\(\int \csc ^6(e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\) [92]

Optimal result

Integrand size = 25, antiderivative size = 209 \[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {\sqrt {b} (3 a+7 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 f}+\frac {b (3 a+7 b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 (a+b) f}-\frac {(3 a+7 b) \cot (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{3 (a+b) f}-\frac {2 \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{3 (a+b) f}-\frac {\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{5 (a+b) f} \] Output:

1/2*b^(1/2)*(3*a+7*b)*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2 
))/f+1/2*b*(3*a+7*b)*tan(f*x+e)*(a+b+b*tan(f*x+e)^2)^(1/2)/(a+b)/f-1/3*(3* 
a+7*b)*cot(f*x+e)*(a+b+b*tan(f*x+e)^2)^(3/2)/(a+b)/f-2/3*cot(f*x+e)^3*(a+b 
+b*tan(f*x+e)^2)^(5/2)/(a+b)/f-1/5*cot(f*x+e)^5*(a+b+b*tan(f*x+e)^2)^(5/2) 
/(a+b)/f
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 8.64 (sec) , antiderivative size = 512, normalized size of antiderivative = 2.45 \[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {\sqrt {2} e^{i (e+f x)} \sqrt {4 b+a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2} \cos ^3(e+f x) \left (-\frac {i \left (16 a^2 \left (1+e^{2 i (e+f x)}\right )^2 \left (1-6 e^{2 i (e+f x)}+16 e^{4 i (e+f x)}-6 e^{6 i (e+f x)}+e^{8 i (e+f x)}\right )+b^2 \left (105-350 e^{2 i (e+f x)}+231 e^{4 i (e+f x)}+412 e^{6 i (e+f x)}+231 e^{8 i (e+f x)}-350 e^{10 i (e+f x)}+105 e^{12 i (e+f x)}\right )+a b \left (115-402 e^{2 i (e+f x)}+317 e^{4 i (e+f x)}+708 e^{6 i (e+f x)}+317 e^{8 i (e+f x)}-402 e^{10 i (e+f x)}+115 e^{12 i (e+f x)}\right )\right )}{(a+b) \left (-1+e^{2 i (e+f x)}\right )^5 \left (1+e^{2 i (e+f x)}\right )^2}-\frac {15 \sqrt {b} (3 a+7 b) \log \left (\frac {-4 \sqrt {b} \left (-1+e^{2 i (e+f x)}\right ) f+4 i \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2} f}{1+e^{2 i (e+f x)}}\right )}{\sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}}\right ) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{15 f (a+2 b+a \cos (2 (e+f x)))^{3/2}} \] Input:

Integrate[Csc[e + f*x]^6*(a + b*Sec[e + f*x]^2)^(3/2),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]^3*(((-I)*(16*a^2*(1 + E^((2*I)*(e + f*x)))^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*(105 - 350*E^((2*I)*(e + f*x)) + 231*E^((4*I 
)*(e + f*x)) + 412*E^((6*I)*(e + f*x)) + 231*E^((8*I)*(e + f*x)) - 350*E^( 
(10*I)*(e + f*x)) + 105*E^((12*I)*(e + f*x))) + a*b*(115 - 402*E^((2*I)*(e 
 + f*x)) + 317*E^((4*I)*(e + f*x)) + 708*E^((6*I)*(e + f*x)) + 317*E^((8*I 
)*(e + f*x)) - 402*E^((10*I)*(e + f*x)) + 115*E^((12*I)*(e + f*x)))))/((a 
+ b)*(-1 + E^((2*I)*(e + f*x)))^5*(1 + E^((2*I)*(e + f*x)))^2) - (15*Sqrt[ 
b]*(3*a + 7*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] 
)*(a + b*Sec[e + f*x]^2)^(3/2))/(15*f*(a + 2*b + a*Cos[2*(e + f*x)])^(3/2) 
)
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.90, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4620, 365, 27, 359, 247, 211, 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.

