\(\int \csc ^4(e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\) [91]

Optimal result

Integrand size = 25, antiderivative size = 172 \[ \int \csc ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {\sqrt {b} (3 a+5 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 f}+\frac {b (3 a+5 b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 (a+b) f}-\frac {(3 a+5 b) \cot (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{3 (a+b) f}-\frac {\cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{3 (a+b) f} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 5.55 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.95 \[ \int \csc ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=-\frac {2 \left (a+b \sec ^2(e+f x)\right )^{3/2} \left (-\left ((a+b) (-2 a-b+a \cos (2 (e+f x))) \csc ^2(e+f x) \operatorname {Hypergeometric2F1}\left (1,2,\frac {1}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right )\right )+8 b \operatorname {Hypergeometric2F1}\left (2,3,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \left (a+b \sec ^2(e+f x)\right )\right ) ((a+b) \csc (e+f x)-a \sin (e+f x))^2 \tan (e+f x)}{3 (a+b)^3 f (a+2 b+a \cos (2 (e+f x)))} \] Input:

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

Output:

(-2*(a + b*Sec[e + f*x]^2)^(3/2)*(-((a + b)*(-2*a - b + a*Cos[2*(e + f*x)] 
)*Csc[e + f*x]^2*Hypergeometric2F1[1, 2, 1/2, -((b*Tan[e + f*x]^2)/(a + b) 
)]) + 8*b*Hypergeometric2F1[2, 3, 3/2, -((b*Tan[e + f*x]^2)/(a + b))]*(a + 
 b*Sec[e + f*x]^2))*((a + b)*Csc[e + f*x] - a*Sin[e + f*x])^2*Tan[e + f*x] 
)/(3*(a + b)^3*f*(a + 2*b + a*Cos[2*(e + f*x)]))
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.90, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4620, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4620

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

\(\Big \downarrow \) 359

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

\(\Big \downarrow \) 247

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(-1/3*(Cot[e + f*x]^3*(a + b + b*Tan[e + f*x]^2)^(5/2))/(a + b) + ((3*a + 
5*b)*(-(Cot[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(3/2)) + 3*b*(((a + b)*Arc 
Tanh[(Sqrt[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 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 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. \(653\) vs. \(2(152)=304\).

Time = 21.83 (sec) , antiderivative size = 654, normalized size of antiderivative = 3.80

method	result	size
default	\(\frac {\left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}} \left (\left (-15 \cos \left (f x +e \right )+15\right ) b^{\frac {5}{2}} \ln \left (\frac {4 \sqrt {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 )+4 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a -4 a -4 b}{\sin \left (f x +e \right )+1}\right ) \cos \left (f x +e \right ) \cot \left (f x +e \right )^{2}+\left (-9 \cos \left (f x +e \right )+9\right ) b^{\frac {3}{2}} \ln \left (\frac {4 \sqrt {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 )+4 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a -4 a -4 b}{\sin \left (f x +e \right )+1}\right ) a \cos \left (f x +e \right ) \cot \left (f x +e \right )^{2}+\left (-15 \cos \left (f x +e \right )+15\right ) b^{\frac {5}{2}} \ln \left (-\frac {4 \left (\sqrt {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 {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-\sin \left (f x +e \right ) a +a +b \right )}{\sin \left (f x +e \right )-1}\right ) \cos \left (f x +e \right ) \cot \left (f x +e \right )^{2}+\left (-9 \cos \left (f x +e \right )+9\right ) b^{\frac {3}{2}} \ln \left (-\frac {4 \left (\sqrt {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 {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-\sin \left (f x +e \right ) a +a +b \right )}{\sin \left (f x +e \right )-1}\right ) a \cos \left (f x +e \right ) \cot \left (f x +e \right )^{2}+\left (8 \cos \left (f x +e \right )^{2}-12\right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b \cot \left (f x +e \right )^{3}+\left (30 \cos \left (f x +e \right )^{4}-40 \cos \left (f x +e \right )^{2}+6\right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2} \cot \left (f x +e \right ) \csc \left (f x +e \right )^{2}\right )}{12 f b \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (b +a \cos \left (f x +e \right )^{2}\right )}\)	\(654\)

Input:

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

Output:

