\(\int \frac {\csc (e+f x)}{(a+b \tan ^2(e+f x))^{5/2}} \, dx\) [143]

Optimal result

Integrand size = 23, antiderivative size = 136 \[ \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{a^{5/2} f}-\frac {b \sec (e+f x)}{3 a (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {(5 a-3 b) b \sec (e+f x)}{3 a^2 (a-b)^2 f \sqrt {a-b+b \sec ^2(e+f x)}} \] Output:

-arctanh(a^(1/2)*sec(f*x+e)/(a-b+b*sec(f*x+e)^2)^(1/2))/a^(5/2)/f-1/3*b*se 
c(f*x+e)/a/(a-b)/f/(a-b+b*sec(f*x+e)^2)^(3/2)-1/3*(5*a-3*b)*b*sec(f*x+e)/a 
^2/(a-b)^2/f/(a-b+b*sec(f*x+e)^2)^(1/2)
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(300\) vs. \(2(136)=272\).

Time = 6.63 (sec) , antiderivative size = 300, normalized size of antiderivative = 2.21 \[ \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\frac {\cos (e+f x) \left (-\frac {2 \sqrt {2} \sqrt {a} b \left (6 a^2+a b-3 b^2+3 \left (2 a^2-3 a b+b^2\right ) \cos (2 (e+f x))\right )}{(a-b)^2 (a+b+(a-b) \cos (2 (e+f x)))^2}+\frac {3 \left (2 \text {arctanh}\left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-\frac {\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{\sqrt {a}}\right )+\log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )\right ) \sec ^2\left (\frac {1}{2} (e+f x)\right )}{\sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^4\left (\frac {1}{2} (e+f x)\right )}}\right ) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{6 a^{5/2} f} \] Input:

Integrate[Csc[e + f*x]/(a + b*Tan[e + f*x]^2)^(5/2),x]
 

Output:

(Cos[e + f*x]*((-2*Sqrt[2]*Sqrt[a]*b*(6*a^2 + a*b - 3*b^2 + 3*(2*a^2 - 3*a 
*b + b^2)*Cos[2*(e + f*x)]))/((a - b)^2*(a + b + (a - b)*Cos[2*(e + f*x)]) 
^2) + (3*(2*ArcTanh[Tan[(e + f*x)/2]^2 - Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*( 
-1 + Tan[(e + f*x)/2]^2)^2]/Sqrt[a]] + Log[a - 2*b - a*Tan[(e + f*x)/2]^2 
+ Sqrt[a]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]])*S 
ec[(e + f*x)/2]^2)/Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[(e + f*x)/2 
]^4])*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(6*a^(5/2)* 
f)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4147, 25, 316, 25, 402, 27, 291, 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.

\(\displaystyle \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (e+f x) \left (a+b \tan (e+f x)^2\right )^{5/2}}dx\)

\(\Big \downarrow \) 4147

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 219

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

Input:

Int[Csc[e + f*x]/(a + b*Tan[e + f*x]^2)^(5/2),x]
 

Output:

(-1/3*(b*Sec[e + f*x])/(a*(a - b)*(a - b + b*Sec[e + f*x]^2)^(3/2)) - ((3* 
(a - b)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/a^ 
(3/2) + ((5*a - 3*b)*b*Sec[e + f*x])/(a*(a - b)*Sqrt[a - b + b*Sec[e + f*x 
]^2]))/(3*a*(a - b)))/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 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 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 316

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d))   Int[(a + b*x^2)^(p + 1)*(c + d*x 
^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x 
], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  ! 
( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, 
 p, q, x]
 

rule 402

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

rule 3042

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

rule 4147

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(3061\) vs. \(2(122)=244\).

Time = 171.67 (sec) , antiderivative size = 3062, normalized size of antiderivative = 22.51

method	result	size
default	\(\text {Expression too large to display}\)	\(3062\)

Input:

int(csc(f*x+e)/(a+b*tan(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/6/f/a^(13/2)/(2*(-b*(a-b))^(1/2)-a+2*b)^2/(a^2-2*a*b+b^2)/(2*(-b*(a-b)) 
^(1/2)+a-2*b)^2*(12*cos(f*x+e)^8*a^(27/2)*b+cos(f*x+e)^6*(-54*cos(f*x+e)^2 
+46)*a^(25/2)*b^2+cos(f*x+e)^4*(-96*cos(f*x+e)^2+66)*sin(f*x+e)^2*a^(23/2) 
*b^3+cos(f*x+e)^2*(-84*cos(f*x+e)^2+42)*sin(f*x+e)^4*a^(21/2)*b^4+(-36*cos 
(f*x+e)^2+10)*sin(f*x+e)^6*a^(19/2)*b^5-6*sin(f*x+e)^8*a^(17/2)*b^6+cos(f* 
x+e)^8*(3*cos(f*x+e)+3)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2) 
^(1/2)*ln(2/a^(1/2)*(a^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+ 
1)^2)^(1/2)*cos(f*x+e)+((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^ 
(1/2)*a^(1/2)-a*cos(f*x+e)+cos(f*x+e)*b+b)/(cos(f*x+e)+1))*a^14+cos(f*x+e) 
^6*(-18*cos(f*x+e)^3-18*cos(f*x+e)^2+12*cos(f*x+e)+12)*((a*cos(f*x+e)^2+b* 
sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*ln(2/a^(1/2)*(a^(1/2)*((a*cos(f*x+e) 
^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)+((a*cos(f*x+e)^2+b*s 
in(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-a*cos(f*x+e)+cos(f*x+e)*b+b)/ 
(cos(f*x+e)+1))*a^13*b+cos(f*x+e)^4*(45*cos(f*x+e)^5+45*cos(f*x+e)^4-60*co 
s(f*x+e)^3-60*cos(f*x+e)^2+18*cos(f*x+e)+18)*((a*cos(f*x+e)^2+b*sin(f*x+e) 
^2)/(cos(f*x+e)+1)^2)^(1/2)*ln(2/a^(1/2)*(a^(1/2)*((a*cos(f*x+e)^2+b*sin(f 
*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)+((a*cos(f*x+e)^2+b*sin(f*x+e)^ 
2)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-a*cos(f*x+e)+cos(f*x+e)*b+b)/(cos(f*x+e 
)+1))*a^12*b^2+cos(f*x+e)^2*(60*cos(f*x+e)^5+60*cos(f*x+e)^4-60*cos(f*x+e) 
^3-60*cos(f*x+e)^2+12*cos(f*x+e)+12)*sin(f*x+e)^2*((a*cos(f*x+e)^2+b*si...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 350 vs. \(2 (122) = 244\).

