\(\int \frac {\tan (e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\) [454]

Optimal result

Integrand size = 23, antiderivative size = 63 \[ \int \frac {\tan (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2} f}-\frac {1}{(a+b) f \sqrt {a+b \sin ^2(e+f x)}} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.09 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.86 \[ \int \frac {\tan (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1-\frac {b \cos ^2(e+f x)}{a+b}\right )}{(a+b) f \sqrt {a+b-b \cos ^2(e+f x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 3673, 61, 73, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3673

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {2 \int \frac {1}{\frac {a+b}{b}-\frac {\sin ^4(e+f x)}{b}}d\sqrt {b \sin ^2(e+f x)+a}}{b (a+b)}-\frac {2}{(a+b) \sqrt {a+b \sin ^2(e+f x)}}}{2 f}\)

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 3042

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

rule 3673

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1316\) vs. \(2(55)=110\).

Time = 0.64 (sec) , antiderivative size = 1317, normalized size of antiderivative = 20.90

method	result	size
default	\(\text {Expression too large to display}\)	\(1317\)

Input:

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

Output:

1/2/a/(a^2*b^2*cos(f*x+e)^4+2*a*b^3*cos(f*x+e)^4+b^4*cos(f*x+e)^4-2*a^3*b* 
cos(f*x+e)^2-6*a^2*b^2*cos(f*x+e)^2-6*a*b^3*cos(f*x+e)^2-2*b^4*cos(f*x+e)^ 
2+a^4+4*a^3*b+6*a^2*b^2+4*a*b^3+b^4)*(-b^3*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^ 
2)^(1/2)+b^2*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(3/2)-(-b*cos(f*x+e)^2+(a*b 
^2+b^3)/b^2)^(3/2)*a^2+(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a^3-2*(a+b- 
b*cos(f*x+e)^2)^(1/2)*a^3+b*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a^2-4* 
b*(a+b-b*cos(f*x+e)^2)^(1/2)*a^2-2*b^2*(a+b-b*cos(f*x+e)^2)^(1/2)*a-b^2*(- 
b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a+(a+b)^(1/2)*ln(2/(sin(f*x+e)-1)*(( 
a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^3+(a+b)^(1/2)*ln( 
2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))* 
a^3+2*b*(a+b)^(1/2)*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^ 
(1/2)-b*sin(f*x+e)+a))*a^2+b^2*(a+b)^(1/2)*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2 
)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a+b^2*(a+b)^(1/2)*ln(2/(1+si 
n(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a+2*b*( 
a+b)^(1/2)*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*s 
in(f*x+e)+a))*a^2+b^2*((a+b)^(1/2)*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b 
*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a+(a+b)^(1/2)*ln(2/(sin(f*x+e)-1)*(( 
a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a+(-b*cos(f*x+e)^2+ 
(a*b^2+b^3)/b^2)^(1/2)*a-b*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)-2*(a+b- 
b*cos(f*x+e)^2)^(1/2)*a)*cos(f*x+e)^4-b*(2*(a+b)^(1/2)*ln(2/(1+sin(f*x+...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 145 vs. \(2 (55) = 110\).

Time = 0.13 (sec) , antiderivative size = 293, normalized size of antiderivative = 4.65 \[ \int \frac {\tan (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\left [\frac {{\left (b \cos \left (f x + e\right )^{2} - a - b\right )} \sqrt {a + b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}}{2 \, {\left ({\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f\right )}}, \frac {{\left (b \cos \left (f x + e\right )^{2} - a - b\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{b \cos \left (f x + e\right )^{2} - a - b}\right ) + \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}}{{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 143 vs. \(2 (55) = 110\).

Time = 0.13 (sec) , antiderivative size = 143, normalized size of antiderivative = 2.27 \[ \int \frac {\tan (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\frac {\operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}}\right )}{{\left (a + b\right )}^{\frac {3}{2}}} - \frac {\operatorname {arsinh}\left (-\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}}\right )}{{\left (a + b\right )}^{\frac {3}{2}}} + \frac {2}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a + \sqrt {b \sin \left (f x + e\right )^{2} + a} b}}{2 \, f} \] Input:

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

Output:

-1/2*(arcsinh(b*sin(f*x + e)/(sqrt(a*b)*(sin(f*x + e) + 1)) - a/(sqrt(a*b) 
*(sin(f*x + e) + 1)))/(a + b)^(3/2) - arcsinh(-b*sin(f*x + e)/(sqrt(a*b)*( 
sin(f*x + e) - 1)) - a/(sqrt(a*b)*(sin(f*x + e) - 1)))/(a + b)^(3/2) + 2/( 
sqrt(b*sin(f*x + e)^2 + a)*a + sqrt(b*sin(f*x + e)^2 + a)*b))/f
                                                                                    
                                                                                    
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 242 vs. \(2 (55) = 110\).

Time = 0.45 (sec) , antiderivative size = 242, normalized size of antiderivative = 3.84 \[ \int \frac {\tan (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\frac {\frac {{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}} + \frac {a^{2} b + 2 \, a b^{2} + b^{3}}{a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}}}{\sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}} + \frac {2 \, \arctan \left (-\frac {\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a} - \sqrt {a}}{2 \, \sqrt {-a - b}}\right )}{{\left (a + b\right )} \sqrt {-a - b}}}{f} \] Input:

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

Output:

-(((a^2*b + 2*a*b^2 + b^3)*tan(1/2*f*x + 1/2*e)^2/(a^3*b + 3*a^2*b^2 + 3*a 
*b^3 + b^4) + (a^2*b + 2*a*b^2 + b^3)/(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)) 
/sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2* 
f*x + 1/2*e)^2 + a) + 2*arctan(-1/2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt 
(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 
 1/2*e)^2 + a) - sqrt(a))/sqrt(-a - b))/((a + b)*sqrt(-a - b)))/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(sin(e + f*x)**2*b + a)*tan(e + f*x))/(sin(e + f*x)**4*b**2 + 2*s 
in(e + f*x)**2*a*b + a**2),x)