\(\int \frac {\csc ^4(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\) [117]

Optimal result

Integrand size = 25, antiderivative size = 115 \[ \int \frac {\csc ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {a b \tan (e+f x)}{(a+b)^3 f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {(3 a-2 b) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 (a+b)^3 f}-\frac {\cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 (a+b)^2 f} \] Output:

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

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.89 \[ \int \frac {\csc ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \left (-2 a (a-3 b)+\left (a^2-2 a b-3 b^2\right ) \csc ^2(e+f x)+(a+b)^2 \csc ^4(e+f x)\right ) \sec ^2(e+f x) \tan (e+f x)}{6 (a+b)^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \] Input:

Integrate[Csc[e + f*x]^4/(a + b*Sec[e + f*x]^2)^(3/2),x]
 

Output:

-1/6*((a + 2*b + a*Cos[2*(e + f*x)])*(-2*a*(a - 3*b) + (a^2 - 2*a*b - 3*b^ 
2)*Csc[e + f*x]^2 + (a + b)^2*Csc[e + f*x]^4)*Sec[e + f*x]^2*Tan[e + f*x]) 
/((a + b)^3*f*(a + b*Sec[e + f*x]^2)^(3/2))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4620, 359, 245, 208}

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.

\(\displaystyle \int \frac {\csc ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (e+f x)^4 \left (a+b \sec (e+f x)^2\right )^{3/2}}dx\)

\(\Big \downarrow \) 4620

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

\(\Big \downarrow \) 359

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

\(\Big \downarrow \) 245

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

\(\Big \downarrow \) 208

\(\displaystyle \frac {\frac {(3 a-b) \left (-\frac {2 b \tan (e+f x)}{(a+b)^2 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\cot (e+f x)}{(a+b) \sqrt {a+b \tan ^2(e+f x)+b}}\right )}{3 (a+b)}-\frac {\cot ^3(e+f x)}{3 (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{f}\)

Input:

Int[Csc[e + f*x]^4/(a + b*Sec[e + f*x]^2)^(3/2),x]
 

Output:

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

Defintions of rubi rules used

rule 208

Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), 
x] /; FreeQ[{a, b}, x]
 

rule 245

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + 
 b*x^2)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + 2*(p + 1) + 1)/(a*(m + 1))) 
   Int[x^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, m, p}, x] && ILtQ[Si 
mplify[(m + 1)/2 + p + 1], 0] && NeQ[m, -1]
 

rule 359

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

rule 3042

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

rule 4620

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

Maple [A] (verified)

Time = 5.02 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.21

method	result	size
default	\(\frac {\left (b +a \cos \left (f x +e \right )^{2}\right ) \left (\left (-6 \cos \left (f x +e \right )^{4}+10 \cos \left (f x +e \right )^{2}\right ) a b -6 a b +\left (2 \cos \left (f x +e \right )^{4}-3 \cos \left (f x +e \right )^{2}\right ) a^{2}+\left (-3 \cos \left (f x +e \right )^{2}+2\right ) b^{2}\right ) \sec \left (f x +e \right )^{3} \csc \left (f x +e \right )^{3}}{3 f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}\)	\(139\)

Input:

int(csc(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/3/f/(a^3+3*a^2*b+3*a*b^2+b^3)*(b+a*cos(f*x+e)^2)*((-6*cos(f*x+e)^4+10*co 
s(f*x+e)^2)*a*b-6*a*b+(2*cos(f*x+e)^4-3*cos(f*x+e)^2)*a^2+(-3*cos(f*x+e)^2 
+2)*b^2)/(a+b*sec(f*x+e)^2)^(3/2)*sec(f*x+e)^3*csc(f*x+e)^3
 

Fricas [A] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.64 \[ \int \frac {\csc ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {{\left (2 \, {\left (a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )^{5} - {\left (3 \, a^{2} - 10 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 2 \, {\left (3 \, a b - b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, {\left ({\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{4} - {\left (a^{4} + 2 \, a^{3} b - 2 \, a b^{3} - b^{4}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} f\right )} \sin \left (f x + e\right )} \] Input:

integrate(csc(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

-1/3*(2*(a^2 - 3*a*b)*cos(f*x + e)^5 - (3*a^2 - 10*a*b + 3*b^2)*cos(f*x + 
e)^3 - 2*(3*a*b - b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + 
 e)^2)/(((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f*cos(f*x + e)^4 - (a^4 + 2*a 
^3*b - 2*a*b^3 - b^4)*f*cos(f*x + e)^2 - (a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^ 
4)*f)*sin(f*x + e))
 

Sympy [F]

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

integrate(csc(f*x+e)**4/(a+b*sec(f*x+e)**2)**(3/2),x)
 

