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

Optimal. Leaf size=548 \[ \frac {\sqrt {-\sqrt {a^2-2 a c+b^2+c^2}+a-c} \tanh ^{-1}\left (\frac {-\sqrt {a^2-2 a c+b^2+c^2}+a+b \tan (d+e x)-c}{\sqrt {2} \sqrt {-\sqrt {a^2-2 a c+b^2+c^2}+a-c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} e \sqrt {a^2-2 a c+b^2+c^2}}-\frac {\sqrt {\sqrt {a^2-2 a c+b^2+c^2}+a-c} \tanh ^{-1}\left (\frac {\sqrt {a^2-2 a c+b^2+c^2}+a+b \tan (d+e x)-c}{\sqrt {2} \sqrt {\sqrt {a^2-2 a c+b^2+c^2}+a-c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} e \sqrt {a^2-2 a c+b^2+c^2}}-\frac {b \left (5 b^2-12 a c\right ) \tanh ^{-1}\left (\frac {b+2 c \tan (d+e x)}{2 \sqrt {c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{16 c^{7/2} e}+\frac {\left (-16 a c+15 b^2-10 b c \tan (d+e x)\right ) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{24 c^3 e}+\frac {b \tanh ^{-1}\left (\frac {b+2 c \tan (d+e x)}{2 \sqrt {c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{2 c^{3/2} e}+\frac {\tan ^2(d+e x) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{3 c e}-\frac {\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{c e} \]

[Out]

1/2*b*arctanh(1/2*(b+2*c*tan(e*x+d))/c^(1/2)/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(1/2))/c^(3/2)/e-1/16*b*(-12*a*c+
5*b^2)*arctanh(1/2*(b+2*c*tan(e*x+d))/c^(1/2)/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(1/2))/c^(7/2)/e+1/2*arctanh(1/2
*(a-c-(a^2-2*a*c+b^2+c^2)^(1/2)+b*tan(e*x+d))*2^(1/2)/(a-c-(a^2-2*a*c+b^2+c^2)^(1/2))^(1/2)/(a+b*tan(e*x+d)+c*
tan(e*x+d)^2)^(1/2))*(a-c-(a^2-2*a*c+b^2+c^2)^(1/2))^(1/2)/e*2^(1/2)/(a^2-2*a*c+b^2+c^2)^(1/2)-1/2*arctanh(1/2
*(a-c+(a^2-2*a*c+b^2+c^2)^(1/2)+b*tan(e*x+d))*2^(1/2)/(a-c+(a^2-2*a*c+b^2+c^2)^(1/2))^(1/2)/(a+b*tan(e*x+d)+c*
tan(e*x+d)^2)^(1/2))*(a-c+(a^2-2*a*c+b^2+c^2)^(1/2))^(1/2)/e*2^(1/2)/(a^2-2*a*c+b^2+c^2)^(1/2)-(a+b*tan(e*x+d)
+c*tan(e*x+d)^2)^(1/2)/c/e+1/3*(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(1/2)*tan(e*x+d)^2/c/e+1/24*(a+b*tan(e*x+d)+c*t
an(e*x+d)^2)^(1/2)*(15*b^2-16*a*c-10*b*c*tan(e*x+d))/c^3/e

