3.4.94 \(\int \tan (x) (a+b \tan ^4(x))^{3/2} \, dx\) [394]

Optimal. Leaf size=126 \[ -\frac {1}{4} \sqrt {b} (3 a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )-\frac {1}{2} (a+b)^{3/2} \tanh ^{-1}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )+\frac {1}{4} \left (2 (a+b)-b \tan ^2(x)\right ) \sqrt {a+b \tan ^4(x)}+\frac {1}{6} \left (a+b \tan ^4(x)\right )^{3/2} \]

[Out]

-1/2*(a+b)^(3/2)*arctanh((a-b*tan(x)^2)/(a+b)^(1/2)/(a+b*tan(x)^4)^(1/2))-1/4*(3*a+2*b)*arctanh(b^(1/2)*tan(x)
^2/(a+b*tan(x)^4)^(1/2))*b^(1/2)+1/4*(a+b*tan(x)^4)^(1/2)*(2*a+2*b-b*tan(x)^2)+1/6*(a+b*tan(x)^4)^(3/2)

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Rubi [A]
time = 0.13, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {3751, 1262, 749, 829, 858, 223, 212, 739} \begin {gather*} \frac {1}{6} \left (a+b \tan ^4(x)\right )^{3/2}+\frac {1}{4} \left (2 (a+b)-b \tan ^2(x)\right ) \sqrt {a+b \tan ^4(x)}-\frac {1}{2} (a+b)^{3/2} \tanh ^{-1}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )-\frac {1}{4} \sqrt {b} (3 a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[x]*(a + b*Tan[x]^4)^(3/2),x]

[Out]

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

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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 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 749

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((a + c*x^2)^p/(e
*(m + 2*p + 1))), x] + Dist[2*(p/(e*(m + 2*p + 1))), Int[(d + e*x)^m*Simp[a*e - c*d*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] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !Ration
alQ[m] || LtQ[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 829

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

Rule 858

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

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

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Mathematica [A]
time = 3.04, size = 166, normalized size = 1.32 \begin {gather*} \frac {1}{12} \left (-6 \sqrt {b} (a+b) \tanh ^{-1}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )-6 (a+b)^{3/2} \tanh ^{-1}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )+\sqrt {a+b \tan ^4(x)} \left (8 a+6 b-3 b \tan ^2(x)+2 b \tan ^4(x)\right )-\frac {3 \sqrt {a} \sqrt {b} \sinh ^{-1}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a}}\right ) \sqrt {a+b \tan ^4(x)}}{\sqrt {1+\frac {b \tan ^4(x)}{a}}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*Sqrt[b]*(a + b)*ArcTanh[(Sqrt[b]*Tan[x]^2)/Sqrt[a + b*Tan[x]^4]] - 6*(a + b)^(3/2)*ArcTanh[(a - b*Tan[x]^2
)/(Sqrt[a + b]*Sqrt[a + b*Tan[x]^4])] + Sqrt[a + b*Tan[x]^4]*(8*a + 6*b - 3*b*Tan[x]^2 + 2*b*Tan[x]^4) - (3*Sq
rt[a]*Sqrt[b]*ArcSinh[(Sqrt[b]*Tan[x]^2)/Sqrt[a]]*Sqrt[a + b*Tan[x]^4])/Sqrt[1 + (b*Tan[x]^4)/a])/12

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(312\) vs. \(2(103)=206\).
time = 0.09, size = 313, normalized size = 2.48

