\(\int \csc ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\) [71]

Optimal result

Integrand size = 25, antiderivative size = 124 \[ \int \csc ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{f}-\frac {(a+2 b) \text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 \sqrt {a+b} f}-\frac {\cot (e+f x) \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 f} \] Output:

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

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.31 \[ \int \csc ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)} \left (2 \sqrt {b} (a+b) \text {arctanh}\left (\frac {\sqrt {a+b-a \sin ^2(e+f x)}}{\sqrt {b}}\right )-\sqrt {a+b} (a+2 b) \text {arctanh}\left (\frac {\sqrt {a+b-a \sin ^2(e+f x)}}{\sqrt {a+b}}\right )-(a+b) \csc ^2(e+f x) \sqrt {a+b-a \sin ^2(e+f x)}\right )}{\sqrt {2} (a+b) f \sqrt {a+2 b+a \cos (2 (e+f x))}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4622, 369, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4622

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

\(\Big \downarrow \) 369

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

\(\Big \downarrow \) 398

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

Defintions of rubi rules used

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 369

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(2* 
b*(p + 1))), x] - Simp[e^2/(2*b*(p + 1))   Int[(e*x)^(m - 2)*(a + b*x^2)^(p 
 + 1)*(c + d*x^2)^(q - 1)*Simp[c*(m - 1) + d*(m + 2*q - 1)*x^2, x], x], x] 
/; FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 0 
] && GtQ[m, 1] && IntBinomialQ[a, b, c, d, e, m, 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. \(1243\) vs. \(2(106)=212\).

Time = 4.07 (sec) , antiderivative size = 1244, normalized size of antiderivative = 10.03

method	result	size
default	\(\text {Expression too large to display}\)	\(1244\)

Input:

int(csc(f*x+e)^3*(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/4/f/(a+b)^(5/2)*((4*cos(f*x+e)-4)*b^(1/2)*(a+b)^(3/2)*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+(-5*cos(f*x+e)+5)*ln(2/(a+b)^(1/2)*((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^2+(-5*cos(f*x+e) 
+5)*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^2+(-4*cos(f*x+e)+4)*ln(2/(a+b)^(1/2)*((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*co 
s(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+ 
(-4*cos(f*x+e)+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^2*b+(4*cos(f*x+e)-4)*b^(3/2)*(a+b)^(3/2 
)*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))+(1-cos(f*x+e))*ln(2/(a+b)^(1/2)*((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*co 
s(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+(- 
2*cos(f*x+e)+2)*ln(2/(a+b)^(1/2)*((a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+co...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (106) = 212\).

Time = 0.23 (sec) , antiderivative size = 907, normalized size of antiderivative = 7.31 \[ \int \csc ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\text {Too large to display} \] Input:

integrate(csc(f*x+e)^3*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

Output:

[1/4*(2*(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 - a - 2*b)*sqrt(a + b)*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)) + 2*((a + b)*cos(f*x + e)^2 - a - b)*sqrt( 
b)*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))/((a + b)*f*cos(f*x + e)^2 - (a 
 + b)*f), 1/2*(((a + 2*b)*cos(f*x + e)^2 - a - 2*b)*sqrt(-a - b)*arctan(sq 
rt(-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)) + (a + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*co 
s(f*x + e) + ((a + b)*cos(f*x + e)^2 - a - b)*sqrt(b)*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))/((a + b)*f*cos(f*x + e)^2 - (a + b)*f), -1/4*(4*((a + 
 b)*cos(f*x + e)^2 - a - 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)) - 2*(a + b)*sq 
rt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) - ((a + 2*b)*cos(f* 
x + e)^2 - a - 2*b)*sqrt(a + b)*log(2*(a*cos(f*x + e)^2 - 2*sqrt(a + b)*sq 
rt((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)))/((a + b)*f*cos(f*x + e)^2 - (a + b)*f), 1/2*(((a + 2*b)*cos 
(f*x + e)^2 - a - 2*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)) - 2*((a ...
 

Sympy [F]

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

integrate(csc(f*x+e)**3*(a+b*sec(f*x+e)**2)**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate(csc(f*x+e)^3*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 578 vs. \(2 (106) = 212\).

Time = 0.63 (sec) , antiderivative size = 578, normalized size of antiderivative = 4.66 \[ \int \csc ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx =\text {Too large to display} \] Input:

integrate(csc(f*x+e)^3*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

1/8*(4*(a + 2*b)*arctan(-(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan( 
1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 
 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))/sqrt(-a - b))/sqrt(-a - b) - 16*b* 
arctan(-1/2*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2 
*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/ 
2*f*x + 1/2*e)^2 + a + b) - sqrt(a + b))/sqrt(-b))/sqrt(-b) + 2*(a + 2*b)* 
log(abs(-(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e) 
^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f 
*x + 1/2*e)^2 + a + b))*(a + b) + sqrt(a + b)*(a - b)))/sqrt(a + b) - sqrt 
(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1 
/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b) - 2*((sqrt(a + b)*tan(1/2*f* 
x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 
2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))*(a - b) 
- (a + b)^(3/2))/((sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x 
 + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b* 
tan(1/2*f*x + 1/2*e)^2 + a + b))^2 - a - b))*sgn(cos(f*x + e))/f
 

Mupad [F(-1)]

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

int((a + b/cos(e + f*x)^2)^(1/2)/sin(e + f*x)^3,x)
 

Output:

int((a + b/cos(e + f*x)^2)^(1/2)/sin(e + f*x)^3, x)
 

Reduce [F]

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

int(csc(f*x+e)^3*(a+b*sec(f*x+e)^2)^(1/2),x)
 

Output:

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