\(\int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) [97]

Optimal result

Integrand size = 25, antiderivative size = 87 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {a \text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 (a+b)^{3/2} f}-\frac {\cot (e+f x) \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) f} \] Output:

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

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.61 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {a \sqrt {a+2 b+a \cos (2 e+2 f x)} \sec (e+f x) \sqrt {a+b-a \sin ^2(e+f x)} \left (\frac {(a+b) \csc ^2(e+f x)}{a}+\frac {\text {arctanh}\left (\sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}\right )}{\sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}\right )}{2 \sqrt {2} (a+b)^2 f \sqrt {a+b \sec ^2(e+f x)}} \] Input:

Integrate[Csc[e + f*x]^3/Sqrt[a + b*Sec[e + f*x]^2],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4622, 373, 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.

\(\displaystyle \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4622

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

\(\Big \downarrow \) 373

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 219

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

Input:

Int[Csc[e + f*x]^3/Sqrt[a + b*Sec[e + f*x]^2],x]
 

Output:

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

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 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 373

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

rule 3042

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

rule 4622

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + ( 
f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Si 
mp[1/(f*ff^m)   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p 
/x^(m + 1)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x 
] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(566\) vs. \(2(75)=150\).

Time = 1.41 (sec) , antiderivative size = 567, normalized size of antiderivative = 6.52

method	result	size
default	\(\frac {\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+2 \sqrt {a +b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-2 \cos \left (f x +e \right ) a +2 b}{\sqrt {a +b}\, \left (1+\cos \left (f x +e \right )\right )}\right ) a^{2} \left (1-\cos \left (f x +e \right )\right )^{2}+\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+2 \sqrt {a +b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-2 \cos \left (f x +e \right ) a +2 b}{\sqrt {a +b}\, \left (1+\cos \left (f x +e \right )\right )}\right ) \left (1-\cos \left (f x +e \right )\right )^{2} a b +\ln \left (-\frac {4 \left (\sqrt {a +b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+\sqrt {a +b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+\cos \left (f x +e \right ) a +b \right )}{-1+\cos \left (f x +e \right )}\right ) a^{2} \left (1-\cos \left (f x +e \right )\right )^{2}+\ln \left (-\frac {4 \left (\sqrt {a +b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+\sqrt {a +b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+\cos \left (f x +e \right ) a +b \right )}{-1+\cos \left (f x +e \right )}\right ) \left (1-\cos \left (f x +e \right )\right )^{2} a b +\left (-2 \cos \left (f x +e \right )+2\right ) \left (a +b \right )^{\frac {3}{2}} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\right )}{2 f \left (a +b \right )^{\frac {5}{2}} \sqrt {a +b \sec \left (f x +e \right )^{2}}\, \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right ) \left (1-\cos \left (f x +e \right )\right )^{2}}\)	\(567\)

Input:

int(csc(f*x+e)^3/(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/2/f/(a+b)^(5/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)/(a+b*sec(f*x 
+e)^2)^(1/2)/((1-cos(f*x+e))^2*csc(f*x+e)^2-1)/(1-cos(f*x+e))^2*(ln(2/(a+b 
)^(1/2)*((a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e 
)+(a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-cos(f*x+e)*a+b)/ 
(1+cos(f*x+e)))*a^2*(1-cos(f*x+e))^2+ln(2/(a+b)^(1/2)*((a+b)^(1/2)*((b+a*c 
os(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+(a+b)^(1/2)*((b+a*cos(f*x+ 
e)^2)/(1+cos(f*x+e))^2)^(1/2)-cos(f*x+e)*a+b)/(1+cos(f*x+e)))*(1-cos(f*x+e 
))^2*a*b+ln(-4*((a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*co 
s(f*x+e)+(a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+cos(f*x+e 
)*a+b)/(-1+cos(f*x+e)))*a^2*(1-cos(f*x+e))^2+ln(-4*((a+b)^(1/2)*((b+a*cos( 
f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+(a+b)^(1/2)*((b+a*cos(f*x+e)^ 
2)/(1+cos(f*x+e))^2)^(1/2)+cos(f*x+e)*a+b)/(-1+cos(f*x+e)))*(1-cos(f*x+e)) 
^2*a*b+(-2*cos(f*x+e)+2)*(a+b)^(3/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2) 
^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 152 vs. \(2 (75) = 150\).

Time = 0.15 (sec) , antiderivative size = 314, normalized size of antiderivative = 3.61 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\left [\frac {2 \, {\left (a + b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + {\left (a \cos \left (f x + e\right )^{2} - a\right )} \sqrt {a + b} \log \left (\frac {2 \, {\left (a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a + b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + 2 \, b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right )}{4 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{2} + 2 \, a b + b^{2}\right )} f\right )}}, \frac {{\left (a \cos \left (f x + e\right )^{2} - a\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-a - b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a \cos \left (f x + e\right )^{2} + b}\right ) + {\left (a + b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{2 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{2} + 2 \, a b + b^{2}\right )} f\right )}}\right ] \] Input:

integrate(csc(f*x+e)^3/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate(csc(f*x+e)**3/(a+b*sec(f*x+e)**2)**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate(csc(f*x+e)^3/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 480 vs. \(2 (75) = 150\).

Time = 0.47 (sec) , antiderivative size = 480, normalized size of antiderivative = 5.52 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx =\text {Too large to display} \] Input:

integrate(csc(f*x+e)^3/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

1/8*(4*a*arctan(-(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x 
+ 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*t 
an(1/2*f*x + 1/2*e)^2 + a + b))/sqrt(-a - b))/((a + b)*sqrt(-a - b)) + 2*a 
*log(abs(-(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e 
)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2* 
f*x + 1/2*e)^2 + a + b))*sqrt(a + b) + a - b))/(a + b)^(3/2) - sqrt(a*tan( 
1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 
 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b)/(a + b) - 2*((sqrt(a + b)*tan(1/2*f 
*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 
 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))*(a - b) 
 - (a + b)^(3/2))/(((sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f 
*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2* 
b*tan(1/2*f*x + 1/2*e)^2 + a + b))^2 - a - b)*(a + b)))/(f*sgn(cos(f*x + e 
)))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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