\(\int \csc ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\) [79]

Optimal result

Integrand size = 25, antiderivative size = 149 \[ \int \csc ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}-\frac {\cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{f}-\frac {2 (5 a+4 b) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{15 (a+b)^2 f}-\frac {\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{5 (a+b) f} \] Output:

b^(1/2)*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/f-cot(f*x+e 
)*(a+b+b*tan(f*x+e)^2)^(1/2)/f-2/15*(5*a+4*b)*cot(f*x+e)^3*(a+b+b*tan(f*x+ 
e)^2)^(3/2)/(a+b)^2/f-1/5*cot(f*x+e)^5*(a+b+b*tan(f*x+e)^2)^(3/2)/(a+b)/f
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 6.94 (sec) , antiderivative size = 422, normalized size of antiderivative = 2.83 \[ \int \csc ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {\sqrt {2} e^{i (e+f x)} \sqrt {4 b+a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2} \cos (e+f x) \left (-\frac {i \left (8 a^2 \left (1-6 e^{2 i (e+f x)}+16 e^{4 i (e+f x)}-6 e^{6 i (e+f x)}+e^{8 i (e+f x)}\right )+b^2 \left (15-80 e^{2 i (e+f x)}+178 e^{4 i (e+f x)}-80 e^{6 i (e+f x)}+15 e^{8 i (e+f x)}\right )+a b \left (25-136 e^{2 i (e+f x)}+318 e^{4 i (e+f x)}-136 e^{6 i (e+f x)}+25 e^{8 i (e+f x)}\right )\right )}{(a+b)^2 \left (-1+e^{2 i (e+f x)}\right )^5}-\frac {15 \sqrt {b} \log \left (\frac {-4 \sqrt {b} \left (-1+e^{2 i (e+f x)}\right ) f+4 i \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2} f}{1+e^{2 i (e+f x)}}\right )}{\sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}}\right ) \sqrt {a+b \sec ^2(e+f x)}}{15 f \sqrt {a+2 b+a \cos (2 e+2 f x)}} \] Input:

Integrate[Csc[e + f*x]^6*Sqrt[a + b*Sec[e + f*x]^2],x]
 

Output:

(Sqrt[2]*E^(I*(e + f*x))*Sqrt[4*b + (a*(1 + E^((2*I)*(e + f*x)))^2)/E^((2* 
I)*(e + f*x))]*Cos[e + f*x]*(((-I)*(8*a^2*(1 - 6*E^((2*I)*(e + f*x)) + 16* 
E^((4*I)*(e + f*x)) - 6*E^((6*I)*(e + f*x)) + E^((8*I)*(e + f*x))) + b^2*( 
15 - 80*E^((2*I)*(e + f*x)) + 178*E^((4*I)*(e + f*x)) - 80*E^((6*I)*(e + f 
*x)) + 15*E^((8*I)*(e + f*x))) + a*b*(25 - 136*E^((2*I)*(e + f*x)) + 318*E 
^((4*I)*(e + f*x)) - 136*E^((6*I)*(e + f*x)) + 25*E^((8*I)*(e + f*x)))))/( 
(a + b)^2*(-1 + E^((2*I)*(e + f*x)))^5) - (15*Sqrt[b]*Log[(-4*Sqrt[b]*(-1 
+ E^((2*I)*(e + f*x)))*f + (4*I)*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^( 
(2*I)*(e + f*x)))^2]*f)/(1 + E^((2*I)*(e + f*x)))])/Sqrt[4*b*E^((2*I)*(e + 
 f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2])*Sqrt[a + b*Sec[e + f*x]^2])/(15*f 
*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]])
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4620, 365, 358, 247, 224, 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.

