3.378 \(\int \sqrt{a+b \sec ^2(e+f x)} \tan (e+f x) \, dx\)

Optimal. Leaf size=54 \[ \frac{\sqrt{a+b \sec ^2(e+f x)}}{f}-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.0647038, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {4139, 266, 50, 63, 208} \[ \frac{\sqrt{a+b \sec ^2(e+f x)}}{f}-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4139

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Sec[e + f*x], x]}, Dist[1/f, Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p)/x
, 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] || IGtQ[p, 0] || IntegersQ[2*n, p])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 50

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

Rule 63

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 208

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

Rubi steps

\begin{align*} \int \sqrt{a+b \sec ^2(e+f x)} \tan (e+f x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+b x^2}}{x} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,\sec ^2(e+f x)\right )}{2 f}\\ &=\frac{\sqrt{a+b \sec ^2(e+f x)}}{f}+\frac{a \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\sec ^2(e+f x)\right )}{2 f}\\ &=\frac{\sqrt{a+b \sec ^2(e+f x)}}{f}+\frac{a \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sec ^2(e+f x)}\right )}{b f}\\ &=-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{f}+\frac{\sqrt{a+b \sec ^2(e+f x)}}{f}\\ \end{align*}

Mathematica [B]  time = 0.486268, size = 119, normalized size = 2.2 \[ \frac{\sqrt{a+b \sec ^2(e+f x)} \left (\sqrt{2} \sqrt{b} \sqrt{\frac{a \cos (2 (e+f x))+a+2 b}{b}}-2 \sqrt{a} \cos (e+f x) \sinh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )\right )}{\sqrt{2} \sqrt{b} f \sqrt{\frac{a \cos (2 (e+f x))+a+2 b}{b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.079, size = 61, normalized size = 1.1 \begin{align*} -{\frac{1}{f}\sqrt{a}\ln \left ({\frac{1}{\sec \left ( fx+e \right ) } \left ( 2\,a+2\,\sqrt{a}\sqrt{a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2}} \right ) } \right ) }+{\frac{1}{f}\sqrt{a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/f*a^(1/2)*ln((2*a+2*a^(1/2)*(a+b*sec(f*x+e)^2)^(1/2))/sec(f*x+e))+(a+b*sec(f*x+e)^2)^(1/2)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sec \left (f x + e\right )^{2} + a} \tan \left (f x + e\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 0.834996, size = 779, normalized size = 14.43 \begin{align*} \left [\frac{\sqrt{a} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} + 256 \, a^{3} b \cos \left (f x + e\right )^{6} + 160 \, a^{2} b^{2} \cos \left (f x + e\right )^{4} + 32 \, a b^{3} \cos \left (f x + e\right )^{2} + b^{4} - 8 \,{\left (16 \, a^{3} \cos \left (f x + e\right )^{8} + 24 \, a^{2} b \cos \left (f x + e\right )^{6} + 10 \, a b^{2} \cos \left (f x + e\right )^{4} + b^{3} \cos \left (f x + e\right )^{2}\right )} \sqrt{a} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}\right ) + 8 \, \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{8 \, f}, \frac{\sqrt{-a} \arctan \left (\frac{{\left (8 \, a^{2} \cos \left (f x + e\right )^{4} + 8 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt{-a} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \,{\left (2 \, a^{3} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b \cos \left (f x + e\right )^{2} + a b^{2}\right )}}\right ) + 4 \, \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(sqrt(a)*log(128*a^4*cos(f*x + e)^8 + 256*a^3*b*cos(f*x + e)^6 + 160*a^2*b^2*cos(f*x + e)^4 + 32*a*b^3*co
s(f*x + e)^2 + b^4 - 8*(16*a^3*cos(f*x + e)^8 + 24*a^2*b*cos(f*x + e)^6 + 10*a*b^2*cos(f*x + e)^4 + b^3*cos(f*
x + e)^2)*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)) + 8*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)
)/f, 1/4*(sqrt(-a)*arctan(1/4*(8*a^2*cos(f*x + e)^4 + 8*a*b*cos(f*x + e)^2 + b^2)*sqrt(-a)*sqrt((a*cos(f*x + e
)^2 + b)/cos(f*x + e)^2)/(2*a^3*cos(f*x + e)^4 + 3*a^2*b*cos(f*x + e)^2 + a*b^2)) + 4*sqrt((a*cos(f*x + e)^2 +
b)/cos(f*x + e)^2))/f]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \sec ^{2}{\left (e + f x \right )}} \tan{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.50058, size = 537, normalized size = 9.94 \begin{align*} \frac{2 \,{\left (\frac{a \arctan \left (-\frac{\sqrt{a + b} \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} + b \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} + 2 \, b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a + b} + \sqrt{a + b}}{2 \, \sqrt{-a}}\right )}{\sqrt{-a}} + \frac{2 \,{\left ({\left (\sqrt{a + b} \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} + b \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} + 2 \, b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a + b}\right )} b + \sqrt{a + b} b\right )}}{{\left (\sqrt{a + b} \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} + b \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} + 2 \, b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a + b}\right )}^{2} - 2 \,{\left (\sqrt{a + b} \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} + b \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} + 2 \, b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a + b}\right )} \sqrt{a + b} + a - 3 \, b}\right )} \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(a*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(-a))/sqrt(-a) + 2*((s
qrt(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))*b + sqrt(a + b)*b)/((sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - s
qrt(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 - 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) + a - 3*b))*sgn(cos(f*x +
e))/f