3.320 \(\int \frac {\tan ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\)

Optimal. Leaf size=95 \[ \frac {\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2 f}-\frac {(a+b) \sqrt {a+b \tan ^2(e+f x)}}{b^2 f}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f \sqrt {a-b}} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3670, 446, 88, 63, 208} \[ \frac {\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2 f}-\frac {(a+b) \sqrt {a+b \tan ^2(e+f x)}}{b^2 f}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f \sqrt {a-b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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]

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 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]))

Rubi steps

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

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Mathematica [A]  time = 2.45, size = 87, normalized size = 0.92 \[ -\frac {\frac {2 \left (2 a-b \tan ^2(e+f x)+3 b\right ) \sqrt {a+b \tan ^2(e+f x)}}{3 b^2}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.55, size = 314, normalized size = 3.31 \[ \left [\frac {3 \, \sqrt {a - b} b^{2} \log \left (-\frac {b^{2} \tan \left (f x + e\right )^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 4 \, {\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) + 4 \, {\left ({\left (a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2} - a b + 3 \, b^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{12 \, {\left (a b^{2} - b^{3}\right )} f}, \frac {3 \, \sqrt {-a + b} b^{2} \arctan \left (\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{b \tan \left (f x + e\right )^{2} + 2 \, a - b}\right ) + 2 \, {\left ({\left (a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2} - a b + 3 \, b^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{6 \, {\left (a b^{2} - b^{3}\right )} f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/12*(3*sqrt(a - b)*b^2*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*((a*b - b^2)*tan(f*x + e)^2 - 2*a^2 - a*b + 3*b^2)*sqrt(b*tan(f*x + e)^2 + a))/((a*b^2 - b^3)*f), 1/6*(3*
sqrt(-a + b)*b^2*arctan(2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)/(b*tan(f*x + e)^2 + 2*a - b)) + 2*((a*b - b^
2)*tan(f*x + e)^2 - 2*a^2 - a*b + 3*b^2)*sqrt(b*tan(f*x + e)^2 + a))/((a*b^2 - b^3)*f)]

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giac [A]  time = 2.00, size = 114, normalized size = 1.20 \[ \frac {\arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a}}{\sqrt {-a + b}}\right )}{\sqrt {-a + b} f} + \frac {{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b^{4} f^{2} - 3 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} a b^{4} f^{2} - 3 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b^{5} f^{2}}{3 \, b^{6} f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

arctan(sqrt(b*tan(f*x + e)^2 + a)/sqrt(-a + b))/(sqrt(-a + b)*f) + 1/3*((b*tan(f*x + e)^2 + a)^(3/2)*b^4*f^2 -
3*sqrt(b*tan(f*x + e)^2 + a)*a*b^4*f^2 - 3*sqrt(b*tan(f*x + e)^2 + a)*b^5*f^2)/(b^6*f^3)

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maple [A]  time = 0.35, size = 111, normalized size = 1.17 \[ \frac {\left (\tan ^{2}\left (f x +e \right )\right ) \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}{3 f b}-\frac {2 a \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}{3 f \,b^{2}}-\frac {\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}{b f}+\frac {\arctan \left (\frac {\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {-a +b}}\right )}{f \sqrt {-a +b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)^5/(a+b*tan(f*x+e)^2)^(1/2),x)

[Out]

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

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 12.88, size = 97, normalized size = 1.02 \[ \frac {{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}}{3\,b^2\,f}-\frac {\mathrm {atanh}\left (\frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{\sqrt {a-b}}\right )}{f\,\sqrt {a-b}}-\left (\frac {2\,a}{b^2\,f}-\frac {a-b}{b^2\,f}\right )\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^5/(a + b*tan(e + f*x)^2)^(1/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{5}{\left (e + f x \right )}}{\sqrt {a + b \tan ^{2}{\left (e + f x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**5/(a+b*tan(f*x+e)**2)**(1/2),x)

[Out]

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

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