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

Optimal. Leaf size=73 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f (a-b)^{3/2}}-\frac{a}{b f (a-b) \sqrt{a+b \tan ^2(e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.132861, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3670, 446, 78, 63, 208} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f (a-b)^{3/2}}-\frac{a}{b f (a-b) \sqrt{a+b \tan ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

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 78

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

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

Mathematica [A]  time = 0.362278, size = 75, normalized size = 1.03 \[ \frac{\frac{a (b-a)}{b \sqrt{a+b \tan ^2(e+f x)}}+\sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f (a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.017, size = 92, normalized size = 1.3 \begin{align*} -{\frac{1}{fb}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}}-{\frac{1}{ \left ( a-b \right ) f}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}}-{\frac{1}{ \left ( a-b \right ) 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(tan(f*x+e)^3/(a+b*tan(f*x+e)^2)^(3/2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.665, size = 829, normalized size = 11.36 \begin{align*} \left [-\frac{{\left (b^{2} \tan \left (f x + e\right )^{2} + a b\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 \, \sqrt{b \tan \left (f x + e\right )^{2} + a}{\left (a^{2} - a b\right )}}{4 \,{\left ({\left (a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} f\right )}}, -\frac{{\left (b^{2} \tan \left (f x + e\right )^{2} + a b\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 \, \sqrt{b \tan \left (f x + e\right )^{2} + a}{\left (a^{2} - a b\right )}}{2 \,{\left ({\left (a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{3}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.37426, size = 103, normalized size = 1.41 \begin{align*} -\frac{\frac{b \arctan \left (\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a}}{\sqrt{-a + b}}\right )}{{\left (a f - b f\right )} \sqrt{-a + b}} + \frac{a}{\sqrt{b \tan \left (f x + e\right )^{2} + a}{\left (a f - b f\right )}}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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