3.299 \(\int \tan ^6(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\)

Optimal. Leaf size=222 \[ \frac {\left (a^3+2 a^2 b+8 a b^2-16 b^3\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{16 b^{5/2} f}-\frac {(a-2 b) (a+4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{16 b^2 f}-\frac {\sqrt {a-b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}+\frac {\tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{6 f}+\frac {(a-6 b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{24 b f} \]

[Out]

1/16*(a^3+2*a^2*b+8*a*b^2-16*b^3)*arctanh(b^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/b^(5/2)/f-arctan((a-b)^
(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))*(a-b)^(1/2)/f-1/16*(a-2*b)*(a+4*b)*(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x
+e)/b^2/f+1/24*(a-6*b)*(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)^3/b/f+1/6*(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)^5/f

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Rubi [A]  time = 0.34, antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3670, 478, 582, 523, 217, 206, 377, 203} \[ \frac {\left (2 a^2 b+a^3+8 a b^2-16 b^3\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{16 b^{5/2} f}-\frac {(a-2 b) (a+4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{16 b^2 f}+\frac {\tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{6 f}+\frac {(a-6 b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{24 b f}-\frac {\sqrt {a-b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^6*Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

-((Sqrt[a - b]*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/f) + ((a^3 + 2*a^2*b + 8*a*b^2 -
16*b^3)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(16*b^(5/2)*f) - ((a - 2*b)*(a + 4*b)*Tan
[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(16*b^2*f) + ((a - 6*b)*Tan[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/(24*b
*f) + (Tan[e + f*x]^5*Sqrt[a + b*Tan[e + f*x]^2])/(6*f)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 478

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(b*(m + n*(p + q) + 1)), x] - Dist[e^n/(b*(m + n*(p +
q) + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[a*c*(m - n + 1) + (a*d*(m - n + 1) - n*q*(b
*c - a*d))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && GtQ[q, 0] &&
GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 582

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(f*g^(n - 1)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*d*(m + n*(p + q
+ 1) + 1)), x] - Dist[g^n/(b*d*(m + n*(p + q + 1) + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*
f*c*(m - n + 1) + (a*f*d*(m + n*q + 1) + b*(f*c*(m + n*p + 1) - e*d*(m + n*(p + q + 1) + 1)))*x^n, x], x], x]
/; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 0] && GtQ[m, n - 1]

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps

\begin {align*} \int \tan ^6(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^6 \sqrt {a+b x^2}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{6 f}-\frac {\operatorname {Subst}\left (\int \frac {x^4 \left (5 a+(-a+6 b) x^2\right )}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{6 f}\\ &=\frac {(a-6 b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{24 b f}+\frac {\tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{6 f}+\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (-3 a (a-6 b)-3 (a-2 b) (a+4 b) x^2\right )}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{24 b f}\\ &=-\frac {(a-2 b) (a+4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{16 b^2 f}+\frac {(a-6 b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{24 b f}+\frac {\tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{6 f}-\frac {\operatorname {Subst}\left (\int \frac {-3 a (a-2 b) (a+4 b)-3 \left (a^3+2 a^2 b+8 a b^2-16 b^3\right ) x^2}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{48 b^2 f}\\ &=-\frac {(a-2 b) (a+4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{16 b^2 f}+\frac {(a-6 b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{24 b f}+\frac {\tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{6 f}-\frac {(a-b) \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}+\frac {\left (a^3+2 a^2 b+8 a b^2-16 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{16 b^2 f}\\ &=-\frac {(a-2 b) (a+4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{16 b^2 f}+\frac {(a-6 b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{24 b f}+\frac {\tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{6 f}-\frac {(a-b) \operatorname {Subst}\left (\int \frac {1}{1-(-a+b) x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}+\frac {\left (a^3+2 a^2 b+8 a b^2-16 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{16 b^2 f}\\ &=-\frac {\sqrt {a-b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}+\frac {\left (a^3+2 a^2 b+8 a b^2-16 b^3\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{16 b^{5/2} f}-\frac {(a-2 b) (a+4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{16 b^2 f}+\frac {(a-6 b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{24 b f}+\frac {\tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{6 f}\\ \end {align*}

