\(\int \csc ^2(e+f x) (a+b \tan ^2(e+f x))^{3/2} \, dx\) [113]

Optimal result

Integrand size = 25, antiderivative size = 100 \[ \int \csc ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\frac {3 a \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 f}+\frac {3 b \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 f}-\frac {\cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{f} \] Output:

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 2.17 (sec) , antiderivative size = 220, normalized size of antiderivative = 2.20 \[ \int \csc ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\frac {\csc (e+f x) \sec ^3(e+f x) \left (-6 a^2-a b+3 b^2-4 \left (2 a^2+b^2\right ) \cos (2 (e+f x))-2 a^2 \cos (4 (e+f x))+a b \cos (4 (e+f x))+b^2 \cos (4 (e+f x))+3 \sqrt {2} a b \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right ) \sin ^2(2 (e+f x))\right )}{8 \sqrt {2} f \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}} \] Input:

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

Output:

(Csc[e + f*x]*Sec[e + f*x]^3*(-6*a^2 - a*b + 3*b^2 - 4*(2*a^2 + b^2)*Cos[2 
*(e + f*x)] - 2*a^2*Cos[4*(e + f*x)] + a*b*Cos[4*(e + f*x)] + b^2*Cos[4*(e 
 + f*x)] + 3*Sqrt[2]*a*b*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + 
f*x]^2)/b]*EllipticF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e 
 + f*x]^2)/b]/Sqrt[2]], 1]*Sin[2*(e + f*x)]^2))/(8*Sqrt[2]*f*Sqrt[(a + b + 
 (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 98, 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, 4146, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4146

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

\(\Big \downarrow \) 247

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(-(Cot[e + f*x]*(a + b*Tan[e + f*x]^2)^(3/2)) + 3*b*((a*ArcTanh[(Sqrt[b]*T 
an[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(2*Sqrt[b]) + (Tan[e + f*x]*Sqrt 
[a + 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 4146

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

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.21

method	result	size
derivativedivides	\(-\frac {\left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {5}{2}}}{f a \tan \left (f x +e \right )}+\frac {b \tan \left (f x +e \right ) \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{f a}+\frac {3 b \tan \left (f x +e \right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}}{2 f}+\frac {3 \sqrt {b}\, a \ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{2 f}\)	\(121\)
default	\(-\frac {\left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {5}{2}}}{f a \tan \left (f x +e \right )}+\frac {b \tan \left (f x +e \right ) \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{f a}+\frac {3 b \tan \left (f x +e \right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}}{2 f}+\frac {3 \sqrt {b}\, a \ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{2 f}\)	\(121\)

Input:

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

Output:

-1/f/a/tan(f*x+e)*(a+b*tan(f*x+e)^2)^(5/2)+1/f/a*b*tan(f*x+e)*(a+b*tan(f*x 
+e)^2)^(3/2)+3/2*b*tan(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2)/f+3/2/f*b^(1/2)*a*l 
n(b^(1/2)*tan(f*x+e)+(a+b*tan(f*x+e)^2)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (86) = 172\).

Time = 0.32 (sec) , antiderivative size = 387, normalized size of antiderivative = 3.87 \[ \int \csc ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\left [\frac {3 \, a \sqrt {b} \cos \left (f x + e\right ) \log \left (\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) \sin \left (f x + e\right ) - 4 \, {\left ({\left (2 \, a + b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{8 \, f \cos \left (f x + e\right ) \sin \left (f x + e\right )}, -\frac {3 \, a \sqrt {-b} \arctan \left (\frac {{\left ({\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left ({\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 2 \, {\left ({\left (2 \, a + b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, f \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.73 \[ \int \csc ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\frac {3 \, a \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right ) + 3 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b \tan \left (f x + e\right ) - \frac {2 \, {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}{\tan \left (f x + e\right )}}{2 \, f} \] Input:

integrate(csc(f*x+e)^2*(a+b*tan(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)) + 3*sqrt(b*tan(f*x + e) 
^2 + a)*b*tan(f*x + e) - 2*(b*tan(f*x + e)^2 + a)^(3/2)/tan(f*x + e))/f
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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