\(\int \csc ^2(e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\) [90]

Optimal result

Integrand size = 25, antiderivative size = 105 \[ \int \csc ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {3 \sqrt {b} (a+b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 f}+\frac {3 b \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 f}-\frac {\cot (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{f} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.18 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.61 \[ \int \csc ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=-\frac {(a+b) \cot (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},2,\frac {1}{2},\frac {b \sin ^2(e+f x)}{a+b-a \sin ^2(e+f x)}\right ) \sqrt {a+b \sec ^2(e+f x)}}{f} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4620, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4620

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

\(\Big \downarrow \) 247

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(-(Cot[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(3/2)) + 3*b*(((a + b)*ArcTanh[ 
(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))/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 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. \(615\) vs. \(2(91)=182\).

Time = 20.87 (sec) , antiderivative size = 616, normalized size of antiderivative = 5.87

method	result	size
default	\(-\frac {\left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}} \left (-3 \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 )^{3} b^{\frac {5}{2}}-3 \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 )^{3} b^{\frac {3}{2}} a -3 \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 )^{3} b^{\frac {5}{2}}-3 \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 )^{3} b^{\frac {3}{2}} a +\left (4 \cos \left (f x +e \right )+4\right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b \cos \left (f x +e \right )^{2} \cot \left (f x +e \right )+\left (6 \cos \left (f x +e \right )^{3}+6 \cos \left (f x +e \right )^{2}-2 \cos \left (f x +e \right )-2\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 )\right )}{4 f b \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (a \cos \left (f x +e \right )^{3}+a \cos \left (f x +e \right )^{2}+\cos \left (f x +e \right ) b +b \right )}\)	\(616\)

Input:

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

Output:

-1/4/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)/(a*cos(f*x+e)^3+a*cos(f*x+e)^2+cos(f*x+e)*b+b)*(-3*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)^3 
*b^(5/2)-3*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)^3*b^(3/2)*a-3*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)^3*b^(5/2)-3* 
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)^3*b^(3/2)*a+(4*cos(f*x+e)+4)*((b+a*cos(f*x+e)^2)/(1+cos 
(f*x+e))^2)^(1/2)*a*b*cos(f*x+e)^2*cot(f*x+e)+(6*cos(f*x+e)^3+6*cos(f*x+e) 
^2-2*cos(f*x+e)-2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2*cot(f*x 
+e))
 

Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 370, normalized size of antiderivative = 3.52 \[ \int \csc ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\left [\frac {3 \, {\left (a + b\right )} \sqrt {b} \cos \left (f x + e\right ) \log \left (\frac {{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) \sin \left (f x + e\right ) - 4 \, {\left ({\left (2 \, a + 3 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{8 \, f \cos \left (f x + e\right ) \sin \left (f x + e\right )}, \frac {3 \, {\left (a + b\right )} \sqrt {-b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, {\left ({\left (2 \, a + 3 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, f \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ] \] Input:

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

Output:

[1/8*(3*(a + b)*sqrt(b)*cos(f*x + e)*log(((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*((2*a + 3*b)*cos(f*x + e)^2 - b)* 
sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e)*sin(f*x + e)) 
, 1/4*(3*(a + b)*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)))*cos(f*x + e)*sin(f*x + e) - 2*((2*a + 3*b)*c 
os(f*x + e)^2 - b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x 
 + e)*sin(f*x + e))]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.93 \[ \int \csc ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {3 \, a \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) + 3 \, b^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) + 3 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} b \tan \left (f x + e\right ) - \frac {2 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}}}{\tan \left (f x + e\right )}}{2 \, f} \] Input:

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

Output:

1/2*(3*a*sqrt(b)*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b)) + 3*b^(3/2)*arcsi 
nh(b*tan(f*x + e)/sqrt((a + b)*b)) + 3*sqrt(b*tan(f*x + e)^2 + a + b)*b*ta 
n(f*x + e) - 2*(b*tan(f*x + e)^2 + a + b)^(3/2)/tan(f*x + e))/f
 

Giac [F]

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

integrate(csc(f*x+e)^2*(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)^2, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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