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Mathematica [C]  time = 6.33, size = 823, normalized size = 3.71 \[ \frac {-\frac {b \left (a^3+2 b a^2-8 b^3\right ) \sqrt {\frac {a+b+(a-b) \cos (2 (e+f x))}{\cos (2 (e+f x))+1}} \sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) F\left (\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right ) \sin ^4(e+f x)}{a (a+b+(a-b) \cos (2 (e+f x)))}-\frac {4 b \left (8 b^3-8 a b^2\right ) \sqrt {\cos (2 (e+f x))+1} \sqrt {\frac {a+b+(a-b) \cos (2 (e+f x))}{\cos (2 (e+f x))+1}} \left (\frac {\sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) F\left (\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right ) \sin ^4(e+f x)}{4 a \sqrt {\cos (2 (e+f x))+1} \sqrt {a+b+(a-b) \cos (2 (e+f x))}}-\frac {\sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \Pi \left (-\frac {b}{a-b};\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right ) \sin ^4(e+f x)}{2 (a-b) \sqrt {\cos (2 (e+f x))+1} \sqrt {a+b+(a-b) \cos (2 (e+f x))}}\right )}{\sqrt {a+b+(a-b) \cos (2 (e+f x))}}}{8 b^2 f}+\frac {\sqrt {\frac {\cos (2 (e+f x)) a+a+b-b \cos (2 (e+f x))}{\cos (2 (e+f x))+1}} \left (\frac {1}{6} \tan (e+f x) \sec ^4(e+f x)+\frac {(a \sin (e+f x)-14 b \sin (e+f x)) \sec ^3(e+f x)}{24 b}+\frac {\left (-3 \sin (e+f x) a^2-8 b \sin (e+f x) a+44 b^2 \sin (e+f x)\right ) \sec (e+f x)}{48 b^2}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^6*Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

(-((b*(a^3 + 2*a^2*b - 8*b^3)*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*Sqrt[-((a*Cot[e
+ f*x]^2)/b)]*Sqrt[-((a*(1 + Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b)]*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Cs
c[e + f*x]^2)/b]*Csc[2*(e + f*x)]*EllipticF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]
/Sqrt[2]], 1]*Sin[e + f*x]^4)/(a*(a + b + (a - b)*Cos[2*(e + f*x)]))) - (4*b*(-8*a*b^2 + 8*b^3)*Sqrt[1 + Cos[2
*(e + f*x)]]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*((Sqrt[-((a*Cot[e + f*x]^2)/b)]*S
qrt[-((a*(1 + Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b)]*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b
]*Csc[2*(e + f*x)]*EllipticF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1]*S
in[e + f*x]^4)/(4*a*Sqrt[1 + Cos[2*(e + f*x)]]*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]]) - (Sqrt[-((a*Cot[e + f*
x]^2)/b)]*Sqrt[-((a*(1 + Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b)]*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e
+ f*x]^2)/b]*Csc[2*(e + f*x)]*EllipticPi[-(b/(a - b)), ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e +
f*x]^2)/b]/Sqrt[2]], 1]*Sin[e + f*x]^4)/(2*(a - b)*Sqrt[1 + Cos[2*(e + f*x)]]*Sqrt[a + b + (a - b)*Cos[2*(e +
f*x)]])))/Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]])/(8*b^2*f) + (Sqrt[(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*(e
+ f*x)])/(1 + Cos[2*(e + f*x)])]*((Sec[e + f*x]^3*(a*Sin[e + f*x] - 14*b*Sin[e + f*x]))/(24*b) + (Sec[e + f*x]
*(-3*a^2*Sin[e + f*x] - 8*a*b*Sin[e + f*x] + 44*b^2*Sin[e + f*x]))/(48*b^2) + (Sec[e + f*x]^4*Tan[e + f*x])/6)
)/f

