3.333 \(\int \frac{\tan ^5(e+f x)}{(a+b \tan ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=98 \[ \frac{a^2}{b^2 f (a-b) \sqrt{a+b \tan ^2(e+f x)}}+\frac{\sqrt{a+b \tan ^2(e+f x)}}{b^2 f}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f (a-b)^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.169312, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3670, 446, 87, 63, 208} \[ \frac{a^2}{b^2 f (a-b) \sqrt{a+b \tan ^2(e+f x)}}+\frac{\sqrt{a+b \tan ^2(e+f x)}}{b^2 f}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f (a-b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^5/(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)) + a^2/((a - b)*b^2*f*Sqrt[a + b*Tan[e + f
*x]^2]) + Sqrt[a + b*Tan[e + f*x]^2]/(b^2*f)

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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

Rule 87

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], ((c + d*x)^n*(e + f*x)^IntegerPart[p])/(a + b*x), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

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 ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^5}{\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{x^2}{(1+x) (a+b x)^{3/2}} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a^2}{(a-b) b (a+b x)^{3/2}}+\frac{1}{b \sqrt{a+b x}}+\frac{1}{(a-b) (1+x) \sqrt{a+b x}}\right ) \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac{a^2}{(a-b) b^2 f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\sqrt{a+b \tan ^2(e+f x)}}{b^2 f}+\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{a^2}{(a-b) b^2 f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\sqrt{a+b \tan ^2(e+f x)}}{b^2 f}+\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{a^2}{(a-b) b^2 f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\sqrt{a+b \tan ^2(e+f x)}}{b^2 f}\\ \end{align*}

Mathematica [C]  time = 0.365129, size = 84, normalized size = 0.86 \[ \frac{b^2 \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{a+b \tan ^2(e+f x)}{a-b}\right )+(a-b) \left (2 a+b \tan ^2(e+f x)+b\right )}{b^2 f (a-b) \sqrt{a+b \tan ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^5/(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.026, size = 141, normalized size = 1.4 \begin{align*}{\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{fb}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}}+2\,{\frac{a}{f{b}^{2}\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}+{\frac{1}{fb}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}}+{\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)^5/(a+b*tan(f*x+e)^2)^(3/2),x)

[Out]

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

________________________________________________________________________________________

Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [B]  time = 2.75476, size = 987, normalized size = 10.07 \begin{align*} \left [-\frac{{\left (b^{3} \tan \left (f x + e\right )^{2} + a b^{2}\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 \,{\left (2 \, a^{3} - 3 \, a^{2} b + a b^{2} +{\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{4 \,{\left ({\left (a^{2} b^{3} - 2 \, a b^{4} + b^{5}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4}\right )} f\right )}}, \frac{{\left (b^{3} \tan \left (f x + e\right )^{2} + a b^{2}\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 \,{\left (2 \, a^{3} - 3 \, a^{2} b + a b^{2} +{\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{2 \,{\left ({\left (a^{2} b^{3} - 2 \, a b^{4} + b^{5}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4}\right )} f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/4*((b^3*tan(f*x + e)^2 + a*b^2)*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*tan(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*(2*a^3 - 3*a^2*b + a*b^2 + (a^2*b - 2*a*b^2 + b^3)*tan(f*x + e)^2)*sqrt(b*tan(f*x
+ e)^2 + a))/((a^2*b^3 - 2*a*b^4 + b^5)*f*tan(f*x + e)^2 + (a^3*b^2 - 2*a^2*b^3 + a*b^4)*f), 1/2*((b^3*tan(f*x
+ e)^2 + a*b^2)*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*(2*a^3 - 3*a^2*b + a*b^2 + (a^2*b - 2*a*b^2 + b^3)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^2*b^3 -
2*a*b^4 + b^5)*f*tan(f*x + e)^2 + (a^3*b^2 - 2*a^2*b^3 + a*b^4)*f)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{5}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**5/(a+b*tan(f*x+e)**2)**(3/2),x)

[Out]

Integral(tan(e + f*x)**5/(a + b*tan(e + f*x)**2)**(3/2), x)

________________________________________________________________________________________

Giac [A]  time = 1.40194, size = 134, normalized size = 1.37 \begin{align*} \frac{a^{2}}{{\left (a b^{2} f - b^{3} f\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}} + \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{\sqrt{b \tan \left (f x + e\right )^{2} + a}}{b^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

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