\(\int \csc ^3(e+f x) (a+b \sin ^2(e+f x))^{3/2} \, dx\) [86]

Optimal result

Integrand size = 25, antiderivative size = 128 \[ \int \csc ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=-\frac {b^{3/2} \arctan \left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{f}-\frac {\sqrt {a} (a+3 b) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{2 f}-\frac {a \sqrt {a+b-b \cos ^2(e+f x)} \cot (e+f x) \csc (e+f x)}{2 f} \] Output:

-b^(3/2)*arctan(b^(1/2)*cos(f*x+e)/(a+b-b*cos(f*x+e)^2)^(1/2))/f-1/2*a^(1/ 
2)*(a+3*b)*arctanh(a^(1/2)*cos(f*x+e)/(a+b-b*cos(f*x+e)^2)^(1/2))/f-1/2*a* 
(a+b-b*cos(f*x+e)^2)^(1/2)*cot(f*x+e)*csc(f*x+e)/f
 

Mathematica [A] (verified)

Time = 1.58 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.15 \[ \int \csc ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=-\frac {2 \sqrt {a} (a+3 b) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \cos (e+f x)}{\sqrt {2 a+b-b \cos (2 (e+f x))}}\right )+\sqrt {2} a \sqrt {2 a+b-b \cos (2 (e+f x))} \cot (e+f x) \csc (e+f x)+4 (-b)^{3/2} \log \left (\sqrt {2} \sqrt {-b} \cos (e+f x)+\sqrt {2 a+b-b \cos (2 (e+f x))}\right )}{4 f} \] Input:

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

Output:

-1/4*(2*Sqrt[a]*(a + 3*b)*ArcTanh[(Sqrt[2]*Sqrt[a]*Cos[e + f*x])/Sqrt[2*a 
+ b - b*Cos[2*(e + f*x)]]] + Sqrt[2]*a*Sqrt[2*a + b - b*Cos[2*(e + f*x)]]* 
Cot[e + f*x]*Csc[e + f*x] + 4*(-b)^(3/2)*Log[Sqrt[2]*Sqrt[-b]*Cos[e + f*x] 
 + Sqrt[2*a + b - b*Cos[2*(e + f*x)]]])/f
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 3665, 315, 25, 398, 224, 216, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3665

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

\(\Big \downarrow \) 315

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 398

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

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

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

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 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 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 315

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), 
x] - Simp[1/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S 
imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) 
*x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 
1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
 

rule 398

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

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. \(286\) vs. \(2(110)=220\).

Time = 0.42 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.24

method	result	size
default	\(\frac {\sqrt {\cos \left (f x +e \right )^{2} \left (a +b \sin \left (f x +e \right )^{2}\right )}\, \left (2 b^{\frac {3}{2}} \arctan \left (\frac {2 b \sin \left (f x +e \right )^{2}+a -b}{2 \sqrt {b}\, \sqrt {\cos \left (f x +e \right )^{2} \left (a +b \sin \left (f x +e \right )^{2}\right )}}\right ) \sin \left (f x +e \right )^{2}-a^{\frac {3}{2}} \ln \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )^{2}+2 \sqrt {a}\, \sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}+a +b}{\sin \left (f x +e \right )^{2}}\right ) \sin \left (f x +e \right )^{2}-3 \sqrt {a}\, b \ln \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )^{2}+2 \sqrt {a}\, \sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}+a +b}{\sin \left (f x +e \right )^{2}}\right ) \sin \left (f x +e \right )^{2}-2 a \sqrt {\cos \left (f x +e \right )^{2} \left (a +b \sin \left (f x +e \right )^{2}\right )}\right )}{4 \sin \left (f x +e \right )^{2} \cos \left (f x +e \right ) \sqrt {a +b \sin \left (f x +e \right )^{2}}\, f}\)	\(287\)

Input:

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

Output:

1/4*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*(2*b^(3/2)*arctan(1/2/b^(1/2)* 
(2*b*sin(f*x+e)^2+a-b)/(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2))*sin(f*x+e) 
^2-a^(3/2)*ln(((a-b)*cos(f*x+e)^2+2*a^(1/2)*(-b*cos(f*x+e)^4+(a+b)*cos(f*x 
+e)^2)^(1/2)+a+b)/sin(f*x+e)^2)*sin(f*x+e)^2-3*a^(1/2)*b*ln(((a-b)*cos(f*x 
+e)^2+2*a^(1/2)*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)+a+b)/sin(f*x+e) 
^2)*sin(f*x+e)^2-2*a*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2))/sin(f*x+e)^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. 279 vs. \(2 (110) = 220\).

Time = 0.40 (sec) , antiderivative size = 1449, normalized size of antiderivative = 11.32 \[ \int \csc ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/8*(4*sqrt(-b*cos(f*x + e)^2 + a + b)*a*cos(f*x + e) + (b*cos(f*x + e)^2 
 - b)*sqrt(-b)*log(128*b^4*cos(f*x + e)^8 - 256*(a*b^3 + b^4)*cos(f*x + e) 
^6 + 160*(a^2*b^2 + 2*a*b^3 + b^4)*cos(f*x + e)^4 + a^4 + 4*a^3*b + 6*a^2* 
b^2 + 4*a*b^3 + b^4 - 32*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*cos(f*x + e)^ 
2 - 8*(16*b^3*cos(f*x + e)^7 - 24*(a*b^2 + b^3)*cos(f*x + e)^5 + 10*(a^2*b 
 + 2*a*b^2 + b^3)*cos(f*x + e)^3 - (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(f*x 
 + e))*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-b)) + ((a + 3*b)*cos(f*x + e) 
^2 - a - 3*b)*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)))/(f*cos(f*x + e)^2 - f), 1/8*(2*((a + 
 3*b)*cos(f*x + e)^2 - a - 3*b)*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))) + 4*sqrt(-b*cos(f*x + e)^2 + a + b)*a*cos(f*x 
+ e) + (b*cos(f*x + e)^2 - b)*sqrt(-b)*log(128*b^4*cos(f*x + e)^8 - 256*(a 
*b^3 + b^4)*cos(f*x + e)^6 + 160*(a^2*b^2 + 2*a*b^3 + b^4)*cos(f*x + e)^4 
+ a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4 - 32*(a^3*b + 3*a^2*b^2 + 3*a* 
b^3 + b^4)*cos(f*x + e)^2 - 8*(16*b^3*cos(f*x + e)^7 - 24*(a*b^2 + b^3)*co 
s(f*x + e)^5 + 10*(a^2*b + 2*a*b^2 + b^3)*cos(f*x + e)^3 - (a^3 + 3*a^2*b 
+ 3*a*b^2 + b^3)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-b)...
 

Sympy [F(-1)]

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

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

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \csc ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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