3.400 \(\int \frac{\tan ^3(x)}{(a+b \tan ^4(x))^{3/2}} \, dx\)

Optimal. Leaf size=71 \[ \frac{\tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{2 (a+b)^{3/2}}-\frac{1-\tan ^2(x)}{2 (a+b) \sqrt{a+b \tan ^4(x)}} \]

[Out]

ArcTanh[(a - b*Tan[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tan[x]^4])]/(2*(a + b)^(3/2)) - (1 - Tan[x]^2)/(2*(a + b)*Sqr
t[a + b*Tan[x]^4])

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Rubi [A]  time = 0.162221, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {3670, 1252, 823, 12, 725, 206} \[ \frac{\tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{2 (a+b)^{3/2}}-\frac{1-\tan ^2(x)}{2 (a+b) \sqrt{a+b \tan ^4(x)}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]^3/(a + b*Tan[x]^4)^(3/2),x]

[Out]

ArcTanh[(a - b*Tan[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tan[x]^4])]/(2*(a + b)^(3/2)) - (1 - Tan[x]^2)/(2*(a + b)*Sqr
t[a + b*Tan[x]^4])

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 1252

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m
- 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\tan ^3(x)}{\left (a+b \tan ^4(x)\right )^{3/2}} \, dx &=\operatorname{Subst}\left (\int \frac{x^3}{\left (1+x^2\right ) \left (a+b x^4\right )^{3/2}} \, dx,x,\tan (x)\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x}{(1+x) \left (a+b x^2\right )^{3/2}} \, dx,x,\tan ^2(x)\right )\\ &=-\frac{1-\tan ^2(x)}{2 (a+b) \sqrt{a+b \tan ^4(x)}}-\frac{\operatorname{Subst}\left (\int \frac{a b}{(1+x) \sqrt{a+b x^2}} \, dx,x,\tan ^2(x)\right )}{2 a b (a+b)}\\ &=-\frac{1-\tan ^2(x)}{2 (a+b) \sqrt{a+b \tan ^4(x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{a+b x^2}} \, dx,x,\tan ^2(x)\right )}{2 (a+b)}\\ &=-\frac{1-\tan ^2(x)}{2 (a+b) \sqrt{a+b \tan ^4(x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{a+b-x^2} \, dx,x,\frac{a-b \tan ^2(x)}{\sqrt{a+b \tan ^4(x)}}\right )}{2 (a+b)}\\ &=\frac{\tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{2 (a+b)^{3/2}}-\frac{1-\tan ^2(x)}{2 (a+b) \sqrt{a+b \tan ^4(x)}}\\ \end{align*}

Mathematica [A]  time = 0.321206, size = 67, normalized size = 0.94 \[ \frac{1}{2} \left (\frac{\tan ^2(x)-1}{(a+b) \sqrt{a+b \tan ^4(x)}}+\frac{\tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{(a+b)^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]^3/(a + b*Tan[x]^4)^(3/2),x]

[Out]

(ArcTanh[(a - b*Tan[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tan[x]^4])]/(a + b)^(3/2) + (-1 + Tan[x]^2)/((a + b)*Sqrt[a
+ b*Tan[x]^4]))/2

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Maple [B]  time = 0.076, size = 267, normalized size = 3.8 \begin{align*}{\frac{ \left ( \tan \left ( x \right ) \right ) ^{2}}{2\,a}{\frac{1}{\sqrt{a+b \left ( \tan \left ( x \right ) \right ) ^{4}}}}}+{\frac{1}{4\,a}\sqrt{b \left ( \left ( \tan \left ( x \right ) \right ) ^{2}+{\frac{1}{b}\sqrt{-ab}} \right ) ^{2}-2\,\sqrt{-ab} \left ( \left ( \tan \left ( x \right ) \right ) ^{2}+{\frac{\sqrt{-ab}}{b}} \right ) } \left ( \sqrt{-ab}-b \right ) ^{-1} \left ( \left ( \tan \left ( x \right ) \right ) ^{2}+{\frac{1}{b}\sqrt{-ab}} \right ) ^{-1}}-{\frac{b}{2}\ln \left ({\frac{1}{1+ \left ( \tan \left ( x \right ) \right ) ^{2}} \left ( 2\,a+2\,b-2\,b \left ( 1+ \left ( \tan \left ( x \right ) \right ) ^{2} \right ) +2\,\sqrt{a+b}\sqrt{b \left ( 1+ \left ( \tan \left ( x \right ) \right ) ^{2} \right ) ^{2}-2\,b \left ( 1+ \left ( \tan \left ( x \right ) \right ) ^{2} \right ) +a+b} \right ) } \right ) \left ( \sqrt{-ab}+b \right ) ^{-1} \left ( \sqrt{-ab}-b \right ) ^{-1}{\frac{1}{\sqrt{a+b}}}}-{\frac{1}{4\,a}\sqrt{b \left ( \left ( \tan \left ( x \right ) \right ) ^{2}-{\frac{1}{b}\sqrt{-ab}} \right ) ^{2}+2\,\sqrt{-ab} \left ( \left ( \tan \left ( x \right ) \right ) ^{2}-{\frac{\sqrt{-ab}}{b}} \right ) } \left ( \sqrt{-ab}+b \right ) ^{-1} \left ( \left ( \tan \left ( x \right ) \right ) ^{2}-{\frac{1}{b}\sqrt{-ab}} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)^3/(a+b*tan(x)^4)^(3/2),x)

