3.27 \(\int \tan ^5(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx\)

Optimal. Leaf size=270 \[ \frac {\left (-4 b c (a-2 c)-8 c^2 (a+2 c)+b^3+2 b^2 c\right ) \tanh ^{-1}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{32 c^{5/2} e}-\frac {\left (2 c (b+2 c) \tan ^2(d+e x)+(b-2 c) (b+4 c)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{16 c^2 e}+\frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{6 c e}-\frac {\sqrt {a-b+c} \tanh ^{-1}\left (\frac {2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e} \]

[Out]

1/32*(b^3+2*b^2*c-4*b*(a-2*c)*c-8*c^2*(a+2*c))*arctanh(1/2*(b+2*c*tan(e*x+d)^2)/c^(1/2)/(a+b*tan(e*x+d)^2+c*ta
n(e*x+d)^4)^(1/2))/c^(5/2)/e-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*ta
n(e*x+d)^4)^(1/2))*(a-b+c)^(1/2)/e-1/16*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)*((b-2*c)*(b+4*c)+2*c*(b+2*c)*t
an(e*x+d)^2)/c^2/e+1/6*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(3/2)/c/e

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Rubi [A]  time = 0.61, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {3700, 1251, 1653, 814, 843, 621, 206, 724} \[ \frac {\left (-4 b c (a-2 c)-8 c^2 (a+2 c)+2 b^2 c+b^3\right ) \tanh ^{-1}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{32 c^{5/2} e}-\frac {\left (2 c (b+2 c) \tan ^2(d+e x)+(b-2 c) (b+4 c)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{16 c^2 e}+\frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{6 c e}-\frac {\sqrt {a-b+c} \tanh ^{-1}\left (\frac {2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e} \]

Antiderivative was successfully verified.

[In]

Int[Tan[d + e*x]^5*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^3 + 2*b^2*c - 4*b*(a - 2*c)*c - 8*c^2*(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])])/(32*c^(5/2)*e) - (((b - 2*c)*(b + 4*c) + 2*c*
(b + 2*c)*Tan[d + e*x]^2)*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])/(16*c^2*e) + (a + b*Tan[d + e*x]^2 +
c*Tan[d + e*x]^4)^(3/2)/(6*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 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 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 1653

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq
, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*e^(q - 1)*(
m + q + 2*p + 1)), x] + Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^
q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q
- 1) - c*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p +
1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

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 \tan ^5(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^5 \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^2 \sqrt {a+b x+c x^2}}{1+x} \, dx,x,\tan ^2(d+e x)\right )}{2 e}\\ &=\frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{6 c e}+\frac {\operatorname {Subst}\left (\int \frac {\left (-\frac {3 b}{2}-\frac {3}{2} (b+2 c) x\right ) \sqrt {a+b x+c x^2}}{1+x} \, dx,x,\tan ^2(d+e x)\right )}{6 c e}\\ &=-\frac {\left ((b-2 c) (b+4 c)+2 c (b+2 c) \tan ^2(d+e x)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{16 c^2 e}+\frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{6 c e}-\frac {\operatorname {Subst}\left (\int \frac {-\frac {3}{4} (b-2 c) \left (b^2-4 a c+4 b c\right )-\frac {3}{4} \left (b^3+2 b^2 c-4 b (a-2 c) c-8 c^2 (a+2 c)\right ) x}{(1+x) \sqrt {a+b x+c x^2}} \, dx,x,\tan ^2(d+e x)\right )}{24 c^2 e}\\ &=-\frac {\left ((b-2 c) (b+4 c)+2 c (b+2 c) \tan ^2(d+e x)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{16 c^2 e}+\frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{6 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^3+2 b^2 c-4 b (a-2 c) c-8 c^2 (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 )}{32 c^2 e}\\ &=-\frac {\left ((b-2 c) (b+4 c)+2 c (b+2 c) \tan ^2(d+e x)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{16 c^2 e}+\frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{6 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^3+2 b^2 c-4 b (a-2 c) c-8 c^2 (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 )}{16 c^2 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^3+2 b^2 c-4 b (a-2 c) c-8 c^2 (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 )}{32 c^{5/2} e}-\frac {\left ((b-2 c) (b+4 c)+2 c (b+2 c) \tan ^2(d+e x)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{16 c^2 e}+\frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{6 c e}\\ \end {align*}

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Mathematica [A]  time = 6.13, size = 467, normalized size = 1.73 \[ \frac {\frac {\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{8 c^{3/2}}-\frac {b \left (\frac {\left (b+2 c \tan ^2(d+e x)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{4 c}-\frac {\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{8 c^{3/2}}\right )}{2 c}+\frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{3 c}-\frac {\left (b+2 c \tan ^2(d+e x)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{4 c}+\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}+\frac {1}{2} \left (\frac {4 (-2 a+2 b-2 c) \sqrt {a-b+c} \tanh ^{-1}\left (\frac {2 a-\left ((2 c-b) \tan ^2(d+e x)\right )-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{4 a-4 b+4 c}-\frac {(2 c-b) \tanh ^{-1}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {c}}\right )}{2 e} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[d + e*x]^5*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4],x]

[Out]

(((b^2 - 4*a*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])])/(
8*c^(3/2)) + (-(((-b + 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])])/Sqrt[c]) + (4*(-2*a + 2*b - 2*c)*Sqrt[a - b + c]*ArcTanh[(2*a - b - (-b + 2*c)*Tan[d + e*x]^2)/(2*S
qrt[a - b + c]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])])/(4*a - 4*b + 4*c))/2 + Sqrt[a + b*Tan[d + e*x]
^2 + c*Tan[d + e*x]^4] - ((b + 2*c*Tan[d + e*x]^2)*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])/(4*c) + (a +
b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4)^(3/2)/(3*c) - (b*(-1/8*((b^2 - 4*a*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])])/c^(3/2) + ((b + 2*c*Tan[d + e*x]^2)*Sqrt[a + b*Tan[
d + e*x]^2 + c*Tan[d + e*x]^4])/(4*c)))/(2*c))/(2*e)