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Rubi [A]  time = 1.09, antiderivative size = 548, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3700, 6725, 640, 621, 206, 742, 779, 1036, 1030, 208} \[ \frac {\sqrt {-\sqrt {a^2-2 a c+b^2+c^2}+a-c} \tanh ^{-1}\left (\frac {-\sqrt {a^2-2 a c+b^2+c^2}+a+b \tan (d+e x)-c}{\sqrt {2} \sqrt {-\sqrt {a^2-2 a c+b^2+c^2}+a-c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} e \sqrt {a^2-2 a c+b^2+c^2}}-\frac {\sqrt {\sqrt {a^2-2 a c+b^2+c^2}+a-c} \tanh ^{-1}\left (\frac {\sqrt {a^2-2 a c+b^2+c^2}+a+b \tan (d+e x)-c}{\sqrt {2} \sqrt {\sqrt {a^2-2 a c+b^2+c^2}+a-c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} e \sqrt {a^2-2 a c+b^2+c^2}}+\frac {\left (-16 a c+15 b^2-10 b c \tan (d+e x)\right ) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{24 c^3 e}-\frac {b \left (5 b^2-12 a c\right ) \tanh ^{-1}\left (\frac {b+2 c \tan (d+e x)}{2 \sqrt {c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{16 c^{7/2} e}+\frac {b \tanh ^{-1}\left (\frac {b+2 c \tan (d+e x)}{2 \sqrt {c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{2 c^{3/2} e}+\frac {\tan ^2(d+e x) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{3 c e}-\frac {\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a - c - Sqrt[a^2 + b^2 - 2*a*c + c^2]]*ArcTanh[(a - c - Sqrt[a^2 + b^2 - 2*a*c + c^2] + b*Tan[d + e*x])/
(Sqrt[2]*Sqrt[a - c - Sqrt[a^2 + b^2 - 2*a*c + c^2]]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])])/(Sqrt[2]*S
qrt[a^2 + b^2 - 2*a*c + c^2]*e) - (Sqrt[a - c + Sqrt[a^2 + b^2 - 2*a*c + c^2]]*ArcTanh[(a - c + Sqrt[a^2 + b^2
- 2*a*c + c^2] + b*Tan[d + e*x])/(Sqrt[2]*Sqrt[a - c + Sqrt[a^2 + b^2 - 2*a*c + c^2]]*Sqrt[a + b*Tan[d + e*x]
+ c*Tan[d + e*x]^2])])/(Sqrt[2]*Sqrt[a^2 + b^2 - 2*a*c + c^2]*e) + (b*ArcTanh[(b + 2*c*Tan[d + e*x])/(2*Sqrt[
c]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])])/(2*c^(3/2)*e) - (b*(5*b^2 - 12*a*c)*ArcTanh[(b + 2*c*Tan[d +
e*x])/(2*Sqrt[c]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])])/(16*c^(7/2)*e) - Sqrt[a + b*Tan[d + e*x] + c*
Tan[d + e*x]^2]/(c*e) + (Tan[d + e*x]^2*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])/(3*c*e) + ((15*b^2 - 16*a
*c - 10*b*c*Tan[d + e*x])*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])/(24*c^3*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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

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 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rule 779

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

Rule 1030

Int[((g_) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[-2*
a*g*h, Subst[Int[1/Simp[2*a^2*g*h*c + a*e*x^2, x], x], x, Simp[a*h - g*c*x, x]/Sqrt[d + e*x + f*x^2]], x] /; F
reeQ[{a, c, d, e, f, g, h}, x] && EqQ[a*h^2*e + 2*g*h*(c*d - a*f) - g^2*c*e, 0]

Rule 1036

Int[((g_.) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q
= Rt[(c*d - a*f)^2 + a*c*e^2, 2]}, Dist[1/(2*q), Int[Simp[-(a*h*e) - g*(c*d - a*f - q) + (h*(c*d - a*f + q) -
g*c*e)*x, x]/((a + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x] - Dist[1/(2*q), Int[Simp[-(a*h*e) - g*(c*d - a*f + q
) + (h*(c*d - a*f - q) - g*c*e)*x, x]/((a + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, c, d, e, f, g,
h}, x] && NeQ[e^2 - 4*d*f, 0] && NegQ[-(a*c)]