method	result	size
derivativedivides	\(\frac {b \left (\tan ^{4}\left (x \right )\right ) \sqrt {a +b \left (\tan ^{4}\left (x \right )\right )}}{6}+\frac {2 a \sqrt {a +b \left (\tan ^{4}\left (x \right )\right )}}{3}-\frac {b \left (\tan ^{2}\left (x \right )\right ) \sqrt {a +b \left (\tan ^{4}\left (x \right )\right )}}{4}-\frac {3 a \sqrt {b}\, \ln \left (\sqrt {b}\, \left (\tan ^{2}\left (x \right )\right )+\sqrt {a +b \left (\tan ^{4}\left (x \right )\right )}\right )}{4}+\frac {b \sqrt {a +b \left (\tan ^{4}\left (x \right )\right )}}{2}-\frac {b^{\frac {3}{2}} \ln \left (\sqrt {b}\, \left (\tan ^{2}\left (x \right )\right )+\sqrt {a +b \left (\tan ^{4}\left (x \right )\right )}\right )}{2}-\frac {\ln \left (\frac {2 a +2 b -2 b \left (1+\tan ^{2}\left (x \right )\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tan ^{2}\left (x \right )\right )^{2}-2 b \left (1+\tan ^{2}\left (x \right )\right )+a +b}}{1+\tan ^{2}\left (x \right )}\right ) a^{2}}{2 \sqrt {a +b}}-\frac {\ln \left (\frac {2 a +2 b -2 b \left (1+\tan ^{2}\left (x \right )\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tan ^{2}\left (x \right )\right )^{2}-2 b \left (1+\tan ^{2}\left (x \right )\right )+a +b}}{1+\tan ^{2}\left (x \right )}\right ) a b}{\sqrt {a +b}}-\frac {\ln \left (\frac {2 a +2 b -2 b \left (1+\tan ^{2}\left (x \right )\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tan ^{2}\left (x \right )\right )^{2}-2 b \left (1+\tan ^{2}\left (x \right )\right )+a +b}}{1+\tan ^{2}\left (x \right )}\right ) b^{2}}{2 \sqrt {a +b}}\)	\(313\)
default	\(\frac {b \left (\tan ^{4}\left (x \right )\right ) \sqrt {a +b \left (\tan ^{4}\left (x \right )\right )}}{6}+\frac {2 a \sqrt {a +b \left (\tan ^{4}\left (x \right )\right )}}{3}-\frac {b \left (\tan ^{2}\left (x \right )\right ) \sqrt {a +b \left (\tan ^{4}\left (x \right )\right )}}{4}-\frac {3 a \sqrt {b}\, \ln \left (\sqrt {b}\, \left (\tan ^{2}\left (x \right )\right )+\sqrt {a +b \left (\tan ^{4}\left (x \right )\right )}\right )}{4}+\frac {b \sqrt {a +b \left (\tan ^{4}\left (x \right )\right )}}{2}-\frac {b^{\frac {3}{2}} \ln \left (\sqrt {b}\, \left (\tan ^{2}\left (x \right )\right )+\sqrt {a +b \left (\tan ^{4}\left (x \right )\right )}\right )}{2}-\frac {\ln \left (\frac {2 a +2 b -2 b \left (1+\tan ^{2}\left (x \right )\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tan ^{2}\left (x \right )\right )^{2}-2 b \left (1+\tan ^{2}\left (x \right )\right )+a +b}}{1+\tan ^{2}\left (x \right )}\right ) a^{2}}{2 \sqrt {a +b}}-\frac {\ln \left (\frac {2 a +2 b -2 b \left (1+\tan ^{2}\left (x \right )\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tan ^{2}\left (x \right )\right )^{2}-2 b \left (1+\tan ^{2}\left (x \right )\right )+a +b}}{1+\tan ^{2}\left (x \right )}\right ) a b}{\sqrt {a +b}}-\frac {\ln \left (\frac {2 a +2 b -2 b \left (1+\tan ^{2}\left (x \right )\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tan ^{2}\left (x \right )\right )^{2}-2 b \left (1+\tan ^{2}\left (x \right )\right )+a +b}}{1+\tan ^{2}\left (x \right )}\right ) b^{2}}{2 \sqrt {a +b}}\)	\(313\)

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]

1/6*b*tan(x)^4*(a+b*tan(x)^4)^(1/2)+2/3*a*(a+b*tan(x)^4)^(1/2)-1/4*b*tan(x)^2*(a+b*tan(x)^4)^(1/2)-3/4*a*b^(1/
2)*ln(b^(1/2)*tan(x)^2+(a+b*tan(x)^4)^(1/2))+1/2*b*(a+b*tan(x)^4)^(1/2)-1/2*b^(3/2)*ln(b^(1/2)*tan(x)^2+(a+b*t
an(x)^4)^(1/2))-1/2/(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))*a^2-1/(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))*a*b-1/2/(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))*b^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((b*tan(x)^4 + a)^(3/2)*tan(x), x)

