3.403 \(\int \frac{\tan ^3(x)}{(a+b \tan ^4(x))^{5/2}} \, dx\)

Optimal. Leaf size=109 \[ -\frac{(b-2 a) \tan ^2(x)+3 a}{6 a (a+b)^2 \sqrt{a+b \tan ^4(x)}}-\frac{1-\tan ^2(x)}{6 (a+b) \left (a+b \tan ^4(x)\right )^{3/2}}+\frac{\tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{2 (a+b)^{5/2}} \]

[Out]

ArcTanh[(a - b*Tan[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tan[x]^4])]/(2*(a + b)^(5/2)) - (1 - Tan[x]^2)/(6*(a + b)*(a
+ b*Tan[x]^4)^(3/2)) - (3*a + (-2*a + b)*Tan[x]^2)/(6*a*(a + b)^2*Sqrt[a + b*Tan[x]^4])

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Rubi [A]  time = 0.227757, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{(b-2 a) \tan ^2(x)+3 a}{6 a (a+b)^2 \sqrt{a+b \tan ^4(x)}}-\frac{1-\tan ^2(x)}{6 (a+b) \left (a+b \tan ^4(x)\right )^{3/2}}+\frac{\tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{2 (a+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]^3/(a + b*Tan[x]^4)^(5/2),x]

[Out]

ArcTanh[(a - b*Tan[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tan[x]^4])]/(2*(a + b)^(5/2)) - (1 - Tan[x]^2)/(6*(a + b)*(a
+ b*Tan[x]^4)^(3/2)) - (3*a + (-2*a + b)*Tan[x]^2)/(6*a*(a + b)^2*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 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 )^{5/2}} \, dx &=\operatorname{Subst}\left (\int \frac{x^3}{\left (1+x^2\right ) \left (a+b x^4\right )^{5/2}} \, dx,x,\tan (x)\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x}{(1+x) \left (a+b x^2\right )^{5/2}} \, dx,x,\tan ^2(x)\right )\\ &=-\frac{1-\tan ^2(x)}{6 (a+b) \left (a+b \tan ^4(x)\right )^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{a b-2 a b x}{(1+x) \left (a+b x^2\right )^{3/2}} \, dx,x,\tan ^2(x)\right )}{6 a b (a+b)}\\ &=-\frac{1-\tan ^2(x)}{6 (a+b) \left (a+b \tan ^4(x)\right )^{3/2}}-\frac{3 a-(2 a-b) \tan ^2(x)}{6 a (a+b)^2 \sqrt{a+b \tan ^4(x)}}+\frac{\operatorname{Subst}\left (\int -\frac{3 a^2 b^2}{(1+x) \sqrt{a+b x^2}} \, dx,x,\tan ^2(x)\right )}{6 a^2 b^2 (a+b)^2}\\ &=-\frac{1-\tan ^2(x)}{6 (a+b) \left (a+b \tan ^4(x)\right )^{3/2}}-\frac{3 a-(2 a-b) \tan ^2(x)}{6 a (a+b)^2 \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)^2}\\ &=-\frac{1-\tan ^2(x)}{6 (a+b) \left (a+b \tan ^4(x)\right )^{3/2}}-\frac{3 a-(2 a-b) \tan ^2(x)}{6 a (a+b)^2 \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)^2}\\ &=\frac{\tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{2 (a+b)^{5/2}}-\frac{1-\tan ^2(x)}{6 (a+b) \left (a+b \tan ^4(x)\right )^{3/2}}-\frac{3 a-(2 a-b) \tan ^2(x)}{6 a (a+b)^2 \sqrt{a+b \tan ^4(x)}}\\ \end{align*}

Mathematica [A]  time = 0.7698, size = 104, normalized size = 0.95 \[ \frac{1}{6} \left (\frac{3 a^2 \tan ^2(x)+b (2 a-b) \tan ^6(x)-3 a b \tan ^4(x)-a (4 a+b)}{a (a+b)^2 \left (a+b \tan ^4(x)\right )^{3/2}}+\frac{3 \tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{(a+b)^{5/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]^3/(a + b*Tan[x]^4)^(5/2),x]

[Out]

((3*ArcTanh[(a - b*Tan[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tan[x]^4])])/(a + b)^(5/2) + (-(a*(4*a + b)) + 3*a^2*Tan[
x]^2 - 3*a*b*Tan[x]^4 + (2*a - b)*b*Tan[x]^6)/(a*(a + b)^2*(a + b*Tan[x]^4)^(3/2)))/6

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Maple [B]  time = 0.093, size = 654, normalized size = 6. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)^3/(a+b*tan(x)^4)^(5/2),x)

[Out]

