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

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

[Out]

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

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Rubi [A]  time = 0.208331, antiderivative size = 126, normalized size of antiderivative = 1., 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 = {3670, 1248, 735, 815, 844, 217, 206, 725} \[ \frac{1}{6} \left (a+b \tan ^4(x)\right )^{3/2}+\frac{1}{4} \left (2 (a+b)-b \tan ^2(x)\right ) \sqrt{a+b \tan ^4(x)}-\frac{1}{2} (a+b)^{3/2} \tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )-\frac{1}{4} \sqrt{b} (3 a+2 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tan ^2(x)}{\sqrt{a+b \tan ^4(x)}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]*(a + b*Tan[x]^4)^(3/2),x]

[Out]

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

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 735

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

Mathematica [A]  time = 4.39079, size = 166, normalized size = 1.32 \[ \frac{1}{12} \left (\sqrt{a+b \tan ^4(x)} \left (8 a+2 b \tan ^4(x)-3 b \tan ^2(x)+6 b\right )-6 (a+b)^{3/2} \tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )-6 \sqrt{b} (a+b) \tanh ^{-1}\left (\frac{\sqrt{b} \tan ^2(x)}{\sqrt{a+b \tan ^4(x)}}\right )-\frac{3 \sqrt{a} \sqrt{b} \sqrt{a+b \tan ^4(x)} \sinh ^{-1}\left (\frac{\sqrt{b} \tan ^2(x)}{\sqrt{a}}\right )}{\sqrt{\frac{b \tan ^4(x)}{a}+1}}\right ) \]

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]  time = 0.046, size = 313, normalized size = 2.5 \begin{align*}{\frac{b \left ( \tan \left ( x \right ) \right ) ^{4}}{6}\sqrt{a+b \left ( \tan \left ( x \right ) \right ) ^{4}}}+{\frac{2\,a}{3}\sqrt{a+b \left ( \tan \left ( x \right ) \right ) ^{4}}}-{\frac{b \left ( \tan \left ( x \right ) \right ) ^{2}}{4}\sqrt{a+b \left ( \tan \left ( x \right ) \right ) ^{4}}}-{\frac{3\,a}{4}\sqrt{b}\ln \left ( \sqrt{b} \left ( \tan \left ( x \right ) \right ) ^{2}+\sqrt{a+b \left ( \tan \left ( x \right ) \right ) ^{4}} \right ) }+{\frac{b}{2}\sqrt{a+b \left ( \tan \left ( x \right ) \right ) ^{4}}}-{\frac{1}{2}{b}^{{\frac{3}{2}}}\ln \left ( \sqrt{b} \left ( \tan \left ( x \right ) \right ) ^{2}+\sqrt{a+b \left ( \tan \left ( x \right ) \right ) ^{4}} \right ) }-{\frac{{a}^{2}}{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 ){\frac{1}{\sqrt{a+b}}}}-{ab\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 ){\frac{1}{\sqrt{a+b}}}}-{\frac{{b}^{2}}{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 ){\frac{1}{\sqrt{a+b}}}} \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]

1/6*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., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (x\right )^{4} + a\right )}^{\frac{3}{2}} \tan \left (x\right )\,{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((b*tan(x)^4 + a)^(3/2)*tan(x), x)

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Fricas [A]  time = 4.23001, size = 1609, normalized size = 12.77 \begin{align*} \text{result too large to display} \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]

[1/8*(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., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan ^{4}{\left (x \right )}\right )^{\frac{3}{2}} \tan{\left (x \right )}\, 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((a + b*tan(x)**4)**(3/2)*tan(x), x)

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

[Out]

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