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

Optimal. Leaf size=69 \[ \frac{1}{f (a-b) \sqrt{a+b \tan ^2(e+f x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f (a-b)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0884195, antiderivative size = 69, 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 = {3670, 444, 51, 63, 208} \[ \frac{1}{f (a-b) \sqrt{a+b \tan ^2(e+f x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f (a-b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

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

Mathematica [C]  time = 0.0734709, size = 56, normalized size = 0.81 \[ -\frac{\text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{a+b \tan ^2(e+f x)}{a-b}\right )}{f (b-a) \sqrt{a+b \tan ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.014, size = 68, normalized size = 1. \begin{align*}{\frac{1}{ \left ( a-b \right ) f}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}}+{\frac{1}{ \left ( a-b \right ) f}\arctan \left ({\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{\frac{1}{\sqrt{-a+b}}}} \right ){\frac{1}{\sqrt{-a+b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 \tan \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*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.74776, size = 790, normalized size = 11.45 \begin{align*} \left [-\frac{{\left (b \tan \left (f x + e\right )^{2} + a\right )} \sqrt{a - b} \log \left (-\frac{b^{2} \tan \left (f x + e\right )^{4} + 2 \,{\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} + 4 \,{\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) - 4 \, \sqrt{b \tan \left (f x + e\right )^{2} + a}{\left (a - b\right )}}{4 \,{\left ({\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} f\right )}}, \frac{{\left (b \tan \left (f x + e\right )^{2} + a\right )} \sqrt{-a + b} \arctan \left (\frac{2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{-a + b}}{b \tan \left (f x + e\right )^{2} + 2 \, a - b}\right ) + 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a}{\left (a - b\right )}}{2 \,{\left ({\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 13.658, size = 56, normalized size = 0.81 \begin{align*} \frac{1}{f \left (a - b\right ) \sqrt{a + b \tan ^{2}{\left (e + f x \right )}}} + \frac{\operatorname{atan}{\left (\frac{\sqrt{a + b \tan ^{2}{\left (e + f x \right )}}}{\sqrt{- a + b}} \right )}}{f \sqrt{- a + b} \left (a - b\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.28866, size = 93, normalized size = 1.35 \begin{align*} \frac{\arctan \left (\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a}}{\sqrt{-a + b}}\right )}{{\left (a f - b f\right )} \sqrt{-a + b}} + \frac{1}{\sqrt{b \tan \left (f x + e\right )^{2} + a}{\left (a f - b f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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