3.5.1 \(\int \frac {\tan (x)}{(a+b \tan ^4(x))^{3/2}} \, dx\) [401]

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

[Out]

-1/2*arctanh((a-b*tan(x)^2)/(a+b)^(1/2)/(a+b*tan(x)^4)^(1/2))/(a+b)^(3/2)+1/2*(a+b*tan(x)^2)/a/(a+b)/(a+b*tan(
x)^4)^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3751, 1262, 755, 12, 739, 212} \begin {gather*} \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}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[x]/(a + b*Tan[x]^4)^(3/2),x]

[Out]

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

Rule 12

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

Rule 212

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

Rule 739

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 755

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 1262

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 3751

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]))

Rubi steps

\begin {align*} \int \frac {\tan (x)}{\left (a+b \tan ^4(x)\right )^{3/2}} \, dx &=\text {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} \text {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 {\text {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 {\text {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 {\text {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*}

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Mathematica [A]
time = 0.20, size = 73, normalized size = 0.99 \begin {gather*} \frac {1}{2} \left (-\frac {\tanh ^{-1}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )}{(a+b)^{3/2}}+\frac {a+b \tan ^2(x)}{a (a+b) \sqrt {a+b \tan ^4(x)}}\right ) \end {gather*}

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(247\) vs. \(2(62)=124\).
time = 0.09, size = 248, normalized size = 3.35

method	result	size
derivativedivides	\(-\frac {\sqrt {b \left (\tan ^{2}\left (x \right )+\frac {\sqrt {-a b}}{b}\right )^{2}-2 \sqrt {-a b}\, \left (\tan ^{2}\left (x \right )+\frac {\sqrt {-a b}}{b}\right )}}{4 \left (\sqrt {-a b}-b \right ) a \left (\tan ^{2}\left (x \right )+\frac {\sqrt {-a b}}{b}\right )}+\frac {\sqrt {b \left (\tan ^{2}\left (x \right )-\frac {\sqrt {-a b}}{b}\right )^{2}+2 \sqrt {-a b}\, \left (\tan ^{2}\left (x \right )-\frac {\sqrt {-a b}}{b}\right )}}{4 \left (\sqrt {-a b}+b \right ) a \left (\tan ^{2}\left (x \right )-\frac {\sqrt {-a b}}{b}\right )}+\frac {b \ln \left (\frac {2 a +2 b -2 b \left (1+\tan ^{2}\left (x \right )\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tan ^{2}\left (x \right )\right )^{2}-2 b \left (1+\tan ^{2}\left (x \right )\right )+a +b}}{1+\tan ^{2}\left (x \right )}\right )}{2 \left (\sqrt {-a b}+b \right ) \left (\sqrt {-a b}-b \right ) \sqrt {a +b}}\)	\(248\)
default	\(-\frac {\sqrt {b \left (\tan ^{2}\left (x \right )+\frac {\sqrt {-a b}}{b}\right )^{2}-2 \sqrt {-a b}\, \left (\tan ^{2}\left (x \right )+\frac {\sqrt {-a b}}{b}\right )}}{4 \left (\sqrt {-a b}-b \right ) a \left (\tan ^{2}\left (x \right )+\frac {\sqrt {-a b}}{b}\right )}+\frac {\sqrt {b \left (\tan ^{2}\left (x \right )-\frac {\sqrt {-a b}}{b}\right )^{2}+2 \sqrt {-a b}\, \left (\tan ^{2}\left (x \right )-\frac {\sqrt {-a b}}{b}\right )}}{4 \left (\sqrt {-a b}+b \right ) a \left (\tan ^{2}\left (x \right )-\frac {\sqrt {-a b}}{b}\right )}+\frac {b \ln \left (\frac {2 a +2 b -2 b \left (1+\tan ^{2}\left (x \right )\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tan ^{2}\left (x \right )\right )^{2}-2 b \left (1+\tan ^{2}\left (x \right )\right )+a +b}}{1+\tan ^{2}\left (x \right )}\right )}{2 \left (\sqrt {-a b}+b \right ) \left (\sqrt {-a b}-b \right ) \sqrt {a +b}}\)	\(248\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)+1/4/((-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+ta
n(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))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 150 vs. \(2 (64) = 128\).
time = 2.68, size = 319, normalized size = 4.31 \begin {gather*} \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 {gather*}

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)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan {\left (x \right )}}{\left (a + b \tan ^{4}{\left (x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

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)

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Giac [A]
time = 0.44, size = 119, normalized size = 1.61 \begin {gather*} \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 {gather*}

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))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {tan}\left (x\right )}{{\left (b\,{\mathrm {tan}\left (x\right )}^4+a\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(a + b*tan(x)^4)^(3/2),x)

[Out]

int(tan(x)/(a + b*tan(x)^4)^(3/2), x)

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