Time = 0.30 (sec) , antiderivative size = 711, normalized size of antiderivative = 5.23 \[ \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(csc(f*x+e)/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="fricas")
 

Output:

[1/6*(3*((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^4 + a^2* 
b^2 - 2*a*b^3 + b^4 + 2*(a^3*b - 3*a^2*b^2 + 3*a*b^3 - b^4)*cos(f*x + e)^2 
)*sqrt(a)*log(-2*((a - b)*cos(f*x + e)^2 - 2*sqrt(a)*sqrt(((a - b)*cos(f*x 
 + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + a + b)/(cos(f*x + e)^2 - 1)) - 
 2*(3*(2*a^3*b - 3*a^2*b^2 + a*b^3)*cos(f*x + e)^3 + (5*a^2*b^2 - 3*a*b^3) 
*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^7 - 
4*a^6*b + 6*a^5*b^2 - 4*a^4*b^3 + a^3*b^4)*f*cos(f*x + e)^4 + 2*(a^6*b - 3 
*a^5*b^2 + 3*a^4*b^3 - a^3*b^4)*f*cos(f*x + e)^2 + (a^5*b^2 - 2*a^4*b^3 + 
a^3*b^4)*f), -1/3*(3*((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x 
+ e)^4 + a^2*b^2 - 2*a*b^3 + b^4 + 2*(a^3*b - 3*a^2*b^2 + 3*a*b^3 - b^4)*c 
os(f*x + e)^2)*sqrt(-a)*arctan(-sqrt(-a)*sqrt(((a - b)*cos(f*x + e)^2 + b) 
/cos(f*x + e)^2)*cos(f*x + e)/((a - b)*cos(f*x + e)^2 + b)) + (3*(2*a^3*b 
- 3*a^2*b^2 + a*b^3)*cos(f*x + e)^3 + (5*a^2*b^2 - 3*a*b^3)*cos(f*x + e))* 
sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^7 - 4*a^6*b + 6*a^5 
*b^2 - 4*a^4*b^3 + a^3*b^4)*f*cos(f*x + e)^4 + 2*(a^6*b - 3*a^5*b^2 + 3*a^ 
4*b^3 - a^3*b^4)*f*cos(f*x + e)^2 + (a^5*b^2 - 2*a^4*b^3 + a^3*b^4)*f)]
 

Sympy [F]

\[ \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\csc {\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(csc(f*x+e)/(a+b*tan(f*x+e)**2)**(5/2),x)
 

Output:

Integral(csc(e + f*x)/(a + b*tan(e + f*x)**2)**(5/2), x)
 

Maxima [F]

\[ \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\csc \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(csc(f*x+e)/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="maxima")
 

Output:

integrate(csc(f*x + e)/(b*tan(f*x + e)^2 + a)^(5/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(csc(f*x+e)/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Degree mismatch inside factorisatio 
n over extensionNot implemented, e.g. for multivariate mod/approx polynomi 
alsError:
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {1}{\sin \left (e+f\,x\right )\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \] Input:

int(1/(sin(e + f*x)*(a + b*tan(e + f*x)^2)^(5/2)),x)
 

Output:

int(1/(sin(e + f*x)*(a + b*tan(e + f*x)^2)^(5/2)), x)
 

Reduce [F]

\[ \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\tan \left (f x +e \right )^{2} b +a}\, \csc \left (f x +e \right )}{\tan \left (f x +e \right )^{6} b^{3}+3 \tan \left (f x +e \right )^{4} a \,b^{2}+3 \tan \left (f x +e \right )^{2} a^{2} b +a^{3}}d x \] Input:

int(csc(f*x+e)/(a+b*tan(f*x+e)^2)^(5/2),x)
 

Output:

int((sqrt(tan(e + f*x)**2*b + a)*csc(e + f*x))/(tan(e + f*x)**6*b**3 + 3*t 
an(e + f*x)**4*a*b**2 + 3*tan(e + f*x)**2*a**2*b + a**3),x)