1/6*(a+b*tan(x)^4)^(1/2)*tan(x)^2*(2*b*tan(x)^4+3*a)/a^2/(b^2*tan(x)^8+2*tan(x)^4*a*b+a^2)-1/24/((-a*b)^(1/2)+
b)/a/(-a*b)^(1/2)/(tan(x)^2-(-a*b)^(1/2)/b)^2*(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)-1/24/((-a*b)^(1/2)+b)/a^2/(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)+1/8*(2*(-a*b)^(1/2)-b)/((-a*b)^(1/2)-b)^2/a^2/(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^2/((-a*b)^(1/2)+b)^2/((-a*b
)^(1/2)-b)^2/(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))-1/8*(2*(-a*b)^(1/2)+b)/((-a*b)^(1/2)+b)^2/a^2/(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)-1/24/((-a*b)^(1/2)-b)/a/(-a*b)^(1/2)/(tan(x)^2+(-a
*b)^(1/2)/b)^2*(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)+1/24/((-a*b)^(1/
2)-b)/a^2/(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{5}{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)^(5/2),x, algorithm="maxima")

[Out]

integrate(tan(x)^3/(b*tan(x)^4 + a)^(5/2), x)

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Fricas [B]  time = 4.04866, size = 1273, normalized size = 11.68 \begin{align*} \left [\frac{3 \,{\left (a b^{2} \tan \left (x\right )^{8} + 2 \, a^{2} b \tan \left (x\right )^{4} + a^{3}\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 \,{\left ({\left (2 \, a^{2} b + a b^{2} - b^{3}\right )} \tan \left (x\right )^{6} - 3 \,{\left (a^{2} b + a b^{2}\right )} \tan \left (x\right )^{4} - 4 \, a^{3} - 5 \, a^{2} b - a b^{2} + 3 \,{\left (a^{3} + a^{2} b\right )} \tan \left (x\right )^{2}\right )} \sqrt{b \tan \left (x\right )^{4} + a}}{12 \,{\left ({\left (a^{4} b^{2} + 3 \, a^{3} b^{3} + 3 \, a^{2} b^{4} + a b^{5}\right )} \tan \left (x\right )^{8} + a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3} + 2 \,{\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4}\right )} \tan \left (x\right )^{4}\right )}}, \frac{3 \,{\left (a b^{2} \tan \left (x\right )^{8} + 2 \, a^{2} b \tan \left (x\right )^{4} + a^{3}\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 ) +{\left ({\left (2 \, a^{2} b + a b^{2} - b^{3}\right )} \tan \left (x\right )^{6} - 3 \,{\left (a^{2} b + a b^{2}\right )} \tan \left (x\right )^{4} - 4 \, a^{3} - 5 \, a^{2} b - a b^{2} + 3 \,{\left (a^{3} + a^{2} b\right )} \tan \left (x\right )^{2}\right )} \sqrt{b \tan \left (x\right )^{4} + a}}{6 \,{\left ({\left (a^{4} b^{2} + 3 \, a^{3} b^{3} + 3 \, a^{2} b^{4} + a b^{5}\right )} \tan \left (x\right )^{8} + a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3} + 2 \,{\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4}\right )} \tan \left (x\right )^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^3/(a+b*tan(x)^4)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(3*(a*b^2*tan(x)^8 + 2*a^2*b*tan(x)^4 + a^3)*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*((2*a^2*b
+ a*b^2 - b^3)*tan(x)^6 - 3*(a^2*b + a*b^2)*tan(x)^4 - 4*a^3 - 5*a^2*b - a*b^2 + 3*(a^3 + a^2*b)*tan(x)^2)*sqr
t(b*tan(x)^4 + a))/((a^4*b^2 + 3*a^3*b^3 + 3*a^2*b^4 + a*b^5)*tan(x)^8 + a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3 +
2*(a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*tan(x)^4), 1/6*(3*(a*b^2*tan(x)^8 + 2*a^2*b*tan(x)^4 + a^3)*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)) + ((2*a^
2*b + a*b^2 - b^3)*tan(x)^6 - 3*(a^2*b + a*b^2)*tan(x)^4 - 4*a^3 - 5*a^2*b - a*b^2 + 3*(a^3 + a^2*b)*tan(x)^2)
*sqrt(b*tan(x)^4 + a))/((a^4*b^2 + 3*a^3*b^3 + 3*a^2*b^4 + a*b^5)*tan(x)^8 + a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b
^3 + 2*(a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*tan(x)^4)]

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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{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)**3/(a+b*tan(x)**4)**(5/2),x)

[Out]

Integral(tan(x)**3/(a + b*tan(x)**4)**(5/2), x)