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fricas [A]  time = 2.10, size = 826, normalized size = 3.72 \[ \left [\frac {48 \, \sqrt {-a + b} b^{3} \log \left (-\frac {{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b} \tan \left (f x + e\right ) - a}{\tan \left (f x + e\right )^{2} + 1}\right ) - 3 \, {\left (a^{3} + 2 \, a^{2} b + 8 \, a b^{2} - 16 \, b^{3}\right )} \sqrt {b} \log \left (2 \, b \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {b} \tan \left (f x + e\right ) + a\right ) + 2 \, {\left (8 \, b^{3} \tan \left (f x + e\right )^{5} + 2 \, {\left (a b^{2} - 6 \, b^{3}\right )} \tan \left (f x + e\right )^{3} - 3 \, {\left (a^{2} b + 2 \, a b^{2} - 8 \, b^{3}\right )} \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{96 \, b^{3} f}, -\frac {96 \, \sqrt {a - b} b^{3} \arctan \left (-\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a}}{\sqrt {a - b} \tan \left (f x + e\right )}\right ) + 3 \, {\left (a^{3} + 2 \, a^{2} b + 8 \, a b^{2} - 16 \, b^{3}\right )} \sqrt {b} \log \left (2 \, b \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {b} \tan \left (f x + e\right ) + a\right ) - 2 \, {\left (8 \, b^{3} \tan \left (f x + e\right )^{5} + 2 \, {\left (a b^{2} - 6 \, b^{3}\right )} \tan \left (f x + e\right )^{3} - 3 \, {\left (a^{2} b + 2 \, a b^{2} - 8 \, b^{3}\right )} \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{96 \, b^{3} f}, \frac {24 \, \sqrt {-a + b} b^{3} \log \left (-\frac {{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b} \tan \left (f x + e\right ) - a}{\tan \left (f x + e\right )^{2} + 1}\right ) - 3 \, {\left (a^{3} + 2 \, a^{2} b + 8 \, a b^{2} - 16 \, b^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-b}}{b \tan \left (f x + e\right )}\right ) + {\left (8 \, b^{3} \tan \left (f x + e\right )^{5} + 2 \, {\left (a b^{2} - 6 \, b^{3}\right )} \tan \left (f x + e\right )^{3} - 3 \, {\left (a^{2} b + 2 \, a b^{2} - 8 \, b^{3}\right )} \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{48 \, b^{3} f}, -\frac {48 \, \sqrt {a - b} b^{3} \arctan \left (-\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a}}{\sqrt {a - b} \tan \left (f x + e\right )}\right ) + 3 \, {\left (a^{3} + 2 \, a^{2} b + 8 \, a b^{2} - 16 \, b^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-b}}{b \tan \left (f x + e\right )}\right ) - {\left (8 \, b^{3} \tan \left (f x + e\right )^{5} + 2 \, {\left (a b^{2} - 6 \, b^{3}\right )} \tan \left (f x + e\right )^{3} - 3 \, {\left (a^{2} b + 2 \, a b^{2} - 8 \, b^{3}\right )} \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{48 \, b^{3} f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)^6,x, algorithm="fricas")

[Out]

[1/96*(48*sqrt(-a + b)*b^3*log(-((a - 2*b)*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)*tan(f*x
+ e) - a)/(tan(f*x + e)^2 + 1)) - 3*(a^3 + 2*a^2*b + 8*a*b^2 - 16*b^3)*sqrt(b)*log(2*b*tan(f*x + e)^2 - 2*sqrt
(b*tan(f*x + e)^2 + a)*sqrt(b)*tan(f*x + e) + a) + 2*(8*b^3*tan(f*x + e)^5 + 2*(a*b^2 - 6*b^3)*tan(f*x + e)^3
- 3*(a^2*b + 2*a*b^2 - 8*b^3)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/(b^3*f), -1/96*(96*sqrt(a - b)*b^3*arc
tan(-sqrt(b*tan(f*x + e)^2 + a)/(sqrt(a - b)*tan(f*x + e))) + 3*(a^3 + 2*a^2*b + 8*a*b^2 - 16*b^3)*sqrt(b)*log
(2*b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(b)*tan(f*x + e) + a) - 2*(8*b^3*tan(f*x + e)^5 + 2*(a*
b^2 - 6*b^3)*tan(f*x + e)^3 - 3*(a^2*b + 2*a*b^2 - 8*b^3)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/(b^3*f), 1
/48*(24*sqrt(-a + b)*b^3*log(-((a - 2*b)*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)*tan(f*x +
e) - a)/(tan(f*x + e)^2 + 1)) - 3*(a^3 + 2*a^2*b + 8*a*b^2 - 16*b^3)*sqrt(-b)*arctan(sqrt(b*tan(f*x + e)^2 + a
)*sqrt(-b)/(b*tan(f*x + e))) + (8*b^3*tan(f*x + e)^5 + 2*(a*b^2 - 6*b^3)*tan(f*x + e)^3 - 3*(a^2*b + 2*a*b^2 -
8*b^3)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/(b^3*f), -1/48*(48*sqrt(a - b)*b^3*arctan(-sqrt(b*tan(f*x +
e)^2 + a)/(sqrt(a - b)*tan(f*x + e))) + 3*(a^3 + 2*a^2*b + 8*a*b^2 - 16*b^3)*sqrt(-b)*arctan(sqrt(b*tan(f*x +
e)^2 + a)*sqrt(-b)/(b*tan(f*x + e))) - (8*b^3*tan(f*x + e)^5 + 2*(a*b^2 - 6*b^3)*tan(f*x + e)^3 - 3*(a^2*b + 2
*a*b^2 - 8*b^3)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/(b^3*f)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \tan \left (f x + e\right )^{2} + a} \tan \left (f x + e\right )^{6}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)^6,x, algorithm="giac")

