3.37 \(\int \frac{\tan ^3(d+e x)}{\sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx\)
Optimal. Leaf size=141 \[ \frac{\tanh ^{-1}\left (\frac{2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt{a-b+c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e \sqrt{a-b+c}}+\frac{\tanh ^{-1}\left (\frac{b+2 c \tan ^2(d+e x)}{2 \sqrt{c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 \sqrt{c} e} \]
[Out]
ArcTanh[(2*a - b + (b - 2*c)*Tan[d + e*x]^2)/(2*Sqrt[a - b + c]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])
]/(2*Sqrt[a - b + c]*e) + ArcTanh[(b + 2*c*Tan[d + e*x]^2)/(2*Sqrt[c]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*
x]^4])]/(2*Sqrt[c]*e)
________________________________________________________________________________________
Rubi [A] time = 0.21029, antiderivative size = 141, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used =
{3700, 1251, 843, 621, 206, 724} \[ \frac{\tanh ^{-1}\left (\frac{2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt{a-b+c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e \sqrt{a-b+c}}+\frac{\tanh ^{-1}\left (\frac{b+2 c \tan ^2(d+e x)}{2 \sqrt{c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 \sqrt{c} e} \]
Antiderivative was successfully verified.
[In]
Int[Tan[d + e*x]^3/Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4],x]
[Out]
ArcTanh[(2*a - b + (b - 2*c)*Tan[d + e*x]^2)/(2*Sqrt[a - b + c]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])
]/(2*Sqrt[a - b + c]*e) + ArcTanh[(b + 2*c*Tan[d + e*x]^2)/(2*Sqrt[c]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*
x]^4])]/(2*Sqrt[c]*e)
Rule 3700
Int[tan[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*((f_.)*tan[(d_.) + (e_.)*(x_)])^(n_.) + (c_.)*((f_.)*tan[(d_.
) + (e_.)*(x_)])^(n2_.))^(p_), x_Symbol] :> Dist[f/e, Subst[Int[((x/f)^m*(a + b*x^n + c*x^(2*n))^p)/(f^2 + x^2
), x], x, f*Tan[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0]
Rule 1251
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]
Rule 843
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&& !IGtQ[m, 0]
Rule 621
Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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 724
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]
Rubi steps
\begin{align*} \int \frac{\tan ^3(d+e x)}{\sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^3}{\left (1+x^2\right ) \sqrt{a+b x^2+c x^4}} \, dx,x,\tan (d+e x)\right )}{e}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x}{(1+x) \sqrt{a+b x+c x^2}} \, dx,x,\tan ^2(d+e x)\right )}{2 e}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x+c x^2}} \, dx,x,\tan ^2(d+e x)\right )}{2 e}-\frac{\operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{a+b x+c x^2}} \, dx,x,\tan ^2(d+e x)\right )}{2 e}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c \tan ^2(d+e x)}{\sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{e}+\frac{\operatorname{Subst}\left (\int \frac{1}{4 a-4 b+4 c-x^2} \, dx,x,\frac{2 a-b-(-b+2 c) \tan ^2(d+e x)}{\sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{e}\\ &=\frac{\tanh ^{-1}\left (\frac{2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt{a-b+c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 \sqrt{a-b+c} e}+\frac{\tanh ^{-1}\left (\frac{b+2 c \tan ^2(d+e x)}{2 \sqrt{c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 \sqrt{c} e}\\ \end{align*}
Mathematica [A] time = 0.236429, size = 136, normalized size = 0.96 \[ \frac{\frac{\tanh ^{-1}\left (\frac{2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt{a-b+c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt{a-b+c}}+\frac{\tanh ^{-1}\left (\frac{b+2 c \tan ^2(d+e x)}{2 \sqrt{c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt{c}}}{2 e} \]
Antiderivative was successfully verified.
