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]

1/2*arctanh(1/2*(b+2*c*tan(e*x+d)^2)/c^(1/2)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2))/e/c^(1/2)+1/2*arctanh(1/
2*(2*a-b+(b-2*c)*tan(e*x+d)^2)/(a-b+c)^(1/2)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2))/e/(a-b+c)^(1/2)

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Rubi [A]  time = 0.21, antiderivative size = 141, normalized size of antiderivative = 1.00, 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 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 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.25, 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)

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fricas [A]  time = 2.20, size = 993, normalized size = 7.04 \[ \text {result too large to display} \]

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)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (e x + d\right )^{3}}{\sqrt {c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a}}\,{d x} \]

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)

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maple [A]  time = 0.44, size = 155, normalized size = 1.10 \[ \frac {\ln \left (\frac {c \left (\tan ^{2}\left (e x +d \right )\right )+\frac {b}{2}}{\sqrt {c}}+\sqrt {a +b \left (\tan ^{2}\left (e x +d \right )\right )+c \left (\tan ^{4}\left (e x +d \right )\right )}\right )}{2 e \sqrt {c}}+\frac {\ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+\tan ^{2}\left (e x +d \right )\right )+2 \sqrt {a -b +c}\, \sqrt {\left (1+\tan ^{2}\left (e x +d \right )\right )^{2} c +\left (b -2 c \right ) \left (1+\tan ^{2}\left (e x +d \right )\right )+a -b +c}}{1+\tan ^{2}\left (e x +d \right )}\right )}{2 e \sqrt {a -b +c}} \]

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((c*tan(e*x+d)^2+1/2*b)/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))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (e x + d\right )^{3}}{\sqrt {c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a}}\,{d x} \]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {tan}\left (d+e\,x\right )}^3}{\sqrt {c\,{\mathrm {tan}\left (d+e\,x\right )}^4+b\,{\mathrm {tan}\left (d+e\,x\right )}^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{3}{\left (d + e x \right )}}{\sqrt {a + b \tan ^{2}{\left (d + e x \right )} + c \tan ^{4}{\left (d + e x \right )}}}\, dx \]

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)

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