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

Optimal. Leaf size=209 \[ -\frac{\left (-4 c (a+2 c)+b^2+4 b c\right ) \tanh ^{-1}\left (\frac{b+2 c \tan ^2(d+e x)}{2 \sqrt{c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{16 c^{3/2} e}+\frac{\left (b+2 c \tan ^2(d+e x)-4 c\right ) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{8 c e}+\frac{\sqrt{a-b+c} \tanh ^{-1}\left (\frac{2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt{a-b+c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e} \]

[Out]

(Sqrt[a - b + c]*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*e) - ((b^2 + 4*b*c - 4*c*(a + 2*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])])/(16*c^(3/2)*e) + ((b - 4*c + 2*c*Tan[d + e*x]^2)*Sqrt[a + b*Tan[d +
e*x]^2 + c*Tan[d + e*x]^4])/(8*c*e)

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Rubi [A]  time = 0.34928, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3700, 1251, 814, 843, 621, 206, 724} \[ -\frac{\left (-4 c (a+2 c)+b^2+4 b c\right ) \tanh ^{-1}\left (\frac{b+2 c \tan ^2(d+e x)}{2 \sqrt{c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{16 c^{3/2} e}+\frac{\left (b+2 c \tan ^2(d+e x)-4 c\right ) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{8 c e}+\frac{\sqrt{a-b+c} \tanh ^{-1}\left (\frac{2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt{a-b+c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e} \]

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]

(Sqrt[a - b + c]*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*e) - ((b^2 + 4*b*c - 4*c*(a + 2*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])])/(16*c^(3/2)*e) + ((b - 4*c + 2*c*Tan[d + e*x]^2)*Sqrt[a + b*Tan[d +
e*x]^2 + c*Tan[d + e*x]^4])/(8*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 814

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*(a + b*x + c*x^
2)^p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*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] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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 \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 \sqrt{a+b x^2+c x^4}}{1+x^2} \, dx,x,\tan (d+e x)\right )}{e}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x \sqrt{a+b x+c x^2}}{1+x} \, dx,x,\tan ^2(d+e x)\right )}{2 e}\\ &=\frac{\left (b-4 c+2 c \tan ^2(d+e x)\right ) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{8 c e}-\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} \left (b^2+4 a c-4 b c\right )+\frac{1}{2} \left (b^2+4 b c-4 c (a+2 c)\right ) x}{(1+x) \sqrt{a+b x+c x^2}} \, dx,x,\tan ^2(d+e x)\right )}{8 c e}\\ &=\frac{\left (b-4 c+2 c \tan ^2(d+e x)\right ) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{8 c e}-\frac{(a-b+c) \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{\left (b^2+4 b c-4 c (a+2 c)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x+c x^2}} \, dx,x,\tan ^2(d+e x)\right )}{16 c e}\\ &=\frac{\left (b-4 c+2 c \tan ^2(d+e x)\right ) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{8 c e}+\frac{(a-b+c) \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{\left (b^2+4 b c-4 c (a+2 c)\right ) \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 )}{8 c e}\\ &=\frac{\sqrt{a-b+c} \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 e}-\frac{\left (b^2+4 b c-4 c (a+2 c)\right ) \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 )}{16 c^{3/2} e}+\frac{\left (b-4 c+2 c \tan ^2(d+e x)\right ) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{8 c e}\\ \end{align*}

Mathematica [A]  time = 1.49379, size = 208, normalized size = 1. \[ \frac{-\left (-4 c (a+2 c)+b^2+4 b c\right ) \tanh ^{-1}\left (\frac{b+2 c \tan ^2(d+e x)}{2 \sqrt{c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )+8 c^{3/2} \sqrt{a-b+c} \tanh ^{-1}\left (\frac{2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt{a-b+c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )+2 \sqrt{c} \left (b+2 c \tan ^2(d+e x)-4 c\right ) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{16 c^{3/2} e} \]

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]

(8*c^(3/2)*Sqrt[a - b + c]*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])] - (b^2 + 4*b*c - 4*c*(a + 2*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])] + 2*Sqrt[c]*(b - 4*c + 2*c*Tan[d + e*x]^2)*Sqrt[a + b*Tan[d + e*x]^
2 + c*Tan[d + e*x]^4])/(16*c^(3/2)*e)

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Maple [B]  time = 0.179, size = 467, normalized size = 2.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/e*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)*tan(e*x+d)^2+1/8/e/c*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)*b+1
/4/e/c^(1/2)*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))*a-1/16/e/c^(3/2)*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))*b^2-1/2/e*((1+tan(e*x+d)^2)^2*c+(b-2*c)*
(1+tan(e*x+d)^2)+a-b+c)^(1/2)-1/4/e*ln((1/2*b-c+c*(1+tan(e*x+d)^2))/c^(1/2)+((1+tan(e*x+d)^2)^2*c+(b-2*c)*(1+t
an(e*x+d)^2)+a-b+c)^(1/2))/c^(1/2)*b+1/2/e*ln((1/2*b-c+c*(1+tan(e*x+d)^2))/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))*c^(1/2)+1/2/e*(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))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a} \tan \left (e x + d\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 21.2282, size = 2952, normalized size = 14.12 \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(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)*tan(e*x+d)^3,x, algorithm="fricas")

[Out]

[1/32*(8*sqrt(a - b + c)*c^2*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)) - (b^2 - 4*(a - b)*c - 8*c^
2)*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) + 4*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*(2*c^2*tan(e*
x + d)^2 + b*c - 4*c^2))/(c^2*e), 1/16*(4*sqrt(a - b + c)*c^2*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)) + (b^2 - 4*(a - b)*c - 8*c^2)*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)) + 2*sqrt(c*tan(e*x + d)^4 + b*tan(e
*x + d)^2 + a)*(2*c^2*tan(e*x + d)^2 + b*c - 4*c^2))/(c^2*e), 1/32*(16*sqrt(-a + b - c)*c^2*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)) - (b^2 - 4*(a - b)*c - 8*c^2)*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) + 4*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*(2*c^2*tan(e*x + d)^2 +
b*c - 4*c^2))/(c^2*e), 1/16*(8*sqrt(-a + b - c)*c^2*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)*t
an(e*x + d)^2 + a^2 - a*b + a*c)) + (b^2 - 4*(a - b)*c - 8*c^2)*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)) + 2*sqr
t(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*(2*c^2*tan(e*x + d)^2 + b*c - 4*c^2))/(c^2*e)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \tan ^{2}{\left (d + e x \right )} + c \tan ^{4}{\left (d + e x \right )}} \tan ^{3}{\left (d + e x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out