)-1/4/a^(3/2)*ln((2*a+(-a+b)*sin(f*x+e)^2+2 
*a^(1/2)*(-(-a-b*sin(f*x+e)^2)*cos(f*x+e)^2)^(1/2))/sin(f*x+e)^2)+3/4*b/a^ 
(5/2)*ln((2*a+(-a+b)*sin(f*x+e)^2+2*a^(1/2)*(-(-a-b*sin(f*x+e)^2)*cos(f*x+ 
e)^2)^(1/2))/sin(f*x+e)^2)-b^2/a^2*cos(f*x+e)^2/(a+b)/(-(-a-b*sin(f*x+e)^2 
)*cos(f*x+e)^2)^(1/2))/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 289 vs. \(2 (118) = 236\).

Time = 0.35 (sec) , antiderivative size = 634, normalized size of antiderivative = 4.73 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate(csc(f*x+e)^3/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

[-1/8*(((a^2*b - 2*a*b^2 - 3*b^3)*cos(f*x + e)^4 + a^3 - a^2*b - 5*a*b^2 - 
 3*b^3 - (a^3 - 7*a*b^2 - 6*b^3)*cos(f*x + e)^2)*sqrt(a)*log(2*((a^2 - 6*a 
*b + b^2)*cos(f*x + e)^4 + 2*(3*a^2 + 2*a*b - b^2)*cos(f*x + e)^2 + 4*((a 
- b)*cos(f*x + e)^3 + (a + b)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b 
)*sqrt(a) + a^2 + 2*a*b + b^2)/(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)) - 
4*((a^2*b + 3*a*b^2)*cos(f*x + e)^3 - (a^3 + 2*a^2*b + 3*a*b^2)*cos(f*x + 
e))*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^4*b + a^3*b^2)*f*cos(f*x + e)^4 - 
 (a^5 + 3*a^4*b + 2*a^3*b^2)*f*cos(f*x + e)^2 + (a^5 + 2*a^4*b + a^3*b^2)* 
f), 1/4*(((a^2*b - 2*a*b^2 - 3*b^3)*cos(f*x + e)^4 + a^3 - a^2*b - 5*a*b^2 
 - 3*b^3 - (a^3 - 7*a*b^2 - 6*b^3)*cos(f*x + e)^2)*sqrt(-a)*arctan(-1/2*(( 
a - b)*cos(f*x + e)^2 + a + b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a)/(a 
*b*cos(f*x + e)^3 - (a^2 + a*b)*cos(f*x + e))) + 2*((a^2*b + 3*a*b^2)*cos( 
f*x + e)^3 - (a^3 + 2*a^2*b + 3*a*b^2)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^ 
2 + a + b))/((a^4*b + a^3*b^2)*f*cos(f*x + e)^4 - (a^5 + 3*a^4*b + 2*a^3*b 
^2)*f*cos(f*x + e)^2 + (a^5 + 2*a^4*b + a^3*b^2)*f)]
 

Sympy [F]

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

integrate(csc(f*x+e)**3/(a+b*sin(f*x+e)**2)**(3/2),x)
 

Output:

Integral(csc(e + f*x)**3/(a + b*sin(e + f*x)**2)**(3/2), x)
 

Maxima [F]

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

integrate(csc(f*x+e)^3/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate(csc(f*x + e)^3/(b*sin(f*x + e)^2 + a)^(3/2), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 563 vs. \(2 (118) = 236\).

Time = 0.74 (sec) , antiderivative size = 563, normalized size of antiderivative = 4.20 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate(csc(f*x+e)^3/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

1/8*((((a^5*b + a^4*b^2)*tan(1/2*f*x + 1/2*e)^2/(a^6*b + a^5*b^2) + 2*(a^5 
*b + 3*a^4*b^2 + 6*a^3*b^3)/(a^6*b + a^5*b^2))*tan(1/2*f*x + 1/2*e)^2 + (a 
^5*b + a^4*b^2 - 8*a^3*b^3)/(a^6*b + a^5*b^2))/sqrt(a*tan(1/2*f*x + 1/2*e) 
^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a) + 4*(a - 
 3*b)*arctan(-(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e 
)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))/sqrt(- 
a))/(sqrt(-a)*a^2) - 2*(a - 3*b)*log(abs((sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - 
 sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2* 
f*x + 1/2*e)^2 + a))*sqrt(a) + a + 2*b))/a^(5/2) + 2*((sqrt(a)*tan(1/2*f*x 
 + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 
 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a + 2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - 
sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f 
*x + 1/2*e)^2 + a))*b + a^(3/2))/(((sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt( 
a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 
1/2*e)^2 + a))^2 - a)*a^2))/f
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {1}{{\sin \left (e+f\,x\right )}^3\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \] Input:

int(1/(sin(e + f*x)^3*(a + b*sin(e + f*x)^2)^(3/2)),x)
 

Output:

int(1/(sin(e + f*x)^3*(a + b*sin(e + f*x)^2)^(3/2)), x)
 

Reduce [F]

\[ \int \frac {\csc ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\sqrt {\sin \left (f x +e \right )^{2} b +a}\, \csc \left (f x +e \right )^{3}}{\sin \left (f x +e \right )^{4} b^{2}+2 \sin \left (f x +e \right )^{2} a b +a^{2}}d x \] Input:

int(csc(f*x+e)^3/(a+b*sin(f*x+e)^2)^(3/2),x)
 

Output:

int((sqrt(sin(e + f*x)**2*b + a)*csc(e + f*x)**3)/(sin(e + f*x)**4*b**2 + 
2*sin(e + f*x)**2*a*b + a**2),x)