[Out]

integrate(sqrt(b*tan(f*x + e)^2 + a)*tan(f*x + e)^6, x)

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maple [B]  time = 0.37, size = 451, normalized size = 2.03 \[ \frac {\left (\tan ^{3}\left (f x +e \right )\right ) \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}{6 f b}-\frac {a \tan \left (f x +e \right ) \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}{8 f \,b^{2}}+\frac {a^{2} \tan \left (f x +e \right ) \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}{16 f \,b^{2}}+\frac {a^{3} \ln \left (\tan \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\right )}{16 f \,b^{\frac {5}{2}}}-\frac {\tan \left (f x +e \right ) \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}{4 f b}+\frac {a \tan \left (f x +e \right ) \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}{8 f b}+\frac {a^{2} \ln \left (\tan \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\right )}{8 f \,b^{\frac {3}{2}}}+\frac {\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\, \tan \left (f x +e \right )}{2 f}+\frac {a \ln \left (\tan \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\right )}{2 f \sqrt {b}}-\frac {\sqrt {b}\, \ln \left (\tan \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\right )}{f}+\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {\left (a -b \right ) b^{2} \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}\right )}{f b \left (a -b \right )}-\frac {a \sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {\left (a -b \right ) b^{2} \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}\right )}{f \,b^{2} \left (a -b \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)^6,x)

[Out]

1/6/f*tan(f*x+e)^3*(a+b*tan(f*x+e)^2)^(3/2)/b-1/8/f/b^2*a*tan(f*x+e)*(a+b*tan(f*x+e)^2)^(3/2)+1/16/f/b^2*a^2*t
an(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2)+1/16/f/b^(5/2)*a^3*ln(tan(f*x+e)*b^(1/2)+(a+b*tan(f*x+e)^2)^(1/2))-1/4/f*ta
n(f*x+e)*(a+b*tan(f*x+e)^2)^(3/2)/b+1/8/f/b*a*tan(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2)+1/8/f/b^(3/2)*a^2*ln(tan(f*x
+e)*b^(1/2)+(a+b*tan(f*x+e)^2)^(1/2))+1/2*(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/f+1/2/f*a/b^(1/2)*ln(tan(f*x+e)*
b^(1/2)+(a+b*tan(f*x+e)^2)^(1/2))-1/f*b^(1/2)*ln(tan(f*x+e)*b^(1/2)+(a+b*tan(f*x+e)^2)^(1/2))+1/f*(b^4*(a-b))^
(1/2)/b/(a-b)*arctan((a-b)*b^2/(b^4*(a-b))^(1/2)/(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e))-1/f*a*(b^4*(a-b))^(1/2)/
b^2/(a-b)*arctan((a-b)*b^2/(b^4*(a-b))^(1/2)/(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \tan \left (f x + e\right )^{2} + a} \tan \left (f x + e\right )^{6}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)^6,x, algorithm="maxima")

[Out]

integrate(sqrt(b*tan(f*x + e)^2 + a)*tan(f*x + e)^6, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\mathrm {tan}\left (e+f\,x\right )}^6\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^6*(a + b*tan(e + f*x)^2)^(1/2),x)

[Out]

int(tan(e + f*x)^6*(a + b*tan(e + f*x)^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b \tan ^{2}{\left (e + f x \right )}} \tan ^{6}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e)**2)**(1/2)*tan(f*x+e)**6,x)

[Out]

Integral(sqrt(a + b*tan(e + f*x)**2)*tan(e + f*x)**6, x)

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