3.389 \(\int \tan ^3(x) \sqrt{a+b \tan ^4(x)} \, dx\)

Optimal. Leaf size=103 \[ -\frac{1}{4} \left (2-\tan ^2(x)\right ) \sqrt{a+b \tan ^4(x)}+\frac{(a+2 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tan ^2(x)}{\sqrt{a+b \tan ^4(x)}}\right )}{4 \sqrt{b}}+\frac{1}{2} \sqrt{a+b} \tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.206797, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.412, Rules used = {3670, 1252, 815, 844, 217, 206, 725} \[ -\frac{1}{4} \left (2-\tan ^2(x)\right ) \sqrt{a+b \tan ^4(x)}+\frac{(a+2 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tan ^2(x)}{\sqrt{a+b \tan ^4(x)}}\right )}{4 \sqrt{b}}+\frac{1}{2} \sqrt{a+b} \tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]^3*Sqrt[a + b*Tan[x]^4],x]

[Out]

((a + 2*b)*ArcTanh[(Sqrt[b]*Tan[x]^2)/Sqrt[a + b*Tan[x]^4]])/(4*Sqrt[b]) + (Sqrt[a + b]*ArcTanh[(a - b*Tan[x]^
2)/(Sqrt[a + b]*Sqrt[a + b*Tan[x]^4])])/2 - ((2 - Tan[x]^2)*Sqrt[a + b*Tan[x]^4])/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 815

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 844

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 217

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 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])

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]

Rubi steps

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

Mathematica [A]  time = 3.67383, size = 145, normalized size = 1.41 \[ \frac{1}{4} \left (\frac{\frac{a^{3/2} \sqrt{\frac{b \tan ^4(x)}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} \tan ^2(x)}{\sqrt{a}}\right )}{\sqrt{b}}+\left (\tan ^2(x)-2\right ) \left (a+b \tan ^4(x)\right )}{\sqrt{a+b \tan ^4(x)}}+2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tan ^2(x)}{\sqrt{a+b \tan ^4(x)}}\right )+2 \sqrt{a+b} \tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]^3*Sqrt[a + b*Tan[x]^4],x]

[Out]

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

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Maple [B]  time = 0.085, size = 181, normalized size = 1.8 \begin{align*}{\frac{ \left ( \tan \left ( x \right ) \right ) ^{2}}{4}\sqrt{a+b \left ( \tan \left ( x \right ) \right ) ^{4}}}+{\frac{a}{4}\ln \left ( \sqrt{b} \left ( \tan \left ( x \right ) \right ) ^{2}+\sqrt{a+b \left ( \tan \left ( x \right ) \right ) ^{4}} \right ){\frac{1}{\sqrt{b}}}}-{\frac{1}{2}\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}}+{\frac{1}{2}\sqrt{b}\ln \left ({(b \left ( 1+ \left ( \tan \left ( x \right ) \right ) ^{2} \right ) -b){\frac{1}{\sqrt{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 ) }+{\frac{1}{2}\sqrt{a+b}\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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(x)^4)^(1/2)*tan(x)^3,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(x)^4)^(1/2)*tan(x)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.67354, size = 1443, normalized size = 14.01 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(x)^4)^(1/2)*tan(x)^3,x, algorithm="fricas")

[Out]

[1/8*((a + 2*b)*sqrt(b)*log(-2*b*tan(x)^4 - 2*sqrt(b*tan(x)^4 + a)*sqrt(b)*tan(x)^2 - a) + 2*sqrt(a + b)*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)*(b*tan(x)^2 - 2*b))/b, -1/4*((a + 2*b)*sqrt(-b)*arctan(
sqrt(b*tan(x)^4 + a)*sqrt(-b)/(b*tan(x)^2)) - sqrt(a + b)*b*log(((a*b + 2*b^2)*tan(x)^4 - 2*a*b*tan(x)^2 - 2*s
qrt(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)) - sqrt(b*tan(x)^4
+ a)*(b*tan(x)^2 - 2*b))/b, 1/8*(4*sqrt(-a - b)*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)) + (a + 2*b)*sqrt(b)*log(-2*b*tan(x)^4 - 2*sqrt(b*tan(x)^4 + a)*sqrt(b)*tan(
x)^2 - a) + 2*sqrt(b*tan(x)^4 + a)*(b*tan(x)^2 - 2*b))/b, 1/4*(2*sqrt(-a - b)*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)) - (a + 2*b)*sqrt(-b)*arctan(sqrt(b*tan(x)^4 +
a)*sqrt(-b)/(b*tan(x)^2)) + sqrt(b*tan(x)^4 + a)*(b*tan(x)^2 - 2*b))/b]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(x)**4)**(1/2)*tan(x)**3,x)

[Out]

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

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Giac [A]  time = 1.22797, size = 144, normalized size = 1.4 \begin{align*} \frac{1}{4} \, \sqrt{b \tan \left (x\right )^{4} + a}{\left (\tan \left (x\right )^{2} - 2\right )} - \frac{{\left (a + b\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}} - \frac{{\left (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 )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(x)^4)^(1/2)*tan(x)^3,x, algorithm="giac")

[Out]

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