3.294 \(\int \tan ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)} \, dx\)

Optimal. Leaf size=88 \[ \frac{\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b f}-\frac{\sqrt{a+b \tan ^2(e+f x)}}{f}+\frac{\sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f} \]

[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*Tan[e
+ f*x]^2)^(3/2)/(3*b*f)

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Rubi [A]  time = 0.114322, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3670, 446, 80, 50, 63, 208} \[ \frac{\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b f}-\frac{\sqrt{a+b \tan ^2(e+f x)}}{f}+\frac{\sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^3*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*Tan[e
+ f*x]^2)^(3/2)/(3*b*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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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

Mathematica [A]  time = 0.35181, size = 82, normalized size = 0.93 \[ \frac{\sqrt{a+b \tan ^2(e+f x)} \left (a+b \tan ^2(e+f x)-3 b\right )+3 b \sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{3 b f} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

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

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Maple [A]  time = 0.027, size = 114, normalized size = 1.3 \begin{align*}{\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)^3,x)

[Out]

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 )^{3}\,{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)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.47136, size = 621, normalized size = 7.06 \begin{align*} \left [\frac{3 \, \sqrt{a - b} 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 (b \tan \left (f x + e\right )^{2} + a - 3 \, b\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{12 \, b f}, -\frac{3 \, \sqrt{-a + b} 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 (b \tan \left (f x + e\right )^{2} + a - 3 \, b\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{6 \, b 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)^3,x, algorithm="fricas")

[Out]

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

[Out]

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

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Giac [A]  time = 1.17739, size = 131, normalized size = 1.49 \begin{align*} -\frac{\frac{3 \,{\left (a b - b^{2}\right )} \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}} f^{2} - 3 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} b f^{2}}{f^{3}}}{3 \, b} \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)^3,x, algorithm="giac")

[Out]

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