\(\int \csc ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\) [77]

Optimal result

Integrand size = 25, antiderivative size = 138 \[ \int \csc ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=-\frac {(3 a-b) (a+b) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{8 a^{3/2} f}-\frac {(3 a+b) \sqrt {a+b-b \cos ^2(e+f x)} \cot (e+f x) \csc (e+f x)}{8 a f}-\frac {\sqrt {a+b-b \cos ^2(e+f x)} \cot (e+f x) \csc ^3(e+f x)}{4 f} \] Output:

-1/8*(3*a-b)*(a+b)*arctanh(a^(1/2)*cos(f*x+e)/(a+b-b*cos(f*x+e)^2)^(1/2))/ 
a^(3/2)/f-1/8*(3*a+b)*(a+b-b*cos(f*x+e)^2)^(1/2)*cot(f*x+e)*csc(f*x+e)/a/f 
-1/4*(a+b-b*cos(f*x+e)^2)^(1/2)*cot(f*x+e)*csc(f*x+e)^3/f
 

Mathematica [A] (verified)

Time = 0.97 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.92 \[ \int \csc ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {\left (-6 a^2-4 a b+2 b^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \cos (e+f x)}{\sqrt {2 a+b-b \cos (2 (e+f x))}}\right )-\sqrt {2} \sqrt {a} \sqrt {2 a+b-b \cos (2 (e+f x))} \cot (e+f x) \csc (e+f x) \left (3 a+b+2 a \csc ^2(e+f x)\right )}{16 a^{3/2} f} \] Input:

Integrate[Csc[e + f*x]^5*Sqrt[a + b*Sin[e + f*x]^2],x]
 

Output:

((-6*a^2 - 4*a*b + 2*b^2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Cos[e + f*x])/Sqrt[2*a 
+ b - b*Cos[2*(e + f*x)]]] - Sqrt[2]*Sqrt[a]*Sqrt[2*a + b - b*Cos[2*(e + f 
*x)]]*Cot[e + f*x]*Csc[e + f*x]*(3*a + b + 2*a*Csc[e + f*x]^2))/(16*a^(3/2 
)*f)
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3665, 296, 292, 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.

\(\displaystyle \int \csc ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {a+b \sin (e+f x)^2}}{\sin (e+f x)^5}dx\)

\(\Big \downarrow \) 3665

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

\(\Big \downarrow \) 296

\(\displaystyle -\frac {\frac {(3 a-b) \int \frac {\sqrt {-b \cos ^2(e+f x)+a+b}}{\left (1-\cos ^2(e+f x)\right )^2}d\cos (e+f x)}{4 a}+\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{4 a \left (1-\cos ^2(e+f x)\right )^2}}{f}\)

\(\Big \downarrow \) 292

\(\displaystyle -\frac {\frac {(3 a-b) \left (\frac {1}{2} (a+b) \int \frac {1}{\left (1-\cos ^2(e+f x)\right ) \sqrt {-b \cos ^2(e+f x)+a+b}}d\cos (e+f x)+\frac {\cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{2 \left (1-\cos ^2(e+f x)\right )}\right )}{4 a}+\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{4 a \left (1-\cos ^2(e+f x)\right )^2}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle -\frac {\frac {(3 a-b) \left (\frac {1}{2} (a+b) \int \frac {1}{1-\frac {a \cos ^2(e+f x)}{-b \cos ^2(e+f x)+a+b}}d\frac {\cos (e+f x)}{\sqrt {-b \cos ^2(e+f x)+a+b}}+\frac {\cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{2 \left (1-\cos ^2(e+f x)\right )}\right )}{4 a}+\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{4 a \left (1-\cos ^2(e+f x)\right )^2}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {\frac {(3 a-b) \left (\frac {(a+b) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{2 \sqrt {a}}+\frac {\cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{2 \left (1-\cos ^2(e+f x)\right )}\right )}{4 a}+\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{4 a \left (1-\cos ^2(e+f x)\right )^2}}{f}\)

Input:

Int[Csc[e + f*x]^5*Sqrt[a + b*Sin[e + f*x]^2],x]
 

Output:

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

Defintions of rubi rules used

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 292

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Si 
mp[(-x)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(2*a*(p + 1))), x] - Simp[c*(q/( 
a*(p + 1)))   Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1), x], x] /; FreeQ[ 
{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && EqQ[2*(p + q + 1) + 1, 0] && Gt 
Q[q, 0] && NeQ[p, -1]
 

rule 296

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[(b*c + 2*(p + 1)*(b*c - a*d))/(2*a*(p + 1)*(b*c - a*d))   Int[ 
(a + b*x^2)^(p + 1)*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, q}, x] && N 
eQ[b*c - a*d, 0] && EqQ[2*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1 
]) && NeQ[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. \(378\) vs. \(2(122)=244\).

Time = 0.45 (sec) , antiderivative size = 379, normalized size of antiderivative = 2.75

