3.417 \(\int \frac{\tan (e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=57 \[ \frac{1}{a f \sqrt{a+b \sec ^2(e+f x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{a^{3/2} f} \]

[Out]

-(ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]]/(a^(3/2)*f)) + 1/(a*f*Sqrt[a + b*Sec[e + f*x]^2])

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Rubi [A]  time = 0.0721779, antiderivative size = 57, 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, 51, 63, 208} \[ \frac{1}{a f \sqrt{a+b \sec ^2(e+f x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{a^{3/2} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]]/(a^(3/2)*f)) + 1/(a*f*Sqrt[a + b*Sec[e + f*x]^2])

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 51

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] - Dist[(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] && NeQ[b*c - a*d, 0] && 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 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 \frac{\tan (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x \left (a+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{x (a+b x)^{3/2}} \, dx,x,\sec ^2(e+f x)\right )}{2 f}\\ &=\frac{1}{a f \sqrt{a+b \sec ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\sec ^2(e+f x)\right )}{2 a f}\\ &=\frac{1}{a f \sqrt{a+b \sec ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sec ^2(e+f x)}\right )}{a b f}\\ &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{a^{3/2} f}+\frac{1}{a f \sqrt{a+b \sec ^2(e+f x)}}\\ \end{align*}

Mathematica [C]  time = 7.15968, size = 382, normalized size = 6.7 \[ \frac{\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b)^{3/2} \left (-\frac{2}{b \sqrt{a \cos (2 (e+f x))+a+2 b}}+\frac{\sqrt{2} e^{i (e+f x)} \sec (e+f x) \sqrt{4 b+a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2} \left (\frac{\sqrt{a} (a+4 b) \left (1+e^{2 i (e+f x)}\right )}{b \left (a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}\right )}+\frac{-2 \log \left (\sqrt{a} \sqrt{a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}}+a e^{2 i (e+f x)}+a+2 b\right )-2 \log \left (\sqrt{a} \sqrt{a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}}+a e^{2 i (e+f x)}+a+2 b e^{2 i (e+f x)}\right )+4 i f x}{\sqrt{a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}}}\right )}{a^{3/2}}\right )}{16 f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])^(3/2)*Sec[e + f*x]^2*(-2/(b*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]) + (Sqrt[2]*E^(
I*(e + f*x))*Sqrt[4*b + (a*(1 + E^((2*I)*(e + f*x)))^2)/E^((2*I)*(e + f*x))]*((Sqrt[a]*(a + 4*b)*(1 + E^((2*I)
*(e + f*x))))/(b*(4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2)) + ((4*I)*f*x - 2*Log[a + 2*b + a*E
^((2*I)*(e + f*x)) + Sqrt[a]*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]] - 2*Log[a + a*E^((
2*I)*(e + f*x)) + 2*b*E^((2*I)*(e + f*x)) + Sqrt[a]*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))
^2]])/Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2])*Sec[e + f*x])/a^(3/2)))/(16*f*(a + b*Sec[
e + f*x]^2)^(3/2))

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Maple [A]  time = 0.056, size = 64, normalized size = 1.1 \begin{align*}{\frac{1}{af}{\frac{1}{\sqrt{a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2}}}}}-{\frac{1}{f}\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 ){a}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a/f/(a+b*sec(f*x+e)^2)^(1/2)-1/f/a^(3/2)*ln((2*a+2*a^(1/2)*(a+b*sec(f*x+e)^2)^(1/2))/sec(f*x+e))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (f x + e\right )}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 0.952492, size = 971, normalized size = 17.04 \begin{align*} \left [\frac{8 \, a \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )^{2} +{\left (a \cos \left (f x + e\right )^{2} + b\right )} \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 \,{\left (a^{3} f \cos \left (f x + e\right )^{2} + a^{2} b f\right )}}, \frac{4 \, a \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )^{2} +{\left (a \cos \left (f x + e\right )^{2} + b\right )} \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 \,{\left (a^{3} f \cos \left (f x + e\right )^{2} + a^{2} b f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 11.0727, size = 53, normalized size = 0.93 \begin{align*} \frac{1}{a f \sqrt{a + b \sec ^{2}{\left (e + f x \right )}}} + \frac{\operatorname{atan}{\left (\frac{\sqrt{a + b \sec ^{2}{\left (e + f x \right )}}}{\sqrt{- a}} \right )}}{a f \sqrt{- a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(a*f*sqrt(a + b*sec(e + f*x)**2)) + atan(sqrt(a + b*sec(e + f*x)**2)/sqrt(-a))/(a*f*sqrt(-a))

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Giac [B]  time = 2.07094, size = 336, normalized size = 5.89 \begin{align*} \frac{\frac{\frac{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2}}{a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )} - \frac{1}{a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )}}{\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}} - \frac{2 \, \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} a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((tan(1/2*f*x + 1/2*e)^2/(a*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)) - 1/(a*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)))/sqrt(a*t
an(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*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)*a
*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)))/f