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

Optimal. Leaf size=74 \[ \frac{a+b \tan ^2(x)}{2 a (a+b) \sqrt{a+b \tan ^4(x)}}-\frac{\tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{2 (a+b)^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.114571, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3670, 1248, 741, 12, 725, 206} \[ \frac{a+b \tan ^2(x)}{2 a (a+b) \sqrt{a+b \tan ^4(x)}}-\frac{\tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{2 (a+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]/(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)) + (a + b*Tan[x]^2)/(2*a*(a + b
)*Sqrt[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 1248

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

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

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

Mathematica [A]  time = 0.27361, size = 73, normalized size = 0.99 \[ \frac{1}{2} \left (\frac{a+b \tan ^2(x)}{a (a+b) \sqrt{a+b \tan ^4(x)}}-\frac{\tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{(a+b)^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]/(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)) + (a + b*Tan[x]^2)/(a*(a + b)*S
qrt[a + b*Tan[x]^4]))/2

________________________________________________________________________________________

Maple [B]  time = 0.056, size = 248, normalized size = 3.4 \begin{align*} -{\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)/(a+b*tan(x)^4)^(3/2),x)

[Out]

-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)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 3.2692, size = 771, normalized size = 10.42 \begin{align*} \left [\frac{{\left (a b \tan \left (x\right )^{4} + a^{2}\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 + b^{2}\right )} \tan \left (x\right )^{2} + a^{2} + a b\right )}}{4 \,{\left ({\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} \tan \left (x\right )^{4} + a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )}}, -\frac{{\left (a b \tan \left (x\right )^{4} + a^{2}\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 + b^{2}\right )} \tan \left (x\right )^{2} + a^{2} + a b\right )}}{2 \,{\left ({\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} \tan \left (x\right )^{4} + a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.24129, size = 161, normalized size = 2.18 \begin{align*} \frac{\frac{{\left (a b + b^{2}\right )} \tan \left (x\right )^{2}}{a^{3} + 2 \, a^{2} b + a b^{2}} + \frac{a^{2} + a b}{a^{3} + 2 \, a^{2} b + a 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)/(a+b*tan(x)^4)^(3/2),x, algorithm="giac")

[Out]

1/2*((a*b + b^2)*tan(x)^2/(a^3 + 2*a^2*b + a*b^2) + (a^2 + a*b)/(a^3 + 2*a^2*b + a*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))