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Fricas [A]
time = 2.91, size = 593, normalized size = 4.71 \begin {gather*} \left [\frac {1}{8} \, {\left (3 \, a + 2 \, b\right )} \sqrt {b} \log \left (-2 \, b \tan \left (x\right )^{4} + 2 \, \sqrt {b \tan \left (x\right )^{4} + a} \sqrt {b} \tan \left (x\right )^{2} - a\right ) + \frac {1}{4} \, {\left (a + b\right )}^{\frac {3}{2}} \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 ) + \frac {1}{12} \, {\left (2 \, b \tan \left (x\right )^{4} - 3 \, b \tan \left (x\right )^{2} + 8 \, a + 6 \, b\right )} \sqrt {b \tan \left (x\right )^{4} + a}, \frac {1}{4} \, {\left (3 \, a + 2 \, b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \tan \left (x\right )^{4} + a} \sqrt {-b}}{b \tan \left (x\right )^{2}}\right ) + \frac {1}{4} \, {\left (a + b\right )}^{\frac {3}{2}} \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 ) + \frac {1}{12} \, {\left (2 \, b \tan \left (x\right )^{4} - 3 \, b \tan \left (x\right )^{2} + 8 \, a + 6 \, b\right )} \sqrt {b \tan \left (x\right )^{4} + a}, -\frac {1}{2} \, {\left (a + b\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 ) + \frac {1}{8} \, {\left (3 \, a + 2 \, b\right )} \sqrt {b} \log \left (-2 \, b \tan \left (x\right )^{4} + 2 \, \sqrt {b \tan \left (x\right )^{4} + a} \sqrt {b} \tan \left (x\right )^{2} - a\right ) + \frac {1}{12} \, {\left (2 \, b \tan \left (x\right )^{4} - 3 \, b \tan \left (x\right )^{2} + 8 \, a + 6 \, b\right )} \sqrt {b \tan \left (x\right )^{4} + a}, -\frac {1}{2} \, {\left (a + b\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 ) + \frac {1}{4} \, {\left (3 \, a + 2 \, b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \tan \left (x\right )^{4} + a} \sqrt {-b}}{b \tan \left (x\right )^{2}}\right ) + \frac {1}{12} \, {\left (2 \, b \tan \left (x\right )^{4} - 3 \, b \tan \left (x\right )^{2} + 8 \, a + 6 \, b\right )} \sqrt {b \tan \left (x\right )^{4} + a}\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]

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

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

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Giac [A]
time = 0.53, size = 138, normalized size = 1.10 \begin {gather*} \frac {1}{4} \, {\left (3 \, a \sqrt {b} + 2 \, b^{\frac {3}{2}}\right )} \log \left ({\left | -\sqrt {b} \tan \left (x\right )^{2} + \sqrt {b \tan \left (x\right )^{4} + a} \right |}\right ) + \frac {1}{12} \, \sqrt {b \tan \left (x\right )^{4} + a} {\left ({\left (2 \, b \tan \left (x\right )^{2} - 3 \, b\right )} \tan \left (x\right )^{2} + \frac {2 \, {\left (4 \, a b + 3 \, b^{2}\right )}}{b}\right )} + \frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} \arctan \left (-\frac {\sqrt {b} \tan \left (x\right )^{2} - \sqrt {b \tan \left (x\right )^{4} + a} + \sqrt {b}}{\sqrt {-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]

1/4*(3*a*sqrt(b) + 2*b^(3/2))*log(abs(-sqrt(b)*tan(x)^2 + sqrt(b*tan(x)^4 + a))) + 1/12*sqrt(b*tan(x)^4 + a)*(
(2*b*tan(x)^2 - 3*b)*tan(x)^2 + 2*(4*a*b + 3*b^2)/b) + (a^2 + 2*a*b + b^2)*arctan(-(sqrt(b)*tan(x)^2 - sqrt(b*
tan(x)^4 + a) + sqrt(b))/sqrt(-a - b))/sqrt(-a - b)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \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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