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 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^5}{\left (1+x^2\right ) \sqrt {a+b x+c x^2}} \, dx,x,\tan (d+e x)\right )}{e}\\ &=\frac {\operatorname {Subst}\left (\int \left (-\frac {x}{\sqrt {a+b x+c x^2}}+\frac {x^3}{\sqrt {a+b x+c x^2}}+\frac {x}{\left (1+x^2\right ) \sqrt {a+b x+c x^2}}\right ) \, dx,x,\tan (d+e x)\right )}{e}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {x}{\sqrt {a+b x+c x^2}} \, dx,x,\tan (d+e x)\right )}{e}+\frac {\operatorname {Subst}\left (\int \frac {x^3}{\sqrt {a+b x+c x^2}} \, dx,x,\tan (d+e x)\right )}{e}+\frac {\operatorname {Subst}\left (\int \frac {x}{\left (1+x^2\right ) \sqrt {a+b x+c x^2}} \, dx,x,\tan (d+e x)\right )}{e}\\ &=-\frac {\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{c e}+\frac {\tan ^2(d+e x) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{3 c e}+\frac {\operatorname {Subst}\left (\int \frac {x \left (-2 a-\frac {5 b x}{2}\right )}{\sqrt {a+b x+c x^2}} \, dx,x,\tan (d+e x)\right )}{3 c e}+\frac {b \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,\tan (d+e x)\right )}{2 c e}-\frac {\operatorname {Subst}\left (\int \frac {-b+\left (a-c-\sqrt {a^2+b^2-2 a c+c^2}\right ) x}{\left (1+x^2\right ) \sqrt {a+b x+c x^2}} \, dx,x,\tan (d+e x)\right )}{2 \sqrt {a^2+b^2-2 a c+c^2} e}+\frac {\operatorname {Subst}\left (\int \frac {-b+\left (a-c+\sqrt {a^2+b^2-2 a c+c^2}\right ) x}{\left (1+x^2\right ) \sqrt {a+b x+c x^2}} \, dx,x,\tan (d+e x)\right )}{2 \sqrt {a^2+b^2-2 a c+c^2} e}\\ &=-\frac {\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{c e}+\frac {\tan ^2(d+e x) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{3 c e}+\frac {\left (15 b^2-16 a c-10 b c \tan (d+e x)\right ) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{24 c^3 e}+\frac {b \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c \tan (d+e x)}{\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{c e}-\frac {\left (b \left (5 b^2-12 a c\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,\tan (d+e x)\right )}{16 c^3 e}-\frac {\left (b \left (a-c-\sqrt {a^2+b^2-2 a c+c^2}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-2 b \left (a-c-\sqrt {a^2+b^2-2 a c+c^2}\right )+b x^2} \, dx,x,\frac {a-c-\sqrt {a^2+b^2-2 a c+c^2}+b \tan (d+e x)}{\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {a^2+b^2-2 a c+c^2} e}+\frac {\left (b \left (a-c+\sqrt {a^2+b^2-2 a c+c^2}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-2 b \left (a-c+\sqrt {a^2+b^2-2 a c+c^2}\right )+b x^2} \, dx,x,\frac {a-c+\sqrt {a^2+b^2-2 a c+c^2}+b \tan (d+e x)}{\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {a^2+b^2-2 a c+c^2} e}\\ &=\frac {\sqrt {a-c-\sqrt {a^2+b^2-2 a c+c^2}} \tanh ^{-1}\left (\frac {a-c-\sqrt {a^2+b^2-2 a c+c^2}+b \tan (d+e x)}{\sqrt {2} \sqrt {a-c-\sqrt {a^2+b^2-2 a c+c^2}} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} \sqrt {a^2+b^2-2 a c+c^2} e}-\frac {\sqrt {a-c+\sqrt {a^2+b^2-2 a c+c^2}} \tanh ^{-1}\left (\frac {a-c+\sqrt {a^2+b^2-2 a c+c^2}+b \tan (d+e x)}{\sqrt {2} \sqrt {a-c+\sqrt {a^2+b^2-2 a c+c^2}} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} \sqrt {a^2+b^2-2 a c+c^2} e}+\frac {b \tanh ^{-1}\left (\frac {b+2 c \tan (d+e x)}{2 \sqrt {c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{2 c^{3/2} e}-\frac {\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{c e}+\frac {\tan ^2(d+e x) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{3 c e}+\frac {\left (15 b^2-16 a c-10 b c \tan (d+e x)\right ) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{24 c^3 e}-\frac {\left (b \left (5 b^2-12 a c\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c \tan (d+e x)}{\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{8 c^3 e}\\ &=\frac {\sqrt {a-c-\sqrt {a^2+b^2-2 a c+c^2}} \tanh ^{-1}\left (\frac {a-c-\sqrt {a^2+b^2-2 a c+c^2}+b \tan (d+e x)}{\sqrt {2} \sqrt {a-c-\sqrt {a^2+b^2-2 a c+c^2}} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} \sqrt {a^2+b^2-2 a c+c^2} e}-\frac {\sqrt {a-c+\sqrt {a^2+b^2-2 a c+c^2}} \tanh ^{-1}\left (\frac {a-c+\sqrt {a^2+b^2-2 a c+c^2}+b \tan (d+e x)}{\sqrt {2} \sqrt {a-c+\sqrt {a^2+b^2-2 a c+c^2}} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} \sqrt {a^2+b^2-2 a c+c^2} e}+\frac {b \tanh ^{-1}\left (\frac {b+2 c \tan (d+e x)}{2 \sqrt {c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{2 c^{3/2} e}-\frac {b \left (5 b^2-12 a c\right ) \tanh ^{-1}\left (\frac {b+2 c \tan (d+e x)}{2 \sqrt {c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{16 c^{7/2} e}-\frac {\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{c e}+\frac {\tan ^2(d+e x) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{3 c e}+\frac {\left (15 b^2-16 a c-10 b c \tan (d+e x)\right ) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{24 c^3 e}\\ \end {align*}

