3.47 \(\int \frac {\tan ^3(d+e x)}{(a+b \tan ^2(d+e x)+c \tan ^4(d+e x))^{3/2}} \, dx\)

Optimal. Leaf size=154 \[ \frac {\tanh ^{-1}\left (\frac {2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e (a-b+c)^{3/2}}-\frac {c (2 a-b) \tan ^2(d+e x)+a (b-2 c)}{e (a-b+c) \left (b^2-4 a c\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \]

[Out]

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

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Rubi [A]  time = 0.28, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3700, 1251, 822, 12, 724, 206} \[ \frac {\tanh ^{-1}\left (\frac {2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e (a-b+c)^{3/2}}-\frac {c (2 a-b) \tan ^2(d+e x)+a (b-2 c)}{e (a-b+c) \left (b^2-4 a c\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[d + e*x]^3/(a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4)^(3/2),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*(a - b + c)^(3/2)*e) - (a*(b - 2*c) + (2*a - b)*c*Tan[d + e*x]^2)/((a - b + c)*(b^2 - 4*a*c)*e*Sqrt[a + b
*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

Rule 822

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x
)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*
c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2
*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d -
b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b,
c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] ||
IntegerQ[p] || IntegersQ[2*m, 2*p])

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

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Mathematica [A]  time = 2.90, size = 155, normalized size = 1.01 \[ \frac {\frac {2 c (2 a-b) \tan ^2(d+e x)+2 a (b-2 c)}{(a-b+c) \left (4 a c-b^2\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}+\frac {\tanh ^{-1}\left (\frac {2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{(a-b+c)^{3/2}}}{2 e} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[d + e*x]^3/(a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4)^(3/2),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]
)]/(a - b + c)^(3/2) + (2*a*(b - 2*c) + 2*(2*a - b)*c*Tan[d + e*x]^2)/((a - b + c)*(-b^2 + 4*a*c)*Sqrt[a + b*T
an[d + e*x]^2 + c*Tan[d + e*x]^4]))/(2*e)

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fricas [B]  time = 1.15, size = 1077, normalized size = 6.99 \[ \text {result too large to display} \]

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)^(3/2),x, algorithm="fricas")

[Out]

[-1/4*(((b^2*c - 4*a*c^2)*tan(e*x + d)^4 + a*b^2 - 4*a^2*c + (b^3 - 4*a*b*c)*tan(e*x + d)^2)*sqrt(a - b + c)*l
og(((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*t
an(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)) - 4*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*(a^2*b
- a*b^2 - 2*a*c^2 + ((2*a - b)*c^2 + (2*a^2 - 3*a*b + b^2)*c)*tan(e*x + d)^2 - (2*a^2 - 3*a*b)*c))/((4*a*c^4 +
(8*a^2 - 8*a*b - b^2)*c^3 + 2*(2*a^3 - 4*a^2*b + a*b^2 + b^3)*c^2 - (a^2*b^2 - 2*a*b^3 + b^4)*c)*e*tan(e*x +
d)^4 - (a^2*b^3 - 2*a*b^4 + b^5 - 4*a*b*c^3 - (8*a^2*b - 8*a*b^2 - b^3)*c^2 - 2*(2*a^3*b - 4*a^2*b^2 + a*b^3 +
b^4)*c)*e*tan(e*x + d)^2 - (a^3*b^2 - 2*a^2*b^3 + a*b^4 - 4*a^2*c^3 - (8*a^3 - 8*a^2*b - a*b^2)*c^2 - 2*(2*a^
4 - 4*a^3*b + a^2*b^2 + a*b^3)*c)*e), -1/2*(((b^2*c - 4*a*c^2)*tan(e*x + d)^4 + a*b^2 - 4*a^2*c + (b^3 - 4*a*b
*c)*tan(e*x + d)^2)*sqrt(-a + b - 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)) - 2*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*(a^2*b - a*b^2 - 2*a*c^2 + ((2*a - b)*c^2
+ (2*a^2 - 3*a*b + b^2)*c)*tan(e*x + d)^2 - (2*a^2 - 3*a*b)*c))/((4*a*c^4 + (8*a^2 - 8*a*b - b^2)*c^3 + 2*(2*a
^3 - 4*a^2*b + a*b^2 + b^3)*c^2 - (a^2*b^2 - 2*a*b^3 + b^4)*c)*e*tan(e*x + d)^4 - (a^2*b^3 - 2*a*b^4 + b^5 - 4
*a*b*c^3 - (8*a^2*b - 8*a*b^2 - b^3)*c^2 - 2*(2*a^3*b - 4*a^2*b^2 + a*b^3 + b^4)*c)*e*tan(e*x + d)^2 - (a^3*b^
2 - 2*a^2*b^3 + a*b^4 - 4*a^2*c^3 - (8*a^3 - 8*a^2*b - a*b^2)*c^2 - 2*(2*a^4 - 4*a^3*b + a^2*b^2 + a*b^3)*c)*e
)]

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.45, size = 508, normalized size = 3.30 \[ \frac {2 c \left (\tan ^{2}\left (e x +d \right )\right )}{e \sqrt {a +b \left (\tan ^{2}\left (e x +d \right )\right )+c \left (\tan ^{4}\left (e x +d \right )\right )}\, \left (4 c a -b^{2}\right )}+\frac {b}{e \sqrt {a +b \left (\tan ^{2}\left (e x +d \right )\right )+c \left (\tan ^{4}\left (e x +d \right )\right )}\, \left (4 c a -b^{2}\right )}+\frac {2 c \sqrt {\left (\tan ^{2}\left (e x +d \right )-\frac {-b +\sqrt {-4 c a +b^{2}}}{2 c}\right )^{2} c +\sqrt {-4 c a +b^{2}}\, \left (\tan ^{2}\left (e x +d \right )-\frac {-b +\sqrt {-4 c a +b^{2}}}{2 c}\right )}}{e \left (\sqrt {-4 c a +b^{2}}-b +2 c \right ) \left (-4 c a +b^{2}\right ) \left (\tan ^{2}\left (e x +d \right )-\frac {\sqrt {-4 c a +b^{2}}}{2 c}+\frac {b}{2 c}\right )}-\frac {2 c \ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+\tan ^{2}\left (e x +d \right )\right )+2 \sqrt {a -b +c}\, \sqrt {\left (1+\tan ^{2}\left (e x +d \right )\right )^{2} c +\left (b -2 c \right ) \left (1+\tan ^{2}\left (e x +d \right )\right )+a -b +c}}{1+\tan ^{2}\left (e x +d \right )}\right )}{e \left (\sqrt {-4 c a +b^{2}}-b +2 c \right ) \left (b -2 c +\sqrt {-4 c a +b^{2}}\right ) \sqrt {a -b +c}}-\frac {2 c \sqrt {\left (\tan ^{2}\left (e x +d \right )+\frac {b +\sqrt {-4 c a +b^{2}}}{2 c}\right )^{2} c -\sqrt {-4 c a +b^{2}}\, \left (\tan ^{2}\left (e x +d \right )+\frac {b +\sqrt {-4 c a +b^{2}}}{2 c}\right )}}{e \left (b -2 c +\sqrt {-4 c a +b^{2}}\right ) \left (-4 c a +b^{2}\right ) \left (\tan ^{2}\left (e x +d \right )+\frac {\sqrt {-4 c a +b^{2}}}{2 c}+\frac {b}{2 c}\right )} \]

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)^(3/2),x)

[Out]

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

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

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)^(3/2),x, algorithm="maxima")

[Out]

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

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

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)^(3/2),x)

[Out]

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

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

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)**(3/2),x)

[Out]

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

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