\(\int \csc (e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\) [83]

Optimal result

Integrand size = 23, antiderivative size = 122 \[ \int \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {\sqrt {b} (3 a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 f}-\frac {(a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{f}+\frac {b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 f} \] Output:

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

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.40 \[ \int \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {\left (2 \sqrt {b} (3 a+2 b) \text {arctanh}\left (\frac {\sqrt {a+b-a \sin ^2(e+f x)}}{\sqrt {b}}\right ) \cos ^2(e+f x)-4 (a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b-a \sin ^2(e+f x)}}{\sqrt {a+b}}\right ) \cos ^2(e+f x)+\sqrt {2} b \sqrt {a+2 b+a \cos (2 (e+f x))}\right ) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 \sqrt {2} f \sqrt {a+2 b+a \cos (2 (e+f x))}} \] Input:

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

Output:

((2*Sqrt[b]*(3*a + 2*b)*ArcTanh[Sqrt[a + b - a*Sin[e + f*x]^2]/Sqrt[b]]*Co 
s[e + f*x]^2 - 4*(a + b)^(3/2)*ArcTanh[Sqrt[a + b - a*Sin[e + f*x]^2]/Sqrt 
[a + b]]*Cos[e + f*x]^2 + Sqrt[2]*b*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]])*Se 
c[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2])/(2*Sqrt[2]*f*Sqrt[a + 2*b + a*Cos[2 
*(e + f*x)]])
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.98, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 4622, 25, 318, 25, 398, 224, 219, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4622

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 318

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 398

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 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 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 318

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

rule 398

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

rule 3042

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

rule 4622

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + ( 
f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Si 
mp[1/(f*ff^m)   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p 
/x^(m + 1)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x 
] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1465\) vs. \(2(104)=208\).

Time = 8.94 (sec) , antiderivative size = 1466, normalized size of antiderivative = 12.02

method	result	size
default	\(\text {Expression too large to display}\)	\(1466\)

Input:

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

Output:

1/2/f/b/(a+b)^(5/2)*(a+b*sec(f*x+e)^2)^(3/2)/((b+a*cos(f*x+e)^2)/(1+cos(f* 
x+e))^2)^(1/2)/(a*cos(f*x+e)^3+a*cos(f*x+e)^2+cos(f*x+e)*b+b)*(2*b^(5/2)*( 
a+b)^(5/2)*ln(4*(b*cot(f*x+e)^2-2*b*cot(f*x+e)*csc(f*x+e)+b*csc(f*x+e)^2+2 
*b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+b)/(cot(f*x+e)^2-2*cs 
c(f*x+e)*cot(f*x+e)+csc(f*x+e)^2-1))*cos(f*x+e)^3+3*b^(3/2)*(a+b)^(5/2)*ln 
(4*(b*cot(f*x+e)^2-2*b*cot(f*x+e)*csc(f*x+e)+b*csc(f*x+e)^2+2*b^(1/2)*((b+ 
a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+b)/(cot(f*x+e)^2-2*csc(f*x+e)*cot( 
f*x+e)+csc(f*x+e)^2-1))*a*cos(f*x+e)^3+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^ 
2)^(1/2)*(a+b)^(5/2)*b^2*(cos(f*x+e)^2+cos(f*x+e))-ln(-4*((a+b)^(1/2)*((b+ 
a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+(a+b)^(1/2)*((b+a*cos(f 
*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+cos(f*x+e)*a+b)/(-1+cos(f*x+e)))*a^4*b*co 
s(f*x+e)^3-4*ln(-4*((a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2 
)*cos(f*x+e)+(a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+cos(f 
*x+e)*a+b)/(-1+cos(f*x+e)))*a^3*b^2*cos(f*x+e)^3-6*ln(-4*((a+b)^(1/2)*((b+ 
a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+(a+b)^(1/2)*((b+a*cos(f 
*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+cos(f*x+e)*a+b)/(-1+cos(f*x+e)))*a^2*b^3* 
cos(f*x+e)^3-4*ln(-4*((a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1 
/2)*cos(f*x+e)+(a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+cos 
(f*x+e)*a+b)/(-1+cos(f*x+e)))*a*b^4*cos(f*x+e)^3-ln(-4*((a+b)^(1/2)*((b+a* 
cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+(a+b)^(1/2)*((b+a*cos(...
 

Fricas [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 755, normalized size of antiderivative = 6.19 \[ \int \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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