[In]
Integrate[Tan[d + e*x]^3/Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4],x]
[Out]
(ArcTanh[(2*a - b + (b - 2*c)*Tan[d + e*x]^2)/(2*Sqrt[a - b + c]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4]
)]/Sqrt[a - b + c] + ArcTanh[(b + 2*c*Tan[d + e*x]^2)/(2*Sqrt[c]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4]
)]/Sqrt[c])/(2*e)
________________________________________________________________________________________
Maple [A] time = 0.189, size = 155, normalized size = 1.1 \begin{align*}{\frac{1}{2\,e}\ln \left ({ \left ({\frac{b}{2}}+c \left ( \tan \left ( ex+d \right ) \right ) ^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{a+b \left ( \tan \left ( ex+d \right ) \right ) ^{2}+c \left ( \tan \left ( ex+d \right ) \right ) ^{4}} \right ){\frac{1}{\sqrt{c}}}}+{\frac{1}{2\,e}\ln \left ({\frac{1}{1+ \left ( \tan \left ( ex+d \right ) \right ) ^{2}} \left ( 2\,a-2\,b+2\,c+ \left ( b-2\,c \right ) \left ( 1+ \left ( \tan \left ( ex+d \right ) \right ) ^{2} \right ) +2\,\sqrt{a-b+c}\sqrt{ \left ( 1+ \left ( \tan \left ( ex+d \right ) \right ) ^{2} \right ) ^{2}c+ \left ( b-2\,c \right ) \left ( 1+ \left ( \tan \left ( ex+d \right ) \right ) ^{2} \right ) +a-b+c} \right ) } \right ){\frac{1}{\sqrt{a-b+c}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(e*x+d)^3/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x)
[Out]
1/2/e*ln((1/2*b+c*tan(e*x+d)^2)/c^(1/2)+(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2))/c^(1/2)+1/2/e/(a-b+c)^(1/2)*l
n((2*a-2*b+2*c+(b-2*c)*(1+tan(e*x+d)^2)+2*(a-b+c)^(1/2)*((1+tan(e*x+d)^2)^2*c+(b-2*c)*(1+tan(e*x+d)^2)+a-b+c)^
(1/2))/(1+tan(e*x+d)^2))
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (e x + d\right )^{3}}{\sqrt{c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(e*x+d)^3/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorithm="maxima")
[Out]
integrate(tan(e*x + d)^3/sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a), x)
________________________________________________________________________________________
Fricas [A] time = 13.274, size = 2450, normalized size = 17.38 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(e*x+d)^3/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorithm="fricas")
[Out]
[1/4*((a - b + c)*sqrt(c)*log(8*c^2*tan(e*x + d)^4 + 8*b*c*tan(e*x + d)^2 + b^2 + 4*sqrt(c*tan(e*x + d)^4 + b*
tan(e*x + d)^2 + a)*(2*c*tan(e*x + d)^2 + b)*sqrt(c) + 4*a*c) + sqrt(a - b + c)*c*log(((b^2 + 4*(a - 2*b)*c +
8*c^2)*tan(e*x + d)^4 + 2*(4*a*b - 3*b^2 - 4*(a - b)*c)*tan(e*x + d)^2 + 4*sqrt(c*tan(e*x + d)^4 + b*tan(e*x +
d)^2 + a)*((b - 2*c)*tan(e*x + d)^2 + 2*a - b)*sqrt(a - b + c) + 8*a^2 - 8*a*b + b^2 + 4*a*c)/(tan(e*x + d)^4
+ 2*tan(e*x + d)^2 + 1)))/(((a - b)*c + c^2)*e), -1/4*(2*(a - b + c)*sqrt(-c)*arctan(1/2*sqrt(c*tan(e*x + d)^
4 + b*tan(e*x + d)^2 + a)*(2*c*tan(e*x + d)^2 + b)*sqrt(-c)/(c^2*tan(e*x + d)^4 + b*c*tan(e*x + d)^2 + a*c)) -
sqrt(a - b + c)*c*log(((b^2 + 4*(a - 2*b)*c + 8*c^2)*tan(e*x + d)^4 + 2*(4*a*b - 3*b^2 - 4*(a - b)*c)*tan(e*x
+ d)^2 + 4*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*((b - 2*c)*tan(e*x + d)^2 + 2*a - b)*sqrt(a - b + c)
+ 8*a^2 - 8*a*b + b^2 + 4*a*c)/(tan(e*x + d)^4 + 2*tan(e*x + d)^2 + 1)))/(((a - b)*c + c^2)*e), 1/4*(2*sqrt(-
a + b - c)*c*arctan(-1/2*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*((b - 2*c)*tan(e*x + d)^2 + 2*a - b)*sq
rt(-a + b - c)/(((a - b)*c + c^2)*tan(e*x + d)^4 + (a*b - b^2 + b*c)*tan(e*x + d)^2 + a^2 - a*b + a*c)) + (a -
b + c)*sqrt(c)*log(8*c^2*tan(e*x + d)^4 + 8*b*c*tan(e*x + d)^2 + b^2 + 4*sqrt(c*tan(e*x + d)^4 + b*tan(e*x +
d)^2 + a)*(2*c*tan(e*x + d)^2 + b)*sqrt(c) + 4*a*c))/(((a - b)*c + c^2)*e), 1/2*(sqrt(-a + b - c)*c*arctan(-1/
2*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*((b - 2*c)*tan(e*x + d)^2 + 2*a - b)*sqrt(-a + b - c)/(((a - b
)*c + c^2)*tan(e*x + d)^4 + (a*b - b^2 + b*c)*tan(e*x + d)^2 + a^2 - a*b + a*c)) - (a - b + c)*sqrt(-c)*arctan
(1/2*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*(2*c*tan(e*x + d)^2 + b)*sqrt(-c)/(c^2*tan(e*x + d)^4 + b*c
*tan(e*x + d)^2 + a*c)))/(((a - b)*c + c^2)*e)]
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{3}{\left (d + e x \right )}}{\sqrt{a + b \tan ^{2}{\left (d + e x \right )} + c \tan ^{4}{\left (d + e x \right )}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(e*x+d)**3/(a+b*tan(e*x+d)**2+c*tan(e*x+d)**4)**(1/2),x)
[Out]
Integral(tan(d + e*x)**3/sqrt(a + b*tan(d + e*x)**2 + c*tan(d + e*x)**4), x)
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (e x + d\right )^{3}}{\sqrt{c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(e*x+d)^3/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorithm="giac")
[Out]
integrate(tan(e*x + d)^3/sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a), x)