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Giac [B]  time = 1.249, size = 806, normalized size = 7.39 \begin{align*} \frac{{\left ({\left (\frac{{\left (2 \, a^{7} b^{2} + 11 \, a^{6} b^{3} + 24 \, a^{5} b^{4} + 25 \, a^{4} b^{5} + 10 \, a^{3} b^{6} - 3 \, a^{2} b^{7} - 4 \, a b^{8} - b^{9}\right )} \tan \left (x\right )^{2}}{a^{9} b + 8 \, a^{8} b^{2} + 28 \, a^{7} b^{3} + 56 \, a^{6} b^{4} + 70 \, a^{5} b^{5} + 56 \, a^{4} b^{6} + 28 \, a^{3} b^{7} + 8 \, a^{2} b^{8} + a b^{9}} - \frac{3 \,{\left (a^{7} b^{2} + 6 \, a^{6} b^{3} + 15 \, a^{5} b^{4} + 20 \, a^{4} b^{5} + 15 \, a^{3} b^{6} + 6 \, a^{2} b^{7} + a b^{8}\right )}}{a^{9} b + 8 \, a^{8} b^{2} + 28 \, a^{7} b^{3} + 56 \, a^{6} b^{4} + 70 \, a^{5} b^{5} + 56 \, a^{4} b^{6} + 28 \, a^{3} b^{7} + 8 \, a^{2} b^{8} + a b^{9}}\right )} \tan \left (x\right )^{2} + \frac{3 \,{\left (a^{8} b + 6 \, a^{7} b^{2} + 15 \, a^{6} b^{3} + 20 \, a^{5} b^{4} + 15 \, a^{4} b^{5} + 6 \, a^{3} b^{6} + a^{2} b^{7}\right )}}{a^{9} b + 8 \, a^{8} b^{2} + 28 \, a^{7} b^{3} + 56 \, a^{6} b^{4} + 70 \, a^{5} b^{5} + 56 \, a^{4} b^{6} + 28 \, a^{3} b^{7} + 8 \, a^{2} b^{8} + a b^{9}}\right )} \tan \left (x\right )^{2} - \frac{4 \, a^{8} b + 25 \, a^{7} b^{2} + 66 \, a^{6} b^{3} + 95 \, a^{5} b^{4} + 80 \, a^{4} b^{5} + 39 \, a^{3} b^{6} + 10 \, a^{2} b^{7} + a b^{8}}{a^{9} b + 8 \, a^{8} b^{2} + 28 \, a^{7} b^{3} + 56 \, a^{6} b^{4} + 70 \, a^{5} b^{5} + 56 \, a^{4} b^{6} + 28 \, a^{3} b^{7} + 8 \, a^{2} b^{8} + a b^{9}}}{6 \,{\left (b \tan \left (x\right )^{4} + a\right )}^{\frac{3}{2}}} + \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^{2} + 2 \, a b + b^{2}\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)^(5/2),x, algorithm="giac")

[Out]

1/6*((((2*a^7*b^2 + 11*a^6*b^3 + 24*a^5*b^4 + 25*a^4*b^5 + 10*a^3*b^6 - 3*a^2*b^7 - 4*a*b^8 - b^9)*tan(x)^2/(a
^9*b + 8*a^8*b^2 + 28*a^7*b^3 + 56*a^6*b^4 + 70*a^5*b^5 + 56*a^4*b^6 + 28*a^3*b^7 + 8*a^2*b^8 + a*b^9) - 3*(a^
7*b^2 + 6*a^6*b^3 + 15*a^5*b^4 + 20*a^4*b^5 + 15*a^3*b^6 + 6*a^2*b^7 + a*b^8)/(a^9*b + 8*a^8*b^2 + 28*a^7*b^3
+ 56*a^6*b^4 + 70*a^5*b^5 + 56*a^4*b^6 + 28*a^3*b^7 + 8*a^2*b^8 + a*b^9))*tan(x)^2 + 3*(a^8*b + 6*a^7*b^2 + 15
*a^6*b^3 + 20*a^5*b^4 + 15*a^4*b^5 + 6*a^3*b^6 + a^2*b^7)/(a^9*b + 8*a^8*b^2 + 28*a^7*b^3 + 56*a^6*b^4 + 70*a^
5*b^5 + 56*a^4*b^6 + 28*a^3*b^7 + 8*a^2*b^8 + a*b^9))*tan(x)^2 - (4*a^8*b + 25*a^7*b^2 + 66*a^6*b^3 + 95*a^5*b
^4 + 80*a^4*b^5 + 39*a^3*b^6 + 10*a^2*b^7 + a*b^8)/(a^9*b + 8*a^8*b^2 + 28*a^7*b^3 + 56*a^6*b^4 + 70*a^5*b^5 +
56*a^4*b^6 + 28*a^3*b^7 + 8*a^2*b^8 + a*b^9))/(b*tan(x)^4 + a)^(3/2) + arctan((sqrt(b)*tan(x)^2 - sqrt(b*tan(
x)^4 + a) + sqrt(b))/sqrt(-a - b))/((a^2 + 2*a*b + b^2)*sqrt(-a - b))