\(\displaystyle \int \csc ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {a+b \sec (e+f x)^2}}{\sin (e+f x)^6}dx\)

\(\Big \downarrow \) 4620

\(\displaystyle \frac {\int \cot ^6(e+f x) \left (\tan ^2(e+f x)+1\right )^2 \sqrt {b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 365

\(\displaystyle \frac {\frac {\int \cot ^4(e+f x) \sqrt {b \tan ^2(e+f x)+a+b} \left (5 (a+b) \tan ^2(e+f x)+2 (5 a+4 b)\right )d\tan (e+f x)}{5 (a+b)}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{5 (a+b)}}{f}\)

\(\Big \downarrow \) 358

\(\displaystyle \frac {\frac {5 (a+b) \int \cot ^2(e+f x) \sqrt {b \tan ^2(e+f x)+a+b}d\tan (e+f x)-\frac {2 (5 a+4 b) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{3 (a+b)}}{5 (a+b)}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{5 (a+b)}}{f}\)

\(\Big \downarrow \) 247

\(\displaystyle \frac {\frac {5 (a+b) \left (b \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)-\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )-\frac {2 (5 a+4 b) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{3 (a+b)}}{5 (a+b)}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{5 (a+b)}}{f}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {5 (a+b) \left (b \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}-\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )-\frac {2 (5 a+4 b) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{3 (a+b)}}{5 (a+b)}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{5 (a+b)}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {5 (a+b) \left (\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )-\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )-\frac {2 (5 a+4 b) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{3 (a+b)}}{5 (a+b)}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{5 (a+b)}}{f}\)

Input:

Int[Csc[e + f*x]^6*Sqrt[a + b*Sec[e + f*x]^2],x]
 

Output:

(-1/5*(Cot[e + f*x]^5*(a + b + b*Tan[e + f*x]^2)^(3/2))/(a + b) + ((-2*(5* 
a + 4*b)*Cot[e + f*x]^3*(a + b + b*Tan[e + f*x]^2)^(3/2))/(3*(a + b)) + 5* 
(a + b)*(Sqrt[b]*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x 
]^2]] - Cot[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2]))/(5*(a + b)))/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 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 247

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

rule 358

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] + S 
imp[d/e^2   Int[(e*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e 
, m, p}, x] && NeQ[b*c - a*d, 0] && EqQ[Simplify[m + 2*p + 3], 0] && NeQ[m, 
 -1]
 

rule 365

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

Leaf count of result is larger than twice the leaf count of optimal. \(902\) vs. \(2(133)=266\).

Time = 19.31 (sec) , antiderivative size = 903, normalized size of antiderivative = 6.06

method	result	size
default	\(\text {Expression too large to display}\)	\(903\)

Input:

int(csc(f*x+e)^6*(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/30/f/(a+b)^2*(sin(f*x+e)^3*(15*cos(f*x+e)-15)*b^(5/2)*ln(4*(b^(1/2)*((b 
+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+ 
e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))+sin(f*x+e) 
^3*(30*cos(f*x+e)-30)*b^(3/2)*ln(4*(b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x 
+e))^2)^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/ 
2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*a+sin(f*x+e)^3*(15*cos(f*x+e)-15)*b^( 
1/2)*ln(4*(b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+ 
b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin 
(f*x+e)+1))*a^2+sin(f*x+e)^3*(15*cos(f*x+e)-15)*b^(5/2)*ln(-4*(b^(1/2)*((b 
+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+ 
e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))+sin(f*x+e) 
^3*(30*cos(f*x+e)-30)*b^(3/2)*ln(-4*(b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f* 
x+e))^2)^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1 
/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))*a+sin(f*x+e)^3*(15*cos(f*x+e)-15)*b^ 
(1/2)*ln(-4*(b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e 
)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(s 
in(f*x+e)-1))*a^2+(16*cos(f*x+e)^4-40*cos(f*x+e)^2+30)*((b+a*cos(f*x+e)^2) 
/(1+cos(f*x+e))^2)^(1/2)*a^2+(50*cos(f*x+e)^4-118*cos(f*x+e)^2+80)*((b+a*c 
os(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b+(30*cos(f*x+e)^4-70*cos(f*x+e)^2+ 
46)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2)*(a+b*sec(f*x+e)^2)...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 312 vs. \(2 (133) = 266\).