[Out]

1/2*tan(x)^2/a/(a+b*tan(x)^4)^(1/2)+1/4/((-a*b)^(1/2)-b)/a/(tan(x)^2+(-a*b)^(1/2)/b)*(b*(tan(x)^2+(-a*b)^(1/2)
/b)^2-2*(-a*b)^(1/2)*(tan(x)^2+(-a*b)^(1/2)/b))^(1/2)-1/2*b/((-a*b)^(1/2)+b)/((-a*b)^(1/2)-b)/(a+b)^(1/2)*ln((
2*a+2*b-2*b*(1+tan(x)^2)+2*(a+b)^(1/2)*(b*(1+tan(x)^2)^2-2*b*(1+tan(x)^2)+a+b)^(1/2))/(1+tan(x)^2))-1/4/((-a*b
)^(1/2)+b)/a/(tan(x)^2-(-a*b)^(1/2)/b)*(b*(tan(x)^2-(-a*b)^(1/2)/b)^2+2*(-a*b)^(1/2)*(tan(x)^2-(-a*b)^(1/2)/b)
)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (x\right )^{3}}{{\left (b \tan \left (x\right )^{4} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^3/(a+b*tan(x)^4)^(3/2),x, algorithm="maxima")

[Out]

integrate(tan(x)^3/(b*tan(x)^4 + a)^(3/2), x)

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Fricas [B]  time = 3.27235, size = 721, normalized size = 10.15 \begin{align*} \left [\frac{{\left (b \tan \left (x\right )^{4} + a\right )} \sqrt{a + b} \log \left (\frac{{\left (a b + 2 \, b^{2}\right )} \tan \left (x\right )^{4} - 2 \, a b \tan \left (x\right )^{2} - 2 \, \sqrt{b \tan \left (x\right )^{4} + a}{\left (b \tan \left (x\right )^{2} - a\right )} \sqrt{a + b} + 2 \, a^{2} + a b}{\tan \left (x\right )^{4} + 2 \, \tan \left (x\right )^{2} + 1}\right ) + 2 \, \sqrt{b \tan \left (x\right )^{4} + a}{\left ({\left (a + b\right )} \tan \left (x\right )^{2} - a - b\right )}}{4 \,{\left ({\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \tan \left (x\right )^{4} + a^{3} + 2 \, a^{2} b + a b^{2}\right )}}, \frac{{\left (b \tan \left (x\right )^{4} + a\right )} \sqrt{-a - b} \arctan \left (\frac{\sqrt{b \tan \left (x\right )^{4} + a}{\left (b \tan \left (x\right )^{2} - a\right )} \sqrt{-a - b}}{{\left (a b + b^{2}\right )} \tan \left (x\right )^{4} + a^{2} + a b}\right ) + \sqrt{b \tan \left (x\right )^{4} + a}{\left ({\left (a + b\right )} \tan \left (x\right )^{2} - a - b\right )}}{2 \,{\left ({\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \tan \left (x\right )^{4} + a^{3} + 2 \, a^{2} b + a b^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^3/(a+b*tan(x)^4)^(3/2),x, algorithm="fricas")

[Out]

[1/4*((b*tan(x)^4 + a)*sqrt(a + b)*log(((a*b + 2*b^2)*tan(x)^4 - 2*a*b*tan(x)^2 - 2*sqrt(b*tan(x)^4 + a)*(b*ta
n(x)^2 - a)*sqrt(a + b) + 2*a^2 + a*b)/(tan(x)^4 + 2*tan(x)^2 + 1)) + 2*sqrt(b*tan(x)^4 + a)*((a + b)*tan(x)^2
- a - b))/((a^2*b + 2*a*b^2 + b^3)*tan(x)^4 + a^3 + 2*a^2*b + a*b^2), 1/2*((b*tan(x)^4 + a)*sqrt(-a - b)*arct
an(sqrt(b*tan(x)^4 + a)*(b*tan(x)^2 - a)*sqrt(-a - b)/((a*b + b^2)*tan(x)^4 + a^2 + a*b)) + sqrt(b*tan(x)^4 +
a)*((a + b)*tan(x)^2 - a - b))/((a^2*b + 2*a*b^2 + b^3)*tan(x)^4 + a^3 + 2*a^2*b + a*b^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{3}{\left (x \right )}}{\left (a + b \tan ^{4}{\left (x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)**3/(a+b*tan(x)**4)**(3/2),x)

[Out]

Integral(tan(x)**3/(a + b*tan(x)**4)**(3/2), x)

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Giac [A]  time = 1.31742, size = 139, normalized size = 1.96 \begin{align*} \frac{\frac{{\left (a + b\right )} \tan \left (x\right )^{2}}{a^{2} + 2 \, a b + b^{2}} - \frac{a + b}{a^{2} + 2 \, a b + b^{2}}}{2 \, \sqrt{b \tan \left (x\right )^{4} + a}} + \frac{\arctan \left (\frac{\sqrt{b} \tan \left (x\right )^{2} - \sqrt{b \tan \left (x\right )^{4} + a} + \sqrt{b}}{\sqrt{-a - b}}\right )}{{\left (a + b\right )} \sqrt{-a - b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^3/(a+b*tan(x)^4)^(3/2),x, algorithm="giac")

[Out]

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