\(\displaystyle \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \sec (e+f x)^2\right )^{3/2}}{\sin (e+f x)^6}dx\)

\(\Big \downarrow \) 4620

\(\displaystyle \frac {\int \cot ^6(e+f x) \left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a+b\right )^{3/2}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 365

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 359

\(\displaystyle \frac {\frac {(3 a+7 b) \int \cot ^2(e+f x) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}d\tan (e+f x)}{3 (a+b)}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{5 (a+b)}-\frac {2 \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{3 (a+b)}}{f}\)

\(\Big \downarrow \) 247

\(\displaystyle \frac {\frac {(3 a+7 b) \left (3 b \int \sqrt {b \tan ^2(e+f x)+a+b}d\tan (e+f x)-\cot (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}\right )}{3 (a+b)}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{5 (a+b)}-\frac {2 \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{3 (a+b)}}{f}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\frac {(3 a+7 b) \left (3 b \left (\frac {1}{2} (a+b) \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )-\cot (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}\right )}{3 (a+b)}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{5 (a+b)}-\frac {2 \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{3 (a+b)}}{f}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {(3 a+7 b) \left (3 b \left (\frac {1}{2} (a+b) \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )-\cot (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}\right )}{3 (a+b)}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{5 (a+b)}-\frac {2 \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{3 (a+b)}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {(3 a+7 b) \left (3 b \left (\frac {(a+b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 \sqrt {b}}+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )-\cot (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}\right )}{3 (a+b)}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{5 (a+b)}-\frac {2 \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{3 (a+b)}}{f}\)

Input:

Int[Csc[e + f*x]^6*(a + b*Sec[e + f*x]^2)^(3/2),x]
 

Output:

((-2*Cot[e + f*x]^3*(a + b + b*Tan[e + f*x]^2)^(5/2))/(3*(a + b)) - (Cot[e 
 + f*x]^5*(a + b + b*Tan[e + f*x]^2)^(5/2))/(5*(a + b)) + ((3*a + 7*b)*(-( 
Cot[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(3/2)) + 3*b*(((a + b)*ArcTanh[(Sq 
rt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(2*Sqrt[b]) + (Tan[e 
+ f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/2)))/(3*(a + b)))/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 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

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 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 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. \(998\) vs. \(2(185)=370\).

Time = 23.00 (sec) , antiderivative size = 999, normalized size of antiderivative = 4.78

method	result	size
default	\(\text {Expression too large to display}\)	\(999\)

Input:

int(csc(f*x+e)^6*(a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/60/f/(a+b)/b*(sin(f*x+e)^3*cos(f*x+e)^2*(105*cos(f*x+e)-105)*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*c 
os(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*b^( 
7/2)+sin(f*x+e)^3*cos(f*x+e)^2*(150*cos(f*x+e)-150)*ln(4*(b^(1/2)*((b+a*co 
s(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))*b^(5/2)*a+sin(f 
*x+e)^3*cos(f*x+e)^2*(45*cos(f*x+e)-45)*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))*b^(3/2)*a^2+sin(f*x+e)^3*co 
s(f*x+e)^2*(105*cos(f*x+e)-105)*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))*b^(7/2)+sin(f*x+e)^3*cos(f*x+e)^2* 
(150*cos(f*x+e)-150)*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))*b^(5/2)*a+sin(f*x+e)^3*cos(f*x+e)^2*(45*cos(f 
*x+e)-45)*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))*b^(3/2)*a^2+cos(f*x+e)^2*(32*cos(f*x+e)^4-80*cos(f*x+e)^ 
2+60)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*b+(230*cos(f*x+e)^6- 
546*cos(f*x+e)^4+370*cos(f*x+e)^2-30)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e)...
 

Fricas [A] (verification not implemented)

Time = 5.78 (sec) , antiderivative size = 682, normalized size of antiderivative = 3.26 \[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx =\text {Too large to display} \] Input:

