\(\int \frac {\tan (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) [512]

Optimal result

Integrand size = 23, antiderivative size = 36 \[ \int \frac {\tan (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b} f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3273, 65, 214} \[ \int \frac {\tan (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{f \sqrt {a+b}} \]

[In]

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

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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 3273

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x]^2, x]}, Dist[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] && IntegerQ[(m - 1)/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{(1-x) \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{2 f} \\ & = \frac {\text {Subst}\left (\int \frac {1}{1+\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sin ^2(e+f x)}\right )}{b f} \\ & = \frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b} f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.06 \[ \int \frac {\tan (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b-b \cos ^2(e+f x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b} f} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(102\) vs. \(2(30)=60\).

Time = 1.37 (sec) , antiderivative size = 103, normalized size of antiderivative = 2.86

method	result	size
default	\(\frac {\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right )+\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right )}{2 \sqrt {a +b}\, f}\)	\(103\)

[In]

int(tan(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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))+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)^(1/2)/f

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 112, normalized size of antiderivative = 3.11 \[ \int \frac {\tan (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\left [\frac {\log \left (\frac {b \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right )}{2 \, \sqrt {a + b} f}, -\frac {\sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{a + b}\right )}{{\left (a + b\right )} f}\right ] \]

[In]

integrate(tan(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

integrate(tan(f*x+e)/(a+b*sin(f*x+e)**2)**(1/2),x)

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (30) = 60\).

Time = 0.32 (sec) , antiderivative size = 106, normalized size of antiderivative = 2.94 \[ \int \frac {\tan (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=-\frac {\frac {\operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}}\right )}{\sqrt {a + b}} - \frac {\operatorname {arsinh}\left (-\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}}\right )}{\sqrt {a + b}}}{2 \, f} \]

[In]

integrate(tan(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(tan(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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