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

Optimal. Leaf size=182 \[ -\frac {(b+2 c) \tanh ^{-1}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{4 c^{3/2} e}+\frac {\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{2 c e}-\frac {\tanh ^{-1}\left (\frac {2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e \sqrt {a-b+c}} \]

[Out]

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

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Rubi [A]  time = 0.35, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3700, 1251, 1653, 843, 621, 206, 724} \[ -\frac {(b+2 c) \tanh ^{-1}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{4 c^{3/2} e}+\frac {\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{2 c e}-\frac {\tanh ^{-1}\left (\frac {2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e \sqrt {a-b+c}} \]

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]

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

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Mathematica [A]  time = 2.49, size = 173, normalized size = 0.95 \[ -\frac {\frac {(b+2 c) \tanh ^{-1}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{c^{3/2}}-\frac {2 \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{c}+\frac {2 \tanh ^{-1}\left (\frac {2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {a-b+c}}}{4 e} \]

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]

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

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fricas [A]  time = 2.92, size = 1226, normalized size = 6.74 \[ \text {result too large to display} \]

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)^(1/2),x, algorithm="fricas")

[Out]

[1/8*(2*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)) + (a*b - b^2 + (2*a - b)*c +
2*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)*((a - b)
*c + c^2))/(((a - b)*c^2 + c^3)*e), 1/4*(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)) + (a*b - b^2 + (2*a - b)*c + 2*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)*((a - b)*c + c^2))/(((a - b)*c^2 + c^3)*e), -1/8*(4*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)) - (a*b - b^2 + (2*a - b)*c + 2*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)*((a - b)*c + c^2
))/(((a - b)*c^2 + c^3)*e), -1/4*(2*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)) - (a*b - b^2 + (2*a - b)*c + 2*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)*((a - b)*c + c^2))/(((a - b)*c^2 + c^3)*e)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (e x + d\right )^{5}}{\sqrt {c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a}}\,{d x} \]

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)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.48, size = 240, normalized size = 1.32 \[ \frac {\sqrt {a +b \left (\tan ^{2}\left (e x +d \right )\right )+c \left (\tan ^{4}\left (e x +d \right )\right )}}{2 c e}-\frac {b \ln \left (\frac {c \left (\tan ^{2}\left (e x +d \right )\right )+\frac {b}{2}}{\sqrt {c}}+\sqrt {a +b \left (\tan ^{2}\left (e x +d \right )\right )+c \left (\tan ^{4}\left (e x +d \right )\right )}\right )}{4 e \,c^{\frac {3}{2}}}-\frac {\ln \left (\frac {c \left (\tan ^{2}\left (e x +d \right )\right )+\frac {b}{2}}{\sqrt {c}}+\sqrt {a +b \left (\tan ^{2}\left (e x +d \right )\right )+c \left (\tan ^{4}\left (e x +d \right )\right )}\right )}{2 e \sqrt {c}}-\frac {\ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+\tan ^{2}\left (e x +d \right )\right )+2 \sqrt {a -b +c}\, \sqrt {\left (1+\tan ^{2}\left (e x +d \right )\right )^{2} c +\left (b -2 c \right ) \left (1+\tan ^{2}\left (e x +d \right )\right )+a -b +c}}{1+\tan ^{2}\left (e x +d \right )}\right )}{2 e \sqrt {a -b +c}} \]

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)^(1/2),x)

[Out]

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {tan}\left (d+e\,x\right )}^5}{\sqrt {c\,{\mathrm {tan}\left (d+e\,x\right )}^4+b\,{\mathrm {tan}\left (d+e\,x\right )}^2+a}} \,d x \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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