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

Optimal. Leaf size=115 \[ \frac {a^2}{3 b^2 f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {a (a-2 b)}{b^2 f (a-b)^2 \sqrt {a+b \tan ^2(e+f x)}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f (a-b)^{5/2}} \]

[Out]

-arctanh((a+b*tan(f*x+e)^2)^(1/2)/(a-b)^(1/2))/(a-b)^(5/2)/f-a*(a-2*b)/(a-b)^2/b^2/f/(a+b*tan(f*x+e)^2)^(1/2)+
1/3*a^2/(a-b)/b^2/f/(a+b*tan(f*x+e)^2)^(3/2)

________________________________________________________________________________________

Rubi [A]  time = 0.20, antiderivative size = 115, 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, 87, 63, 208} \[ \frac {a^2}{3 b^2 f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {a (a-2 b)}{b^2 f (a-b)^2 \sqrt {a+b \tan ^2(e+f x)}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f (a-b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a - b]]/((a - b)^(5/2)*f)) + a^2/(3*(a - b)*b^2*f*(a + b*Tan[e + f*x
]^2)^(3/2)) - (a*(a - 2*b))/((a - b)^2*b^2*f*Sqrt[a + b*Tan[e + f*x]^2])

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

________________________________________________________________________________________

Mathematica [C]  time = 0.46, size = 91, normalized size = 0.79 \[ \frac {b^2 \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {b \tan ^2(e+f x)+a}{a-b}\right )-(a-b) \left (2 a+3 b \tan ^2(e+f x)-b\right )}{3 b^2 f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Tan[e + f*x]^2)/(a - b)] - (a - b)*(2*a - b + 3*b*Tan[e + f*x]^2)
)/(3*(a - b)*b^2*f*(a + b*Tan[e + f*x]^2)^(3/2))

________________________________________________________________________________________

fricas [B]  time = 0.55, size = 608, normalized size = 5.29 \[ \left [\frac {3 \, {\left (b^{4} \tan \left (f x + e\right )^{4} + 2 \, a b^{3} \tan \left (f x + e\right )^{2} + a^{2} b^{2}\right )} \sqrt {a - b} \log \left (-\frac {b^{2} \tan \left (f x + e\right )^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 4 \, {\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) - 4 \, {\left (2 \, a^{4} - 7 \, a^{3} b + 5 \, a^{2} b^{2} + 3 \, {\left (a^{3} b - 3 \, a^{2} b^{2} + 2 \, a b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{12 \, {\left ({\left (a^{3} b^{4} - 3 \, a^{2} b^{5} + 3 \, a b^{6} - b^{7}\right )} f \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b^{3} - 3 \, a^{3} b^{4} + 3 \, a^{2} b^{5} - a b^{6}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{5} b^{2} - 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} - a^{2} b^{5}\right )} f\right )}}, \frac {3 \, {\left (b^{4} \tan \left (f x + e\right )^{4} + 2 \, a b^{3} \tan \left (f x + e\right )^{2} + a^{2} b^{2}\right )} \sqrt {-a + b} \arctan \left (\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{b \tan \left (f x + e\right )^{2} + 2 \, a - b}\right ) - 2 \, {\left (2 \, a^{4} - 7 \, a^{3} b + 5 \, a^{2} b^{2} + 3 \, {\left (a^{3} b - 3 \, a^{2} b^{2} + 2 \, a b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{6 \, {\left ({\left (a^{3} b^{4} - 3 \, a^{2} b^{5} + 3 \, a b^{6} - b^{7}\right )} f \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b^{3} - 3 \, a^{3} b^{4} + 3 \, a^{2} b^{5} - a b^{6}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{5} b^{2} - 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} - a^{2} b^{5}\right )} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(3*(b^4*tan(f*x + e)^4 + 2*a*b^3*tan(f*x + e)^2 + a^2*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^4 - 7*a^3*b + 5*a^2*b^2 + 3*(a^3*b - 3*a^2*b^2 +
2*a*b^3)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^3*b^4 - 3*a^2*b^5 + 3*a*b^6 - b^7)*f*tan(f*x + e)^4
+ 2*(a^4*b^3 - 3*a^3*b^4 + 3*a^2*b^5 - a*b^6)*f*tan(f*x + e)^2 + (a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 - a^2*b^5)*f
), 1/6*(3*(b^4*tan(f*x + e)^4 + 2*a*b^3*tan(f*x + e)^2 + a^2*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^4 - 7*a^3*b + 5*a^2*b^2 + 3*(a^3*b - 3*a^2*b^2 + 2*a*
b^3)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^3*b^4 - 3*a^2*b^5 + 3*a*b^6 - b^7)*f*tan(f*x + e)^4 + 2*(
a^4*b^3 - 3*a^3*b^4 + 3*a^2*b^5 - a*b^6)*f*tan(f*x + e)^2 + (a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 - a^2*b^5)*f)]

________________________________________________________________________________________

giac [A]  time = 4.22, size = 137, normalized size = 1.19 \[ \frac {\arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a}}{\sqrt {-a + b}}\right )}{{\left (a^{2} f - 2 \, a b f + b^{2} f\right )} \sqrt {-a + b}} - \frac {3 \, {\left (b \tan \left (f x + e\right )^{2} + a\right )} a^{2} - a^{3} - 6 \, {\left (b \tan \left (f x + e\right )^{2} + a\right )} a b + a^{2} b}{3 \, {\left (a^{2} b^{2} f - 2 \, a b^{3} f + b^{4} f\right )} {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.34, size = 169, normalized size = 1.47 \[ -\frac {\tan ^{2}\left (f x +e \right )}{f b \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}-\frac {2 a}{3 f \,b^{2} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}+\frac {1}{3 f b \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}+\frac {1}{3 \left (a -b \right ) f \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}+\frac {1}{\left (a -b \right )^{2} f \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}+\frac {\arctan \left (\frac {\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {-a +b}}\right )}{f \left (a -b \right )^{2} \sqrt {-a +b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is b-a positive or negative?

________________________________________________________________________________________

mupad [B]  time = 16.03, size = 148, normalized size = 1.29 \[ \frac {\frac {a^2}{3\,\left (a-b\right )}+\frac {\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )\,\left (2\,a\,b-a^2\right )}{{\left (a-b\right )}^2}}{b^2\,f\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}}+\frac {\mathrm {atan}\left (\frac {a^2\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,1{}\mathrm {i}+b^2\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,1{}\mathrm {i}-a\,b\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,2{}\mathrm {i}}{{\left (a-b\right )}^{5/2}}\right )\,1{}\mathrm {i}}{f\,{\left (a-b\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((a^2*(a + b*tan(e + f*x)^2)^(1/2)*1i + b^2*(a + b*tan(e + f*x)^2)^(1/2)*1i - a*b*(a + b*tan(e + f*x)^2)^
(1/2)*2i)/(a - b)^(5/2))*1i)/(f*(a - b)^(5/2)) + (a^2/(3*(a - b)) + ((a + b*tan(e + f*x)^2)*(2*a*b - a^2))/(a
- b)^2)/(b^2*f*(a + b*tan(e + f*x)^2)^(3/2))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________