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

Optimal. Leaf size=117 \[ \frac{\left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}-\frac{(a+b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2 f}+\frac{\sqrt{a+b \tan ^2(e+f x)}}{f}-\frac{\sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.146467, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3670, 446, 88, 50, 63, 208} \[ \frac{\left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}-\frac{(a+b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2 f}+\frac{\sqrt{a+b \tan ^2(e+f x)}}{f}-\frac{\sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[a - b]*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a - b]])/f) + Sqrt[a + b*Tan[e + f*x]^2]/f - ((a + b)*(
a + b*Tan[e + f*x]^2)^(3/2))/(3*b^2*f) + (a + b*Tan[e + f*x]^2)^(5/2)/(5*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 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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 \tan ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^5 \sqrt{a+b x^2}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^2 \sqrt{a+b x}}{1+x} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{(-a-b) \sqrt{a+b x}}{b}+\frac{\sqrt{a+b x}}{1+x}+\frac{(a+b x)^{3/2}}{b}\right ) \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=-\frac{(a+b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2 f}+\frac{\left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{1+x} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac{\sqrt{a+b \tan ^2(e+f x)}}{f}-\frac{(a+b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2 f}+\frac{\left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac{(a-b) \operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac{\sqrt{a+b \tan ^2(e+f x)}}{f}-\frac{(a+b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2 f}+\frac{\left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac{(a-b) \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{\sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f}+\frac{\sqrt{a+b \tan ^2(e+f x)}}{f}-\frac{(a+b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2 f}+\frac{\left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}\\ \end{align*}

Mathematica [A]  time = 1.35958, size = 109, normalized size = 0.93 \[ \frac{\frac{\sqrt{a+b \tan ^2(e+f x)} \left (-2 a^2+b (a-5 b) \tan ^2(e+f x)-5 a b+3 b^2 \tan ^4(e+f x)+15 b^2\right )}{b^2}-15 \sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{15 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*Sqrt[a - b]*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a - b]] + (Sqrt[a + b*Tan[e + f*x]^2]*(-2*a^2 - 5*a*b
+ 15*b^2 + (a - 5*b)*b*Tan[e + f*x]^2 + 3*b^2*Tan[e + f*x]^4))/b^2)/(15*f)

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Maple [A]  time = 0.046, size = 166, normalized size = 1.4 \begin{align*}{\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{5\,fb} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{2\,a}{15\,f{b}^{2}} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{1}{3\,fb} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{1}{f}\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}-{\frac{b}{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}}}}+{\frac{a}{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((a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)^5,x)

[Out]

1/5/f*tan(f*x+e)^2*(a+b*tan(f*x+e)^2)^(3/2)/b-2/15/f*a/b^2*(a+b*tan(f*x+e)^2)^(3/2)-1/3*(a+b*tan(f*x+e)^2)^(3/
2)/b/f+(a+b*tan(f*x+e)^2)^(1/2)/f-1/f*b/(-a+b)^(1/2)*arctan((a+b*tan(f*x+e)^2)^(1/2)/(-a+b)^(1/2))+1/f*a/(-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 \sqrt{b \tan \left (f x + e\right )^{2} + a} \tan \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.72676, size = 770, normalized size = 6.58 \begin{align*} \left [\frac{15 \, \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 (3 \, b^{2} \tan \left (f x + e\right )^{4} +{\left (a b - 5 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2} - 5 \, a b + 15 \, b^{2}\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{60 \, b^{2} f}, \frac{15 \, \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 (3 \, b^{2} \tan \left (f x + e\right )^{4} +{\left (a b - 5 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2} - 5 \, a b + 15 \, b^{2}\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{30 \, b^{2} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/60*(15*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*(3*b^2*tan(f*x + e)^4 + (a*b - 5*b^2)*tan(f*x + e)^2 - 2*a^2 - 5*a*b + 15*b^2)*sqrt(b*tan(f*x + e)^2 + a
))/(b^2*f), 1/30*(15*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*(3*b^2*tan(f*x + e)^4 + (a*b - 5*b^2)*tan(f*x + e)^2 - 2*a^2 - 5*a*b + 15*b^2)*sqrt(b*tan(f*x + e)^
2 + a))/(b^2*f)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.20562, size = 193, normalized size = 1.65 \begin{align*} \frac{{\left (a - b\right )} \arctan \left (\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a}}{\sqrt{-a + b}}\right )}{\sqrt{-a + b} f} + \frac{3 \,{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}} b^{8} f^{4} - 5 \,{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} a b^{8} f^{4} - 5 \,{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} b^{9} f^{4} + 15 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} b^{10} f^{4}}{15 \, b^{10} f^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a - b)*arctan(sqrt(b*tan(f*x + e)^2 + a)/sqrt(-a + b))/(sqrt(-a + b)*f) + 1/15*(3*(b*tan(f*x + e)^2 + a)^(5/2
)*b^8*f^4 - 5*(b*tan(f*x + e)^2 + a)^(3/2)*a*b^8*f^4 - 5*(b*tan(f*x + e)^2 + a)^(3/2)*b^9*f^4 + 15*sqrt(b*tan(
f*x + e)^2 + a)*b^10*f^4)/(b^10*f^5)