\(\int \cot ^4(e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\) [400]

Optimal result

Integrand size = 25, antiderivative size = 112 \[ \int \cot ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}+\frac {(3 a-b) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 f}-\frac {(a+b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{3 f} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.25 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.89 \[ \int \cot ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=-\frac {2 (a+b) \cot ^3(e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{2},-\frac {1}{2},\frac {a \sin ^2(e+f x)}{a+b}\right ) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 f (a+2 b+a \cos (2 (e+f x))) \sqrt {\frac {a+b-a \sin ^2(e+f x)}{a+b}}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.98, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4629, 2075, 376, 25, 445, 27, 291, 216}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4629

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

\(\Big \downarrow \) 2075

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

\(\Big \downarrow \) 376

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 445

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[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 376

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

rule 445

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

rule 2075

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa 
ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi 
alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] &&  ! 
BinomialMatchQ[{u, v}, x]
 

rule 3042

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

rule 4629

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f 
_.)*(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim 
p[ff/f   Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2*x^2 
)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && Inte 
gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(249\) vs. \(2(98)=196\).

Time = 3.66 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.23

method	result	size
default	\(-\frac {\left (\left (3 \cos \left (f x +e \right )-3\right ) \sin \left (f x +e \right ) a^{2} \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right )+\left (4 \cos \left (f x +e \right )^{2}-3\right ) \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a +\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {-a}\, b \right ) \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}} \cot \left (f x +e \right )^{3}}{3 f \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (b +a \cos \left (f x +e \right )^{2}\right )}\)	\(250\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (98) = 196\).

Time = 0.65 (sec) , antiderivative size = 597, normalized size of antiderivative = 5.33 \[ \int \cot ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/24*(3*(a*cos(f*x + e)^2 - a)*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256* 
(a^4 - a^3*b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + 
 e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 
 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 - 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - 
a^2*b)*cos(f*x + e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a 
^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e)^ 
2 + b)/cos(f*x + e)^2)*sin(f*x + e))*sin(f*x + e) + 8*(4*a*cos(f*x + e)^3 
- (3*a - b)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((f 
*cos(f*x + e)^2 - f)*sin(f*x + e)), -1/12*(3*(a*cos(f*x + e)^2 - a)*sqrt(a 
)*arctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^3 + (a^2 - 
 6*a*b + b^2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + 
e)^2)/((2*a^3*cos(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e 
)^2)*sin(f*x + e)))*sin(f*x + e) - 4*(4*a*cos(f*x + e)^3 - (3*a - b)*cos(f 
*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((f*cos(f*x + e)^2 - 
 f)*sin(f*x + e))]
 

Sympy [F(-1)]

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

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

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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