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fricas [A]  time = 8.73, size = 1405, normalized size = 5.20 \[ \text {result too large to display} \]

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)^5,x, algorithm="fricas")

[Out]

[1/192*(48*sqrt(a - b + c)*c^3*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)*sqr
t(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)) - 3*(b^3 - 8*(a - b)*c^2
- 16*c^3 - 2*(2*a*b - b^2)*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) + 4*(8*c^3*tan(e*x + d)^4 - 3*b^2*c
+ 2*(4*a - 3*b)*c^2 + 24*c^3 + 2*(b*c^2 - 6*c^3)*tan(e*x + d)^2)*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)
)/(c^3*e), 1/96*(24*sqrt(a - b + c)*c^3*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)) - 3*(b^3 - 8*(a
- b)*c^2 - 16*c^3 - 2*(2*a*b - b^2)*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)) + 2*(8*c^3*tan(e*x + d)^4 - 3*b^
2*c + 2*(4*a - 3*b)*c^2 + 24*c^3 + 2*(b*c^2 - 6*c^3)*tan(e*x + d)^2)*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2
+ a))/(c^3*e), -1/192*(96*sqrt(-a + b - c)*c^3*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)) + 3*(b^3 - 8*(a - b)*c^2 - 16*c^3 - 2*(2*a*b - b^2)*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) - 4*(8*c^3*tan(e*x + d)^4 - 3*b^2*c + 2*(4*a - 3*b)*c^2 + 24*c^3 + 2*(b*c^2 - 6*c^3)*tan(e*x
+ d)^2)*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a))/(c^3*e), -1/96*(48*sqrt(-a + b - c)*c^3*arctan(-1/2*sqr
t(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)) + 3*(b^3 - 8*(a - b)*c^2 - 16*c^3
- 2*(2*a*b - b^2)*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)) - 2*(8*c^3*tan(e*x + d)^4 - 3*b^2*c + 2*(4*a - 3*b
)*c^2 + 24*c^3 + 2*(b*c^2 - 6*c^3)*tan(e*x + d)^2)*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a))/(c^3*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((a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)*tan(e*x+d)^5,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.67, size = 684, normalized size = 2.53 \[ \frac {\left (a +b \left (\tan ^{2}\left (e x +d \right )\right )+c \left (\tan ^{4}\left (e x +d \right )\right )\right )^{\frac {3}{2}}}{6 c e}-\frac {b \sqrt {a +b \left (\tan ^{2}\left (e x +d \right )\right )+c \left (\tan ^{4}\left (e x +d \right )\right )}\, \left (\tan ^{2}\left (e x +d \right )\right )}{8 e c}-\frac {b^{2} \sqrt {a +b \left (\tan ^{2}\left (e x +d \right )\right )+c \left (\tan ^{4}\left (e x +d \right )\right )}}{16 e \,c^{2}}-\frac {b \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 ) a}{8 e \,c^{\frac {3}{2}}}+\frac {b^{3} \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 )}{32 e \,c^{\frac {5}{2}}}-\frac {\sqrt {a +b \left (\tan ^{2}\left (e x +d \right )\right )+c \left (\tan ^{4}\left (e x +d \right )\right )}\, \left (\tan ^{2}\left (e x +d \right )\right )}{4 e}-\frac {\sqrt {a +b \left (\tan ^{2}\left (e x +d \right )\right )+c \left (\tan ^{4}\left (e x +d \right )\right )}\, b}{8 e c}-\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 ) a}{4 e \sqrt {c}}+\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 ) b^{2}}{16 e \,c^{\frac {3}{2}}}+\frac {\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}}{2 e}+\frac {\ln \left (\frac {\frac {b}{2}-c +\left (1+\tan ^{2}\left (e x +d \right )\right ) c}{\sqrt {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}\right ) b}{4 e \sqrt {c}}-\frac {\ln \left (\frac {\frac {b}{2}-c +\left (1+\tan ^{2}\left (e x +d \right )\right ) c}{\sqrt {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}\right ) \sqrt {c}}{2 e}-\frac {\sqrt {a -b +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 )}{2 e} \]

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)^5,x)

[Out]

1/6*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(3/2)/c/e-1/8/e*b/c*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)*tan(e*x+d)^2
-1/16/e*b^2/c^2*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)-1/8/e*b/c^(3/2)*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))*a+1/32/e*b^3/c^(5/2)*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))-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((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))
*a+1/16/e/c^(3/2)*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))*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+(1+tan(e*x+d)^2)*c)/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)*b-1/2/e*ln((1/2*b-c+(1+tan(e*x+d)^2)*c)/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 = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a} \tan \left (e x + d\right )^{5}\,{d x} \]

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)^5,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\mathrm {tan}\left (d+e\,x\right )}^5\,\sqrt {c\,{\mathrm {tan}\left (d+e\,x\right )}^4+b\,{\mathrm {tan}\left (d+e\,x\right )}^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d + e*x)^5*(a + b*tan(d + e*x)^2 + c*tan(d + e*x)^4)^(1/2),x)

[Out]

int(tan(d + e*x)^5*(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 \[ \int \sqrt {a + b \tan ^{2}{\left (d + e x \right )} + c \tan ^{4}{\left (d + e x \right )}} \tan ^{5}{\left (d + e x \right )}\, dx \]

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)**5,x)

[Out]

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

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