Time = 1.12 (sec) , antiderivative size = 656, normalized size of antiderivative = 4.40 \[ \int \csc ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx =\text {Too large to display} \] Input:

integrate(csc(f*x+e)^6*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

Output:

[1/60*(15*((a^2 + 2*a*b + b^2)*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*cos( 
f*x + e)^2 + a^2 + 2*a*b + b^2)*sqrt(b)*log(((a^2 - 6*a*b + b^2)*cos(f*x + 
 e)^4 + 8*(a*b - b^2)*cos(f*x + e)^2 + 4*((a - b)*cos(f*x + e)^3 + 2*b*cos 
(f*x + e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e 
) + 8*b^2)/cos(f*x + e)^4)*sin(f*x + e) - 4*((8*a^2 + 25*a*b + 15*b^2)*cos 
(f*x + e)^5 - (20*a^2 + 59*a*b + 35*b^2)*cos(f*x + e)^3 + (15*a^2 + 40*a*b 
 + 23*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(((a 
^2 + 2*a*b + b^2)*f*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*f*cos(f*x + e)^ 
2 + (a^2 + 2*a*b + b^2)*f)*sin(f*x + e)), 1/30*(15*((a^2 + 2*a*b + b^2)*co 
s(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2)*s 
qrt(-b)*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*s 
qrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a*b*cos(f*x + e)^2 + b^2)*sin 
(f*x + e)))*sin(f*x + e) - 2*((8*a^2 + 25*a*b + 15*b^2)*cos(f*x + e)^5 - ( 
20*a^2 + 59*a*b + 35*b^2)*cos(f*x + e)^3 + (15*a^2 + 40*a*b + 23*b^2)*cos( 
f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(((a^2 + 2*a*b + b^ 
2)*f*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*f*cos(f*x + e)^2 + (a^2 + 2*a* 
b + b^2)*f)*sin(f*x + e))]
 

Sympy [F(-1)]

Timed out. \[ \int \csc ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\text {Timed out} \] Input:

integrate(csc(f*x+e)**6*(a+b*sec(f*x+e)**2)**(1/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.96 \[ \int \csc ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {15 \, \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) - \frac {15 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b}}{\tan \left (f x + e\right )} - \frac {10 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}}}{{\left (a + b\right )} \tan \left (f x + e\right )^{3}} + \frac {2 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b}{{\left (a + b\right )}^{2} \tan \left (f x + e\right )^{3}} - \frac {3 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}}}{{\left (a + b\right )} \tan \left (f x + e\right )^{5}}}{15 \, f} \] Input:

integrate(csc(f*x+e)^6*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

Output:

1/15*(15*sqrt(b)*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b)) - 15*sqrt(b*tan(f 
*x + e)^2 + a + b)/tan(f*x + e) - 10*(b*tan(f*x + e)^2 + a + b)^(3/2)/((a 
+ b)*tan(f*x + e)^3) + 2*(b*tan(f*x + e)^2 + a + b)^(3/2)*b/((a + b)^2*tan 
(f*x + e)^3) - 3*(b*tan(f*x + e)^2 + a + b)^(3/2)/((a + b)*tan(f*x + e)^5) 
)/f
 

Giac [F]

\[ \int \csc ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int { \sqrt {b \sec \left (f x + e\right )^{2} + a} \csc \left (f x + e\right )^{6} \,d x } \] Input:

integrate(csc(f*x+e)^6*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

integrate(sqrt(b*sec(f*x + e)^2 + a)*csc(f*x + e)^6, x)
 

Mupad [F(-1)]

Timed out. \[ \int \csc ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int \frac {\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}}{{\sin \left (e+f\,x\right )}^6} \,d x \] Input:

int((a + b/cos(e + f*x)^2)^(1/2)/sin(e + f*x)^6,x)
 

Output:

int((a + b/cos(e + f*x)^2)^(1/2)/sin(e + f*x)^6, x)
 

Reduce [F]

\[ \int \csc ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \csc \left (f x +e \right )^{6}d x \] Input:

int(csc(f*x+e)^6*(a+b*sec(f*x+e)^2)^(1/2),x)
 

Output:

int(sqrt(sec(e + f*x)**2*b + a)*csc(e + f*x)**6,x)