Output:

Integral(csc(e + f*x)**4/(a + b*sec(e + f*x)**2)**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.36 \[ \int \frac {\csc ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\frac {6 \, b \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2}} - \frac {8 \, b^{2} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{3}} + \frac {3}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \tan \left (f x + e\right )} - \frac {4 \, b}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2} \tan \left (f x + e\right )} + \frac {1}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \tan \left (f x + e\right )^{3}}}{3 \, f} \] Input:

integrate(csc(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

-1/3*(6*b*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^2) - 8*b^2* 
tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^3) + 3/(sqrt(b*tan(f* 
x + e)^2 + a + b)*(a + b)*tan(f*x + e)) - 4*b/(sqrt(b*tan(f*x + e)^2 + a + 
 b)*(a + b)^2*tan(f*x + e)) + 1/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)*ta 
n(f*x + e)^3))/f
 

Giac [F]

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

integrate(csc(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

integrate(csc(f*x + e)^4/(b*sec(f*x + e)^2 + a)^(3/2), x)
 

Mupad [B] (verification not implemented)

Time = 32.87 (sec) , antiderivative size = 124682, normalized size of antiderivative = 1084.19 \[ \int \frac {\csc ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

int(1/(sin(e + f*x)^4*(a + b/cos(e + f*x)^2)^(3/2)),x)
 

Output:

(a^2*(a + b/(((cos(2*f*x) - sin(2*f*x)*1i)*(cos(2*e) - sin(2*e)*1i))/4 + ( 
(cos(2*f*x) + sin(2*f*x)*1i)*(cos(2*e) + sin(2*e)*1i))/4 + 1/2))^(1/2)*128 
i)/(3*(32*a^4*f + 32*b^4*f + 192*a^2*b^2*f + 128*a*b^3*f + 128*a^3*b*f - 3 
2*a^4*f*(cos(2*f*x) + sin(2*f*x)*1i)*(cos(2*e) + sin(2*e)*1i) - 32*a^4*f*( 
cos(4*f*x) + sin(4*f*x)*1i)*(cos(4*e) + sin(4*e)*1i) + 32*a^4*f*(cos(6*f*x 
) + sin(6*f*x)*1i)*(cos(6*e) + sin(6*e)*1i) - 32*b^4*f*(cos(2*f*x) + sin(2 
*f*x)*1i)*(cos(2*e) + sin(2*e)*1i) - 32*b^4*f*(cos(4*f*x) + sin(4*f*x)*1i) 
*(cos(4*e) + sin(4*e)*1i) + 32*b^4*f*(cos(6*f*x) + sin(6*f*x)*1i)*(cos(6*e 
) + sin(6*e)*1i) - 128*a*b^3*f*(cos(2*f*x) + sin(2*f*x)*1i)*(cos(2*e) + si 
n(2*e)*1i) - 128*a^3*b*f*(cos(2*f*x) + sin(2*f*x)*1i)*(cos(2*e) + sin(2*e) 
*1i) - 128*a*b^3*f*(cos(4*f*x) + sin(4*f*x)*1i)*(cos(4*e) + sin(4*e)*1i) - 
 128*a^3*b*f*(cos(4*f*x) + sin(4*f*x)*1i)*(cos(4*e) + sin(4*e)*1i) + 128*a 
*b^3*f*(cos(6*f*x) + sin(6*f*x)*1i)*(cos(6*e) + sin(6*e)*1i) + 128*a^3*b*f 
*(cos(6*f*x) + sin(6*f*x)*1i)*(cos(6*e) + sin(6*e)*1i) - 192*a^2*b^2*f*(co 
s(2*f*x) + sin(2*f*x)*1i)*(cos(2*e) + sin(2*e)*1i) - 192*a^2*b^2*f*(cos(4* 
f*x) + sin(4*f*x)*1i)*(cos(4*e) + sin(4*e)*1i) + 192*a^2*b^2*f*(cos(6*f*x) 
 + sin(6*f*x)*1i)*(cos(6*e) + sin(6*e)*1i))) - (5*a*(a + b/(((cos(2*f*x) - 
 sin(2*f*x)*1i)*(cos(2*e) - sin(2*e)*1i))/4 + ((cos(2*f*x) + sin(2*f*x)*1i 
)*(cos(2*e) + sin(2*e)*1i))/4 + 1/2))^(1/2))/(3*(a^3*f*1i + b^3*f*1i + a*b 
^2*f*3i + a^2*b*f*3i - a^3*f*(cos(4*f*x) + sin(4*f*x)*1i)*(cos(4*e) + s...
 

Reduce [F]

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

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

Output:

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