1/12/f/b*(a+b*sec(f*x+e)^2)^(3/2)/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1 
/2)/(b+a*cos(f*x+e)^2)*((-15*cos(f*x+e)+15)*b^(5/2)*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))*cos(f*x+e)*cot( 
f*x+e)^2+(-9*cos(f*x+e)+9)*b^(3/2)*ln(4*(b^(1/2)*((b+a*cos(f*x+e)^2)/(1+co 
s(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*cos(f*x+e)*cot(f*x+e)^2+(-15*c 
os(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))*cos(f*x+e)*cot(f*x+e)^2+(-9*cos(f*x+e)+9)*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)/(s 
in(f*x+e)-1))*a*cos(f*x+e)*cot(f*x+e)^2+(8*cos(f*x+e)^2-12)*((b+a*cos(f*x+ 
e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b*cot(f*x+e)^3+(30*cos(f*x+e)^4-40*cos(f*x 
+e)^2+6)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2*cot(f*x+e)*csc(f* 
x+e)^2)
 

Fricas [A] (verification not implemented)

Time = 1.24 (sec) , antiderivative size = 472, normalized size of antiderivative = 2.74 \[ \int \csc ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\left [\frac {3 \, {\left ({\left (3 \, a + 5 \, b\right )} \cos \left (f x + e\right )^{3} - {\left (3 \, a + 5 \, b\right )} \cos \left (f x + e\right )\right )} \sqrt {b} \log \left (\frac {{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (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} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) \sin \left (f x + e\right ) - 4 \, {\left ({\left (4 \, a + 15 \, b\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a + 10 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{24 \, {\left (f \cos \left (f x + e\right )^{3} - f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}, \frac {3 \, {\left ({\left (3 \, a + 5 \, b\right )} \cos \left (f x + e\right )^{3} - {\left (3 \, a + 5 \, b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \, {\left ({\left (4 \, a + 15 \, b\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a + 10 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{12 \, {\left (f \cos \left (f x + e\right )^{3} - f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}\right ] \] Input:

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

Output:

[1/24*(3*((3*a + 5*b)*cos(f*x + e)^3 - (3*a + 5*b)*cos(f*x + e))*sqrt(b)*l 
og(((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*((4*a + 15*b)*cos(f*x + e)^4 - 2*(3*a + 10*b)*cos(f*x + e)^2 + 3*b)*sqr 
t((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((f*cos(f*x + e)^3 - f*cos(f*x + 
 e))*sin(f*x + e)), 1/12*(3*((3*a + 5*b)*cos(f*x + e)^3 - (3*a + 5*b)*cos( 
f*x + e))*sqrt(-b)*arctan(-1/2*((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)/((a*b*cos(f*x + e)^2 
 + b^2)*sin(f*x + e)))*sin(f*x + e) - 2*((4*a + 15*b)*cos(f*x + e)^4 - 2*( 
3*a + 10*b)*cos(f*x + e)^2 + 3*b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e) 
^2))/((f*cos(f*x + e)^3 - f*cos(f*x + e))*sin(f*x + e))]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.41 \[ \int \csc ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {9 \, a \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) + \frac {6 \, a b^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{a + b} + 9 \, b^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) + \frac {6 \, b^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{a + b} + 9 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} b \tan \left (f x + e\right ) + \frac {6 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} b^{2} \tan \left (f x + e\right )}{a + b} - \frac {6 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}}}{\tan \left (f x + e\right )} - \frac {4 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b}{{\left (a + b\right )} \tan \left (f x + e\right )} - \frac {2 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {5}{2}}}{{\left (a + b\right )} \tan \left (f x + e\right )^{3}}}{6 \, f} \] Input:

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

Output:

1/6*(9*a*sqrt(b)*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b)) + 6*a*b^(3/2)*arc 
sinh(b*tan(f*x + e)/sqrt((a + b)*b))/(a + b) + 9*b^(3/2)*arcsinh(b*tan(f*x 
 + e)/sqrt((a + b)*b)) + 6*b^(5/2)*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b)) 
/(a + b) + 9*sqrt(b*tan(f*x + e)^2 + a + b)*b*tan(f*x + e) + 6*sqrt(b*tan( 
f*x + e)^2 + a + b)*b^2*tan(f*x + e)/(a + b) - 6*(b*tan(f*x + e)^2 + a + b 
)^(3/2)/tan(f*x + e) - 4*(b*tan(f*x + e)^2 + a + b)^(3/2)*b/((a + b)*tan(f 
*x + e)) - 2*(b*tan(f*x + e)^2 + a + b)^(5/2)/((a + b)*tan(f*x + e)^3))/f
 

Giac [F]

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

integrate(csc(f*x+e)^4*(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)^4, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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