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Mathematica [C]  time = 6.14, size = 456, normalized size = 0.83 \[ \frac {\frac {\frac {\left (9 a b c-\frac {15 b^3}{4}\right ) \tanh ^{-1}\left (\frac {b+2 c \tan (d+e x)}{2 \sqrt {c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{4 c^{5/2}}+\frac {\left (-4 a c+\frac {15 b^2}{4}-\frac {5}{2} b c \tan (d+e x)\right ) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{2 c^2}}{3 c}+\frac {b \tanh ^{-1}\left (\frac {b+2 c \tan (d+e x)}{2 \sqrt {c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{2 c^{3/2}}+\frac {\tan ^2(d+e x) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{3 c}-\frac {\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{c}-\frac {2 \sqrt {a+i b-c} \tanh ^{-1}\left (\frac {2 a-(-b-2 i c) \tan (d+e x)+i b}{2 \sqrt {a+i b-c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{4 a+4 i b-4 c}-\frac {2 \sqrt {a-i b-c} \tanh ^{-1}\left (\frac {2 a-(-b+2 i c) \tan (d+e x)-i b}{2 \sqrt {a-i b-c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{4 a-4 i b-4 c}}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*Sqrt[a + I*b - c]*ArcTanh[(2*a + I*b - (-b - (2*I)*c)*Tan[d + e*x])/(2*Sqrt[a + I*b - c]*Sqrt[a + b*Tan[d
+ e*x] + c*Tan[d + e*x]^2])])/(4*a + (4*I)*b - 4*c) - (2*Sqrt[a - I*b - c]*ArcTanh[(2*a - I*b - (-b + (2*I)*c
)*Tan[d + e*x])/(2*Sqrt[a - I*b - c]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])])/(4*a - (4*I)*b - 4*c) + (b
*ArcTanh[(b + 2*c*Tan[d + e*x])/(2*Sqrt[c]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])])/(2*c^(3/2)) - Sqrt[a
+ b*Tan[d + e*x] + c*Tan[d + e*x]^2]/c + (Tan[d + e*x]^2*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])/(3*c) +
((((-15*b^3)/4 + 9*a*b*c)*ArcTanh[(b + 2*c*Tan[d + e*x])/(2*Sqrt[c]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^
2])])/(4*c^(5/2)) + (((15*b^2)/4 - 4*a*c - (5*b*c*Tan[d + e*x])/2)*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2]
)/(2*c^2))/(3*c))/e

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fricas [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)^5/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(e*x+d)^5/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep^2-1)]Discontinuities at zeroes of t_nostep^2-1 were not checkedWarning, integration of abs or sign
assumes constant sign by intervals (correct if the argument is real):Check [abs(t_nostep^2-1)]Evaluation time:
3.36Not invertible Error: Bad Argument Value

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maple [B]  time = 0.84, size = 9581348, normalized size = 17484.21 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [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)^5/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(1/2),x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{5}{\left (d + e x \right )}}{\sqrt {a + b \tan {\left (d + e x \right )} + c \tan ^{2}{\left (d + e x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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