integrate(csc(f*x+e)^6*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

[1/120*(15*((3*a^2 + 10*a*b + 7*b^2)*cos(f*x + e)^5 - 2*(3*a^2 + 10*a*b + 
7*b^2)*cos(f*x + e)^3 + (3*a^2 + 10*a*b + 7*b^2)*cos(f*x + e))*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 
*((16*a^2 + 115*a*b + 105*b^2)*cos(f*x + e)^6 - (40*a^2 + 273*a*b + 245*b^ 
2)*cos(f*x + e)^4 + (30*a^2 + 185*a*b + 161*b^2)*cos(f*x + e)^2 - 15*a*b - 
 15*b^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(((a + b)*f*cos(f*x 
+ e)^5 - 2*(a + b)*f*cos(f*x + e)^3 + (a + b)*f*cos(f*x + e))*sin(f*x + e) 
), 1/60*(15*((3*a^2 + 10*a*b + 7*b^2)*cos(f*x + e)^5 - 2*(3*a^2 + 10*a*b + 
 7*b^2)*cos(f*x + e)^3 + (3*a^2 + 10*a*b + 7*b^2)*cos(f*x + e))*sqrt(-b)*a 
rctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sqrt((a*co 
s(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*((16*a^2 + 115*a*b + 105*b^2)*cos(f*x + e)^6 - (40*a^2 
 + 273*a*b + 245*b^2)*cos(f*x + e)^4 + (30*a^2 + 185*a*b + 161*b^2)*cos(f* 
x + e)^2 - 15*a*b - 15*b^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/( 
((a + b)*f*cos(f*x + e)^5 - 2*(a + b)*f*cos(f*x + e)^3 + (a + b)*f*cos(f*x 
 + e))*sin(f*x + e))]
 

Sympy [F(-1)]

Timed out. \[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \] Input:

integrate(csc(f*x+e)**6*(a+b*sec(f*x+e)**2)**(3/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.31 \[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {45 \, a \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) + \frac {60 \, a b^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{a + b} + 45 \, b^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) + \frac {60 \, b^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{a + b} + 45 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} b \tan \left (f x + e\right ) + \frac {60 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} b^{2} \tan \left (f x + e\right )}{a + b} - \frac {30 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}}}{\tan \left (f x + e\right )} - \frac {40 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b}{{\left (a + b\right )} \tan \left (f x + e\right )} - \frac {20 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {5}{2}}}{{\left (a + b\right )} \tan \left (f x + e\right )^{3}} - \frac {6 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {5}{2}}}{{\left (a + b\right )} \tan \left (f x + e\right )^{5}}}{30 \, f} \] Input:

integrate(csc(f*x+e)^6*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

1/30*(45*a*sqrt(b)*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b)) + 60*a*b^(3/2)* 
arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/(a + b) + 45*b^(3/2)*arcsinh(b*tan 
(f*x + e)/sqrt((a + b)*b)) + 60*b^(5/2)*arcsinh(b*tan(f*x + e)/sqrt((a + b 
)*b))/(a + b) + 45*sqrt(b*tan(f*x + e)^2 + a + b)*b*tan(f*x + e) + 60*sqrt 
(b*tan(f*x + e)^2 + a + b)*b^2*tan(f*x + e)/(a + b) - 30*(b*tan(f*x + e)^2 
 + a + b)^(3/2)/tan(f*x + e) - 40*(b*tan(f*x + e)^2 + a + b)^(3/2)*b/((a + 
 b)*tan(f*x + e)) - 20*(b*tan(f*x + e)^2 + a + b)^(5/2)/((a + b)*tan(f*x + 
 e)^3) - 6*(b*tan(f*x + e)^2 + a + b)^(5/2)/((a + b)*tan(f*x + e)^5))/f
 

Giac [F]

\[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \csc \left (f x + e\right )^{6} \,d x } \] Input:

integrate(csc(f*x+e)^6*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^(3/2)*csc(f*x + e)^6, x)
 

Mupad [F(-1)]

Timed out. \[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int \frac {{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}}{{\sin \left (e+f\,x\right )}^6} \,d x \] Input:

int((a + b/cos(e + f*x)^2)^(3/2)/sin(e + f*x)^6,x)
 

Output:

int((a + b/cos(e + f*x)^2)^(3/2)/sin(e + f*x)^6, x)
 

Reduce [F]

\[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\left (\int \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \csc \left (f x +e \right )^{6} \sec \left (f x +e \right )^{2}d x \right ) b +\left (\int \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \csc \left (f x +e \right )^{6}d x \right ) a \] Input:

int(csc(f*x+e)^6*(a+b*sec(f*x+e)^2)^(3/2),x)
 

Output:

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