method	result	size
default	\(-\frac {\sqrt {\cos \left (f x +e \right )^{2} \left (a +b \sin \left (f x +e \right )^{2}\right )}\, \left (3 a^{3} \ln \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )^{2}+2 \sqrt {a}\, \sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}+a +b}{\sin \left (f x +e \right )^{2}}\right ) \sin \left (f x +e \right )^{4}+2 b \ln \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )^{2}+2 \sqrt {a}\, \sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}+a +b}{\sin \left (f x +e \right )^{2}}\right ) \sin \left (f x +e \right )^{4} a^{2}-\ln \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )^{2}+2 \sqrt {a}\, \sqrt {-b \cos \left (f x +e \right )^{4}+\left (a +b \right ) \cos \left (f x +e \right )^{2}}+a +b}{\sin \left (f x +e \right )^{2}}\right ) b^{2} \sin \left (f x +e \right )^{4} a +6 \sin \left (f x +e \right )^{2} a^{\frac {5}{2}} \sqrt {\cos \left (f x +e \right )^{2} \left (a +b \sin \left (f x +e \right )^{2}\right )}+2 \sin \left (f x +e \right )^{2} a^{\frac {3}{2}} \sqrt {\cos \left (f x +e \right )^{2} \left (a +b \sin \left (f x +e \right )^{2}\right )}\, b +4 \sqrt {\cos \left (f x +e \right )^{2} \left (a +b \sin \left (f x +e \right )^{2}\right )}\, a^{\frac {5}{2}}\right )}{16 \sin \left (f x +e \right )^{4} a^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {a +b \sin \left (f x +e \right )^{2}}\, f}\)	\(379\)

Input:

int(csc(f*x+e)^5*(a+b*sin(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/16*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*(3*a^3*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)^4+2*b*ln(((a-b)*cos(f*x+e)^2+2*a^(1/2)*(-b*cos(f*x+e)^4+(a+b)*c 
os(f*x+e)^2)^(1/2)+a+b)/sin(f*x+e)^2)*sin(f*x+e)^4*a^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 
)*b^2*sin(f*x+e)^4*a+6*sin(f*x+e)^2*a^(5/2)*(cos(f*x+e)^2*(a+b*sin(f*x+e)^ 
2))^(1/2)+2*sin(f*x+e)^2*a^(3/2)*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*b 
+4*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*a^(5/2))/sin(f*x+e)^4/a^(5/2)/c 
os(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 520, normalized size of antiderivative = 3.77 \[ \int \csc ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\left [-\frac {{\left ({\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} + 2 \, a b - b^{2}\right )} \sqrt {a} \log \left (\frac {2 \, {\left ({\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + {\left (a + b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} + a^{2} + 2 \, a b + b^{2}\right )}}{\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1}\right ) - 4 \, {\left ({\left (3 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - {\left (5 \, a^{2} + a b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{32 \, {\left (a^{2} f \cos \left (f x + e\right )^{4} - 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f\right )}}, \frac {{\left ({\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} + 2 \, a b - b^{2}\right )} \sqrt {-a} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + a + b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{2 \, {\left (a b \cos \left (f x + e\right )^{3} - {\left (a^{2} + a b\right )} \cos \left (f x + e\right )\right )}}\right ) + 2 \, {\left ({\left (3 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - {\left (5 \, a^{2} + a b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{16 \, {\left (a^{2} f \cos \left (f x + e\right )^{4} - 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f\right )}}\right ] \] Input:

integrate(csc(f*x+e)^5*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

Output:

[-1/32*(((3*a^2 + 2*a*b - b^2)*cos(f*x + e)^4 - 2*(3*a^2 + 2*a*b - b^2)*co 
s(f*x + e)^2 + 3*a^2 + 2*a*b - b^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*((3*a^2 + a 
*b)*cos(f*x + e)^3 - (5*a^2 + a*b)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + 
a + b))/(a^2*f*cos(f*x + e)^4 - 2*a^2*f*cos(f*x + e)^2 + a^2*f), 1/16*(((3 
*a^2 + 2*a*b - b^2)*cos(f*x + e)^4 - 2*(3*a^2 + 2*a*b - b^2)*cos(f*x + e)^ 
2 + 3*a^2 + 2*a*b - b^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*((3*a^2 + a*b)*cos(f*x + e)^3 - (5*a^2 + a*b)*cos 
(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b))/(a^2*f*cos(f*x + e)^4 - 2*a^2* 
f*cos(f*x + e)^2 + a^2*f)]
 

Sympy [F]

\[ \int \csc ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int \sqrt {a + b \sin ^{2}{\left (e + f x \right )}} \csc ^{5}{\left (e + f x \right )}\, dx \] Input:

integrate(csc(f*x+e)**5*(a+b*sin(f*x+e)**2)**(1/2),x)
 

Output:

Integral(sqrt(a + b*sin(e + f*x)**2)*csc(e + f*x)**5, x)
 

Maxima [F]

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

integrate(csc(f*x+e)^5*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(b*sin(f*x + e)^2 + a)*csc(f*x + e)^5, x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 914 vs. \(2 (122) = 244\).

Time = 0.65 (sec) , antiderivative size = 914, normalized size of antiderivative = 6.62 \[ \int \csc ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\text {Too large to display} \] Input:

integrate(csc(f*x+e)^5*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

1/64*(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)*(tan(1/2*f*x + 1/2*e)^2 + (7*a + 2*b)/a) + 8*(3*a 
^2 + 2*a*b - b^2)*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) - 4*(3*a^(5/2) + 2*a^(3/2)*b - sqrt(a)*b^2)*l 
og(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))*a - a^(3/2) 
- 2*sqrt(a)*b))/a^2 + 4*(4*(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))^3*a^2 + 8*(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))^3*a 
*b + 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))^3*b^2 + 5*(sq 
rt(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^(5/2) + 4*(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^(3/2)*b - 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^3 - 4*(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...
 

Mupad [F(-1)]

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

int((a + b*sin(e + f*x)^2)^(1/2)/sin(e + f*x)^5,x)
 

Output:

int((a + b*sin(e + f*x)^2)^(1/2)/sin(e + f*x)^5, x)
 

Reduce [F]

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

int(csc(f*x+e)^5*(a+b*sin(f*x+e)^2)^(1/2),x)
 

Output:

int(sqrt(sin(e + f*x)**2*b + a)*csc(e + f*x)**5,x)