\(\int \frac {\tan ^5(e+f x)}{(a+b \tan ^2(e+f x))^{3/2}} \, dx\) [333]

Optimal result

Integrand size = 25, antiderivative size = 98 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{(a-b)^{3/2} f}+\frac {a^2}{(a-b) b^2 f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\sqrt {a+b \tan ^2(e+f x)}}{b^2 f} \]

[Out]

-arctanh((a+b*tan(f*x+e)^2)^(1/2)/(a-b)^(1/2))/(a-b)^(3/2)/f+a^2/(a-b)/b^2/f/(a+b*tan(f*x+e)^2)^(1/2)+(a+b*tan
(f*x+e)^2)^(1/2)/b^2/f

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3751, 457, 89, 65, 214} \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\frac {a^2}{b^2 f (a-b) \sqrt {a+b \tan ^2(e+f x)}}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f (a-b)^{3/2}}+\frac {\sqrt {a+b \tan ^2(e+f x)}}{b^2 f} \]

[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 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 89

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 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 457

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 3751

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]))

Rubi steps \begin{align*} \text {integral}& = \frac {\text {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 {\text {Subst}\left (\int \frac {x^2}{(1+x) (a+b x)^{3/2}} \, dx,x,\tan ^2(e+f x)\right )}{2 f} \\ & = \frac {\text {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 {\text {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 {\text {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 {\text {arctanh}\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] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.42 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.86 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\frac {b^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan ^2(e+f x)}{a-b}\right )+(a-b) \left (2 a+b+b \tan ^2(e+f x)\right )}{(a-b) b^2 f \sqrt {a+b \tan ^2(e+f x)}} \]

[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] (verified)

Time = 0.07 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.33

method	result	size
derivativedivides	\(\frac {\frac {\tan \left (f x +e \right )^{2}}{b \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {2 a}{b^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {1}{b \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {1}{\left (a -b \right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {\arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\left (a -b \right ) \sqrt {-a +b}}}{f}\)	\(130\)
default	\(\frac {\frac {\tan \left (f x +e \right )^{2}}{b \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {2 a}{b^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {1}{b \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {1}{\left (a -b \right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {\arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\left (a -b \right ) \sqrt {-a +b}}}{f}\)	\(130\)

[In]

int(tan(f*x+e)^5/(a+b*tan(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (88) = 176\).

Time = 0.33 (sec) , antiderivative size = 432, normalized size of antiderivative = 4.41 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\left [-\frac {{\left (b^{3} \tan \left (f x + e\right )^{2} + a b^{2}\right )} \sqrt {a - b} \log \left (-\frac {b^{2} \tan \left (f x + e\right )^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} + 4 \, {\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) - 4 \, {\left (2 \, a^{3} - 3 \, a^{2} b + a b^{2} + {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{4 \, {\left ({\left (a^{2} b^{3} - 2 \, a b^{4} + b^{5}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4}\right )} f\right )}}, \frac {{\left (b^{3} \tan \left (f x + e\right )^{2} + a b^{2}\right )} \sqrt {-a + b} \arctan \left (\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{b \tan \left (f x + e\right )^{2} + 2 \, a - b}\right ) + 2 \, {\left (2 \, a^{3} - 3 \, a^{2} b + a b^{2} + {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{2 \, {\left ({\left (a^{2} b^{3} - 2 \, a b^{4} + b^{5}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4}\right )} f\right )}}\right ] \]

[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]

\[ \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\tan ^{5}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\tan \left (f x + e\right )^{5}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F(-1)]

Timed out. \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 14.11 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.14 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{b^2\,f}+\frac {a^2}{b^2\,f\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\left (a-b\right )}+\frac {\mathrm {atan}\left (\frac {a\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,1{}\mathrm {i}-b\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,1{}\mathrm {i}}{{\left (a-b\right )}^{3/2}}\right )\,1{}\mathrm {i}}{f\,{\left (a-b\right )}^{3/2}} \]

[In]

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

[Out]

(a + b*tan(e + f*x)^2)^(1/2)/(b^2*f) + (atan((a*(a + b*tan(e + f*x)^2)^(1/2)*1i - b*(a + b*tan(e + f*x)^2)^(1/
2)*1i)/(a - b)^(3/2))*1i)/(f*(a - b)^(3/2)) + a^2/(b^2*f*(a + b*tan(e + f*x)^2)^(1/2)*(a - b))