3.46 \(\int \frac{\tan ^5(d+e x)}{(a+b \tan ^2(d+e x)+c \tan ^4(d+e x))^{3/2}} \, dx\)

Optimal. Leaf size=159 \[ \frac{(b (a-b)+2 a c) \tan ^2(d+e x)+a (2 a-b)}{e (a-b+c) \left (b^2-4 a c\right ) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac{\tanh ^{-1}\left (\frac{2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt{a-b+c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e (a-b+c)^{3/2}} \]

[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*(2*a - b) + ((a - b)*b + 2*a*c)*Tan[d + e*x]^2)/((a - b + c)*(b^2 - 4*a*c)*e*S
qrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])

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Rubi [A]  time = 0.393696, antiderivative size = 159, normalized size of antiderivative = 1., 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, 1646, 12, 724, 206} \[ \frac{(b (a-b)+2 a c) \tan ^2(d+e x)+a (2 a-b)}{e (a-b+c) \left (b^2-4 a c\right ) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac{\tanh ^{-1}\left (\frac{2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt{a-b+c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e (a-b+c)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[d + e*x]^5/(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*(2*a - b) + ((a - b)*b + 2*a*c)*Tan[d + e*x]^2)/((a - b + c)*(b^2 - 4*a*c)*e*S
qrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])

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 1646

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = Polynomi
alQuotient[(d + e*x)^m*Pq, a + b*x + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2,
x], x, 0], g = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2
*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(d
+ e*x)^m*(a + b*x + c*x^2)^(p + 1)*ExpandToSum[((p + 1)*(b^2 - 4*a*c)*Q)/(d + e*x)^m - ((2*p + 3)*(2*c*f - b*
g))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]

Rule 12

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

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 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])

Rubi steps

\begin{align*} \int \frac{\tan ^5(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^5}{\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^2}{(1+x) \left (a+b x+c x^2\right )^{3/2}} \, dx,x,\tan ^2(d+e x)\right )}{2 e}\\ &=\frac{a (2 a-b)+((a-b) b+2 a 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 (a-b+c) (1+x) \sqrt{a+b x+c x^2}} \, dx,x,\tan ^2(d+e x)\right )}{\left (b^2-4 a c\right ) e}\\ &=\frac{a (2 a-b)+((a-b) b+2 a 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 (2 a-b)+((a-b) b+2 a 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 (2 a-b)+((a-b) b+2 a 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*}

Mathematica [C]  time = 35.6431, size = 57597, normalized size = 362.25 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Tan[d + e*x]^5/(a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4)^(3/2),x]

[Out]

Result too large to show

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Maple [B]  time = 0.165, size = 601, normalized size = 3.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e*x+d)^5/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(3/2),x)

[Out]

-1/e/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)/(4*a*c-b^2)*b*tan(e*x+d)^2-2/e/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^
(1/2)/(4*a*c-b^2)*a-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)/(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/((-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/(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., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (e x + d\right )^{5}}{{\left (c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(e*x+d)^5/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(3/2),x, algorithm="maxima")

[Out]

integrate(tan(e*x + d)^5/(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)^(3/2), x)

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Fricas [B]  time = 6.31012, size = 2421, normalized size = 15.23 \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)^5/(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)*(2*a^3
- 3*a^2*b + a*b^2 + (a^2*b - 2*a*b^2 + b^3 + 2*a*c^2 + (2*a^2 - a*b - b^2)*c)*tan(e*x + d)^2 + (2*a^2 - 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)*(2*a^3 - 3*a^2*b + a*b^2 + (a
^2*b - 2*a*b^2 + b^3 + 2*a*c^2 + (2*a^2 - a*b - b^2)*c)*tan(e*x + d)^2 + (2*a^2 - 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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{5}{\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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(e*x+d)**5/(a+b*tan(e*x+d)**2+c*tan(e*x+d)**4)**(3/2),x)

[Out]

Integral(tan(d + e*x)**5/(a + b*tan(d + e*x)**2 + c*tan(d + e*x)**4)**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (e x + d\right )^{5}}{{\left (c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(e*x+d)^5/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(3/2),x, algorithm="giac")

[Out]

integrate(tan(e*x + d)^5/(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)^(3/2), x)